Unit 2 Energy and Momentum - classconnect

Unit 2 Energy and MomentumARE YOU READY?(Pages 174–175)Knowledge and Understanding1. Some forms of energy are the chemical potential energy stored in the gasoline, kinetic energy, gravitational potentialenergy, thermal energy (heat from the motor), and sound.2. (a) mass(b) energy(c) force(d) power3. joule – James Joulenewton – Isaac Newtonwatt – James Watt4. The work you do on the box depends on its mass, m, the gravitational field strength, g, and the vertical height it is raised,∆y.Inquiry and Communication29.5 − 25.15. (a)× 100% = 14.9%29.5The percentage of initial kinetic energy lost is 14.9%.(b) The energy did not cease to exist, but was converted into other forms (likely heat and sound).6. (a) The measure the efficiency of this incline, you would need to know the distance up the ramp, the force applied, themass of the sled and child and the vertical height. If you knew the angle that the hill was compared to the horizontal,you could calculate the vertical height from the length up the slope using trigonometry.(b) For 100% efficiency, there could be no friction between the sled and the ground.7. The conservation of energy refers to the inability to create or destroy energy during an interaction. Conserving energy isthe reduction of our use of energy converted from fossil fuels and other sources to allow them to last longer.Making Connections8. (a) The large mass of the bus help to reduce injury because the acceleration will be small. The bumpers and body panelsalso help to absorb energy to minimize the energy that will be absorbed by passengers.(b) We could increase the safety of buses by adding more padding around passenger compartments, seat belts, and perhapseven air bags.(c) School busses currently have excellent safety ratings, and the added features in part (b) are expensive to install.9. (a) The ski jumper will acquire gravitational potential energy as he is transported to the top of the hill. This gravitationalenergy will be converted to kinetic energy as the jumper slides down the hill, and then reconverted back into somegravitational energy as he jumps from the end of the ramp. All of the energy will be converted to kinetic as he reachesthe bottom of the hill (if we ignore the loss due to friction).(b) The law of conservation of energy is what governs the energy transformations.(c) The main safety feature of this sport is the careful construction of the hill so that the ski jumper does not experience alarge normal force from landing. Also the protection of the head by a helmet and suits minimize injury in case of a fall.Math Skills10. (a) W ∝∆ d(b)(c)(d)EP2∝ x1P ∝t∝ vEK2Copyright © 2003 Nelson Unit 2 Are You Ready? 231

The graphs for (a)–(d) are shown below.11. (a) x-components: F K , F A cos αy-components: F N , F A sin α, F g(b) x-components: F K , F A cos α, F g sin βy-components: F N , F A sin α, F g cos β12.Technical Skills and Safety13. (a) Place the air pucks on the table while it is running to see if they regularly move in one direction.(b) The determination of speeds depends on the assumption that the speeds are constant. For a table that is not level, acomponent of the force of gravity will be accelerating the objects and changing their speed and direction.(c) Other sources of error can be vibration from motion, and air currents from a location near a ventilation duct for theschool environmental controls. You must also check equipment to make sure there are no rough edges that might catch.14. Most sensors are electronic and have a shock hazard so the electrical equipment should be kept away from water. This isalso wise to protect the equipment. All experiments should be run slowly in a trial form to make sure that any electricalconnectors or cables do not interfere with data collection or damage the electronic devices.CHAPTER 4 WORK AND ENERGYReflect on Your Learning(Page 176)1. A compression spring is designed to be at its maximum length when there is no force, and resist being squeezed together.An extension spring is designed to be at its minimum length when there is no force, and resist being stretched. Someexamples of compression springs are the springs in a car and a retractable ballpoint pen. Some examples of extensionsprings are bungee cords and storm door springs.2. During the bounce of the ball, most of the kinetic energy is stored as elastic potential energy to be converted back intokinetic energy after the bounce. Some energy is “lost” (converted) to unwanted forms such as heat and sound.3. A grandfather clock uses the gravitational potential energy to operate a pendulum with a constant period.4. (a) The work done on each cart is the same.(b) The work done on each box would be the same since it only depends on the size of the component of force in thedirection of motion and the displacement. For each box, these values are still the same.(c) If F 2 is not enough to overcome the frictional force, then the box will not move and no work will be done.Copyright © 2003 Nelson Unit 2 Are You Ready? 232

The graphs for (a)–(d) are shown below.11. (a) x-components: F K , F A cos αy-components: F N , F A sin α, F g(b) x-components: F K , F A cos α, F g sin βy-components: F N , F A sin α, F g cos β12.Technical Skills and Safety13. (a) Place the air pucks on the table while it is running to see if they regularly move in one direction.(b) The determination of speeds depends on the assumption that the speeds are constant. For a table that is not level, acomponent of the force of gravity will be accelerating the objects and changing their speed and direction.(c) Other sources of error can be vibration from motion, and air currents from a location near a ventilation duct for theschool environmental controls. You must also check equipment to make sure there are no rough edges that might catch.14. Most sensors are electronic and have a shock hazard so the electrical equipment should be kept away from water. This isalso wise to protect the equipment. All experiments should be run slowly in a trial form to make sure that any electricalconnectors or cables do not interfere with data collection or damage the electronic devices.CHAPTER 4 WORK AND ENERGYReflect on Your Learning(Page 176)1. A compression spring is designed to be at its maximum length when there is no force, and resist being squeezed together.An extension spring is designed to be at its minimum length when there is no force, and resist being stretched. Someexamples of compression springs are the springs in a car and a retractable ballpoint pen. Some examples of extensionsprings are bungee cords and storm door springs.2. During the bounce of the ball, most of the kinetic energy is stored as elastic potential energy to be converted back intokinetic energy after the bounce. Some energy is “lost” (converted) to unwanted forms such as heat and sound.3. A grandfather clock uses the gravitational potential energy to operate a pendulum with a constant period.4. (a) The work done on each cart is the same.(b) The work done on each box would be the same since it only depends on the size of the component of force in thedirection of motion and the displacement. For each box, these values are still the same.(c) If F 2 is not enough to overcome the frictional force, then the box will not move and no work will be done.Copyright © 2003 Nelson Unit 2 Are You Ready? 232

5.6. The shock absorber would be a spring with a large force constant to absorb lots of energy quickly, combined with a slowair release to prevent too much bouncing. The size would be suited to fit into the length of the forks. The forces involvedwould mean the choice of a strong material, such as steel, for strength.Try This Activity: Which Ball Wins?(Page 177)(a) The ball on track Y will win because it converts its gravitational energy into kinetic energy more quickly, and will speedup more quickly than the ball on track X.(b) The total energy of the balls remains constant down the track. The energy is just transformed from one type to another.4.1 WORK DONE BY A CONSTANT FORCEPRACTICE(Pages 181–182)Understanding Concepts1. F 1 will do more work than F 2 because the component of F 2 in the direction of motion is smaller than F 1 .2. No. The force of kinetic friction is always acting opposite to the direction of motion. Since negative work is alwaysopposite the direction of motion, the kinetic friction will always do negative work.3. Yes. The force of gravity can move an object toward itself, and therefore does positive work on that object.4. m = 2.75 kg(a) ∆d = 1.37 mW = ?W = ( Fcos θ ) ∆d= ( mg cos θ ) ∆d= (2.75 kg)(9.80 N/kg)(cos 0 ° )(1.37 m)W = 36.9 JThe work done to move the plant 1.37 m up is 36.9 J.(b) ∆y = 1.07 mµ K = 0.549W = ?First we must calculate the normal force acting on the potted plant:Σ Fy= may= 0FN− Fg= 0FN= Fg= mg= (2.75 kg)(9.80 N/kg)F = 26.95 NNCopyright © 2003 Nelson Chapter 4 Work and Energy 233

5.6. The shock absorber would be a spring with a large force constant to absorb lots of energy quickly, combined with a slowair release to prevent too much bouncing. The size would be suited to fit into the length of the forks. The forces involvedwould mean the choice of a strong material, such as steel, for strength.Try This Activity: Which Ball Wins?(Page 177)(a) The ball on track Y will win because it converts its gravitational energy into kinetic energy more quickly, and will speedup more quickly than the ball on track X.(b) The total energy of the balls remains constant down the track. The energy is just transformed from one type to another.4.1 WORK DONE BY A CONSTANT FORCEPRACTICE(Pages 181–182)Understanding Concepts1. F 1 will do more work than F 2 because the component of F 2 in the direction of motion is smaller than F 1 .2. No. The force of kinetic friction is always acting opposite to the direction of motion. Since negative work is alwaysopposite the direction of motion, the kinetic friction will always do negative work.3. Yes. The force of gravity can move an object toward itself, and therefore does positive work on that object.4. m = 2.75 kg(a) ∆d = 1.37 mW = ?W = ( Fcos θ ) ∆d= ( mg cos θ ) ∆d= (2.75 kg)(9.80 N/kg)(cos 0 ° )(1.37 m)W = 36.9 JThe work done to move the plant 1.37 m up is 36.9 J.(b) ∆y = 1.07 mµ K = 0.549W = ?First we must calculate the normal force acting on the potted plant:Σ Fy= may= 0FN− Fg= 0FN= Fg= mg= (2.75 kg)(9.80 N/kg)F = 26.95 NNCopyright © 2003 Nelson Chapter 4 Work and Energy 233

Let F A be the applied force to move the potted plant horizontally:Σ Fx= max= 0FA− FK= 0FA= FK= µ FN= (0.549)(26.95 N)FA= 14.796 NTo calculate the work done on the potted plant:W = ( FAcos θ ) ∆d= (14.796 N)(cos0 ° )(1.07 m)W = 15.8 JThe work done to move the plant 1.07 m across the shelf is 15.8 J.5. m = 24.5 kgF = 14.2 N [22.5° below the horizontal]∆d = 14.8 mW = ?We only need to consider the component of force in the direction of motion:W = ( Fcos θ ) ∆d= (14.2 N)(cos 22.5 ° )(14.8 m)W = 194 JThe work done by the force is 194 J.6. FT= 12.5 N [19.5 ° above the horizontal]W = 225 J∆d = ?W = ( Fcos θ ) ∆dW∆ d =F cosθ225 J=(12.5 N)(cos19.5 ° )∆ d = 19.1 mThe toboggan moves 19.1 m.7. (a) We will calculate the area using the formula for a rectangleA=lw= (4.0 N)(2.0 m)A = 8.0 JThe area represents the work done on the object.(b) First calculate the area of the second portion of the graph:A=lw= ( −4.0 N)(6.0 m −2.0 m)A =−8.0 JThe total work is 8.0 + (–8.0) = 0.0 J(c) One situation could be pushing a box across a table and pulling it back.Applying Inquiry Skills8. You could set the pen on the paper and pull the paper across the desk. The force of static friction between the paper andthe pen is in the direction of motion, doing positive work on the pen.Copyright © 2003 Nelson Chapter 4 Work and Energy 234

Understanding Concepts9. Four different situations are:• a book sitting on a desk (∆d = 0)• a student carrying a book at a constant speed (θ = 90º)• a teacher whirling a putty pat in a circle at the end of a string• a toy car travelling in a circular path10. (a) A box being pulled by string at an angle involves the forward component doing positive work, and the verticalcomponent doing zero work.(b) A box being pulled up a ramp involves the parallel component of gravity doing negative work, and the perpendicularcomponent of gravity doing zero work.Section 4.1 Questions(Page 183)Understanding Concepts1. The everyday use of the word “work” is different from the usage in physics when it is referring to employment, or a dutyto perform. “Working” as a teacher involves very little physical work. The physics definition of work means the transferof energy to an object to move it a certain distance. Many types of employment or daily activities involve physical work.For example, the sentence “Loading the cement bags onto the truck was a lot of work,” uses the word “work” similar tothe physics definition of work.2. The centripetal force is always directed toward the centre of the circle, and is by definition perpendicular to the motion ofthe object. The 90º angle means that work is not done on the object by the centripetal force.3. As you push on a wall, you are exerting a force, which involves the use of energy. Even though no physical work is beingdone, your muscles are still burning your body’s fuel, causing you to become tired.4. Assuming the classroom to be 4 m tall, and the student to have a mass of 70 kg:W = ( Fcos θ ) ∆d= ( mg cosθ) ∆d= (70 N)(9.80 N/kg)(cos 0 ° )(4.0 m)3W = 3.0×10 JIt would take about 3.0 10 3 J, or 3.0 kJ of work to climb the ladder.5. F A= 75 N [22° below the horizontal]F T= 75 N [32° above the horizontal]Copyright © 2003 Nelson Chapter 4 Work and Energy 235

(a) The work done by the boy (W B ):WB= ( FTcos θ ) ∆d= (75 N)(cos32 ° )(13 m)W = 826.8 JBThe work done by the girl (W G ):WG= ( FAcos θ ) ∆d= (75 N)(cos 22 ° )(13 m)W = 904.0 JGThe total work done:Wtotal = WB + WG= 828.8 J + 904.0 J3Wtotal= 1.73×10 JThe total amount of work done on the crate is 1.7 10 –3 J.(b) The crate is moving at a constant speed, so it is not gaining any energy. This means that the crate will have the sameamount of energy before and after the move, so it must have work done on it opposite the direction of motion.Therefore, the work done on the crate by the floor is –1.7 10 3 J.6. W = 9.65 10 2 J∆d = 45.3 mF = 24.1 N [parallel to the handle of the sleigh]θ = ?W = ( Fcos θ ) ∆dWcosθ=F ∆ d−1⎛ W ⎞θ = cos ⎜ ⎟⎝ F∆d⎠−1⎛ 965 J ⎞= cos ⎜ ⎟⎝(24.1 N)(45.3 m) ⎠θ = 27.9°The angle between the snowy surface and the handle is 27.9º.7. (a) ∆d = 38 mm = 66 kgF A= 58 N [18° above the horizontal]F N = ?µ K = ?First we must calculate the normal force:Σ F = ma = 0F + F sin18°− F = 0N A gyF = F − F sin18°N g ANy= mg − F sin18°A= (66 kg)(9.80 N/kg) − (58 N)sin18°F = 628.88 NCopyright © 2003 Nelson Chapter 4 Work and Energy 236

We then calculate the force of gravity:Σ Fx= max= 0FAcos18°− FK= 0FK= FAcos18°= (58 N) cos18°F = 55.161 NKThe coefficient of kinetic friction is:FKµ K =FN55.161 N=628.88 Nµ K = 0.088The normal force is 6.3 10 2 N, and the coefficient of kinetic friction between the toboggan and the snow is 0.088.(b) W = ?W = ( Fcos θ ) ∆d= (55.161 N)(cos180 ° )(38 m)3W =− 2.1×10 JThe work done by kinetic friction is –2.1 10 3 J.(c) The normal force, the gravitational force, and the vertical component of the applied force do no work on the toboggan.(d) ∆d = 25 mW = ?W = ( Fcos θ ) ∆d= (58 N)(cos18 ° )(25 m)3W = 1.4×10 JThe work done by the parent is 1.4 10 3 J.Applying Inquiry Skills8. As shown below, the graph indicates that the work done on a object is positive for angles less than 90º, zero for anglesequal to 90º, and negative for angles between 90º and 180º.Making Connections9. Work done by friction has the effect of heating up the environment. Most forms of energy usually end up as thermalenergy.4.2 KINETIC ENERGY AND THE WORK–ENERGY THEOREMPRACTICE(Pages 186–187)Understanding Concepts1. The kinetic energy of a moving object is related to both the mass and the velocity. If the mass of the truck is large enough,a slow moving truck can have more kinetic energy than a fast moving car.Copyright © 2003 Nelson Chapter 4 Work and Energy 237

We then calculate the force of gravity:Σ Fx= max= 0FAcos18°− FK= 0FK= FAcos18°= (58 N) cos18°F = 55.161 NKThe coefficient of kinetic friction is:FKµ K =FN55.161 N=628.88 Nµ K = 0.088The normal force is 6.3 10 2 N, and the coefficient of kinetic friction between the toboggan and the snow is 0.088.(b) W = ?W = ( Fcos θ ) ∆d= (55.161 N)(cos180 ° )(38 m)3W =− 2.1×10 JThe work done by kinetic friction is –2.1 10 3 J.(c) The normal force, the gravitational force, and the vertical component of the applied force do no work on the toboggan.(d) ∆d = 25 mW = ?W = ( Fcos θ ) ∆d= (58 N)(cos18 ° )(25 m)3W = 1.4×10 JThe work done by the parent is 1.4 10 3 J.Applying Inquiry Skills8. As shown below, the graph indicates that the work done on a object is positive for angles less than 90º, zero for anglesequal to 90º, and negative for angles between 90º and 180º.Making Connections9. Work done by friction has the effect of heating up the environment. Most forms of energy usually end up as thermalenergy.4.2 KINETIC ENERGY AND THE WORK–ENERGY THEOREMPRACTICE(Pages 186–187)Understanding Concepts1. The kinetic energy of a moving object is related to both the mass and the velocity. If the mass of the truck is large enough,a slow moving truck can have more kinetic energy than a fast moving car.Copyright © 2003 Nelson Chapter 4 Work and Energy 237

2. The kinetic energy is proportional to the speed, so if the speed increases by(a) 2, the kinetic energy will increase by a factor of 2 2 , or 4(b) 3, the kinetic energy will increase by a factor of 3 2 , or 9(c) 37%, the kinetic energy will increase by a factor of 1.37 2 , or 1.93. Assume a 75-kg student running at 8.0 m/s:1 2EK= mv21 (75 kg)(8.0 m/s)2=23EK= 2.4×10 JThe kinetic energy at maximum speed is 2.4 kJ.4. m = 45 g = 4.5 10 –2 kgv i = 0 m/sv f = 43 m/s(a) W = ?W =∆E1 2 2= mv (f−vi)2= × −2W = 41.6025 J, or42 JThe work done by the club is 42 J.(b) ∆d = 2.0 cm = 2.0 10 –2 mF = ?W = F∆dWF = ∆ d41.6025 J=−22.0×10 m3F = 2.1×10 NThe average force exerted by the club is 2.1 × 10 3 N.5. m = 27 g = 2.7 10 –1 kgF = 95 N∆d = 31 cm = 3.1 10 –1 mv f = ?K2 2(( ) ( ) )1 (4.5 10−2kg) 43 m/s 0 m/sW =∆E1 2F∆ d = mvf(since the initial speed is zero)22 2F∆dvf=m2F∆dvf=m−12(95 N)(3.1×10 m)=−12.7×10 kgvf= 47 m/sThe final speed of the arrow is 47 m/s.6. m = 4.55 10 4 kgv i = 1.22 10 4 m/sF = 3.85 10 5 N∆d = 2.45 10 6 mv f = ?KCopyright © 2003 Nelson Chapter 4 Work and Energy 238

W =∆EK1 2 1 2F∆ d = mvf− mvi2 21 2 1 2mvf= F∆ d + mvi2 22F∆d2vf= + vim5 62(3.85× 10 N)(2.45×10 m)= + (1.22 × 10 m/s)44.55×10 kg4vf= 1.38×10 m/sThe final speed of the probe is 1.38 10 4 m/s.7. m = 28.0 kgF A= 95.6 N [35° above the horizontal]F K = 75.5 Nv i = 0 m/s∆d = 0.750 mv f = ?4 2The total work done on the box will become kinetic energy. Since the initial speed is zero:1 2W = mvf21 2FA cos35.0°∆ d + FK cos180°∆ d = mvf22 2∆dvf = ( FA cos35.0°+ FKcos180°)m2∆dvf = ( FA cos35.0°+ FKcos180°)m2(0.750 m)= (95.6 N)(0.81952) + (75.5 N)( − 1)20.8 kgv = 0.45 m/sThe final speed of the box is 0.45 m/s.8. W = 1.47(cos 38º ) = 1.16The toboggan would have increased its kinetic energy by 16%.Applying Inquiry Skills9. W = F∆d1 2E= N⋅mK = mv2kg ⋅m = ⋅2 m⎛m⎞= kg ⋅s⎜ ⎟⎝ s ⎠2kg ⋅m2W =kg ⋅ m2sEK =2sThe base units for both are the same.Making Connections10. m = 6.85 10 3 kgv A = 2.81 10 3 m/sv B = 8.38 10 3 m/sW = ?2f( )Copyright © 2003 Nelson Chapter 4 Work and Energy 239

(a) The work done is equal to the change in kinetic energyW =∆EK1 2 1 2= mvB− mv2 A21 ( 2 2= mv ) B −vA21= (6.85 × 10 kg) (8.38 × 10 m/s) − (2.81 × 10 m/s)211W = 2.13×10 J(b) The work done by Earth to move the satellite from A to B is 2.13 × 10 11 J.W =∆EK( )3 3 2 3 21 2 1 2= mvA− mvB2 21 ( 2 2= mv ) A −vB21= (6.85 × 10 kg) (2.81 × 10 m/s) − (8.38 × 10 m/s)211W =− 2.13×10 JThe work done by Earth to move the satellite from B to A is –2.13 × 10 11 J.( )3 3 2 3 2Section 4.2 Questions(Page 188)Understanding Concepts1. The first doubling will require much less energy than the second doubling of the speed. This can clearly be shown using:1 2 2W = m ( vf−vi)22 2f − iW ∝v vTo go from v to 2v:2 2f iW = v −v2 2= (2 v) −( v)2 2= 4v−vW = 3v2To go from 2v to 4v:2 2f i2 2W = v −v= (4 v) −(2 v)2 2= 16v−4v2W = 11vThe first doubling of speed will require work proportional to 3 times the square of the original speed. The seconddoubling will require work proportional to 11 times the square of the original speed.Copyright © 2003 Nelson Chapter 4 Work and Energy 240

2. m = 1.50 10 3 kgv = 18.0 m/s [E]E K = ?EK1 2= mv21= (1.5 × 10 kg)(18.0 m/s)25= 2.43×10 JEKThe kinetic energy of the car is 2.43 10 5 J.3. (a) v = 1.150 × 18.0 = 20.7 m/sE K = ?3 21 2EK= mv21 (1.5 103 kg)(20.7 m/s)2= ×25EK= 3.21×10 JThe new kinetic energy of the car is 3.21 10 5 J.(b) The increase in E K is:3.21×10 J= 1.3252.43×10 JWe can verify this with (1.15) 2 = 1.32This represents an increase in the kinetic energy of 32%.(c) W = ?W =∆EK5 5= 3.21× 10 J − 2.43×10 J4W = 7.8×10 JThe work done to speed up the car was 7.8 10 4 J.4. m = 55 kgE K = 3.3 10 3 Jv = ?1 2EK= mv22EKv =m32(3.3×10 J)=55 kgv = 11 m/sThe speed of the sprinter is 11 m/s.5. v = 12 m/sE K = 43 Jm = ?1 2EK= mv22EKm =2v2(43 J)=2(12 m/s)m = 0.60 kgThe mass of the basketball is 0.60 kg.5Copyright © 2003 Nelson Chapter 4 Work and Energy 241

6. m = 0.353 kg∆d = 89.3 cm = 0.893 m(a) W = ?W = ( mgcos θ ) ∆d= (0.353 kg)(9.80 N/kg)(cos0 ° )(0.893 m)W = 3.0892 JThe work done by gravity is 3.09 J.(b) Using the work energy theorem, W =∆ EK.Since v i = 0:1 2W = mvf22Wvf=mvf==2(3.0892 J)0.353 kg4.18 m/sThe speed of the plate just before it hits the floor is 4.18 m/s.7. m = 61 kgθ = 23°F K = 72 Nv i = 3.5 m/s∆d = 62 mv f = ?The component of gravity along the slope is mg sin 23º. Using the work energy theorem:1 2 1 2mg sin 23 ° (cos0 ° ) ∆ d + FK(cos180 ° ) ∆ d = mvf − mvi2 21 2 1 2mvf = mg sin 23 ° (1) ∆ d + FK( −1)∆ d + mvi2 22 2FK∆d2vf= 2gsin23°∆d − + vim2FK∆d2vf= 2gsin23°∆d − + vim2(72 N)(62 m)= 2(9.8 m/s )(sin 23 ° )(62 m) − + (3.5 m/s)61 kgvf= 18 m/sThe speed of the skier after travelling 62 m downhill is 18 m/s.8. m = 55.2 kg∆d = 4.18 mµ K = 0.27v i = ?2 2Copyright © 2003 Nelson Chapter 4 Work and Energy 242

Using the FBD to calculate F N ,Σ F = ma = 0FNy− mg = 0FFNN= mgy= (55.2 kg)(9.80 N/kg)= 540.96 NTo calculate F K :FFK K NK= µ F= (0.27)(540.96 N)= 146.06 NUsing the work-energy theorem:W =∆EK1 2 2FK(cos180 °∆ ) d = m( vf −vi)2Since v f = 0,1 2FK(cos180 °∆ ) d =− mvi2− 2 FK(cos180 ° ) ∆dvi=m−2(146.06 N)( −1)(4.18 m)=55.2 kgvi= 4.7 m/sThe initial speed of the skater was 4.7 m/s.Applying Inquiry Skills9. (a)Car Speed (m/s) Car Energy (J) Truck Speed (m/s) Truck Energy (J)10.0 6.0 × 10 4 10.0 2.5 × 10 720.0 2.4 × 10 5 20.0 1.0 × 10 830.0 5.4 × 10 5 30.0 2.2 × 10 840.0 9.6 × 10 5 40.0 4.0 × 10 8(b)To calculate the energy of the car and the truck, use the equationTo convert tonnes to kilograms, multiply by 1000:m = 1.2 t = 1.2 10 3 kg (for the car)m = 5.0 10 2 t = 5.0 10 5 kg (for the truck)EK1 2= mv .2Copyright © 2003 Nelson Chapter 4 Work and Energy 243

(c) A vehicle with a much larger mass has a much larger kinetic energy. As the speed of a vehicle increase, the kineticenergy increases proportional to the square of the speed.Making Connections10. (a) The kinetic energy is used to permanently deform the object as energy is transformed into heat.(b) The kinetic energy of a vehicle is transmitted (at least in part) to the occupants of the vehicle, and anything it contacts.The high kinetic energy of a fast moving vehicle is more than sufficient to damage the human body beyond its limit,causing death.4.3 GRAVITATIONAL POTENTIAL ENERGY AT EARTH’S SURFACEPRACTICE(Page 191)Understanding Concepts1. The total work done by gravity is zero. The work on the way down is positive and is equal to the work done on the wayup, which is negative. At the end, the pen has not lost or gained gravitational potential energy. Alternatively, you couldargue that because the ∆d = 0 (25 cm – 25 cm), the work done must be equal to zero.2. m = 62.5 kg∆y = 346 m (using the ground as y = 0)E g = ?∆ Eg= mg∆y= (62.5 kg)(9.80 N/kg)(346 m)5∆ E = 2.12×10 JgRelative to the ground, the gravitational potential energy is 2.12 10 5 J.3. m = 58.2 g = 5.82 10 –2 kg∆y = 1.55 m(a) ∆E g = ?At 1.55 m above the court:∆ E = mg∆yAt the court height:∆ E = mg∆ygg−2= (5.82×10 m)(9.80 N/kg)(1.55 m)∆ E = 0.884 Jg−2= (5.82×10 m)(9.80 N/kg)(0.00 m)∆ Eg= 0.00 JThe gravitational potential energy when the ball is above the court is 0.884 J, and as it strikes the court surface is0.00 J.(b) W = ?W = ( Fcos θ ) ∆d= ( mg cos θ ) ∆d−2= (5.82× 10 kg)(9.80 N/kg)(cos 0 ° )(1.55 m)W = 0.884 JAt the instant the ball strikes the court surface, the force of gravity has done 0.884 J of work on the ball.(c) The work done in (b) is equal to the change in kinetic energy of the ball.4. m = 68.5 kgm = 2.56 km = 2.56 10 3 mθ = 13.9°E g = ?Copyright © 2003 Nelson Chapter 4 Work and Energy 244

(c) A vehicle with a much larger mass has a much larger kinetic energy. As the speed of a vehicle increase, the kineticenergy increases proportional to the square of the speed.Making Connections10. (a) The kinetic energy is used to permanently deform the object as energy is transformed into heat.(b) The kinetic energy of a vehicle is transmitted (at least in part) to the occupants of the vehicle, and anything it contacts.The high kinetic energy of a fast moving vehicle is more than sufficient to damage the human body beyond its limit,causing death.4.3 GRAVITATIONAL POTENTIAL ENERGY AT EARTH’S SURFACEPRACTICE(Page 191)Understanding Concepts1. The total work done by gravity is zero. The work on the way down is positive and is equal to the work done on the wayup, which is negative. At the end, the pen has not lost or gained gravitational potential energy. Alternatively, you couldargue that because the ∆d = 0 (25 cm – 25 cm), the work done must be equal to zero.2. m = 62.5 kg∆y = 346 m (using the ground as y = 0)E g = ?∆ Eg= mg∆y= (62.5 kg)(9.80 N/kg)(346 m)5∆ E = 2.12×10 JgRelative to the ground, the gravitational potential energy is 2.12 10 5 J.3. m = 58.2 g = 5.82 10 –2 kg∆y = 1.55 m(a) ∆E g = ?At 1.55 m above the court:∆ E = mg∆yAt the court height:∆ E = mg∆ygg−2= (5.82×10 m)(9.80 N/kg)(1.55 m)∆ E = 0.884 Jg−2= (5.82×10 m)(9.80 N/kg)(0.00 m)∆ Eg= 0.00 JThe gravitational potential energy when the ball is above the court is 0.884 J, and as it strikes the court surface is0.00 J.(b) W = ?W = ( Fcos θ ) ∆d= ( mg cos θ ) ∆d−2= (5.82× 10 kg)(9.80 N/kg)(cos 0 ° )(1.55 m)W = 0.884 JAt the instant the ball strikes the court surface, the force of gravity has done 0.884 J of work on the ball.(c) The work done in (b) is equal to the change in kinetic energy of the ball.4. m = 68.5 kgm = 2.56 km = 2.56 10 3 mθ = 13.9°E g = ?Copyright © 2003 Nelson Chapter 4 Work and Energy 244

First, calculate the vertical lift:∆ysin13.9°=32.56×10 m−3∆ y = 2.56× 10 m(sin13.9 ° )∆ y = 614.98 mThen calculate the gravitational potential energy:∆ Eg= mg∆y= (68.5 kg)(9.80 N/kg)(614.98 m)5∆ E = 4.13×10 JgThe skier’s gravitational potential energy at the top of the mountain is 4.13 10 5 J.5. ∆y = 2.36 m∆E g = –1.65 10 3 Jm = ?The ground is 2.36 m down from the pole, therefore:∆ Eg= mg∆yEgm = ∆g ∆ y31.65 10 J=− ×(9.80 N/kg)( −2.36 m)m = 71.3 kgThe mass of the jumper is 71.3 kg.6. (a) The first coin does not need to be lifted, so no work is done on it. Each successive coin will need to be raised one moreycoin thickness, t, than the previous. The thickness of each coin will be t = . The work done on each coin will beNequal to its increase in gravitational potential energy.WT = W1+ W2 + +WN= mg∆ y1+ mg∆ y2 + + mg∆yN= mg( ∆ y1+∆ y2 + +∆yN)= mg(0 + t + 2 t + Nt)= mgt(0 + 1+ 2 + N)mgyWT= (1+ 2 + +N)NThe sum of an arithmetic series is:⎛t1+ tn⎞Sn= n ⎜2 ⎟⎝ ⎠For the series (1 + 2 + ··· + N):1Sn= N ⎜ ⎛ + N⎟⎞⎝ 2 ⎠Copyright © 2003 Nelson Chapter 4 Work and Energy 245

Substituting in the original equation to find the amount of work that must be done on the last coin:mgy ⎛1+N ⎞WT= N⎜ ⎟N ⎝ 2 ⎠⎛1+N ⎞WT= mgy⎜ ⎟⎝ 2 ⎠(b) The energy stored is equal to the work done, therefore:⎛1+N ⎞∆ Eg= mgy⎜ ⎟⎝ 2 ⎠Applying Inquiry Skills7. The units for gravitational potential energy:⎛ m ⎞mg∆ y = ( kg) ⎜ m2 ⎟⎝s⎠2 2= kg ⋅m /sThe units for work:F∆ d =( N)( m)⎛ m ⎞= ⎜ kg2 ⎟⎝ s ⎠( m)2 2= kg ⋅m /sThe units for kinetic energy:1 2 ⎛m⎞mv = ( kg)⎜ ⎟2 ⎝ s ⎠2 2= kg ⋅m /sTherefore, all three units are the same.2Making Connections8. ∆E g = 6.1 10 9 J(a) ∆y = ?Assume 920 students in the school. Assume an average mass of 70.0 kg per student.m = 70.0 kg 920 = 6.44 10 4 kg∆ Eg= mg∆y∆Eg∆ y =mg96.1×10 J=4(6.44×10 kg)(9.80 N/kg)4∆ y = 9.472×10 mThe energy from one barrel of oil could raise the students 9.472 10 4 m above ground level.(b) There are 158.987 L in a barrel of oil96.1×10 J7= 3.8×10 J/L158.987 LThere are 3.8 10 7 J stored in each litre of oil.Copyright © 2003 Nelson Chapter 4 Work and Energy 246

PRACTICE(Page 193)Understanding Concepts9. (a) The Sun’s radiant energy is converted to thermal energy in the water, which is then converted into gravitationalpotential energy as the water rises. The gravitational energy converts into kinetic energy as the water falls and turns theturbine. The kinetic energy of the turbine is converted into electrical energy by the generator.(b) A run-of-the-river generating station does not dam the water to create a large vertical drop in a short area, but ratheruses the natural drop of the land over a certain distance, and diverts part of the water to flow down this path.Making Connections(c) The main difference students will find is that most run-of-the-river generating stations in Canada are much smaller.(d) (i) rainfall, rivers, and glacier/mountain snow melting(ii) Bhutan relies heavily on its environment for exports from logging and energy.(e) possible points: cost, limited suitable locations, low population density, long-term effectsSection 4.3 Questions(Page 194)Understanding Concepts1. As the construction worker raises the wood, the wood’s gravitational energy increases.2. m = 63 kg∆y = 3.4 m(a) On Earth, g = 9.80 N/kg∆E g = ?∆ Eg= mg∆y= (63 kg)(9.80 N/kg)(3.4 m)3∆ E = 2.1×10 JgThe astronaut’s gravitational potential energy is 2.1 10 3 J.(b) On the Moon, g = 1.6 N/kg∆E g = ?∆ Eg= mg∆y= (63 kg)(1.6 N/kg)(3.4 m)2∆ E = 3.4×10 JgThe astronaut’s gravitational potential energy is 3.4 10 2 J.3. m = 125 g = 0.125 kg∆y = 3.50 m(a) ∆E g = ? (of the pear relative to the ground)The pear on the branch:∆ Eg= mg∆y= (0.125 kg)(9.80 N/kg)(3.50 m)∆ E = 4.29 JgThe pear at ground level:∆ Eg= mg∆y= (0.125 kg)(9.80 N/kg)(0.00 m)∆ Eg= 0.00 JThe gravitational potential energy of the pear on the branch relative to the ground is 4.29 J. The gravitational potentialenergy of the pear on the ground relative to the ground is 0.00 J.Copyright © 2003 Nelson Chapter 4 Work and Energy 247

(b) ∆E g = ? (of the pear relative to the branch)The pear on the branch:∆ Eg= mg∆y= (0.125 kg)(9.80 N/kg)(0.00 m)∆ E = 0.00 JgThe pear on the ground:∆ Eg= mg∆y= (0.125 kg)(9.80 N/kg)( −3.50 m)∆ E = −4.29 JgThe gravitational potential energy of the pear on the branch relative to the branch is 0.00 J. The gravitational potentialenergy of the pear on the ground relative to the branch is –4.29 J.4. m = 0.15 kg∆E g = 22 J∆y = ?∆ Eg= mg∆y∆Eg∆ y =mg22 J=(0.15 kg)(9.80 N/kg)∆ y = 15 mThe ball’s maximum height is 15 m above the point where it was hit.5. Let the subscript g represent gravity, and W represent the weightlifter.m = 15 kg∆y = 66 cm = 0.66(a) W g = ?Wg= ( Fcos θ ) ∆d= ( mg cos θ ) ∆d= (15 kg)( − 9.80 N/m)(cos180 ° )(0.66 m)W =−97 JgThe amount of work done by gravity on the mass is –97 J.(b) W W = ?WW= ( Fcos θ ) ∆d= ( mg cos θ ) ∆d= (15 kg)(9.80 N/kg)(cos0 ° )(0.66 m)WW= 97 JThe amount of work done by the weightlifter on the mass is 97 J.(c) ∆E g = ?∆ Eg= mg∆y= (15 kg)(9.80 N/kg)(0.66 m)∆ E = 97 JgThe gravitational potential energy of the mass increases by 97 J.Copyright © 2003 Nelson Chapter 4 Work and Energy 248

Applying Inquiry Skills6.Making Connections7. (a) V = 32.8 km 3 = 3.28 × 10 10 m 3∆y = 23.1 mρ = 1.00 10 3 kg/m 3∆E g = ?First we must determine the mass of the water:m = ρV3 3 10 3= (1.00 × 10 kg/m )(3.28 × 10 m )13m = 3.28×10 kgTo calculate the gravitational potential energy:∆ E = mg∆ygg13= (3.28×10 kg)(9.80 N/kg)(23.1 m)15∆ E = 7.43×10 JThe gravitational potential energy of the lake relative to the turbines is 7.43 10 15 J.157.43×10 J(b) This value is approximately= 6.52 times the annual energy output of the Chukha plant in Bhutan.151.14×10 J4.4 THE LAW OF CONSERVATION OF ENERGYPRACTICE(Page 197)Understanding Concepts1. The force has done negative work on the ball. The speed of the ball has decreased, which corresponds to a decrease inkinetic energy. This can only be done with negative work.2. If the losses of energy are the same, the only source of kinetic energy is the conversion of gravitational potential. Massdoesn’t matter (it cancels out in the equation), so they will both acquire the same final velocity.3. ∆y = 59.4 mv i = 0.0 m/sv f = ?Copyright © 2003 Nelson Chapter 4 Work and Energy 249

Applying Inquiry Skills6.Making Connections7. (a) V = 32.8 km 3 = 3.28 × 10 10 m 3∆y = 23.1 mρ = 1.00 10 3 kg/m 3∆E g = ?First we must determine the mass of the water:m = ρV3 3 10 3= (1.00 × 10 kg/m )(3.28 × 10 m )13m = 3.28×10 kgTo calculate the gravitational potential energy:∆ E = mg∆ygg13= (3.28×10 kg)(9.80 N/kg)(23.1 m)15∆ E = 7.43×10 JThe gravitational potential energy of the lake relative to the turbines is 7.43 10 15 J.157.43×10 J(b) This value is approximately= 6.52 times the annual energy output of the Chukha plant in Bhutan.151.14×10 J4.4 THE LAW OF CONSERVATION OF ENERGYPRACTICE(Page 197)Understanding Concepts1. The force has done negative work on the ball. The speed of the ball has decreased, which corresponds to a decrease inkinetic energy. This can only be done with negative work.2. If the losses of energy are the same, the only source of kinetic energy is the conversion of gravitational potential. Massdoesn’t matter (it cancels out in the equation), so they will both acquire the same final velocity.3. ∆y = 59.4 mv i = 0.0 m/sv f = ?Copyright © 2003 Nelson Chapter 4 Work and Energy 249

Using conservation of energy, we will have no kinetic energy at the top, and no gravitational energy at the bottom.E = ET1T2mgy /1/2v = 2gyv21 = mvff 1f= 2(9.80 m/s )(59.4 m)= 34.1 m/sTo convert to km/h:⎛ m ⎞⎛3600 s ⎞⎛ 1 km ⎞⎜34.1 ⎟⎜ ⎟⎜ ⎟ = 123 km/h⎝ s ⎠⎝ h ⎠⎝1000 m ⎠The roller coaster reaches a maximum speed of 123 km/h at the bottom of the hill.4. v i = v 1 = 9.7 m/s∆y = 4.2 mv f = v 2 = ?We will use y = 0 at the point of contact on the hill.ET1= ET21 2 1 2mv1 + mgy1 = mv2 2+ mgy222 2v1 + 2gy1 = v2 + 2gy22 2v = v + 2gy −2gySince y 2 = 0, 2gy 2 = 0, therefore:22 1 1 222=1+ 212 2v v gyv2= (9.7 m/s) + 2(9.80 m/s )(4.2 m)= 13 m/sThe skier’s speed upon touching the hillside is 13 m/s.5. ∆y = 4.4 10 2 mv 2 = 93 m/sv 1 = ?E = ET1T21 1mv mgy mv mgy2 2v gy v gy2 21+1=2+22 21 + 2 1 = 2 + 2 22 2v1 = v2 + 2gy2 −2gy12v1 = v2 + 2 g( y2 − y1)2 2= (93 m/s) + 2(9.8 m/s )(0 −440 m)v = 5.0 m/sThe speed of the water at the top of the waterfall is 5.0 m/s.6. v 1 = 9.7 m/s∆y = 4.7 mv 2 = ?1Copyright © 2003 Nelson Chapter 4 Work and Energy 250

ET1T21 1mv mgy mv mgy2 2v gy v gy2 21+1=2+22 21 + 2 1 = 2 + 2 22 2v2 = v1 + 2gy1−2gy22v2 = v1 + 2 g( y1−y2)2 2v2= EThe cyclist crests the hill at a speed of 1.4 m/s.7. v 2 = ?= (9.7 m/s) + 2(9.8 m/s )(0 −4.7 m)= 1.4 m/sFirst determine how high the pendulum is vertically raised:2 2 2h + 24.5 cm = 85.5 cm1h + h = 85.5 cm1 2112 2h = 85.5 cm −24.5 cmh = 81.915 cmh = 85.5 cm −h2 1= 85.5 cm −81.915 cmh2= 3.585 cmUsing the value h 2 = 3.585 cm, or 0.03585 m, calculate the maximum speed:E = ET1T21 1mv mgy mv mgy2 2since v = 02 21+1=2+2121=2+22gy v 2gy22 = 2 1 −22v gy gyv = 2 g( y − y )2 1 22= 2(9.80 m/s )(0.03585 m −0)v2= 0.838 m/sThe maximum speed of the pendulum bob is 0.838 m/s.Copyright © 2003 Nelson Chapter 4 Work and Energy 251

Applying Inquiry Skills8.∆y (m) ∆E g (J) E K (J) E T (J)8.00 392 0.00 3926.00 294 98.0 3924.00 196 196 3922.00 98.0 294 3920.00 0.00 392 392Making Connections9. A wrecking ball works as a pendulum that is slowly pulled back, increasing its gravitational energy. When it is released,the gravitational potential energy is converted into kinetic energy which is used to destroy buildings.PRACTICE(Page 200)Understanding Concepts10. (a) The energy supplied becomes sound and thermal energy through friction.(b) The energy supplied still produces sound and thermal energy, but some is also converted into kinetic energy.11. F K = 67 N∆d = 3.5 m(a) W = ?W = ( Fcos θ ) ∆d= (67 N)(cos180 ° )(3.5 m)2W =− 2.3×10 JThe amount of work done by friction is –2.3 10 2 J.(b) E th = ?Eth= FK∆d= (67 N)(3.5 m)2Eth= 2.3×10 JThe amount of thermal energy produced is 2.3 10 2 J.12. E th = 0.620 JF K = 0.83 N∆d = ?Copyright © 2003 Nelson Chapter 4 Work and Energy 252

The 0.620 J of energy comes from the work done by friction, therefore:Eth= FK∆dEth∆ d =FK0.620 J=0.83 N∆ d = 0.75 mThe plate slides 0.75 m.13. m = 22.0 kgF = 98 NF K = 87 Nv i = 0.0 m/s∆d = 1.2 mv f = ?W = Eth+ Ek1 2( Fcos θ ) ∆ d = FK∆ d + mv222 ( Fcos θ ) ∆d −FK∆dv2=1m2( Fcos θ ) ∆d −FK∆dv2=0.5m(98 N)(cos 0 ° )(1.2 m) −87 N(1.2 m)=0.5(22.0 kg)v2= 1.1 m/sThe speed of the cabinet after it moves 1.2 m is 1.1 m/s.14. m = 0.057 kg∆d = 25 cm = 0.25 mv f = 5.7 cm/s = 5.7 10 –2 m/sF K = 0.15 Nv i = ?EK1 = Eth + EK21 2 1 2mv1 = FK∆ d + mv22 21 2FK∆ d + mv22v21 =1m21 2FK∆ d + mv2v21 =0.5m1(0.15 N)(0.25 m) + (0.057 kg)(5.7×10 m/s)=20.5(0.057 kg)v1= 1.1 m/sThe initial speed of the pen is 1.1 m/s.−2 2Applying Inquiry Skills15. (a) The law of conservation of energy could be verified by observing how close to its original position the pendulumwould return after a swing.Copyright © 2003 Nelson Chapter 4 Work and Energy 253

(b) Some sources of error would be loss of energy due to friction at the point of attachment and air friction of the movingpendulum. Some error may be observed if the string is somewhat elastic.Making Connections16. The oil circulation system is an attempt to minimize the frictional forces between the moving parts of the engine, andthereby reduce the loss (and damage) due to the thermal energy caused by friction. The water (coolant) circulation is usedto absorb thermal energy from the engine and dissipate it rapidly into the air to prevent damage from overheating.Section 4.4 Questions(Pages 201–202)Understanding Concepts1. Roller coasters are gravity rides that have an initial input of gravitational potential energy that is converted to kinetic (andback into gravitational) energy throughout the ride. To give them this initial energy, they must be pulled up the largest hillat the beginning.2. m = 0.052 kg∆y = 11 cm = 0.11 my = 0(a) ∆E g = ?∆ E = mg∆yg= (0.052 kg)(9.80 m/s )(0.11 m)∆ E = 0.056 JgThe initial gravitational potential energy of the egg’s contents is 0.056 J.(b) ∆E g = ?∆ E = mg∆yg= (0.052 kg)(9.80 m/s )(0.0 m)∆ E = 0.0 JgThe final gravitational potential energy of the egg’s contents is 0.0 J.(c) ∆E g = 0.0 – 0.056 = –0.056 JThe change in gravitational potential energy as the egg’s contents fall is –0.056 J.(d) E K = ?v = ?The kinetic energy will be equal to the loss of gravitational potential, so E K = 0.056 J.1 2EK= mv22EKv =m2(0.056 J)=0.052 kgv = 1.5 m/sThe kinetic energy is 0.056 J. The speed of the egg’s contents just before hitting the pan is 1.5 m/s.3. ∆y = 1.2 mv f = 9.9 m/sv i = ?22Copyright © 2003 Nelson Chapter 4 Work and Energy 254

ET1T21 1mv mgy mv mgy2 2v gy v gy2 21+1=2+22 21 + 2 1 = 2 + 2 22 2v1 = v2 + 2gy2 −2gy12v1 = v2 + 2 g( y2 − y1)2 21= EThe initial speed of the ball was 11 m/s.4. v 1 = 3.74 m/s∆y = 8.74 mm = 7.12 10 4 kg(a) ∆E g = ?∆ E = mg∆ygg= (9.9 m/s) + 2(9.8 m/s )(1.2 m −0 m)v = 11 m/s4 2= (7.12×10 kg)(9.80 m/s )(8.74 m)6∆ E = 6.10×10 JThe gravitational potential energy of the mass of water at the top of the waterfall is 6.10 10 6 J.(b) v 2 = ?E = ET1T21 1mv mgy mv mgy2 2v gy v gy2 21 + 1 = 2 + 22 21 + 2 1 = 2 + 2 22 2v2 = v1 + 2gy1 −2gy22v2 = v1 + 2 g( y1−y2)2 2= (3.74 m/s) + 2(9.80 m/s )(8.74 m −0 m)v2= 13.6 m/sThe speed of the water at the bottom of the waterfall is 13.6 m/s.5. (a)First, determine the vertical height above the bottom of the swing:h1cos 48°=3.7 mh = 3.7 m cos 48°1h = 2.476 m1( )Copyright © 2003 Nelson Chapter 4 Work and Energy 255

h1+ h2= 3.7 mh2 = 3.7 m −h1= 3.7 m −2.476 mh = 1.224 mNow we can calculate the acrobat’s speed:ET1= ET21 2 1 2mv + mgy = mv + mgy2 2Since v 1 = 0:21 1 2 221 = 2 + 222 = 2 1−222gy v 2gyv gy gyv = 2 g( y − y )v2 1 222= 2(9.80 m/s )(1.224 m −0 m)= 4.9 m/sThe acrobat’s speed at the bottom of the swing is 4.9 m/s.(b) Due to conservation of energy, the maximum height on the other side is equal to the starting height of the acrobat.6. m = 55 kg∆d = 3.7 mF K = 41.5 Nv 1 = 65.7 cm/s = 0.657 m/sv 2 = 7.19 m/sφ = ?Relating φ to h:hsinφ=11.7h = 11.7sinφUsing conservation of energy:T1T2K1 g K2 th1 2 1 2mv1 + mgh = mv2 + FK∆d2 21 1mgh = mv + F ∆d − mv2 21 (2 2mv2 − v1 ) + FK∆dh =2mgsinφ= 0.30054φ = 17.5°The angle is 17.5°.E= EE +∆ E = E + E2 22 K 12 2(( ) ( ) )1 (55.0 kg) 7.19 m/s − 0.657 m/s + (41.5 N)(11.7 m)11.7sinφ= 22(55.0 kg)(9.80 m/s )Copyright © 2003 Nelson Chapter 4 Work and Energy 256

7. v 1 = 0.0 m/sv 2 = 6.8 m/sSince the skateboarder starts even with the height of the centre of the circle, the total vertical drop at the bottom will beequal to the radius of the circle.r = y 1 = ?ET1= ET21 1mv mgy mv mgy2 22 2v1 + 2gy1 = v2 + 2gy2Since v 1 = 0, and y 2 = 0:22gy1 = v22v2y1=2g26.8 m/s=22(9.8 m/s )y = 2.4 mThe radius of the half-pipe is 2.4 m.8. m = 55 g = 5.5 10 –2 kgv 1 = 1.9 m/s∆d = 54 cm = 0.54 mv 2 = 0.0 m/s(a) µ K = ?First, solve for F K :ET1= ET2E = E12 21+1=2+2KKth1 2mv = FK∆d22mvFK=2 ∆ d×=F−2 2(5.5 10 kg)(1.9 m/s)2(0.54 m)= 0.1838 NNow calculate the normal force:Σ Fy= may= 0FN− mg = 0FN= mg−2 2= (5.5×10 kg)(9.80 m/s )F = 0.539 NSolve for µ K :FKµ K =FN0.1838 N=0.539 Nµ K = 0.34The coefficient of kinetic friction is 0.34.NCopyright © 2003 Nelson Chapter 4 Work and Energy 257

(b) µ K =?First, solve for the acceleration:2 2v = v + 2a∆dfi2 2f − viva =2∆d=( 1.9 m/s) − ( 0 m/s)2 22(0.54 m)a = 3.3426 m/s2Using the FBD, calculate the magnitude of kinetic friction:Σ Fx= maxFK= max−2 2= (5.5×10 kg)(3.3426 m/s )F = 0.1838 NKCalculate the normal force:Σ Fy= may= 0FN− mg = 0FN= mg−2 2= (5.5×10 kg)(9.80 m/s )F = 0.539 NCalculate the coefficient of kinetic friction:FKµ K =F0.1838 N=0.539 Nµ K = 0.34The coefficient of kinetic friction is 0.34.(c) The kinetic energy is converted into thermal energy.9. v i = 85 km/h = 23.61 m/sv f = 0.0 m/s∆d = 47 mF K = 7.4 10 3 N(a) E th = ?Eth= FK∆d3= (7.4×10 N)(47 m)5Eth= 3.5×10 JThe amount of thermal energy produced is 3.5 10 5 J.(b) Before the skid, the thermal energy was kinetic energy.NNCopyright © 2003 Nelson Chapter 4 Work and Energy 258

(c) m = ?ET1= ET2EK= Eth1 2mv = FK∆d22FK∆dm =2v32(7.4×10 N)(47 m)=( 23.61 m/s)m = 1.2×10 kgThe mass of the car is 1.2 10 3 kg.(d) µ K = ?First calculate the normal force:Σ Fy= may= 0FN− mg = 0FN= mg3 2= (1.248×10 kg)(9.80 m/s )F = 12 228To calculate the coefficient of kinetic friction:FKµK=FN37.4×10 J=12228 Jµ K = 0.61The coefficient of kinetic friction is 0.61.10. m = 22 kg∆d = 2.5 mθ = 44°F K = 79 N(a) W = ?W = ( FKcos θ ) ∆d= (79 N)(cos180 ° )(2.5 m)2W =− 2.0×10 JThe work done by friction is –2.0 10 2 J.(b) E K = ?N32Copyright © 2003 Nelson Chapter 4 Work and Energy 259

First, determine the vertical drop:hsin 44°=2.5h = 2.5sin 44°h = 1.737 mTo calculate the final kinetic energy:ET1= ET2∆ Eg = EK + Ethmg∆ y = EK+ FK∆dEK= mg∆y−FK∆d= (9.8 N/kg)(22 kg)(1.737 m) −(79 N)(2.5 m)2EK= 1.8×10 JThe box’s final kinetic energy is 1.8 10 2 J.(c) E th = ?Eth= FK∆d= (79 N)(2.5 m)2Eth= 2.0×10 JThe thermal energy produced is 2.0 10 2 J.Applying Inquiry Skills11. (a) The loss of gravitational potential energy from the first lift to the second is equal to the amount of thermal energyproduced.(b) The equation needed would be:ET1= ET2∆ Eg1 = ∆ Eg2 + Ethmgy1 = mgy2 + EthEth = mgy1 −mgy2Eth = mg( y1 − y2)(c) You could determine the height at each point by standing a known distance from the base of the ride and measure theangle up for each height. Using simple trigonometric ratios, you could estimate the amount of thermal energyproduced.Making Connections12. A trebuchet is very accurate. The raised mass supplies a fixed amount of gravitational potential energy. This potentialenergy is converted into the kinetic energy of the projectile.4.5 ELASTIC POTENTIAL ENERGY AND SIMPLE HARMONIC MOTIONPRACTICE(Pages 206–207)Understanding Concepts1. The higher the spring constant, the more force is needed to stretch the spring the same amount. For this reason, spring Awould be more difficult to stretch than spring B.2. The spring would exert a southward force on you.3. k = 25 N/m(a) x = 16 cm = 0.16 mFx= kx= (25 N/m)(0.16 m)F = 4.0 NxCopyright © 2003 Nelson Chapter 4 Work and Energy 260

First, determine the vertical drop:hsin 44°=2.5h = 2.5sin 44°h = 1.737 mTo calculate the final kinetic energy:ET1= ET2∆ Eg = EK + Ethmg∆ y = EK+ FK∆dEK= mg∆y−FK∆d= (9.8 N/kg)(22 kg)(1.737 m) −(79 N)(2.5 m)2EK= 1.8×10 JThe box’s final kinetic energy is 1.8 10 2 J.(c) E th = ?Eth= FK∆d= (79 N)(2.5 m)2Eth= 2.0×10 JThe thermal energy produced is 2.0 10 2 J.Applying Inquiry Skills11. (a) The loss of gravitational potential energy from the first lift to the second is equal to the amount of thermal energyproduced.(b) The equation needed would be:ET1= ET2∆ Eg1 = ∆ Eg2 + Ethmgy1 = mgy2 + EthEth = mgy1 −mgy2Eth = mg( y1 − y2)(c) You could determine the height at each point by standing a known distance from the base of the ride and measure theangle up for each height. Using simple trigonometric ratios, you could estimate the amount of thermal energyproduced.Making Connections12. A trebuchet is very accurate. The raised mass supplies a fixed amount of gravitational potential energy. This potentialenergy is converted into the kinetic energy of the projectile.4.5 ELASTIC POTENTIAL ENERGY AND SIMPLE HARMONIC MOTIONPRACTICE(Pages 206–207)Understanding Concepts1. The higher the spring constant, the more force is needed to stretch the spring the same amount. For this reason, spring Awould be more difficult to stretch than spring B.2. The spring would exert a southward force on you.3. k = 25 N/m(a) x = 16 cm = 0.16 mFx= kx= (25 N/m)(0.16 m)F = 4.0 NxCopyright © 2003 Nelson Chapter 4 Work and Energy 260

x = 32 cm = 0.32 mFx= kx= (25 n/m)(0.32 m)Fx= 8.0 NThe magnitude of force would be 4.0 N for a stretch of 16 cm, and 8.0 N for a stretch of 32 cm.(b) x = 16 cm = 0.16mFx=−kx=−(25 N/m)(0.16 m)Fx=−4.0 Nx = 32 cm = 0.32 mFx=−kx=−(25 N/m)(0.32 m)Fx=−8.0 NThe magnitudes of the forces are 4.0 N and 8.0 N, respectively.4. k = 3.2 10 2 N/mx = 2.0 cm = 2.0 10 –2 mFx=−kx2 −2=− (3.2× 10 N/m)(2.0×10 m)Fx=−6.4 NThe magnitude of the force applied by the air is 6.4 N.5. m = 1.37 kgk = 5.20 10 2 N/m(a) x = ?Σ Fy= ma = 0Fx− mg = 0kx = mgmgx =k(1.37 kg)(9.80 N/kg)=25.20×10 N/mx = 0.0258 mThe spring stretches 0.0258 m.(b) x = 1.59 cm = 0.0159 mΣ F y= ?Σ Fy= Fx−mg= kx −mgΣ Fy2= (5.20× 10 N/m)(0.0159 m) −(1.37 kg)(9.80 N/kg)= −5.16 NThe net force on the fish is 5.16 N [down].(c) x = 2.05 cm = 0.0205 ma = ?Σ Fy= maFx− mg = makx − mga =m2(5.20× 10 N/m)(0.0205 m) −(1.37 kg)(9.80 N/kg)=1.37 kg2a =−2.02 m/sThe acceleration of the fish is 2.02 m/s 2 [down]Copyright © 2003 Nelson Chapter 4 Work and Energy 261

Applying Inquiry Skills6. (a)(b) The slope of the line is negative.Making Connections7. (a)Mass (kg) Stretch (m)1.00 0.1222.00 0.2453.00 0.3684.00 0.4905.00 0.6126.00 0.7357.00 0.8588.00 0.980(b)(c) The mass value may not be correct if the value of g is different then where it was calibrated. The weight value wouldbe accurate anywhere (even on the Moon).Copyright © 2003 Nelson Chapter 4 Work and Energy 262

PRACTICE(Page 211)Understanding Concepts8. (a) The graph shown is the force applied by the spring. Since the spring is being stretched to the right (positive x), the forceexerted by the spring will be to the left (negative). This is what is shown on the graph.(b) The force constant is the slope of the graph. Since the equation F x = kx relates the force exerted by the spring, we mustchange the direction (i.e., the sign) of the force,−F k = xx−− ( 15 N)=0.40 mk = 38 N/mThe force constant of the spring is 38 N/m.(c) x = 35 cm = 0.35 mThe energy stored is the area between the curve and the x-axis.1Ee= A=bh21= (0.35 m)(13 N)2E = 2.3 JeThe elastic potential energy is 2.3 J.9. k = 9.0 10 3 N/m(a) x = 1.0 cm = 0.010 mE e = ?1 2Ee= kx21= (9.0 × 10 N/m)(0.010 m)2E = 0.45 Je3 2The elastic potential energy stored by the spring is 0.45 J.(b) x = –2.0 cm = –0.020 mE e = ?1 2Ee= kx21 (9.0 103 N/m)( 0.020 m)2= × −2E = 1.8 JeThe elastic potential energy stored in the spring is 1.8 J.10. m = 7.8 g = 0.0078 kgk = 3.5 10 2 N/mx = –4.5 cm = –0.045 m(a) E e = ?1 2Ee= kx21 (3.5 102 N/m)( 0.045 kg)2= × −2E = 0.35 JeThe elastic potential energy of the spring is 0.35 J.Copyright © 2003 Nelson Chapter 4 Work and Energy 263

(b) v = ?ET1= ET2Ee= EK1 2 1 2kx = mv2 22kxv =m2 2(3.5× 10 N/m)( −0.045 m)=0.0078 kgv = 9.5 m/sThe speed of the dart as it leaves the toy is 9.5 m/s.11. m = 3.5 10 –3 kgk = 9.5 N/m(a) ∆y = 5.7 cm = 0.057 mx = ?ET1= ET2Ee=∆Eg1 2kx = mg∆y22mg∆yx =k−32(3.5×10 kg)(9.8 N/kg)(0.057 m)=9.5 N/mx = 0.020 mThe spring must be compressed by 0.020 m, or 2.0 cm.(b) If friction was not negligible, the amount of compression would need to be increased. The energy supplied by thecompressed spring would now become both gravitational potential and thermal energy. In order to have the same finalgravitational energy, the spring would need to be compressed more.12. m = 0.20 kgk = 55 N/m(a) ∆y = 1.5 cm = 0.015 mv = ?ET1= ET2∆ E = E + Eg K e1 1mg∆ y = mv + kx2 22 2mv = 2mg∆y −kxv =2 22mg∆y−kxm2(0.20 kg)(9.8 N/kg)(0.015 m) − (55 N/m)(0.015 m)=0.20 kgv = 0.48 m/sThe speed of the mass is 0.48 m/s.22Copyright © 2003 Nelson Chapter 4 Work and Energy 264

(b) For the fall, ∆y = x. At maximum fall, there will be no kinetic energy.ET1= ET2∆ Eg= Ee1 2mg∆ y = kx22kx −2mg∆ y = 02( 55)x − 2(0.20)(9.8) x = 02( 55) x − ( 3.92)x = 0x(55x− 3.92) = 055x− 3.92 = 0 or x = 03.92x =55x = 0.071m or x = 0 mThe value x = 0 refers to the moment of release, so the maximum stretch will be 0.071 m.13. k = 12 N/m∆y = 93.0 cm = 0.93 mm = 8.3 10 –3 kgx = 4.0 cm = 0.040 m∆d = ?Analyze as a projectile motion question. First, determine horizontal speed at launch:ET1= ET2Ee= EK1 2 1 2kx = mv2 22kxv =m2(12 N/m)( −0.040 m)=−38.3×10 kgv = 1.521 m/sChoosing down as positive, the time for the marble to drop 0.93 m vertically is:1 2∆ y = vi∆ t+ a ( ∆ t )2Since v i = 0:1∆ y = a ( ∆t)22∆y∆ t =a2(0.93 m)=29.8 m/s∆ t = 0.4357 sThe horizontal distance travelled during this time is:∆ d = v∆t= (1.521 m/s)(0.4357 s)∆ d = 0.66 mThe marble travels 0.66 m horizontally before hitting the floor.2Copyright © 2003 Nelson Chapter 4 Work and Energy 265

Applying Inquiry Skills14. (a) The measurement needed would be the mass of the largest friend.(b) Assuming a largest mass of 115 kg,Σ Fy= ma = 0Fx− mg = 0kx = mgmgk =x(115 kg)(9.80 N/kg)=0.75 m3k = 1.5×10 N/mThe approximate force constant is 1.5 10 3 N/m.Making Connections15. m = 2.0 10 2 µg = 2.0 10 –7 kg∆y = 65 mm = 0.065 mx = 75 cm = 0.75 mE e = ?ET1= ET2Ee= x( ∆Eg)= xmg ( ∆y)−7= 0.75 m(2.0×10 kg)(9.8 N/kg)(0.065 m)−8Ee= 9.6×10 JThe initial quantity of elastic potential energy is 9.6 10 –8 J.PRACTICE(Pages 214–215)Understanding Concepts16. (a) The maximum displacement from the rest position will be at the top and bottom of the bounce.(b) The speed is a maximum at the rest position.(c) The speed will be zero at the top and the bottom of the bounce.(d) The acceleration will be a maximum at the top and the bottom of the bounce.(e) The acceleration will be zero at the rest position.17. T = ?f = ?(a) number of vibrations = 12t = 48 stotal timeT =number of complete vibrations48=12T = 4.0 sffnumber of complete vibrations=total time12=48= 0.25 HzCopyright © 2003 Nelson Chapter 4 Work and Energy 266

Alternatively, you could calculate frequency by using the equation:1f =T1=4.0 sf = 0.25 HzThe period is 4.0 s, and the frequency is 0.25 Hz.(b) number of vibrations = 210t = 1 min = 60 stotal timeT =number of complete vibrations60=210T = 0.29 snumber of complete vibrationsf =total time210=60f = 3.5 HzThe period is 0.29 s, and the frequency is 3.5 Hz.(c) number of vibrations = 2200t = 5.0 stotal timeT =number of complete vibrations5.0=2200−3T = 2.3×10 snumber of complete vibrationsf =total time2200=5.02f = 4.4×10 HzThe period is 2.3 10 –3 s, and the frequency is 4.4 10 2 Hz.18. m = 0.25 kgA = 8.5 cmk = 1.4 10 2 N/m(a) d = ?During each cycle, the mass moves 4 amplitudes, or 4A. In 5 cycles, the mass will move:d = 5×4A= 5×4(8.5 cm)2d = 1.7×10 cmThe mass moves 1.7 10 2 cm in the first five cycles.(b) T = ?T = 2πmk0.25 kg= 2π21.4×10 N/mT = 0.27 sThe period of vibration is 0.27 s.Copyright © 2003 Nelson Chapter 4 Work and Energy 267

19. m = 0.10 kgf = 2.5 Hzk = ?1 kf =2πmk2πf =m2 2k = 4πf m2 2= 4 π (2.5 Hz) (0.10 kg)k = 25 N/mThe force constant of the spring is 25 N/m.20. k = 1.4 10 2 N/mT = 0.85 sm = ?T = 2πmkT m=2πk2T km =24π2 2(0.85 s) (1.4×10 N/m)=24πm = 2.6 kgThe mass would have to be 2.6 kg.Applying Inquiry Skills21. Examining the base SI units for each:xaxam=⎛ m ⎞⎜ 2 ⎟⎝s⎠2= s= sTherefore, they are dimensionally equivalent.mkmkkg=⎛ N ⎞⎜ ⎟⎝m⎠kg ⋅m=⎛kg ⋅m⎞⎜ 2 ⎟⎝ s ⎠== ss2Making Connections22. number of vibrations = 6.0t = 8.0 s(a) k = ?First we must calculate the frequency:number of complete vibrationsf =total time6.0=8.0 sf = 0.75 HzCopyright © 2003 Nelson Chapter 4 Work and Energy 268

Using a mass of 75 kg:1 kf =2πmk2πf =m2 2k = 4πf m2 2= 4 π (0.75 Hz) (75 kg)3k = 1.7×10 N/mThe force constant is 1.7 10 3 N/m.(b) No, you are not undergoing SHM. When you leave the trampoline, there is a period of time when it is not exerting anyforce on you. SHM requires that the force be proportional to the displacement.PRACTICE(Pages 217–218)Understanding Concepts23. (a) The speed is zero at lengths of 12 cm and 38 cm.(b) The maximum speed will be at the rest position. The rest position will be halfway between the minimum and maximumextensions:12 + 38= 25 cm2(c) The amplitude is 38 – 25 = 13 cm.24. E e = 5.64 Jm = 0.128 kgk = 244 N/m(a) x = ?The maximum energy is constant, and all elastic potential at either end of the system, at the maximum amplitude.1 2Ee= kx22Eex =k2(5.64 J)=244 N/mx = 0.215 mThe amplitude of the vibration is 0.215 m.(b) v = ?Approach 1: All of the energy will be kinetic as it passes through the rest position.2EK= mv2EKv =m2(5.64 J)=0.128 kgv = 9.39 m/sCopyright © 2003 Nelson Chapter 4 Work and Energy 269

Approach 2: All of the energy stored in the spring at maximum compression will be converted to kinetic energy.E = ET1T21 1kx = mv2 22kxv =m2 2(244 N/m)(0.215 m)=0.128 kgv = 9.39 m/sThe speed is 9.39 m/s regardless of the approach used.(c) x 2 = 15.5 cm = 0.155 mv = ?E = ET1T21 1 1kx mv kx2 2 22 2 2max = + 22 22 kxmax − kx22 2max− x2kx ( )v = 6.51 m/sThe speed of the mass is 6.51 m/s.25. x = 0.18 mm = 58 g = 0.058 kgk = 36 N/m(a) E e = ?v = ?v=v ==mm22 2(( ) − ( ) )(244 N/m) 0.215 m 0.155 m0.128 kgMaximum energy during maximum stretch/compression of the spring:1 2Emax= kx21 (36 N/m)(0.18 m)2=2E = 0.58 JmaxThis will all be kinetic energy at the rest position.1 2EK= mv22EKv =m2(0.58 J)=0.058 kgv = 4.5 m/sThe maximum energy of the system is 0.58 J. The maximum speed of the mass is 4.5 m/s.Copyright © 2003 Nelson Chapter 4 Work and Energy 270

(b) x = ?Double the energy would be:E′ max= 2×Ee1 2= 2×kx 2= (36 N/m)(0.18 m)E′ = 1.1664 JmaxThis is all stored as elastic potential energy at the full amplitude.1 2Ee= kx22Eex =k2(1.1664 J)=36 N/mx = 0.25 mThe amplitude of vibration required would be 0.25 m.(c) v = ?1 2EK= mv22EKv =m2(1.1664 J)=0.058 kgv = 6.3 m/sThe maximum speed of the mass is 6.3 m/s.26. For the maximum speed, all of the elastic energy will be converted to kinetic:E = ET1T21 1kx = mv2 2v =2kxmv = xkm2 22For a SHM system, x = A, therefore:f2πf1=2πk=mkmSubstituting above:kv = x m= A(2 π f)v = 2πfACopyright © 2003 Nelson Chapter 4 Work and Energy 271

Applying Inquiry Skills27. (a)⎛ N ⎞k⎜ ⎟2 2 m (A x ) (m2 m2− =⎝ ⎠− )mkg⎛kg ⋅m⎞⎜ 2 ⎟ms ⋅ 2=⎝ ⎠(m )kg2m=2s= m/sThe dimensions are m/s.(b) This expression is to calculate the speed of an object at a location during SHM.Making Connections28. (a) Tuning fork prongs have slow damping to produce a long tone.(b) The voltmeter needle has fast damping to stabilize the reading quickly.(c) A guitar string has slow damping to play the note as long as possible.(d) Saloon doors have medium damping to keep the doors from swinging too much, but they still swing back and forth.(e) The string of the bow has fast damping to prevent dangerous vibration.Section 4.5 Questions(Pages 218–219)Understanding Concepts1. When the two students pull on either end of the spring, it will not stretch as much as when it is pulled by both studentswhile attached to the wall. When they both pull on it from the same side, the wall pulls back with equal force. When theypull from opposite ends, the force will be half as much, and the stretch will also be half as much.2. The elastic potential energy is the same when the spring is stretched or compressed 2.0 cm. The amount of energy storedonly depends on the magnitude of the distortion, not the direction.3. Harmonic means that it is regularly repeated, symmetrical motion.14. (a) Period is inversely proportional to the frequency, T ∝ .f(b) The acceleration is directly proportional to the displacement, a∝ x.1(c) The period is inversely proportional to square root of the force constant, T ∝ .k(d) The maximum speed is directly proportional to the amplitude, v∝ A.5. m = 62 kgk = 2.4 10 3 N/mx = ?Σ Fy= ma = 06Fx− mg = 06kx= mgmgx =6k(62 kg)(9.8 N/kg)=36(2.4×10 N/m)x = 0.042 mThe compression of each spring is 0.042 m.Copyright © 2003 Nelson Chapter 4 Work and Energy 272

6. k = 78 N/mx = 2.3 cm = 0.023 mF x = ?The magnitude of force is 1.8 N.7. x 1 = 1.85 cmF x1 = 85.5 Nx 2 = 4.95 cm = 0.0495 mF x2 = ?First we must calculate k:F = kxFk =Solve for x 2 :Fx= kx= (78 N/m)(0.023 m)F = 1.8 NFFxx1 1=k =x1x185.5 N0.0185 m4621.6 N/mx2 2x2The force required is 229 N.8. m = 97 kgk = 2.2 103 N/ma = 0.45 m/s 2x = ?= kx= (4621.6 N/m)(0.0495 m)= 229 NIgnoring friction to calculate the applied force:Σ Fx= maFA= ma2= (97 kg)(0.45 m/s )F = 43.65 NThis force is the force exerted on the spring:Fx= kxFxx =k43.65 N=32.2×10 N/mx = 0.020 mThe spring stretches 0.020 m, or 2.0 10 –2 m.9. m = 289 g = 0.289 kgk = 18.7 N/m(a) x = 10.0 cm = 0.100 mΣ F y= ?a = ?ACopyright © 2003 Nelson Chapter 4 Work and Energy 273

To calculate the net force:Σ Fy= Fx−mg= kx −mg= (18.7 N/m)(0.100 m) −(0.289 kg)(9.80 N/kg)Σ F = −0.962 NyTo calculate acceleration:Σ Fy= maΣFya =m−0.962 N=0.289 kg2a =−3.33 m/sThe net force is 0.962 N [down], and the acceleration is 3.33 m/s 2 [down].(b) x = ?Σ Fy= ma = 0Fx− mg = 0kx = mgmgx =k(0.289 kg)(9.80 N/kg)=18.7 N/mx = 0.151 mThe spring will be stretched by 0.151 m.10. m = 64.5 kg∆y = 48.0 m –12.5 m = 35.5 mk = 65.5 N/mx = 35.5 m – 10.1 m = 25.4 m.v = ?∆ E = E + Eg e K1 1mg∆ y = kx + mv2 22 2mv = 2mg∆y −kxv =2 22mg∆y−kxm2(64.5 kg)(9.80 N/kg)(35.5 m) − (65.5 N/m)(25.4 m)=64.5 kgv = 6.37 m/sThe jumper’s speed at a height of 12.5 m above the water is 6.37 m/s.11. F x = 8.6 Nx = 9.4 cm = 0.094 m(a) k = ?Fx= kxFxk =x8.6 N=0.094 mk = 91 N/mThe force constant of the spring is 91 N/m.22Copyright © 2003 Nelson Chapter 4 Work and Energy 274

(b) E e = ?1 2Ee= kx21 (91.49 N/m)(0.094 m)2=2Ee= 0.40 JThe maximum energy of the spring is 0.40 J.12. x = 1.0 10 –7 mE e = 1.0 10–13 Jk = ?1 2Ee= kx22Eek =2x−132(1.0 × 10 J)=−7 2(1.0 × 10 m)1k = 2.0×10 N/mThe force constant is 2.0 10 1 N/m.13. m = 22 kgθ = 29°k = 8.9 10 2 N/mx = 0.30 md = ?To calculate the total vertical drop, ∆y:ET1= ET2∆ E = Ege1 2mg∆ y = kx22kx∆ y =2mg(8.9×10 N/m)(0.30 m)=2(22 kg)(9.8 N/kg)∆ y = 0.1858 mTo calculate the distance:∆ysinθ=d∆yd =sinθ0.1858 m=sin 29°d = 0.38 mThe crate slides 0.38 m along the ramp.14. m = 0.20 kgk = 28 N/m∆y = ?2 2Copyright © 2003 Nelson Chapter 4 Work and Energy 275

For fall, ∆y = x. At maximum fall, there will be no kinetic energy.ET1= ET2∆ Eg= Ee1 2mg∆ y = kx22kx −2mg∆ y = 0228x− 2(0.20)(9.8) x = 0228x− 3.92x= 0x(28x− 3.92) = 028x− 3.92 = 0 or x = 03.92x =28x = 0.14 m or x = 0Since x = 0 refers to the moment of release, the maximum stretch will be 0.14 m.Applying Inquiry Skills15. (a) (i) Line A (total energy is constant)(ii) Line B (no kinetic energy at maximum stretch)(iii) Line C (maximum elastic potential energy a maximum stretch)(b) By observation from the graph, A = 10.0 cm = 0.100 m(c) At the end, the total energy of 5.0 J is all elastic potential energy,1 2Ee= kx22Eek =2x2(5.0 J)=2(0.100 m)3k = 1.0×10 N/mThe force constant of the spring is 1.0 10 3 N/m.(d) Maximum kinetic energy is 5.0 J when there is no elastic potential energy.1 2EK= mv22EKv =m2(5.0 J)=0.12 mv = 9.1 m/sThe maximum speed of the mass is 9.1 m/s.16. (a) If the spring were cut in two, it would take more force to stretch the spring the same amount because each coil wouldneed to be moved twice as far. This will cause the force constant to double for the remaining two half springs. If a masshung from two identical springs attached together caused them to stretch 36 cm, than each of them would stretch halfof that amount. Since each bears the same weight, the force constant of each would only allow them to stretch 18 cmunder the same force, indicating a force constant that is twice as great.(b) Answers will vary.Making Connections17. m = 5.5 10 2 kgnumber of vibrations = 6.0t = 3.5 sk = ?Copyright © 2003 Nelson Chapter 4 Work and Energy 276

Calculate the frequency:number of complete vibrationsf =total time6.0=3.5 sf = 1.7143 HzTo calculate the force constant:1 kf =2πmk2πf =m2 2k = 4πf m2 2 2= 4 π (1.7143 Hz) (5.5×10 kg)4k = 6.4×10 N/mThe force constant of each spring is 6.4 10 4 N/m.18. a = 25gf = 8.9 HzA = ?AT = 2πa1 A= 2πf a1 A=2πf aaA =2 24πf225(9.8 m/s )=2 24 π (8.9 Hz)A = 0.078 mThe minimum amplitude is 0.078 m, or 7.8 10 –2 m.CHAPTER 4 LAB ACTIVITIESActivity 4.4.1: Applying the Law of Conservation of Energy(Page 220)(a) Energy conversions would be gravitational potential to kinetic to move the hands.(b) Different types include water clocks, hour-glass clocks, and pendulum clocks.(c) – (f) Answers will vary based on the students’ choice of design.Investigation 4.5.1: Testing Real Springs(Pages 220–221)Questions(i) By graphing the stretch as a function of the applied force we can learn the relationship is linear.(ii) The force constant of two combined springs is always less than the force constant for either spring individually.Copyright © 2003 Nelson Chapter 4 Work and Energy 277

Calculate the frequency:number of complete vibrationsf =total time6.0=3.5 sf = 1.7143 HzTo calculate the force constant:1 kf =2πmk2πf =m2 2k = 4πf m2 2 2= 4 π (1.7143 Hz) (5.5×10 kg)4k = 6.4×10 N/mThe force constant of each spring is 6.4 10 4 N/m.18. a = 25gf = 8.9 HzA = ?AT = 2πa1 A= 2πf a1 A=2πf aaA =2 24πf225(9.8 m/s )=2 24 π (8.9 Hz)A = 0.078 mThe minimum amplitude is 0.078 m, or 7.8 10 –2 m.CHAPTER 4 LAB ACTIVITIESActivity 4.4.1: Applying the Law of Conservation of Energy(Page 220)(a) Energy conversions would be gravitational potential to kinetic to move the hands.(b) Different types include water clocks, hour-glass clocks, and pendulum clocks.(c) – (f) Answers will vary based on the students’ choice of design.Investigation 4.5.1: Testing Real Springs(Pages 220–221)Questions(i) By graphing the stretch as a function of the applied force we can learn the relationship is linear.(ii) The force constant of two combined springs is always less than the force constant for either spring individually.Copyright © 2003 Nelson Chapter 4 Work and Energy 277

Calculate the frequency:number of complete vibrationsf =total time6.0=3.5 sf = 1.7143 HzTo calculate the force constant:1 kf =2πmk2πf =m2 2k = 4πf m2 2 2= 4 π (1.7143 Hz) (5.5×10 kg)4k = 6.4×10 N/mThe force constant of each spring is 6.4 10 4 N/m.18. a = 25gf = 8.9 HzA = ?AT = 2πa1 A= 2πf a1 A=2πf aaA =2 24πf225(9.8 m/s )=2 24 π (8.9 Hz)A = 0.078 mThe minimum amplitude is 0.078 m, or 7.8 10 –2 m.CHAPTER 4 LAB ACTIVITIESActivity 4.4.1: Applying the Law of Conservation of Energy(Page 220)(a) Energy conversions would be gravitational potential to kinetic to move the hands.(b) Different types include water clocks, hour-glass clocks, and pendulum clocks.(c) – (f) Answers will vary based on the students’ choice of design.Investigation 4.5.1: Testing Real Springs(Pages 220–221)Questions(i) By graphing the stretch as a function of the applied force we can learn the relationship is linear.(ii) The force constant of two combined springs is always less than the force constant for either spring individually.Copyright © 2003 Nelson Chapter 4 Work and Energy 277

Hypothesis(a) The graph of F x versus x should be a straight line with a positive slope.kk 1 2(b) The combined force constant will be .k + kPrediction(c)1 2(d)ktotalkk 1 2=k + k1 2Procedure1 – 3. See sample data table below.Mass(g)Force(N)StretchSpring 1(cm)StretchSpring 2(cm)StretchSpring 3(cm)StretchSpring 1 + Spring 2(cm)0 0 0 0 0 050 0.49 1.10 0.72 0.56 1.82100 0.98 2.22 1.43 1.11 3.63150 1.47 3.29 2.14 1.66 5.44200 1.96 4.43 2.86 2.22 7.26Analysis(e)(f)Spring 1riseslope = run1.47 N − 0.49 N=0.0329 m − 0.0110 mslope = 44.5 N/mSpring 2riseslope = run1.47 N − 0.49 N=0.0214 m − 0.0072 mslope = 69.0 N/mSpring 3riseslope =run1.47 N − 0.49 N=0.0166 m − 0.0056 mslope = 89.1 N/mThese slopes represent the force constant for the spring.Copyright © 2003 Nelson Chapter 4 Work and Energy 278

(g) The extrapolation is not feasible because the spring is not capable of supporting that much mass.(h) When two springs are connected linearly, each spring must support the full weight (if we ignore the mass of the lowerspring), so the total stretch will be equal to the sum of the individual stretches, (i.e., x total = x 1 + x 2 ).F = ktotalxtotal= ktotal ( x1 + x2)⎛ F F ⎞= ktotal⎜ + ⎟⎝k1 k2⎠⎛ 1 1 ⎞F = Fktotal⎜ + ⎟⎝k1 k2⎠⎛k2 + k1⎞1 = ktotal⎜ ⎟⎝ kk1 2 ⎠kk1 2ktotal=k + k1 2Evaluation(i) Answers will vary based on the hypothesis and predictions made.(j) Sources of error could be improper zero for measuring, parallax, bend in support apparatus, and permanent distortion ofthe spring. These can be avoided by careful calibration at lab setup, caution to avoid parallax during measurement, andrigid apparatus for the support stand. The value of g in the area, as well as the accuracy of the stamped value of themasses can contribute to error.Synthesis(k) To calculate the slope of the line and have it represent the force constant, the force had to be plotted on the vertical axis.(l) A real spring will heat up when stretched and has an elastic limit beyond which it will not act as a Hooke’s law spring anylonger. Ideal springs do not have these limitations.Investigation 4.5.2: Analyzing Forces and Energies in a Mass-Spring System(Pages 222–223)Questions(i) The action of a real mass–spring system supports the law of conservation of energy.(ii) The damping is slow for a real vertical mass–spring system.Hypothesis(a) A real vibrating spring will undergo damped harmonic motion. The presence of friction within the spring and within theair will transform some of the kinetic and elastic potential energy into thermal energy.(b) The damping of a real spring will likely be very slow.Prediction(c)Copyright © 2003 Nelson Chapter 4 Work and Energy 279

(g) The extrapolation is not feasible because the spring is not capable of supporting that much mass.(h) When two springs are connected linearly, each spring must support the full weight (if we ignore the mass of the lowerspring), so the total stretch will be equal to the sum of the individual stretches, (i.e., x total = x 1 + x 2 ).F = ktotalxtotal= ktotal ( x1 + x2)⎛ F F ⎞= ktotal⎜ + ⎟⎝k1 k2⎠⎛ 1 1 ⎞F = Fktotal⎜ + ⎟⎝k1 k2⎠⎛k2 + k1⎞1 = ktotal⎜ ⎟⎝ kk1 2 ⎠kk1 2ktotal=k + k1 2Evaluation(i) Answers will vary based on the hypothesis and predictions made.(j) Sources of error could be improper zero for measuring, parallax, bend in support apparatus, and permanent distortion ofthe spring. These can be avoided by careful calibration at lab setup, caution to avoid parallax during measurement, andrigid apparatus for the support stand. The value of g in the area, as well as the accuracy of the stamped value of themasses can contribute to error.Synthesis(k) To calculate the slope of the line and have it represent the force constant, the force had to be plotted on the vertical axis.(l) A real spring will heat up when stretched and has an elastic limit beyond which it will not act as a Hooke’s law spring anylonger. Ideal springs do not have these limitations.Investigation 4.5.2: Analyzing Forces and Energies in a Mass-Spring System(Pages 222–223)Questions(i) The action of a real mass–spring system supports the law of conservation of energy.(ii) The damping is slow for a real vertical mass–spring system.Hypothesis(a) A real vibrating spring will undergo damped harmonic motion. The presence of friction within the spring and within theair will transform some of the kinetic and elastic potential energy into thermal energy.(b) The damping of a real spring will likely be very slow.Prediction(c)Copyright © 2003 Nelson Chapter 4 Work and Energy 279

(d)Procedure1. Suspend a mass from the spring and measure the stretch (do several times to get a precise value).mgk =x(0.200 kg)(9.80 N/kg)=0.0126 mk = 156 N/m2. Using a 200.0-g mass, maximum stretch is 2.51 cm.3. Maximum vertical displacement is 2.51 cm, 2.36 cm, 2.22 cm, 2.08 cm, 1.96 cm, 1.84 cm, 1.73 cm, 1.63 cm, 1.53 cm,1.44 cm.Total time for 10 cycles is 2.25 s.Analysis(e)x (cm)∆y (cm)∆E g (J)E e (J)E K (J)E T (J)top middle bottom0 1.26 2.512.51 1.26 00.0492 0.0246 00 0.0124 0.04910 0.0122 00.0492 0.0492 0.0491Copyright © 2003 Nelson Chapter 4 Work and Energy 280

(f)(g)EKv ==v =1= mv22EmK22(0.0122 J)0.200 kg0.349 m/s(h) The energy is lost to thermal energy through friction.(i) (i) The action of a real mass system supports the law of conservation of energy for short time intervals (one bounce), butseems to refute it for longer time intervals (several bounces).(ii) A real spring system slowly damps the motion of a vertically vibrating system.Evaluation(j) Answers will vary depending on the hypotheses and predictions made.(k) The largest source of error is difficulty reading the location of a moving object with accuracy. A slow motion film, or aspring with a lower force constant could reduce the speed.Synthesis(l) A stiff spring will undergo faster damping because the object will move faster. The higher speed will increase the amountof air friction in each cycle, damping the motion more quickly.Activity 4.5.1: Achieving a Smooth and Safe Ride(Page 223)(a) Energy will be converted into elastic potential energy and then into thermal energy, and the spring undergoes dampedharmonic motion.(b) - capable of absorbing large quantities of energy- small amounts of energy are absorbed quickly to provide a smooth ride- often the spring has a changing spring constant to accomplish the above ideas- for safety, shock absorber must be capable of damping violent motion quickly- shock absorber must be able to damp both small and large stored energies(c) The system uses a spring to absorb the energy, and then the shock absorber damps the motion quickly to prevent dangerbounce in the vehicle.(d) Lower friction designs last longer because they do not produce as much heat. The heat can break down the seals andincrease the chances of warping.(e) Answers will vary based on the resources students use.(f) Answers will vary based on the initial hypothesis made.Copyright © 2003 Nelson Chapter 4 Work and Energy 281

(f)(g)EKv ==v =1= mv22EmK22(0.0122 J)0.200 kg0.349 m/s(h) The energy is lost to thermal energy through friction.(i) (i) The action of a real mass system supports the law of conservation of energy for short time intervals (one bounce), butseems to refute it for longer time intervals (several bounces).(ii) A real spring system slowly damps the motion of a vertically vibrating system.Evaluation(j) Answers will vary depending on the hypotheses and predictions made.(k) The largest source of error is difficulty reading the location of a moving object with accuracy. A slow motion film, or aspring with a lower force constant could reduce the speed.Synthesis(l) A stiff spring will undergo faster damping because the object will move faster. The higher speed will increase the amountof air friction in each cycle, damping the motion more quickly.Activity 4.5.1: Achieving a Smooth and Safe Ride(Page 223)(a) Energy will be converted into elastic potential energy and then into thermal energy, and the spring undergoes dampedharmonic motion.(b) - capable of absorbing large quantities of energy- small amounts of energy are absorbed quickly to provide a smooth ride- often the spring has a changing spring constant to accomplish the above ideas- for safety, shock absorber must be capable of damping violent motion quickly- shock absorber must be able to damp both small and large stored energies(c) The system uses a spring to absorb the energy, and then the shock absorber damps the motion quickly to prevent dangerbounce in the vehicle.(d) Lower friction designs last longer because they do not produce as much heat. The heat can break down the seals andincrease the chances of warping.(e) Answers will vary based on the resources students use.(f) Answers will vary based on the initial hypothesis made.Copyright © 2003 Nelson Chapter 4 Work and Energy 281

CHAPTER 4 SUMMARYMake a Summary(Page 224)1. Determine the spring constant by suspending a known mass (and, therefore, weight) from it and measuring the stretch.mgk =x(0.100 kg)(9.80 N/kg)=0.0356 mk = 27.5 N/mUse projectile analysis with the known launch angle and range to determine the launch speed. For this problem, useθ = 34º and ∆d x = 2.6 m, and ∆d y = 0.0 m.In horizontal direction:∆ d = v ∆tIn vertical direction:xx2.6 = ( vcos34 ° ) ∆t2.6v∆ t = (Equation 1)cos34°1 2∆ dy= vyi∆ t+ a ( ∆t)210 = ( vsin34 ° ) ∆ t+ ( −9.8)( ∆t)20 =∆ tv ( sin34°−4.9 ∆t)2Ignore the case where ∆t = 0 since it refers to the launch:0= vsin34°−4.9∆tvsin 34°∆ t = (Equation 2)4.9Substitute Equation 2 into Equation 1:⎛vsin 34°⎞ 2.6v⎜⎟ =⎝ 4.9 ⎠ cos34°v =( )2.6 4.9(sin 34 ° )(cos34 ° )v = 5.2 m/sCopyright © 2003 Nelson Chapter 4 Work and Energy 282

Use conservation of energy to determine the amount of stretch required in the spring. Assume the spring mass is 18 g.ET1= ET2E =∆Eeg1 1kx = mv2 2mvx =k2 222(0.018 kg)(5.2 m/s)=27.5 N/mx = 0.13 m2. The height could be determined by analyzing the vertical component of velocity.2 2vf= vi+ 2a∆d2 2vf− vi∆ d =2a2 2(0 m/s) − (5.2 m/s)=22( −9.8 m/s )∆ d = 1.4 m3. The three forms of friction would use up some of the elastic potential energy intended to launch the spring. Tocompensate, the spring would need to be stretched an extra amount.4. The energy to stretch the string would come from food you ate. The energy stored in the spring would be converted tokinetic energy as it left. The kinetic energy would be partially converted into gravitational potential as it rises, and thanreconverted back to kinetic as it lands. As it lands on the floor and slides to a stop, the kinetic energy is converted intosound and thermal energy.5. The spring would cause a mass to move up and down with decreasing amplitude until it came to rest.6. The concepts and equations can be used to design and manufacture vehicles, beds, and a wide variety of other products.CHAPTER 4 SELF QUIZ(Page 225)True/False1. T2. F The work done by gravity is zero.3. F The work you do on the backpack is negative.4. F The gravitational potential energy decreases in proportion to the distance fallen.5. F This does not refute the law of conservation of energy because some energy is converted into other forms, such as heat(thermal energy) in the ball and the floor.6. T7. T8. F Maximum speed occurs at the equilibrium position, but elastic potential energy is at a minimum at the point ofminimum extension of the spring.9. F A long damping time would not be appropriate for a bathroom scale. It would be appropriate for a “jolly-jumper” toy.Multiple Choice10. (c)11. (c)12. (e)13. (d)14. (e)15. (a)16. (d) W = ( Fcos θ ) ∆ d= ( mg sin θ )(cos180 ° ) LW =−mgLsinθCopyright © 2003 Nelson Chapter 4 Work and Energy 283

Use conservation of energy to determine the amount of stretch required in the spring. Assume the spring mass is 18 g.ET1= ET2E =∆Eeg1 1kx = mv2 2mvx =k2 222(0.018 kg)(5.2 m/s)=27.5 N/mx = 0.13 m2. The height could be determined by analyzing the vertical component of velocity.2 2vf= vi+ 2a∆d2 2vf− vi∆ d =2a2 2(0 m/s) − (5.2 m/s)=22( −9.8 m/s )∆ d = 1.4 m3. The three forms of friction would use up some of the elastic potential energy intended to launch the spring. Tocompensate, the spring would need to be stretched an extra amount.4. The energy to stretch the string would come from food you ate. The energy stored in the spring would be converted tokinetic energy as it left. The kinetic energy would be partially converted into gravitational potential as it rises, and thanreconverted back to kinetic as it lands. As it lands on the floor and slides to a stop, the kinetic energy is converted intosound and thermal energy.5. The spring would cause a mass to move up and down with decreasing amplitude until it came to rest.6. The concepts and equations can be used to design and manufacture vehicles, beds, and a wide variety of other products.CHAPTER 4 SELF QUIZ(Page 225)True/False1. T2. F The work done by gravity is zero.3. F The work you do on the backpack is negative.4. F The gravitational potential energy decreases in proportion to the distance fallen.5. F This does not refute the law of conservation of energy because some energy is converted into other forms, such as heat(thermal energy) in the ball and the floor.6. T7. T8. F Maximum speed occurs at the equilibrium position, but elastic potential energy is at a minimum at the point ofminimum extension of the spring.9. F A long damping time would not be appropriate for a bathroom scale. It would be appropriate for a “jolly-jumper” toy.Multiple Choice10. (c)11. (c)12. (e)13. (d)14. (e)15. (a)16. (d) W = ( Fcos θ ) ∆ d= ( mg sin θ )(cos180 ° ) LW =−mgLsinθCopyright © 2003 Nelson Chapter 4 Work and Energy 283

CHAPTER 4 REVIEW(Pages 226–229)Understanding Concepts1. (a) No work is done because the force is perpendicular to the displacement.(b) The work is negative because the student is exerting a force opposite the direction of motion.(c) The work is negative because gravity is exerting a force opposite the direction of motion.(d) Assuming a level roadway, the work done is zero because the force is perpendicular to the displacement.(e) No work is done because the electrical force is perpendicular to the displacement.(f) No work is done because the tension is perpendicular to the displacement.2. The force must be applied perpendicular to the object for it to do no work on the object.3. The normal force can do work on an object. For example, when you jump, you push down on the ground and the normalforce pushes up on you and accelerates you up, giving you kinetic energy.4. (a) No work is being done on the swimmer because the balancing forces forward and backward produce no motion.(b) Work is done on the student to speed him up, but after that, there is no work being done on the student. Once thestudent reaches the speed of the current, the only force is the upward buoyant force, which is perpendicular to thewaters surface. Technically, a small amount of work is being done by gravity as they whole river/student system ispulled closer to the earth.5. (a) The velocity will be changing because Newton’s second law states that if a net force acts on an object, it will accelerate(i.e., change its velocity).(b) It is possible the speed is constant if the particle is travelling in a circle.(c) The kinetic energy is proportional to the square of the speed and the mass. Since both can remain constant under theseconditions, the kinetic energy may be constant.6. Agree. In the absence of friction (including air resistance), all of the gravitational potential energy will be converted intokinetic energy. The mass will cancel in each term.7. (a) Damped vibrations are useful in the suspension of a vehicle.(b) Damped vibrations are not useful in a pendulum clock.8. It is not possible to have a motion that is not damped. Such a device would be a perpetual motion machine that cannotexist. The force of friction within the system cannot be avoided.9. m = 0.425 kg∆y = 11.8 m(a) W = ?W = ( Fcos θ ) ∆d= ( mg cos θ ) ∆y2= (0.425 kg)(9.80 m/s )(cos180 ° )(11.8 m)W =−49.1 JThe work gravity does on the ball on the way up is –49.1 J.(b) W = ?W = ( Fcos θ ) ∆d= ( mg cos θ ) ∆y2= (0.425 kg)(9.80 m/s )(cos 0 ° )(11.8 m)W = 49.1 JThe work gravity does on the ball on the way down is 49.1 J.10. F = 9.3 NW = 87 J∆d = 11 mθ = ?Copyright © 2003 Nelson Chapter 4 Work and Energy 284

W = ( Fcos θ ) ∆dWcosθ=F ∆ d−1⎛ W ⎞θ = cos ⎜ ⎟⎝ F∆d⎠−1⎛ 87 J ⎞= cos ⎜ ⎟⎝(9.3 N)(11 m) ⎠θ = 32°The angle between the applied force and the horizontal is 32°.11. Let the subscript C represent the child, and TO represent the toboggan.m C = 25.6 kgm TO = 4.81 kg∆y = 27.3 m(a) W C = ?First, find the actual distance:27.3sinφ= ∆ d27.3∆ d =sinφThe applied force, F A , will be equal to the component of gravity down the hill, mgsin φWC= ( Fcos θ ) ∆d= ( mg sin φ)(cos θ)∆d⎛ 27.3 ⎞= ( mg sin φ)(cos θ)⎜ ⎟⎝sinφ⎠= ( mg cos θ )(27.3)= (4.81 kg)(9.80 N/kg)(cos 0 ° )(27.3)3WC= 1.29×10 JThe total work done by the child is 1.29 10 3 J.(b) From part (a), the angle doesn’t matter, therefore W = 1.29 × 10 3 J.(c) The total work on the child and toboggan during the slide will be equal to the work done to take them up the hill. Usingthe equation derived in part (a),WT= ( mgcos θ )(27.3)= (25.6 kg + 4.81 kg)(9.80 N/kg)(cos0 ° )(27.3)3WT= 8.14×10 JThe total work on the child and the toboggan is 8.14 10 3 J.12. m = 73 kgθ = 9.3°v i = 4.2 m/s∆d = ?Copyright © 2003 Nelson Chapter 4 Work and Energy 285

The relation between the distance along the slope and the vertical height is:∆ysin 9.3°= ∆ d∆ y =∆ dsin 9.3°To calculate the change in distance:E = ET11 2mv = mg∆y22v = 2 g( ∆ dsin9.3 ° )2v∆ d =2 g(sin9.3 ° )2(4.2 m/s)=2(9.8 m/s)(sin 9.3 ° )∆ d = 5.6 mThe skier would travel 5.6 m along the hill before stopping.13. An increase of 50% is equivalent of multiplying by 1.5:1 2mvE2K2= 2E 1K12mv1221.50EK1 v2=2EK1 v12( v1+ 2.00)1.5 =2v12 21.5v1 = v1 + 4v1+ 420.5v−4v− 4 = 01 1T2−− ( 4) ± ( −4) −4(0.5)( −4)v1=2(0.5)4 ± 4.899=1v1= 8.90 m/s, or −0.899 m/sSince the negative value is not admissible, the original speed of the object was 8.90 m/s.14. m = 7.0 10 9 kg∆y = 36 m(a) ∆E g = ?∆ E = mg∆ygg9= (7.0×10 kg)(9.8 N/kg)(36 m)12∆ E = 2.5×10 JThe gravitational potential energy is 2.5 10 12 J.(b) The work done by on the pyramid by one person in 20 years is:⎛5× 10 6 J⎞⎛40 d⎞( ) ( )0.20 ⎜ ⎟⎜ ⎟ 20 a = 8×10 J/persond a⎝ ⎠⎝ ⎠To calculate the total number of people:2.5×103= 3×10 people88×10There were 3 10 3 workers involved.1228Copyright © 2003 Nelson Chapter 4 Work and Energy 286

15. m = 45 kg∆d = 66 cm = 0.66 m(a) W = ?W = ( Fcos θ ) ∆d= ( mg cos θ ) ∆d= (45 kg)(9.80 N/kg)(cos180 ° )(0.66 m)2W =− 2.9×10 JThe work done by gravity on the mass is –2.9 10 2 J.(b) W = ?W = ( Fcos θ ) ∆d= ( mg cos θ ) ∆d= (45 kg)(9.80 N/kg)(cos 0 ° )(0.66 m)2W = 2.9×10 JThe work done by the weightlifter on the mass is 2.9 10 2 J.(c) ∆E g = ?∆ Eg= mg∆y= (45 kg)(9.80 N/kg)(0.66 m)2∆ E = 2.9×10 JgThe change in gravitational potential energy is 2.9 10 2 J.16. m = 47 g = 0.047 kg(a) ∆y = 4.3 mv 2 = ?ET1= ET21 2 2mv 11+ mgy1 = mv2 + mgy22 2Since v 1 = 0:22gy1 = v2 + 2gy22v2 = 2gy1−2gy2v = 2 g( y − y )v2 1 222= 2(9.8 m/s )(4.3 m)= 9.2 m/sThe speed of the stick just before it hits the ground is 9.2 m/s.(b) If air resistance was included, the answer in part (a) would be slightly smaller. The energy from gravity would beshared between the thermal energy and the kinetic energy.17. y 1 – 27 mv 1 = 18 m/sθ = 37°(a) v 2 = ?E = ET1T21 1mv mgy mv mgy2 2v gy v gy2 21 + 1 = 2 + 22 21 + 2 1 = 2 + 2 22 2v2 = v1 + 2gy1 −2gy22v2 = v1 + 2 g( y1−y2)2 2= (18 m/s) + 2(9.8 m/s )(27 m −0 m)v2= 29 m/sThe speed of the stick just before it hits the ground is 29 m/s.(b) The angle the stick is thrown at does not appear in the equation, so the answer will still be 29 m/s.Copyright © 2003 Nelson Chapter 4 Work and Energy 287

18. F = 1.5 10 2 N [22° below the horizontal]m = 18 kg∆d = 1.6 mµ k = 0.55(a) F N = ?F K = ?To calculate the normal force:Σ F y= 0FN− FAsin 22°− mg = 0FN= FAsin 22°+mg2= (1.5× 10 N)(sin 22 ° ) + (18 kg)(9.8 N/kg)= 232.6 N2F = 2.3×10 NNTo calculate the force of friction:FK = µKFN= (0.55)(232.6 N)= 127.92FK= 1.3×10 NThe normal force on the box is 2.3 10 2 N. The force of friction on the box is 1.3 10 2 N.(b) v = ?If the box is starting from rest, there is no initial kinetic energy.W = E + EthK1 2( Fcos θ ) ∆ d = FK∆ d + mv22mv = 2( F cos θ ) ∆d −2FK∆d2( Fcos θ ) ∆d −2FK∆dv =m22(1.5 × 10 N)(cos 22 ° )(cos0 ° )(1.6 m) −2(127.9 N)(1.6 m)=18 kgv = 1.4 m/sThe final speed of the box is 1.4 m/s.(c) E th = ?Eth= FK∆d= (127.9 N)(1.6 m)2Eth= 2.0×10 JThe amount of thermal energy produced is 2.0 10 2 J.Copyright © 2003 Nelson Chapter 4 Work and Energy 288

19. m = 1.2 10 3 kgv 1 = 9.5 10 3 m/sF = 9.2 10 4 N∆d = 86 km = 86 10 3 mv 2 = ?W =∆EK1 2 1 2F∆ d = mvf− mvi2 21 2 1 2mvf= F∆ d + mvi2 22F∆d2vf= + vim4 32(9.2× 10 N)(86×10 m)= + (9.5×10 m/s)31.2×10 kg4vf= 1.0×10 m/sThe final speed of the probe is 1.0 10 4 m/s.20. y 1 = 1.15 my 2 = 4.75 mv 1 = ?ET1= ET2E + E = EK1 g1 g21 2mv1 + mgy1 = mgy222v1 = 2( gy2 −gy1)v = 2 g( y − y )1 2 12= 2(9.80 m/s )(4.75 m −1.15 m)v1= 8.40 m/sThe speed with which the gymnast leaves the trampoline is 8.40 m/s.21. The work in the area under the graph. Consider the area in three parts:W = A + A + Atriangle 1 rectangle triangle 23 2= 1 bh 11 1+ lw+b2h22 21 1= (1.0 m)(12.0 N) + (2.0 m − 1.0 m)(12.0 N) + (6.0 m −2.0 m)(12.0 N)2 2W = 42 JThe person does 42 J of work.22. x = 0.418 mF = 1.00 10 2 N(a) k = ?Fx= kxFxk =x21.00×10 N=0.418 mk = 239 N/mThe force constant of the spring is 239 N/m.(b) x = 0.150 mF x = ?Copyright © 2003 Nelson Chapter 4 Work and Energy 289

Fx= kx= (239 N/m)(0.150 m)Fx= 35.9 NThe force required to stretch the spring is 35.9 N.(c) To stretch it 0.150 m:1 2Ee= kx21 (239 N/m)(0.150 m)2=2E = 2.69 JeTo compress it 0.300 m:1 2Ee= kx21 (239 N/m)( 0.300 m)2= −2Ee= 10.8 JThe work required is 2.69 J to stretch it, and 10.8 J to compress it.23. k = 22 N/mm = 7.5 10 –3 kgF K = 4.2 10 –2 Nx = 3.5 cm = 0.035 m∆d = ?ET1= ET2Ee= Eth1 2kx = FK∆d22kx∆ d =2FK2(22 N/m)(0.035 m)=−22(4.2×10 N)∆ d = 0.32 mThe eraser will slide 0.32 m along the desk.24. k = 75 N/mA = 0.15 mv = 1.7 m/sx = 0.12 mm = ?For SHM, all of the energy is E e when at the maximum amplitude, A:ET1= ET2E = E + Ee max e K1 1 1kA = kx + mv2 2 22 2 2mv = kA −kx2 2k( A − x )m =2v2 2 22 2(( ) − ( ) )75 N/m 0.15 m 0.12 m=2(1.7 m/s)m = 0.21 kgThe mass of the block is 0.21 kg.Copyright © 2003 Nelson Chapter 4 Work and Energy 290

25. m = 0.42 kgk = 38 N/mA = 5.3 cm = 0.053 m(a) E e = ?Maximum energy occurs at full amplitude.1 2Ee= kx21= (38 N/m)(0.053 m)2E = 0.053 JeThe maximum energy of the mass-spring system is 0.053 J.(b) v = ?Maximum speed occurs when there is no elastic potential energy:ET1= ET2E = Ee maxK1 1kA = mv2 22 2mv = kA2 2v =38 N/m(0.053 m)=0.42 kgv = 0.50 m/sThe maximum speed of the mass is 0.50 m/s.(c) x = 4.0 cm = 0.040 mv = ?ET1= ET2E = E + E2kAme max e K1 1 1kA = kx + mv2 2 22 2 2mv = kA −kx2 2 2v =2 2k( A − x )mv = 0.33 m/sThe speed of the mass is 0.33 m/s.(d) x = 4.0 cm = 0.040 mE T = ?E = E + E=T e K222 2(( ) − ( ) )38 N/m 0.053 m 0.040 m0.42 kg1 2 1 2= kx + mv2 21 1= (38 N/m)(0.040 m) + (0.42 kg)(0.33 m/s)2 2ET= 0.053 JThe total energy is 0.053 J. The results are the same.2 2Copyright © 2003 Nelson Chapter 4 Work and Energy 291

Applying Inquiry Skills26. (a) One text weighs 2.0 kg and has a thickness of 3.6 cm. The first one doesn’t need to be raised at all.W = Eg1 + Eg2 + Eg3 + Eg4 + Eg5= mgy1+ mgy2 + mgy3 + mgy4 + mgy5= mg( y1+ y2 + y3 + y4 + y5)= (2.0 kg)(9.8 N/kg)(0 m + 0.036 m + 0.072 m + 0.108 m + 0.144 m)W = 7.1 JYou would have to do 7.1 J of work.(b) Some errors would be determining the mass of the text. Also, the thickness may compress the books on the bottom.Not all texts may have the same mass.27. The calculator would need to know the distance the box was pushed.28.29. The tractor seat would need to have a strong spring to absorb large bumps, and a shock absorber to prevent “launching”the driver. There would have to be smaller springs on top of that to absorb small vibrations. Damping would be importantto prevent resonance.Making Connections30. Roller coasters are often shut down in high winds because of the loss of energy that may occur due to increased airresistance. Cold can cause parts to shrink and increase frictional forces beyond safe limits.31. (a) The ball on track B will arrive first. Shortly after the start, the vertical drop of the ball on track B causes an increase inthe speed, which it will enjoy for the majority of the race. At the end, it will slow down to the same speed that the ballon track A has just accelerated to.(b) Racing cyclists use this in a variety of ways. The sprinters stay high on the track until ready to make a break for it,converting all of their stored gravitational energy into kinetic. Team racers use the change to minimize the work of therider. When the leader is ready to give up his spot, he rides up the hill a bit, converting some of his kinetic energy intogravitational potential energy, reducing his speed. Once the last team mate has passed, he can drop down again, gainingthe gravitational potential energy back as kinetic energy without needing to supply it from his own body power.32. (a) v 1 = 0 m/sv 2 = ?E = ET1T21 1mv mgy mv mgy2 2v gy v gy2 21+1=2+22 21 + 2 1 = 2 + 2 22v2 = 2gy1−2gy2v = 2 g( y − y )2 1 22= 2(9.80 m/s )(37.8 m −17.8 m)v2= 19.8 m/sThe speed of the coaster at position C is 19.8 m/s.Copyright © 2003 Nelson Chapter 4 Work and Energy 292

(b) v 1 = 5.00 m/sv 2 = ?ET1= ET21 1mv mgy mv mgy2 2v gy v gy2 21 + 1 = 2 + 22 21 + 2 1 = 2 + 2 22 2v2 = v1 + 2gy1−2gy22v2 = v1 + 2 g( y1−y2)2 2= (5.00 m/s) + 2(9.80 m/s )(37.8 m −17.8 m)v2= 20.4 m/sThe speed of the coaster at position C is 20.4 m/s.33. (a) The mass is not necessary because it cancels out of the mathematical equations.(b) You might expect the speed would be 5.0 m/s more at the second point, but the reality is that the small kinetic energysupplied by having a speed going over the first hill is negligible compared with the large amount of gravitationalpotential energy.34. a = 12 g [upward](a) F A = ?Σ Fy= maFA− mg = maF = ma+mg= ma ( + g)= m(0.12 g+g)FA= 1.12mgThe force required is 1.12mg.(b) W = ?W = ( Fcos θ ) ∆d= (1.12 mg)(cos 0 ° )( ∆y)W = 1.12mg∆yThe work done is 1.12mg∆y.35. m = 1.5 kgk = 2.1 10 3 N/m∆y = 0.37 mx = ?ACopyright © 2003 Nelson Chapter 4 Work and Energy 293

Using the quadratic equation:ET1= ET21 2mg∆ y = kx21(1.5)(9.8)(0.37 + x) = (2.1×10 ) x225.439 + 14.7x= 1050x21050x−14.7x− 5.439 = 0−− ( 14.7) ± ( −14.7) −4(1050)( −5.439)x =2(1050)14.7 ± 151.9=2100= 0.079 m or x = −0.065 m (negative value inadmissable)x = 0.079 mThe maximum distance the spring is compressed is 0.079 m.36. m = 0.55 g = 5.5 10 –4 kg∆d = 95 cm = 0.95 m∆d x = 3.7 mE e = ?23 2First we must calculate the time required for the vertical drop (v i = 0):1 2∆ d = vi∆ t+ a ( ∆t)21 ( )2∆ d = a ∆t22∆d∆ t =a2( −0.95 m)=2−9.8 m/s∆ t = 0.44 sTo calculate the horizontal speed:∆ dx= vx∆t∆dxvx= ∆ t3.7 m=0.44 sv = 8.4 m/sxWe know that initial kinetic energy came from elastic potential energy, therefore:ET1= ET2E = EEeeK1 2= mv21= (5.5 × 10 kg)(8.4 m/s)2= 0.019 J−4 2The elastic potential energy stored was 0.019 J.Copyright © 2003 Nelson Chapter 4 Work and Energy 294

37. y 1 = 16 my 2 = 9.0 mH = ?First we must solve for the launch speed if v 1 = 0:E = EResolve vector components:T1T21 1mv mgy mv mgy2 222gy1 = v2 + 2gy22v2 = 2gy1 −2gy2v = 2 g( y − y )2 21+1=2+2v2 1 222= 2(9.8 m/s )(16 m −9.0 m)= 11.71 m/sAs just clearing the wall, the horizontal component will be the same, and the vertical component must produce a 30º angleto the vertical.Copyright © 2003 Nelson Chapter 4 Work and Energy 295

To calculate the vertical speed:v2cos 45°tan 30°=vyv2cos 45°vy=tan 30°( 11.71 m/s)cos 45°=tan 30°v = 14.35 m/s [down]yDetermine how far the skier has dropped from the launch point:2 2vf= vi+ 2a∆d2 2vf− vi∆ d =2a2( −14.35 m/s) − 11.71 m/s sin 45°=22( −9.8 m/s )∆ d = −7.0 mTherefore, the wall must be 9.0 – 7.0 = 2.0 m tall.38. y 1 = 2.5 mv 1 = 9.0 m/sy 2 = 3.0 mv 2 = ?E = ET1T21 1mv mgy mv mgy2 2v gy v gy2 21+1=2+22(( )( ) )2 21 + 2 1 = 2 + 2 22 2v2 = v1 + 2gy1−2gy22v2 = v1 + 2 g( y1−y2)2 2v2= (9.0 m/s) + 2(9.8 m/s )(2.5 m −3.0 m)= 8.4 m/sThe speed of the ball when it “swishes” through the hoop is 8.4 m/s.39. g = 2.0v 1 = 6.0y 2 = 5.0S = ?Solve for the launch speed at the top of the ramp where y 1 = 0:E = ET1T21 1mv mgy mv mgy2 2v v gy2 21+1=2+22 21 + 0= 2 + 2 22 2v2 = v1 −2gy22v2 = v1 −2gy22= (6.0) −2(2.0)(5.0)v = 4.0 units2Copyright © 2003 Nelson Chapter 4 Work and Energy 296

Analyze projectile motion to find the change in time:2( ∆t) −2∆t− 5=01 2∆ dy= vyi∆ t+ a ( ∆t)21− 5.0 = (4.0sin 30 ° ) ∆ t+ ( −2.0)( ∆t)22− 5=2 ∆t−( ∆t)2Using the quadratic equation:To calculate horizontal range (S):S = v ∆t−− ( 2) ± ( −2) −4(1)( −5)∆ t =2(1)22±4.9=2= 3.45 units or −1.45 units (dismiss negative answer)∆ t = 3.45 units= (4.0cos30 ° )(3.45)S = 12 unitsThe shuttle lands a distance of 12 units from the ramp.xCopyright © 2003 Nelson Chapter 4 Work and Energy 297

CHAPTER 5 MOMENTUM AND COLLISIONSReflect on Your Learning(Page 230)1. (a) The expression refers to once you start playing well, it is easier to keep playing well (or scoring, or winning).(b) The physics meaning refers to a specific quantity. The everyday use of the word momentum can mean continuing to dowell and inertia.2. (a) The momentum of car A is less than the momentum of car B.(b) The momentum of bicycle and rider A is less than the momentum of bicycle and rider B.(c) The momentum of the large truck A is greater than the momentum of car B.3. (a) The phrase “follow-through” refers to continue to swing the racket/club even after contact with the ball is made.This allows the racket/club to be in contact with the ball for a longer period of time.(b) “Follow-through” affects the momentum (or change in it).4. (a) One technique that could be used is to break the total area up into eight different rectangles. The height of each wouldbe an estimate. Each rectangle’s area could be determined and the sum would give the total area.kg ⋅mkg ⋅m(b) F⋅ t = Ns ⋅ = × s=2s s5. The car should be designed with crunch zones to absorb the energy. If elastic bumpers were used, the car would reboundand be able to impact other objects.Try This Activity: Predicting the Bounces(Page 231)(a) Prediction: The balls will bounce to the same height.(b) Ball A and ball B both seem to be made from the same material. They possess approximately equal density and areapproximately the same size. Since the balls appear to be identical, they will possess the same amount of elasticity andwill bounce to the same height.(c) The balls did not bounce to the same height. Our prediction and hypothesis were not supported by the evidence.(d) Much of the kinetic energy of one ball is conserved in the bounce. The kinetic energy of the other ball is not conserved.It is transformed into other forms. This ball loses kinetic energy in the bounce and does not bounce to the same height.Bales of hay placed on the sides of ski-racing runs transform the kinetic energy of the skier into other forms, slowing theskier down and reducing the chance for bodily harm.5.1 MOMENTUM AND IMPULSEPRACTICE(Page 233)Understanding Concepts1. (a) m = 7.0 × 10 3 kgv = 7.9 m/sp = ? p = mv3= (7.0×10 kg)(7.9 m/s)4p = 5.5× 10 kg ⋅m/s [E]The momentum of the African elephant is45.5× 10 kg ⋅ m/s [E] .Copyright © 2003 Nelson Chapter 5 Momentum and Collisions 299

CHAPTER 5 MOMENTUM AND COLLISIONSReflect on Your Learning(Page 230)1. (a) The expression refers to once you start playing well, it is easier to keep playing well (or scoring, or winning).(b) The physics meaning refers to a specific quantity. The everyday use of the word momentum can mean continuing to dowell and inertia.2. (a) The momentum of car A is less than the momentum of car B.(b) The momentum of bicycle and rider A is less than the momentum of bicycle and rider B.(c) The momentum of the large truck A is greater than the momentum of car B.3. (a) The phrase “follow-through” refers to continue to swing the racket/club even after contact with the ball is made.This allows the racket/club to be in contact with the ball for a longer period of time.(b) “Follow-through” affects the momentum (or change in it).4. (a) One technique that could be used is to break the total area up into eight different rectangles. The height of each wouldbe an estimate. Each rectangle’s area could be determined and the sum would give the total area.kg ⋅mkg ⋅m(b) F⋅ t = Ns ⋅ = × s=2s s5. The car should be designed with crunch zones to absorb the energy. If elastic bumpers were used, the car would reboundand be able to impact other objects.Try This Activity: Predicting the Bounces(Page 231)(a) Prediction: The balls will bounce to the same height.(b) Ball A and ball B both seem to be made from the same material. They possess approximately equal density and areapproximately the same size. Since the balls appear to be identical, they will possess the same amount of elasticity andwill bounce to the same height.(c) The balls did not bounce to the same height. Our prediction and hypothesis were not supported by the evidence.(d) Much of the kinetic energy of one ball is conserved in the bounce. The kinetic energy of the other ball is not conserved.It is transformed into other forms. This ball loses kinetic energy in the bounce and does not bounce to the same height.Bales of hay placed on the sides of ski-racing runs transform the kinetic energy of the skier into other forms, slowing theskier down and reducing the chance for bodily harm.5.1 MOMENTUM AND IMPULSEPRACTICE(Page 233)Understanding Concepts1. (a) m = 7.0 × 10 3 kgv = 7.9 m/sp = ? p = mv3= (7.0×10 kg)(7.9 m/s)4p = 5.5× 10 kg ⋅m/s [E]The momentum of the African elephant is45.5× 10 kg ⋅ m/s [E] .Copyright © 2003 Nelson Chapter 5 Momentum and Collisions 299

(b) m = 19 kgv = 726 m/sp = ? p = mv= (19 kg)(26 m/s)2p = 4.9× 10 kg ⋅m/s [S]The momentum of the mute swan is(c) m = 9.1 × 10 –31 kgv = 1.0 × 10 7 m/sp = ?24.9× 10 kg ⋅ m/s [S] . p = mv−31 7= (9.1× 10 kg)(1.0×10 m/s)−24p = 9.1× 10 kg ⋅m/s [forward]The momentum of the electron is 9.1× 10 kg ⋅ m/s [forward] .2. m = 405 kg3p = 5.02× 10 kg ⋅m/sv = ? p = mv pv =m3−245.02× 10 kg ⋅m/s=405 kgv = 12.4 m/s [W]The velocity of the craft is 12.4 m/s [W] .3. p = 4.5 kg ⋅m/sv = 9.0 × 10 2 m/sm = ? p = mvpm = v4.5 kg ⋅ m/s=29.0×10 m/s−3m = 5.0×10 kgThe mass of the bullet is 5.0× 10 kg .4. (a) Assuming a mass of 58 kg and a top speed of 8.0 m/s.p = mv= (58 kg)(8.0 m/s)2p = 4.6× 10 kg ⋅m/s−32The magnitude of the momentum of an average student would be 4.6× 10 kg ⋅ m/s .(b) Assume a typical compact car to have a mass of 1200 kg.p = mvpv =m24.6× 10 kg ⋅m/s=1200 kgv = 0.39 m/sA typical compact car would have to travel at a velocity of 0.39 m/s to achieve the same momentum.300 Unit 2 Energy and Momentum Copyright © 2003 Nelson

PRACTICE(Page 237)Understanding Concepts5. impulse change in momentumF∆ t = Ns ⋅∆ p = m( ∆v)kg ⋅m⎛m⎞= ×2 s= kg⎜ ⎟s⎝ s ⎠kg ⋅mkg ⋅mF∆ t =∆ p =ssTherefore, the units of impulse and change in momentum are equivalent.6. (a) m = 0.065 kgv iy = –3.8 m/sp =i y?ppiyiy= mviy= (0.065 kg)( −3.8 m/s)=−0.25 kg ⋅m/sThe momentum of the snowball before hitting the ground is –0.25 kg⋅m/s.(b) v fy = 0.0 m/sp =f y?ppfyfy= mvfy= (0.065 kg)(0.0 m/s)= 0.0 kg ⋅m/sThe momentum of the snowball after hitting the ground is 0.0 kg⋅m/s.(c) ∆ py = pfy − piy= 0.0 kg ⋅m/s −( −0.25 kg ⋅m/s)∆ p y= 0.25 kg ⋅m/sThe change in momentum is 0.25 kg⋅m/s.37. (a) Σ F x= 4.8×10 N∆ t = 3.5 simpulse = ΣFx∆ t = ?3ΣFx∆ t = (4.8×10 N)(3.5 s)4ΣFx∆ t = 1.7× 10 N⋅s [W]4The impulse on the truck over this time interval is 1.7× 10 N ⋅ s [W] .4(b) p = 5.8× 10 kg ⋅ m/sixp = ?fxΣF ∆ t = ∆ pxΣFx∆ t = pfx− pixp =ΣF ∆ t+ppxfx x ix4 4fxThe final momentum of the truck is= 1.7× 10 N⋅ s+ 5.8× 10 kg⋅m/s4= 7.5× 10 kg ⋅m/s [W]47.5× 10 kg ⋅ m/s [W] .Copyright © 2003 Nelson Chapter 5 Momentum and Collisions 301

8. m = 0.27 kgv ix = 2.7 m/sv fx = 0 m/sΣ F x= 33 N [W]∆t = ?Choosing the initial direction of the ball (east) as positive:ΣFx∆ t = ∆pxΣFx∆ t = m( vfx−vix)mv (fx− vix)∆ t =ΣFx0.27 kg(0 m/s − 2.7 m/s)=−33 N∆ t = 0.022 s, or 22 msThe ball is in contact with the net for 22 ms.9. v = 75.5 m/s [11.1° below the horizontal]5m = 1.24×10 kgp x = ?p y = ?p = p= mv5= (1.24×10 kg)(75.5 m/s)6p = 9.362× 10 kg ⋅m/spxcos11.1°=pppxx= pcos11.1°66= (9.362× 10 kg ⋅ m/s)(cos11.1 ° )= 9.19× 10 kg ⋅m/spysin11.1°=pp = psin11.1°pyy6= (9.362× 10 kg ⋅ m/s)(sin11.1 ° )6= 1.80× 10 kg ⋅m/sThe horizontal component of the plane’s momentum is61.80× 10 kg ⋅ m/s .69.19× 10 kg ⋅ m/s and the vertical component isApplying Inquiry Skills10. (a) Using the area of a triangle:1impulse = A=bh21= (0.5 s)(4.0 N)2impulse = 1.0 N ⋅s [E]The impulse imparted is 1.0 N⋅s [E].302 Unit 2 Energy and Momentum Copyright © 2003 Nelson

(b) Estimate the shape from 0 s – 0.2 s to be a triangle.Estimate the shape from 0.2 s – 0.4 s to be a trapezoid.Estimate the shape from 0.4 s – 1.0 s to be a trapezoid.The sum of all three is the total impulse:1 ⎛h1+ h2 ⎞ ⎛h1′ + h2′⎞impulse = A = bh +2⎜2⎟b + ⎜2⎟b′⎝ ⎠ ⎝ ⎠1 ⎛30 N + 40 N ⎞ ⎛40 N + 60 N ⎞= (0.2 s)(30 N) + ⎜ ⎟(0.4 s − 0.2 s) + ⎜ ⎟(1.0 s −0.4 s)2 ⎝ 2 ⎠ ⎝ 2 ⎠1impulse = 4.0× 10 N ⋅s [S]The impulse imparted is 4.0 × 10 1 N⋅s [S].Making Connections11. (a) Padded gloves increase the duration of time that the moving hand (punch) comes to rest. This causes a decrease in thesize of the force applied to the head (and the hand), reducing fractures.(b) By “rolling with a punch,” a boxer increases the contact time, and reduces the applied force as the punch is applied.Section 5.1 Questions(Page 238)Understanding Concepts1. The net force on an object is the rate of change in momentum.∆pΣ F = ∆ t2. Impulse is most useful when there is only a single net force acting on an object.3. (a) Σ F = 24 N [E]∆ t = 3.2 sΣF ∆t= ?ΣF∆ t = (24 N)(3.2 s)ΣF∆ t = 77 N ⋅s [E]The impulse exerted is 77 N ⋅ s [E] .2(b) Σ F = 1.2×10 N [forward]∆ t = 9.1 msΣF ∆t= ?2 −3ΣF∆ t = (1.2 × 10 N)(9.1×10 s)ΣF∆ t = 1.1 N ⋅s [forward]The impulse exerted is 1.1 N ⋅ s [forward] .(c) m = 12 kgg = 9.8 m/s 2 [down]∆ t = 3.0 sΣF ∆t= ?ΣF∆ t = mg∆t2= (12 kg)(9.8 m/s )(3.0 s)2ΣF∆ t = 3.5× 10 N ⋅s [down]The impulse exerted is23.5× 10 N ⋅ s [down] .Copyright © 2003 Nelson Chapter 5 Momentum and Collisions 303

(d) Solve for the area under the graph. Estimate the geometric shapes shown.ΣF∆ t = A + A + A1 2 31 ⎛1.0 N + 3.7 N ⎞ ⎛3.9 N + 5.0 N ⎞= (0.010 s)(0.81 N) + ⎜ ⎟(0.020 s − 0.010 s) + ⎜ ⎟(0.040 s −0.020 s)2 ⎝ 2 ⎠ ⎝ 2 ⎠ΣF∆ t = 0.12 N ⋅s [S]The impulse exerted is 0.12 N ⋅ s [S] .4. msystem= 41 kg + 21 kg = 62 kg∆ t = 2.0 sΣ Fx= 75 N [W]vix= 0 m/sv = ?5.fxΣFx∆ t = ∆pxΣFx∆ t = mvfx −mvixmv =ΣF ∆ t + mvfx x ixΣFx∆ t+mvixvfx=mΣFx∆t= + vixm(75 N)(2.0 s)= + 0 m/s(41 kg + 21 kg)vfx= 2.4 m/s [W]The final velocity of the cart and the child will be 2.4 m/s [W].3m = 1.1×10 kgvix= 22 m/svfx= 0 m/s∆ t = 1.5 sΣ F = ?x304 Unit 2 Energy and Momentum Copyright © 2003 Nelson

ΣF ∆ t = ∆pThe average force required to stop the car is6. (a) m = 0.17 kgvix= 2.1 m/s [right]vfx= 1.8 m/s [left]∆ p = ?xxxΣFx∆ t = m( vfx −vix)mv (fx− vix)Σ Fx=∆t3(1.1× 10 kg)(0 m/s −22 m/s)=1.5 s4 4Σ F =− 1.6× 10 N [E], or 1.6×10 N [W]x41.6× 10 N [W] .(b)Choose right as the positive direction.∆ p = m( v −v)x fx ix= (0.17 kg)( −1.8 m/s −2.1 m/s)=−0.66 kg ⋅m/s [right]∆ p = 0.66 kg ⋅m/s [left]xThe change in momentum of the ball is 0.66 kg ⋅ m/s [left] .ΣF∆ t = ?xThe impulse is equal to the change in momentum, but the units are written in a different form.ΣFx∆ t = m( vfx −vix)= (0.17 kg)( −1.8 m/s −2.1 m/s)=−0.66 N ⋅s [right]ΣF∆ t = 0.66 N ⋅s [left]xThe impulse given to the ball by the cushion is 0.66 N ⋅ s [left] .7. (a) m = 0.16 kgvfx= 11 m/svix= 18 m/s∆ p = ?x∆ px = m( vfx −vix)= (0.16 kg)(11 m/s −18 m/s)=−1.1 kg ⋅m/s [forward]∆ p = 1.1 kg ⋅m/s [backward]xThe change in momentum of the puck is 1.1 kg ⋅ m/s [backward] .(b) ΣF∆ t = ?xΣFx∆ t = m( vfx −vix)= (0.16 kg)(11 m/s −18 m/s)=−1.1 N ⋅s [forward]ΣF∆ t = 1.1 N ⋅s [backward]xThe impulse exerted by the snow on the puck is 1.1 N ⋅ s [backward] .Copyright © 2003 Nelson Chapter 5 Momentum and Collisions 305

(c)∆ t = 2.5 sF =K?ΣFx∆ t = m( vfx −vix)mv (fx− vix)FK=∆t(0.16 kg)(11 m/s −18 m/s)=2.5 s=−0.45 N [forward]FK= 0.45 N [backward]The average frictional force exerted by the snow on the puck is 0.45 N [backward].8. m = 0.50 kg∆ t = 1.5 sΣ Fx= 1.4 N [W]vix= 2.4 m/sv = ?fxChoose east as the positive direction.ΣFx∆ t = ∆pxΣFx∆ t = mvfx −mvixmvfx=ΣFx∆ t + mvixΣFx∆ t+mvixvfx =mΣFx∆t= + vixm( −1.4 N)(1.5 s)= + 2.4 m/s(0.50 kg)=−1.8 m/s [E]vfx= 1.8 m/s [W]The final velocity of the disk is 1.8 m/s [W].9. m = 2.0 kg∆ t = 0.50 sΣ Fx= 6.0 N [N]vfx= 4.5 m/sv = ?ixChoose north as the positive direction.ΣFx∆ t = ∆pxΣFx∆ t = mvfx −mvixmvix = mvfx −ΣFx∆tmvfx−ΣFx∆tvix=mΣFx∆t= vfx−m(6.0 N)(0.50 s)= 4.5 m/s −(2.0 kg)vix= 3.0 m/s [N]The initial velocity of the skateboard is 3.0 m/s [N].306 Unit 2 Energy and Momentum Copyright © 2003 Nelson

10. (a) m = 0.61 kgvix= 9.6 m/s [down]vfx= 8.5 m/s [up]∆ p = ?xChoose up as positive.∆ px = m( vfx −vix)= (0.61 kg)(8.5 m/s −( −9.6 m/s))∆ px= 11 kg ⋅m/s [up]The change in momentum of the basketball is 11 kg ⋅ m/s [up] .−3(b) ∆ t = 6.5 ms = 6.5×10 sF =floor?Choose up as positive.ΣFx∆ t = m( vfx −vix)mv (fx− vix)Ffloor=∆t(0.61 kg)(8.5 m/s −( −9.6 m/s))=−36.5×10 s3F = 1.7×10 N [up]floorThe average force exerted on the basketball by the floor is 1.7× 10 N [up] .11. The force of gravity can be ignored in question 10 because it is so small compared to the force exerted by the floor(approximately 6.0 N compared with 1700 N).Applying Inquiry Skills12. By using a high-speed digital camera, take rapid photos during the swing. For each athlete, compare the distance that theracket is in contact with the ball. The longer the distance, the greater the follow-through.3Copyright © 2003 Nelson Chapter 5 Momentum and Collisions 307

Making Connections13. The change in momentum when you land is the same for either type of landing. When you bend your knees, the time thatthe force is applied is longer, so the required force is less. This smaller force is less likely to cause pain.14. (a) The analyst could determine the initial speed of the bumper component from its skid marks. To estimate the initialspeeds of each vehicle, you would need to know the mass of each vehicle and how far each one skidded after thecollision. Using this information, you can also determine approximate coefficients of friction and solve for speedsimmediately after the collision. This would provide enough information to calculate the change in momenta of eachvehicle.(b) Answers will vary.5.2 CONSERVATION OF MOMENTUM IN ONE DIMENSIONPRACTICE(Pages 243–244)Understanding Concepts1. The net force on the system must be zero for momentum to be conserved.2. For the centre of mass of the system, the statement is equivalent to Newton’s first law. With no net force acting on thesystem, the centre of mass of the system will not experience any change in velocity. However, the individual parts of thesystem will undergo various changes in speed and direction.3. (a) Earth will exert a downward force on the hairbrush.(b) The hairbrush will exert an upward force on Earth.(c) The forces in (a) and (b) are the same in magnitude.(d) The net force of the system containing Earth and the hairbrush is zero.(e) Momentum of this system will be conserved.(f) Earth will move up as the hairbrush falls down.(g) Choose up as positive.m1= 0.0598 kg24m2= 5.98×10 kgv1y = 10 m/s [down]v = ?2 ypy= p′y0 = mv + mvmv1 1yv2y=−mv2 y1 1y2 2y2(0.0598 kg)( −10 m/s)=−245.98×10 kg−25= 1×10 m/s [up]−25Earth’s speed at this time is 1× 10 m/s [up].4. m1= 45 kgm2= 33 kgv1x= 1.9 m/s [E]v = ?2x308 Unit 2 Energy and Momentum Copyright © 2003 Nelson

Making Connections13. The change in momentum when you land is the same for either type of landing. When you bend your knees, the time thatthe force is applied is longer, so the required force is less. This smaller force is less likely to cause pain.14. (a) The analyst could determine the initial speed of the bumper component from its skid marks. To estimate the initialspeeds of each vehicle, you would need to know the mass of each vehicle and how far each one skidded after thecollision. Using this information, you can also determine approximate coefficients of friction and solve for speedsimmediately after the collision. This would provide enough information to calculate the change in momenta of eachvehicle.(b) Answers will vary.5.2 CONSERVATION OF MOMENTUM IN ONE DIMENSIONPRACTICE(Pages 243–244)Understanding Concepts1. The net force on the system must be zero for momentum to be conserved.2. For the centre of mass of the system, the statement is equivalent to Newton’s first law. With no net force acting on thesystem, the centre of mass of the system will not experience any change in velocity. However, the individual parts of thesystem will undergo various changes in speed and direction.3. (a) Earth will exert a downward force on the hairbrush.(b) The hairbrush will exert an upward force on Earth.(c) The forces in (a) and (b) are the same in magnitude.(d) The net force of the system containing Earth and the hairbrush is zero.(e) Momentum of this system will be conserved.(f) Earth will move up as the hairbrush falls down.(g) Choose up as positive.m1= 0.0598 kg24m2= 5.98×10 kgv1y = 10 m/s [down]v = ?2 ypy= p′y0 = mv + mvmv1 1yv2y=−mv2 y1 1y2 2y2(0.0598 kg)( −10 m/s)=−245.98×10 kg−25= 1×10 m/s [up]−25Earth’s speed at this time is 1× 10 m/s [up].4. m1= 45 kgm2= 33 kgv1x= 1.9 m/s [E]v = ?2x308 Unit 2 Energy and Momentum Copyright © 2003 Nelson

Choose east as positive.px= p′x0 = mv + m vmv1 1xv2x=−m1 1x2 2x(45 kg)(1.9 m/s)=−33 kg=−2.6 m/s [E]v2x = 2.6 m/s [W]The velocity of the raft relative to the water is 2.6 m/s [W].5. m1 = 56.9 kgv1x = 3.28 m/sv2x =−3.69 m/sm2= ?px= p′x0 = mv1 1x+ m2v2xmv1 1xm2=−v(56.9 kg)(3.28 m/s)=−−3.69 m/sm2= 50.6 kgThe mass of the other skater is 50.6 kg.6. m1 = 11 kgm2= 24 kgv1x= 95 m/sv = ?7.2xpx= p′x22x0 = mv + m vmv1 1xv2x=−m1 1x2 2x(11 kg)(95 m/s)=−24 kgv2x=−44 m/sThe speed of the 24-kg piece is 44 m/s in the opposite direction.4m1= 1.37×10 kg4m2= 1.12×10 kgv1= 20.0 km/h [N]v1′ = v2′ = v′= 18.3 m/s (when coupled)v = ?22Copyright © 2003 Nelson Chapter 5 Momentum and Collisions 309

Since v′ = v′ = v′,1 2mv + mv = mv′ + mv′1 1 2 2 1 1 2 2mv1 1+ m2v2= ( m1+m2)v′mv2 2= ( m1+ m2)v′−mv1 1( m1+ m2)v′−mv1 1v2=m24 4 4(1.37 × 10 kg + 1.12× 10 kg)(18.3 km/h) − (1.37×10 kg)(20.0 km/h)=41.12×10 kgv = 16.2 km/h [N]The initial velocity of the second car is 16.2 km/h [N].8. m1= 0.15 kgm2= 0.045 kgv1= 56 m/sv2= 0v2′ = 67 m/sv′ = ?12mv + mv = mv′ + mv′1 1 2 2 1 1 2 2mv 1 1+ 0 = mv 1 1′ + mv 2 2′mv 1 1′ = mv 1 1−mv2 2′mv1 1−m2v′2v1′ =m1mv 2 2′= v1−m(0.045 kg)(67 m/s)= 56 m/s −0.15 kgv1′ = 36 m/sThe head of the driver has a speed of 36 m/s immediately after the collision.1Applying Inquiry Skills9. (a) Cart A moves toward the stationary cart B. This causes the two carts to collide, imparting speed to cart B. There is nodata to tell what happens to cart A during the last part of the collision and after.(b) Cart B has no speed, so the system momentum will be equal to the momentum of cart A.psystem= pA= mAvA= (0.40 kg)(0.60 m/s)p = 0.24 kg ⋅m/s [E]systemThe momentum of the system of carts before the collision is 0.24 kg ⋅ m/s [E] .(c) mA= 0.40 kgmB= 0.80 kgvA= 0.60 m/s [E]vB= 0vB′ = 0.35 m/s [E]v′ = ?A310 Unit 2 Energy and Momentum Copyright © 2003 Nelson

(d)Choosing east as positive.p = p′m v + m v = m v′ + m v′A A B B A A B Bm v + 0 = m v′ + m v′A A A A B Bm v′ = m v −m v′A A A A B BmAvA − mBv′Bv′ A=mAAmv′B B= vA−mA(0.80 kg)(0.35 m/s)= 0.60 m/s −0.40 kgv′ =−0.10 m/s [E], or 0.10 m/s [W]The velocity of cart A after the collision is 0.10 m/s [W].Making Connections10. Assuming the astronaut has something to throw, he could throw it away from the spaceship. This would cause him tomove toward the ship.Section 5.2 Questions(Pages 244–245)Understanding Concepts1. (a) F The forces are the same (Newton’s third law).(b) F The magnitude of the changes in momenta will be equal.(c) T(d) F The magnitude of the changes in momenta will be equal.2. Yes. An example would be two dynamics carts on a track that have had a spring bumper explosion.3. (a) Momentum is conserved. There is no net external force of the system of objects.(b) Momentum is conserved. There is no net external force of the system of objects.(c) Momentum is not conserved. There is an external force applied by the stove on the pan.4. It is not possible to react against the vacuum of space. A rocket ship works by pushing exhaust gasses out the back, andthe exhaust gases push back on the rocket ship. There is not involvement of the air (or not).Copyright © 2003 Nelson Chapter 5 Momentum and Collisions 311

5. m1= 57 kgm2= 27 kgv1 = v2= v = 3.2 m/s [forward] (when together)v1′ = 3.8 m/s [forward]v′ = ?2Choose forward as positive.mv1 1+ m2v2= mv′ 1 1+mv′2 2( m1+ m2)v = mv1 1′ + m2v2′mv 2 2′ = ( m1+ m2)v−mv1 1′( m1+ m2)v−mv1 1′v2′ =m2(57 kg + 27 kg)(3.2 m/s) −(57 kg)(3.8 m/s)=27 kgv′ 2= 1.9 m/s [forward]The final velocity of the cart is 1.9 m/s [forward].6. m1= 65 kg + 35 kg = 100 kg (treating the hiker and raft as a single object)m2= 19 kgv1= 0v2= 0v′ 1= 1.1 m/s [S]v′ = ?7.2Choose south as positive.p = p′mv + mv = mv′ + mv′1 1 2 2 1 1 2 20 = mv′ 1 1+m2v′2mv′1 1v′ 2=−m22(100 kg)(1.1 m/s)=−19 kgv′ =−5.8 m/s [S], or 5.8 m/s [N]The hiker threw the backpack with a velocity of 5.8 m/s [N] relative to the water.3m1= 1.13×10 kg3m2= 1.25×10 kgv1= 25.7 m/s [E]v2= 13.8 m/s [W]v′ = v′ = v′= ? (when coupled)1 2Choose east as positive.mv 1 1+ m2v2= mv 1 1′ + mv 2 2′mv 1 1+ m2v2= ( m1+m2)v′mv 1 1+m2v2v′ =m1+m23 3(1.13× 10 kg)(25.7 m/s) + (1.25× 10 kg)( −13.8 m/s)=3 31.13× 10 kg + 1.25×10 kgv′ = 4.95 m/s [E]The velocity of the system after the collision is 4.95 m/s [E].312 Unit 2 Energy and Momentum Copyright © 2003 Nelson

8. (a) For the first vehicle:3m1= 1.13×10 kgv1= 25.7 m/s [E]v′ 1= v′ 2= v′= 4.95 m/s [E] (when coupled)∆ p = ?∆ p = p2 − p1= mv 1′ −mv1 1= m1( v′−v1)3= (1.13 × 10 kg)(4.95 m/s −25.7 m/s)4∆ p = − 2.34× 10 kg ⋅m/s [E]For the second vehicle:3m2= 1.25×10 kgv2= 13.8 m/s [W]v1′ = v2′ = v′= 4.95 m/s [E] (when coupled)∆ p = ?∆ p = p2 − p1= mv 2′ −mv2 2= m2( v′−v2)3= (1.25 × 10 kg)(4.95 m/s −( −13.8 m/s))4∆ p = 2.34× 10 kg ⋅m/s [E](b) These two quantities are equal in magnitude and opposite in direction.(c) The total change in momentum of the two-automobile system is zero.9. mL = 89 kgvQ= 0vL= 5.2 m/svQ′ = vL′ = v′= 2.7 m/sm = ?Qm v + m v = m v′ + m v′Q Q L L Q Q L L0 + mvL L= mv′ Q+ mv′LmvQ′ = mvL L−mvL′mL( vL− v′)mQ=v′(89 kg)(5.2 m/s − 2.7 m/s)=2.7 m/sm = 82 kgThe mass of the quarterback is 82 kg.10. Choose original direction of 0.25-kg ball as positive.m1= 0.18 kgm2= 0.25 kgv1=−2.5 m/sv2= 1.7 m/sv2′ =−0.10 m/sv′ = ?1QCopyright © 2003 Nelson Chapter 5 Momentum and Collisions 313

mv1 1+ mv2 2= mv1 1′ + mv2 2′mv′ = mv + m v −mv′1 1 1 1 2 2 2 2mv1 1+ m2( v2−v2′)v1′ =m1(0.18 kg)( − 2.5 m/s) + (0.25 kg)(1.7 m/s −( −0.10 m/s))=0.18 kgv′ 1= 0 m/sThe velocity of the 0.18-kg ball after the collision is 0 m/s.Applying Inquiry Skills11. Set up the carts as shown in the text on a track. Attach a spark timing recorder to each cart to record location for each timeinterval. Turn on the timers, and release the carts. Analyze each tape close to the collision to determine the speed of eachcart. Mass each cart and calculate the momentum of each cart after the collision. Check to see if each one adds up to zero(as it should).12. (a) Since the car and SUV came to an immediate halt at the location of the crash, the total momentum of the system waszero, before and after the collision. That would mean that the total momentum before and after would have the samemagnitude.mvC C= mvS S(but mS=2 mC)mvC C=(2 mC)vSmv C C=2mvC SvC= 2vSThis data indicates the car was travelling at twice the speed of the SUV.(b) If both drivers had the same speed, there would have been momentum after the collision in the direction of the originalmotion of the SUV.5.3 ELASTIC AND INELASTIC COLLISIONSTry This Activity: Newton’s Cradle(Page 248)(a) Each collision is successive. Assuming an elastic collision,mv + mv = mv′ + m v′(but v = 0 and all masses are equal)and1 1 2 2 1 1 2 2 2mv + 0 = mv′ + mv′1 1 2v = v′ + v′1 1 2v′ = v −v′2 1 1(Equation 1)+ = ′ + ′ = 0 and all masses are equal)1 2 1 2 1 2 1 2mv 1 1 mv 2 2 mv2 2 2 1 1 m2 2v2 (but v2mv mv′ mv′2 2 21+ 0 =1+2v v′ v′2 2 21 = 1 + 2v′ v v′2 2 22 = 1 − 1Substitute Equation 1 into Equation 2:( v − v′ ) = v −v′2 2 21 1 1 1v v v′ v′ v v′2 2 2 21− 21 1+ 1=1−12/ v′ − 2/′ = 021vv1 1v′ ( v′− v ) = 01 1 1v′ = 0 or v′− v = 01 1 1(Equation 2)If v′ 1− v1 = 0, then v′ 1= v1 (i.e., the speed of the ball after the collision is unchanged). This is not possible, so v′1= 0.The final speed of the first ball is zero after the collision.314 Unit 2 Energy and Momentum Copyright © 2003 Nelson

mv1 1+ mv2 2= mv1 1′ + mv2 2′mv′ = mv + m v −mv′1 1 1 1 2 2 2 2mv1 1+ m2( v2−v2′)v1′ =m1(0.18 kg)( − 2.5 m/s) + (0.25 kg)(1.7 m/s −( −0.10 m/s))=0.18 kgv′ 1= 0 m/sThe velocity of the 0.18-kg ball after the collision is 0 m/s.Applying Inquiry Skills11. Set up the carts as shown in the text on a track. Attach a spark timing recorder to each cart to record location for each timeinterval. Turn on the timers, and release the carts. Analyze each tape close to the collision to determine the speed of eachcart. Mass each cart and calculate the momentum of each cart after the collision. Check to see if each one adds up to zero(as it should).12. (a) Since the car and SUV came to an immediate halt at the location of the crash, the total momentum of the system waszero, before and after the collision. That would mean that the total momentum before and after would have the samemagnitude.mvC C= mvS S(but mS=2 mC)mvC C=(2 mC)vSmv C C=2mvC SvC= 2vSThis data indicates the car was travelling at twice the speed of the SUV.(b) If both drivers had the same speed, there would have been momentum after the collision in the direction of the originalmotion of the SUV.5.3 ELASTIC AND INELASTIC COLLISIONSTry This Activity: Newton’s Cradle(Page 248)(a) Each collision is successive. Assuming an elastic collision,mv + mv = mv′ + m v′(but v = 0 and all masses are equal)and1 1 2 2 1 1 2 2 2mv + 0 = mv′ + mv′1 1 2v = v′ + v′1 1 2v′ = v −v′2 1 1(Equation 1)+ = ′ + ′ = 0 and all masses are equal)1 2 1 2 1 2 1 2mv 1 1 mv 2 2 mv2 2 2 1 1 m2 2v2 (but v2mv mv′ mv′2 2 21+ 0 =1+2v v′ v′2 2 21 = 1 + 2v′ v v′2 2 22 = 1 − 1Substitute Equation 1 into Equation 2:( v − v′ ) = v −v′2 2 21 1 1 1v v v′ v′ v v′2 2 2 21− 21 1+ 1=1−12/ v′ − 2/′ = 021vv1 1v′ ( v′− v ) = 01 1 1v′ = 0 or v′− v = 01 1 1(Equation 2)If v′ 1− v1 = 0, then v′ 1= v1 (i.e., the speed of the ball after the collision is unchanged). This is not possible, so v′1= 0.The final speed of the first ball is zero after the collision.314 Unit 2 Energy and Momentum Copyright © 2003 Nelson

Substituting back into Equation 1 gives:v′ 2= v1−v′1= v1−0v′ 2= v1Each sphere collision will leave the previous one stationary as the next one moves on. Only one sphere can move out fromthe end.(b) Similarly, only one sphere can end up with all of the kinetic energy.(c) According to the above calculations, each sphere will “pass” its momentum onto the next. When the last sphere receivesthe momentum, and therefore, speed, it will rise to the same vertical height as the original sphere was dropped.(d) If two spheres are used, then two spheres will move out from the other side. If three spheres are used, then three sphereswill move out from the other side.PRACTICE(Page 248)Understanding Concepts1. All of the original kinetic energy is transformed to other forms if both objects come to rest after the collision.2. Yes, we can conclude the collision is completely inelastic. After a “hit-and-stick” collision, no energy is stored as elasticpotential to be returned to either object at the end of the collision.3. A head on collision is very dangerous because of the high relative velocity between the vehicles and the large (and rapid)change in speed for each one. These large accelerations produce large forces that are capable of inflicting serious damageon a human body.Applying Inquiry Skills4. If the ball tends to return to shape quickly when squeezed, it will have a more elastic collision than one that returns to itsoriginal shape more slowly.5. (a)(b)Copyright © 2003 Nelson Chapter 5 Momentum and Collisions 315

Making Connections6. (a) The impact force is reduced by the soft interior because it takes the same impulse, and causes the interaction to lastlonger. The larger time interval of collision means that a smaller force is applied. A hard interior would have a shortduration of collision, and a higher force.(b) If a helmet does not fit properly, the force applied to the head is not evenly distributed to the whole head. The smallerdistribution area means the force would be applied to a smaller area, increasing the pressure to that portion of the head.(c) Once a helmet is involved in a collision, it should be replaced. One way a helmet is designed to reduce injury is toabsorb some of the impact by breaking. This is only able to help the wearer one time. In future collisions, the safety ofthe wearer could be compromised.7. Some possible answers:Flexible• absorb energy• reduce deceleration of train (and passengers)Rigid• prevent car from collapsing and injuring passengers located at either end• less likely to have debris flying about the interior (due to crumpling car)PRACTICE(Pages 251–252)Understanding Concepts8. The larger truck would have the larger momentum.2mvEK=22 2mv=2m2( mv)=2m2pEK=2mp = 2mEKp = 2 m E and p = 2m Esmall small K small large large K largeSince E = E ,K smallK largepsmall = 2 msmallEK and plarge = 2mlargeEKThe mass of the larger vehicle is larger, and the momentum will be too.9. (a) All objects have mass. If it also has a velocity, then it will have both momentum and kinetic energy.(b) For an isolated system of two objects, it is possible to have a momentum of zero. One example would be two cartsreleased from an internal spring “explosion”. The total momentum is zero, but both of the individual parts have kineticenergy. If the system of objects has momentum, then at least one of them is moving, and there will be kinetic energy. Itis not possible to have momentum and zero kinetic energy at the same time.10. m1= 0.15 kgm2= 0.15 kgv1= 22 m/s [N]v2= 22 m/s [S]v′ = v′ = v′= ? (when coupled)1 2316 Unit 2 Energy and Momentum Copyright © 2003 Nelson

Choose north as positive.mv + mv = mv′ + mv′1 1 2 2 1 1 2 2mv + m v = ( m + m ) v′1 1 2 2 1 2mv + m vv′ =m + m1 1 2 21 2(0.15 kg)(22 m/s) + (0.15 kg)( −22 m/s)=0.15 kg + 0.15 m/sv′ = 0 m/sThe velocity of the two-ball system after the collision is 0 m/s.11. Conservation of Momentummv + mv = mv′ + m v′(but v = 0 and the masses are equal)Conservation of Energy1 1 2 2 1 1 2 2 2mv + 0 = mv′ + mv′1 1 2v = v′ + v′1 1 2v′ = v −v′2 1 1(Equation 1)+ = ′ + ′ = 0 and the masses are equal)1 2 1 2 1 2 1 2mv 1 1 mv 2 2 mv2 2 2 1 1 m2 2v2 (but v22 2 2mv1 + 0 = mv′ 1+ mv′22 2 2v1 = v1′ + v2′2 2 2v2′ = v1 −v1′(Equation 2)Substitute Equation 1 into Equation 2:2 2 2( v1− v′ 1)= v1 −v′12 2 2 2v1 − 2v1v′ 1+ v′ 1= v1 −v′122/ v′ 1− 2/vv′1 1= 0v′ 1( v1′− v1) = 0v′ = 0 or v′− v = 01 1 1If v1′ − v1 = 0, then v1′ = v1 (i.e., the speed of the proton after the collision is unchanged). This is not possible, so v1′= 0.The final speed of the first proton is zero after the collision.Substituting back into Equation 1 gives:v2′ = v1 −v1′= v1−0= v1v2′ = 815 m/sThe final velocity of the second proton after the collision is 815 m/s in the direction of the initial velocity.12. For an inelastic collision, the vehicles stick together.4m1= 1.3×10 kg3m2= 1.1×10 kgv1= 90 km/h [N]v2= 30 km/h [N]v′ = v′ = v′= ? (when coupled)1 2Copyright © 2003 Nelson Chapter 5 Momentum and Collisions 317

13.Choose north as positive.mv + mv = mv′ + m v′1 1 2 2 1 1 2 2when coupled, v′ = v′ = v′1 2mv + mv = ( m + m ) v′1 1 2 2 1 2mv + mvv′ =m + m1 1 2 21 24 3(1.3 × 10 kg)(90 km/h) + (1.1×10 kg)(30 km/h)=4 31.3× 10 kg + 1.1×10 kgv′ = 85 km/h [N]The velocity of the vehicles is 85 km/h [N] just after the collision.4m = 1.3×10 kgtctm = 1.1×10 kgc3v = 90 km/h [N] = 25 m/s [N]v = 30 km/h [N] = 8.333 m/s [N]E = E + EET K truck K carT1 2 1 2= mv t t + mv c c2 21 1= (1.3 × 10 kg)(25 m/s) + (1.1×10 kg)(8.333 m/s)2 26= 4.1×10 J4 2 3 2The final velocity of the vehicles after the collision is 85.319 km/h [N] = 23.7 m/s [N].E′ = E′ + E′T K truck K car1 2 1 2= mv t t′ + mv c c′2 21 1= (1.3× 10 kg)(23.7 m/s) + (1.1×10 kg)(23.7 m/s)2 26E′ = 4.0×10 JT4 2 3 2The decrease in kinetic energy is 4.1 × 10 6 – 4.1 × 10 6 = 1 × 10 5 J.14. Choose the original direction of motion as the positive direction.−26mO= 5.31×10 kg−26mN= 4.65×10 kgvO= 02v′ O= 4.81×10 m/sv′ N=−34.1 m/svN= ?mNvN + mOvO = mNv′ N+ mOv′OmNvN + 0 = mNvN′ + mOvO′mNvN′ + mOvO′vN=mNmOvO′= vN′+mN−26 2(5.31× 10 kg)(4.81×10 m/s)=− 34.1 m/s +−264.65×10 kg2v = 5.15×10 m/sThe initial speed of the nitrogen molecule wasN25.15× 10 m/s .318 Unit 2 Energy and Momentum Copyright © 2003 Nelson

Applying Inquiry Skills15. (a) Line A represents the total momentum because the total momentum in the system is constant.(b) The collision is an inelastic one. If the collision were elastic, the total kinetic energy before and after would have beenthe same.Making Connections16. (a) The rubber bullet would have the elastic collision and the lead bullet would have the inelastic collision. (b) ∆ pT = −∆pR mTvT2− mTvT1=−( mRvR2−mRvR1)v T1= 0, and v R2= −vR1 mvT T2− 0 = mvR R1−mR( −vR1) mTvT2= mRvR2+ mRvR1 vR1=vR2 mT vT2 = mR vR2 + mR vR1 p′ = p′+ pT R R ∆ p =−∆pmv mv mv mv v mv T T2− 0 = mv L L2−mvL L2 mT vT2 = mL vL1 −mL vL1 p′ = p − p′T L T T2− T T1= − L L2+ L L1 T1=0T L LThe rubber bullet transfers more momentum to the target.(c) Rubbers bullets are preferred in crowd control because they are less likely to kill or permanently injure any of thecrowd and they impart a larger backward impulse on the crowd.Section 5.3 Questions(Page 253)Understanding Concepts1. (a) It is not possible for both objects to be at rest. If they were both at rest, the initial momentum of the first object wouldhave violated the law of conservation of t momentum.(b) It is possible for the first object to be at rest after the collision. One example is a curling stone that strikes another andthen stops moving.2. The does not violate the law of conservation of momentum for the system which contains the earth and the snowball.The earth exerts a net external force on the tree/snowball system. This external force negates the conservation ofmomentum for that system.3. There momentums will only be the same if they have the same mass. The relationship between momentum and kineticenergy is2mvEK=22 2mv=2m2( mv)=2m2pEK=2mp = 2mEKWhen two objects have the same kinetic energy, the object with the larger mass will always have a larger momentum.Copyright © 2003 Nelson Chapter 5 Momentum and Collisions 319

4. Conservation of Momentummv + mv = mv′ + m v′(but v = 0)1 1 2 2 1 1 2 2 2mv + 0 = mv′ + m v′1 1 1 1 2 2mv − mv′ = m v′1 1 1 1 2 2m ( v − v′ ) = m v′(Equation 1)1 1 1 2 2Conservation of Energy1 2 1 2 1 2 1 2mv + mv = mv′ + m v′(but v = 0)2 1 1 2 2 2 2 1 1 2 2 2 2mv mv′ m v′2 2 21 1+ 0 =1 1+2 2mv mv′ m v′2 2 21 1− 1 1=2 2m ( v v′ ) m v′(Equation 2)2 2 21 1 − 1 = 2 2Divide Equation 2 by Equation 1:2 2 2m1( v1 − v′ 1) m2v′2=m ( v − v′ ) m v′1 1 1 2 2( v1 + v1′ )( v1 −v1′)= v2′( v − v′)1 1v′ = v + v′ (Equation 3) or v′ = v′−v(Equation 4)2 1 1 1 2 1Substitute Equations 3 and 4 back into the original conservation of momentum equation:mv + 0 = mv′ + mv′1 1 1 1 2 2mv = mv′ + m ( v + v′)1 1 1 1 2 1 1mv = mv′ + m v + m v′1 1 1 1 2 1 2 1mv′ + mv′= mv −mv1 1 2 1 1 1 2 1v′ ( m + m ) = ( m −m ) v1 1 2 1 2 1⎛m− m ⎞v′ = ⎜ ⎟v⎝ ⎠1 21 1m1+m2⎛0.022 kg − 0.027 kg ⎞= ⎜ ⎟ (3.5 m/s)⎝0.022 kg + 0.027 kg ⎠v′ =− 0.36 m/s [forward], or 0.36 m/s [backward]1The velocity of the 22-g superball after the collision is 0.36 m/s [backward] .mv + 0 = mv′ + mv′1 1 1 1 2 2mv = m( v′ − v)+ m v′1 1 1 2 1 2 2mv = mv′ − mv + m v′1 1 1 2 1 1 2 2mv′ + mv′= 2mv1 2 2 2 1 1v′ ( m + m ) = 2mv2 1 2 1 1⎛ 2m⎞v′ = ⎜ ⎟v⎝ ⎠12 1m1+m2⎛ 2(0.022 kg) ⎞= ⎜ ⎟(3.5 m/s)⎝0.022 kg + 0.027 kg ⎠v′ = 3.1 m/s [forward]2The velocity of the 27-g superball after the collision is 3.1 m/s [forward] .320 Unit 2 Energy and Momentum Copyright © 2003 Nelson

5. Using the equations derived in question 4 ,⎛m1−m2⎞v′ = ⎜ ⎟v⎝m1+m2⎠1 ⎛m−m2⎞v1 = ⎜ ⎟v13 ⎝m+m2⎠m+ m2 = 3( m−m2)m+ m2 = 3m−3m24m2= 2m1m2= m26. m = 66 kgg = 9.8 m/s∆ y = 25 mv = ?121 1Use conservation of energy to solve for the speed of the moving skier at the bottom of the hill.E = E′T1T1 2mg∆ y = mv12v = 2g∆y= 2(9.8 m/s )(25 m)v1= 22.14 m/sm2= 72 kgv1= 22.14 m/sv2= 0v1′ = v2′ = v′= ? (when coupled)mv1 1+ mv2 2= mv′ 1 1+m2v′2v2 = 0, and when coupled, v′ 1= v′ 2= v′mv1 1+ 0 = ( m1+m2)v′mv1 1v′ =m + m21 2(66 kg)(22.14 m/s)=66 kg + 72 kgv′ = 11 m/sThe speed of the two-skier system immediately after the collision is 11 m/s.Applying Inquiry Skills7. (a) The collision takes place very quickly. The duration of time that the bullet and block are interacting, the strings arevertical, and only balance the gravitational forces. During the collision (not during the swing), momentum is conserved.(b) Let V be the speed of the bullet and block combination after the collision.mv1 1+ mv2 2= mv′ 1 1+ mv′ 2 2(but v2= 0 and when coupled, v′ 1= v′2=V)mv + 0 = ( m + M ) VmvV =m+M(c) The law of conservation of energy can be used to relate the maximum vertical height to the speed just after collision.Copyright © 2003 Nelson Chapter 5 Momentum and Collisions 321

(d) ET = E′T1 2mg∆ y = mv21 2gh = V22Vh =2g2⎛ mv ⎞⎜ ⎟m+M=⎝ ⎠2g2 2mvh =22 gm ( + M)2 2mv(e) h =22 gm ( + M)2 2 2mv = 2 ghm ( + M)22 gh( m + M )v =2m2⎛m+M ⎞= 2gh⎜ ⎟⎝ m ⎠⎛m+M ⎞v = 2gh⎜ ⎟⎝ m ⎠(f) m = 87 g = 0.0087 kgM = 5.212 kg2g = 9.8 m/sh = 6.2 cm = 0.062 mv = ?⎛m+M ⎞v = 2gh⎜ ⎟⎝ m ⎠2 ⎛0.0087 kg + 5.212 kg ⎞= 2(9.8 m/s )(0.062 m) ⎜ ⎟⎝ 0.0087 kg ⎠2v = 6.6×10 m/s2The initial speed of the bullet was 6.6× 10 m/s .(g) Some of the sources of error would be friction of the moving parts of the gun and loss of energy to thermal energy inthe spring. The frictional draw of the catch mechanism would use some of the energy that should be converted intogravitational potential.8. There are a number of advantages of a crumple zone. One is that it converts a significant part of the kinetic energy of thevehicle into thermal energy as the steel become permanently deformed. The crumple zone also causes the vehicle to slowdown over a greater distance. This greater distance increases the length of time the vehicle is slowing down, and theaverage force to stop the vehicle (and its passengers) decreases.9. Most meteorites burn up in the atmosphere. Larges known one is about 60 metric tons, and the next known one is about30 metric tons. Large collisions have rarely (ever 300 million years or so) and can cause devastating climactic change.322 Unit 2 Energy and Momentum Copyright © 2003 Nelson

5.4 CONSERVATION OF MOMENTUM IN TWO DIMENSIONSPRACTICE(Pages 257–258)Understanding Concepts1.2. (a) ms= 52 kgmc= 26 kgvs= 1.2 m/s [W]vc= 1.2 m/s [S]vs′ = ?v′ = ?cCopyright © 2003 Nelson Chapter 5 Momentum and Collisions 323

Calculate the momenta: p = m vsss s= (52 kg)(1.2 m/s)p = 62.4 kg ⋅m/s [S] ps + pc = 93.6 kg ⋅m/s [S] p′ = m v′sss s= (52 kg)(1.0 m/s)p′ = 52 kg ⋅m/s [W]pcc= m vc c= (26 kg)(1.2 m/s)p = 31.2 kg ⋅m/s [S]Measure length and angle to get:2p′c= 1.1× 10 kg ⋅ m/s [61 ° S of E] pc′ = mcvc′ pc′vc′ =mcc21.1× 10 kg ⋅m/s=26 kgv′ = 4.1 m/s [61 ° S of E]The approximate final velocity of the cart is 4.1 m/s [61 ° S of E] .(b) Using the diagram from part (a), the total momentum after is 2 2p′ = p′+ p + ppc s s c2 2= (52 kg ⋅ m/s) + (93.6 kg ⋅m/s)1.1 10 kg m/s2c′ = × ⋅ ps+ pctanθ= p′s ⎛−1ps+ pc⎞θ = tan⎜ p ⎟⎝′s ⎠−1⎛93.6⎞= tan ⎜ ⎟⎝ 52 ⎠θ = 61°2So, p′c= 1.1× 10 kg ⋅ m/s [61 ° S of E].3. Diagram is not to scale. Dotted line represents the direction “after.”A completely inelastic collision means they will stick together.123m1= 1.4×10 kg3m2= 1.3×10 kgv1= 45 km/h [S] = 12.5 m/s [S]v2= 39 km/h [E] = 10.83 m/s [E]v′ = ?324 Unit 2 Energy and Momentum Copyright © 2003 Nelson

pp1 1 11pp= m v3= (1.4 × 10 kg)(12.5 m/s)= 1.75× 10 kg ⋅m/s2 2 21= m v43= (1.3 × 10 kg)(10.83 m/s)= 1.408× 10 kg ⋅m/s p p p2 2′12=1+2p412′ = × ⋅44 2 4 2= (1.75× 10 kg ⋅ m/s) + (1.408× 10 kg ⋅m/s)p′ = m v′p12′v12′ =m2.246 10 kg m/s12 12 1212122.246× 10 kg ⋅m/s=3 31.3× 10 kg + 1.4×10 kgv′ = 8.3 m/sp1tanθ= p2⎛1p1⎞−θ = tan⎜ p ⎟⎝ 2 ⎠4−1⎛ 1.75× 10 kg ⋅m/s⎞= tan ⎜ 4 ⎟⎝1.408× 10 kg ⋅m/s⎠θ = 51°The final speed of the cars is 8.3 m/s [51º S of E], or 3.0 × 10 1 km/h [51º S of E].4Copyright © 2003 Nelson Chapter 5 Momentum and Collisions 325

4. (a) Diagram not to scale.(b)Bv0 = mv −mvByBByAyABy0 = v −v= vAysin 31.1°= vsin 48.9°Ayv sin 48.9°= v sin 31.1°vmv = mv + mvA1 AxBxv = v + vA1 AxBx2.25 = v cos31.1°+ v cos 48.9°AAAB⎛sin 31.1°⎞2.25 = vAcos31.1°+ ⎜ vA⎟cos 48.9°⎝sin 48.9°⎠2.25 = 1.307vv= 1.72 m/sAvvBBsin 31.1°= vAsin 48.9°sin 31.1 °= (1.72 m/s)sin 48.9°= 1.18 m/sThe velocities of the balls are 1.18 m/s at 48.9° and 1.72 m/s at 31.1°.326 Unit 2 Energy and Momentum Copyright © 2003 Nelson

(c)(d)EEKT==12121KT =2mv2A1m(2.25)m(5.06)2E′ = mv + mvE1 2 1 2KT 2 A 2 B121KT′ =22 2A vB2 21= mv (2+ )= m(1.72 + 1.18 )m(4.35)The total kinetic energy after is less than the total kinetic energy before, so the collision is not elastic.5. (a) The residual nucleus will move in the opposite direction of the combined momentums of the two particles so the totalmomentum is still zero. (Note: Diagram is not to scale.)−21pn= 4.8× 10 kg ⋅m/s [S]−21pe= 9.0× 10 kg ⋅m/s [E]θ = ?ptanθ=p−1 ⎛ pn⎞θ = tan ⎜ ⎟⎝ pe⎠−21−1⎛4.8× 10 kg ⋅m/s⎞= tan ⎜ −21⎟⎝9.0× 10 kg ⋅m/s⎠θ = 28°The nucleus will move in the direction 28º N of W.(b) pnucleus= ?2 2p = p + ppnenucleus n e−21 2 −21 2nucleus= (4.8× 10 kg ⋅ m/s) + (9.0× 10 kg ⋅m/s)−20= 1.0× 10 kg ⋅m/s−20The magnitude of the nucleus’s momentum is 1.0× 10 kg ⋅ m/s .Copyright © 2003 Nelson Chapter 5 Momentum and Collisions 327

(c)−25m nucleus= 3.6×10 kgvnucleus= ?p = m vpnucleusvnucleus=mvvnucleus nucleus nucleusnucleusnucleusnucleus−20The recoil velocity of the nucleus is1.0× 10 kg ⋅m/s=−253.6×10 kg4= 2.8×10 m/s4= 2.8× 10 m/s [28 ° N of W]42.8× 10 m/s [28 ° N of W] .Applying Inquiry Skills6. (a) The black car was travelling faster at the moment of impact. There was a larger component of momentum in theoriginal direction of the motion of the black car because there is less deviation in its path than the path of the white car.(b) The investigator could determine more precise information by measuring the angle of deviation for the cars and thetotal distance that the combined cars slid.Making Connections7. Answers will vary depending on the type of equipment chosen. Most likely there will be some padding included in it thatwill reduce the force required by increasing the time of collision.Section 5.4 Questions(Pages 258–239)Understanding Concepts1.328 Unit 2 Energy and Momentum Copyright © 2003 Nelson

2.−26m Li= 1.2×10 kgv′ = 0.40 km/s [54 ° to original direction of motion of neutron]Liv′ = 2.5 km/snp′ = ?Lip′ = ?nθ = ?Diagram is not to scale.p′ = m v′pLi Li Li−26= (1.2 × 10 kg)(0.40 km/s)−27Li′ = × ⋅4.8 10 kg km/sp′ = m v′pn n n−27= (1.7×10 kg)(2.5 km/s)−27n′ = × ⋅4.25 10 kg km/sIn the y-direction:py= p′y0 = p′ Liy− p′ny0= pLi′ sin54°−pn′sinθpLi′ sin 54°sinθ=pn′−1 ⎛ p′Lisin 54°⎞θ = sin ⎜ ⎟⎝ p′n ⎠−27−1⎛(4.8× 10 kg ⋅ km/s)(sin 54 ° ) ⎞= sin ⎜ −27⎟⎝ 4.25× 10 kg ⋅km/s⎠θ = 66 ° [from original direction of motion of the neutron]The neutron is now travelling at 66 ° from its original direction of motion.3. Diagram is not to scale.Copyright © 2003 Nelson Chapter 5 Momentum and Collisions 329

m1= 71 kgv1= 2.3 m/s [12 ° N of E]v2= 1.9 kg ⋅ m/s [52 ° S of W]m = ?2p1 1 11= mv= (71 kg)(2.3 m/s)p = 163.3 kg ⋅m/spp= m v= m2(1.9 kg ⋅m/s)= 1.9 m kg ⋅m/s2 2 22 2Using the sine law:sin101° sin 39°=163.3 1.9m2⎛ sin 39° ⎞⎛163.3 kg ⋅m/s⎞m2= ⎜ ⎟⎜ ⎟⎝1.9 kg ⋅ m/s ⎠⎝ sin101°⎠m2= 55 kgThe mass of the second skater is 55 kg.4. Diagram not to scale.p1x1 1xp1x= mv= (0.50 kg)(2.0 m/s)= 1.0 kg ⋅m/sp′ = mv′1 1 11= (0.50 kg)(1.5 m/s)p′ = 0.75 kg ⋅m/sp′ = m v′2 2 2= (0.30 kg) v′p′ = 0.30 v′kg ⋅m/s2 222p = p′ + p′1x 1x 2xp = p′ cos30°+p′cosθ1x1 21.0 = (0.75)cos30 °+ (0.30 v′)cosθv′ cosθ= 1.168 (Equation 1)220 = p′ − p′1y2y0= p′ sin30°−p′sinθ1 20 = (0.75)sin 30 °−(0.30 v′)sinθv′ sinθ= 1.25 (Equation 2)2Divide Equation 2 by Equation 1:v2′sinθ1.25=v′cosθ1.1682tanθ= 1.070−1θ = tan (1.070)θ = 47°330 Unit 2 Energy and Momentum Copyright © 2003 Nelson

Substitute back into Equation 2:v′ 2sinθ= 1.251.25v′ 2=sinθ1.25=sin 47°v′ 2= 1.7 m/sThe velocity of the second ball after collision is 1.7 m/s [47º S of E].Applying Inquiry Skills5. (a) Referring to the diagram:Ignore points A3 and B3 for calculations because they are during the collision.Measure between points A1 and A2, A4 and A5, and divide by 0.50 s to calculate velocities.Repeat for Puck B (using two values for the after calculation because you have them).Measure angles before and after for velocities. 2.00vA = = 4.0 cm/s [45 ° N of E]0.50 1.9vB = = 3.8 cm/s [38 ° W of N]0.50 1.80v′A = = 3.6 cm/s [N]0.501.3+1.4vB′ = = 2.7 cm/s [43 ° E of N]0.50 + 0.50Copyright © 2003 Nelson Chapter 5 Momentum and Collisions 331

Let the x direction be the east-west direction.px= p′xmAvAx − mBvBx = mAvA′ x+ mBvB′xvA′ x= 0mvA Ax − mvB Bx = 0 + mv′B Bxmv′ B Bx + mvB Bx = mvA AxmB( v′ Bx + vB x)= mAvAxmAvAxmB=vB′x + vBx(0.32 kg)(4.0cos 45 ° )=2.7sin 43°+ 3.8sin 38°= 0.2165 kgm = 0.22 kg(b) mA = 0.32 kgmB= 0.2165 kgvA= 0.040 m/svB= 0.038 m/sE = ?K lostTB1 2 1 2ET = mAvA + mBvB2 21 1= (0.32 kg)(0.040 m/s) + (0.2165 kg)(0.038 m/s)2 2−4E = 4.123×10 JT2 21 2 1 2E′ T= mAv′ A+ mBv′B2 21 1= (0.32 kg)(0.036 m/s) + (0.2165 kg)(0.027 m/s)2 2−4E′ = 2.863×10 J−4 −42 2EK lost = 4.123× 10 J − 2.863×10 J−4EK lost= 1.3×10 J−4The total kinetic energy lost is 1.3× 10 J .(c) This is an inelastic collision.(d) The most likely source of error is measuring the angles from the diagram.Making Connections6. (a) Some possible points:• The perception of a safe vehicle is a common advertising feature of most vehicles.• The idea of protecting loved ones and yourself in a collision is socially desirable.• Insurance companies offer better rates for vehicles with high safety performance in collisions.(b) Some points are:• seat belts• roll bars (roll safety)• bumpers• crumple zones• air bag• anti-lock braking systems• traction control (4WD, AWD, tires)332 Unit 2 Energy and Momentum Copyright © 2003 Nelson

(c) Any of the ones listed in (b).(d) Could compare safety ratings and insurance prices, cost of vehicle (often safer vehicles or options are more expensive),and fuel economy.CHAPTER 5 LAB ACTIVITIESInvestigation 5.2.1: Analyzing One-Dimensional Collisions(Pages 260–262)QuestionObserving two objects before, during, and after a collision allows the verification of conservation of momentum and kineticenergy theory.Prediction(a) For all three categories, the total momentum of the system will always be the same.(b) Category I: The total kinetic energy before and after will be the same. During the collision there will be some loss ofkinetic energy as it is stored as some other form in the repulsive device.Category II: The total kinetic energy before the collision will be zero. It will be a maximum at the end of the explosion,increasing as the explosion takes place.Category III: The total kinetic energy will be a maximum before, decreasing throughout the collision and a minimum atthe end.Hypothesis(c) The velocity of both carts before and after the collision (an adhesive collision would be the easiest) can be determinedusing the ticker-tape timer. Using the known mass of the carts, the conservation of momentum can be used to calculate theunknown mass.Collisionm 1Before the CollisionAfter the CollisionTotal p(kg⋅m/s)Total E K(J)v 1 m 2 v 2 m 1 v 1 ’ m 2 v 2 ’(kg) (m/s) (kg) (m/s) (kg) (m/s) (kg) (m/s)before after before afterI (a) 0.50 0.19 0.50 0.0 0.50 0.0 0.50 0.17 0.095 0.085 0.0090 0.0072 20I (b) 1.0 0.18 0.50 0.0 1.0 0.060 0.50 0.22 0.18 0.17 0.016 0.014 12I (c) 0.50 0.15 0.50 –0.13 0.50 –0.12 0.50 0.14 0.010 0.010 0.0098 0.0085 13II (a) 0.50 0.0 0.50 0.0 0.50 –0.22 0.50 0.21 0.0 –0.0050 0.0 0.023 —II (b) 1.0 0.0 0.50 0.0 1.0 –0.091 0.50 0.27 0.0 0.044 0.0 0.022 —II (c) 0.50 0.0 0.50 0.0 0.50 –0.10 0.50 0.12 0.0 0.010 0.0 0.0061 —III (a) 0.50 0.24 0.50 0.0 0.50 0.11 0.50 0.11 0.12 0.11 0.014 0.0060 57III (b) 1.0 0.20 0.50 0.0 1.0 0.12 0.50 0.12 0.20 0.18 0.020 0.011 45III (c) 0.50 0.18 0.50 –0.17 0.50 0.0 0.50 0.0 0.0050 0.0 0.015 0.0 100IV 0.50 0.22 m 2 0.0 0.50 0.088 m 2 0.088 — — — — —Lossin E K(%)Analysis(d) For each collision, the total momentum before and after was essentially the same.(e) For Category I collisions, there was very little loss of kinetic energy during the collisions. Category II collisions did notstart with any, but received the kinetic energy from the spring bumper. Category III collisions lost a significant amount ofkinetic energy during the collisions.(f) Category I collisions were all elastic (within experimental error). Category III collisions were all completely inelastic.(g) During Category I collisions, there was a temporary loss of kinetic energy that reappeared after the collision.Copyright © 2003 Nelson Chapter 5 Momentum and Collisions 333

(c) Any of the ones listed in (b).(d) Could compare safety ratings and insurance prices, cost of vehicle (often safer vehicles or options are more expensive),and fuel economy.CHAPTER 5 LAB ACTIVITIESInvestigation 5.2.1: Analyzing One-Dimensional Collisions(Pages 260–262)QuestionObserving two objects before, during, and after a collision allows the verification of conservation of momentum and kineticenergy theory.Prediction(a) For all three categories, the total momentum of the system will always be the same.(b) Category I: The total kinetic energy before and after will be the same. During the collision there will be some loss ofkinetic energy as it is stored as some other form in the repulsive device.Category II: The total kinetic energy before the collision will be zero. It will be a maximum at the end of the explosion,increasing as the explosion takes place.Category III: The total kinetic energy will be a maximum before, decreasing throughout the collision and a minimum atthe end.Hypothesis(c) The velocity of both carts before and after the collision (an adhesive collision would be the easiest) can be determinedusing the ticker-tape timer. Using the known mass of the carts, the conservation of momentum can be used to calculate theunknown mass.Collisionm 1Before the CollisionAfter the CollisionTotal p(kg⋅m/s)Total E K(J)v 1 m 2 v 2 m 1 v 1 ’ m 2 v 2 ’(kg) (m/s) (kg) (m/s) (kg) (m/s) (kg) (m/s)before after before afterI (a) 0.50 0.19 0.50 0.0 0.50 0.0 0.50 0.17 0.095 0.085 0.0090 0.0072 20I (b) 1.0 0.18 0.50 0.0 1.0 0.060 0.50 0.22 0.18 0.17 0.016 0.014 12I (c) 0.50 0.15 0.50 –0.13 0.50 –0.12 0.50 0.14 0.010 0.010 0.0098 0.0085 13II (a) 0.50 0.0 0.50 0.0 0.50 –0.22 0.50 0.21 0.0 –0.0050 0.0 0.023 —II (b) 1.0 0.0 0.50 0.0 1.0 –0.091 0.50 0.27 0.0 0.044 0.0 0.022 —II (c) 0.50 0.0 0.50 0.0 0.50 –0.10 0.50 0.12 0.0 0.010 0.0 0.0061 —III (a) 0.50 0.24 0.50 0.0 0.50 0.11 0.50 0.11 0.12 0.11 0.014 0.0060 57III (b) 1.0 0.20 0.50 0.0 1.0 0.12 0.50 0.12 0.20 0.18 0.020 0.011 45III (c) 0.50 0.18 0.50 –0.17 0.50 0.0 0.50 0.0 0.0050 0.0 0.015 0.0 100IV 0.50 0.22 m 2 0.0 0.50 0.088 m 2 0.088 — — — — —Lossin E K(%)Analysis(d) For each collision, the total momentum before and after was essentially the same.(e) For Category I collisions, there was very little loss of kinetic energy during the collisions. Category II collisions did notstart with any, but received the kinetic energy from the spring bumper. Category III collisions lost a significant amount ofkinetic energy during the collisions.(f) Category I collisions were all elastic (within experimental error). Category III collisions were all completely inelastic.(g) During Category I collisions, there was a temporary loss of kinetic energy that reappeared after the collision.Copyright © 2003 Nelson Chapter 5 Momentum and Collisions 333

+ = ′ + ′mv1 1−mv1 1′=v′(h) mv1 10 mv1 1mv2 2mm222(0.50 kg)(0.22 m/s) − (0.50 kg)(0.088 m/s)=0.088 m/s= 0.75 kgBut m = m + m2 cart unknown mass, thereforem = m −mmunknown mass 2 cartunknown mass= 0.75 kg −0.50 kg= 0.25 kgEvaluation(i) The predictions and hypothesis were correct to within experimental errors.(j) Sources of error during the experiment include:• There was friction between the cart and the track.• There was friction between the tape and the timing recorder.• It is difficult to measure the exact distance between the ticker tape timer dots because the distance is so small.Synthesis(k) It is important to distinguish between scalar and vector quantities because a system momentum of zero may still involvedangerous speeds or motions.(l) Using electronic values, the refuse bag could be tossed to a stationary astronaut. After catching the bag, the velocities andknown mass of the astronaut could be used to calculate the mass of the refuse bag.(m) The friction pads would provide a source of external force to the system. This would increase the loss of kinetic energybefore and after the collision and change the total momentum before and after as well.Investigation 5.3.1: Analyzing Two-Dimensional Collisions(Pages 262–265)QuestionThe laws of conservation of momentum and kinetic energy can be verified through two-dimensional collisions.Prediction(a) The kinetic energy and the total momentum will be the same before and after the collisions.(b) The change in momentum for each puck will be the same as the change in momentum of the other puck.Hypothesis(c) I expect that the two pucks will stick together and move along a straight path that will bisect the initial angle of collision.(d) The change in momentum of puck A will be equal to the change in momentum of puck B. The velocities can memeasured, to the mass can be calculated.(e) Category I and Category II pA + pB = p′ A+ p′B mvA + mvB = mvA′ + mvB′ v + v = v′ + v′A B A B334 Unit 2 Energy and Momentum Copyright © 2003 Nelson

+ = ′ + ′mv1 1−mv1 1′=v′(h) mv1 10 mv1 1mv2 2mm222(0.50 kg)(0.22 m/s) − (0.50 kg)(0.088 m/s)=0.088 m/s= 0.75 kgBut m = m + m2 cart unknown mass, thereforem = m −mmunknown mass 2 cartunknown mass= 0.75 kg −0.50 kg= 0.25 kgEvaluation(i) The predictions and hypothesis were correct to within experimental errors.(j) Sources of error during the experiment include:• There was friction between the cart and the track.• There was friction between the tape and the timing recorder.• It is difficult to measure the exact distance between the ticker tape timer dots because the distance is so small.Synthesis(k) It is important to distinguish between scalar and vector quantities because a system momentum of zero may still involvedangerous speeds or motions.(l) Using electronic values, the refuse bag could be tossed to a stationary astronaut. After catching the bag, the velocities andknown mass of the astronaut could be used to calculate the mass of the refuse bag.(m) The friction pads would provide a source of external force to the system. This would increase the loss of kinetic energybefore and after the collision and change the total momentum before and after as well.Investigation 5.3.1: Analyzing Two-Dimensional Collisions(Pages 262–265)QuestionThe laws of conservation of momentum and kinetic energy can be verified through two-dimensional collisions.Prediction(a) The kinetic energy and the total momentum will be the same before and after the collisions.(b) The change in momentum for each puck will be the same as the change in momentum of the other puck.Hypothesis(c) I expect that the two pucks will stick together and move along a straight path that will bisect the initial angle of collision.(d) The change in momentum of puck A will be equal to the change in momentum of puck B. The velocities can memeasured, to the mass can be calculated.(e) Category I and Category II pA + pB = p′ A+ p′B mvA + mvB = mvA′ + mvB′ v + v = v′ + v′A B A B334 Unit 2 Energy and Momentum Copyright © 2003 Nelson

Category III pA + pB = pA′ + pB′ mvA + mvB = mvA′ + mvB′ v′ A= v′ B(call this v′) vA+ vB= v′ + v′ v + v = 2v′ABCategory IV pA + pB = p′ A+ p′B mAvA + mBvB = mAvA′ + mBvB′ mvB B− mvB B′ = mvA A′−mvA A mB( vB − vB′ ) = mA( vA′−vA) mA( vA′ − vA)mB= ( vB− v′B)(f) Mark off the four markers to get a known distance over a known time (the distance between each pair of dots should beapproximately the same). Measure the angles. The sum of the two velocity vectors before should be equal to the sum ofthe velocity vectors after for Category I.(g) For category II, the change in each vector should be the same magnitude and opposite direction.(h) Category III should have the two velocity vectors before be equal to the twice the combined after velocity.(i) For Category IV, subtract the two velocity vectors for each puck and substitute the values into the above equation.(j) Measure each length of velocity vectors before and after. Substitute with the mass into the kinetic energy equation andcompare the total values before and after. Any lost energy went into thermal energy.Evaluation(l) Some possible sources of error:• The table surface may not be flat.• There may be some frictional or air current forces acting on the pucks.• The air supply line to the pucks may exert some lateral forces on the pucks.• Masses of the pucks may not be identical.• Frequency of the spark timer may not be perfectly uniform.Synthesis(m) This would apply to all of the categories because all of them have the total momentum conserved.(n) The momentum would only be conserved if you considered both pucks.(o) It is wish to avoid dots during the impulse and collision because there is a changing speed during these interactions. Theanalysis of this experiment is not designed to include a puck that is accelerating.(p) It is better to have steel barriers because they will absorb some energy and prevent the cars from bouncing off as much asa rubber barrier would.CHAPTER 5 SELF QUIZ(Pages 267–268)True/False1. F The impulse is the same in magnitude and is in the same direction.2. T3. F You have increased the amount of time the same force is applied.4. F The total momentum before and after is the same.5. T6. T7. F The kinetic energies will be different.8. F The momentum is conserved in the snowball-earth system, even though the collision is completely inelastic.9. T10. F The final kinetic energy can be greater (e.g., an exploding cart system).Copyright © 2003 Nelson Chapter 5 Momentum and Collisions 335

Category III pA + pB = pA′ + pB′ mvA + mvB = mvA′ + mvB′ v′ A= v′ B(call this v′) vA+ vB= v′ + v′ v + v = 2v′ABCategory IV pA + pB = p′ A+ p′B mAvA + mBvB = mAvA′ + mBvB′ mvB B− mvB B′ = mvA A′−mvA A mB( vB − vB′ ) = mA( vA′−vA) mA( vA′ − vA)mB= ( vB− v′B)(f) Mark off the four markers to get a known distance over a known time (the distance between each pair of dots should beapproximately the same). Measure the angles. The sum of the two velocity vectors before should be equal to the sum ofthe velocity vectors after for Category I.(g) For category II, the change in each vector should be the same magnitude and opposite direction.(h) Category III should have the two velocity vectors before be equal to the twice the combined after velocity.(i) For Category IV, subtract the two velocity vectors for each puck and substitute the values into the above equation.(j) Measure each length of velocity vectors before and after. Substitute with the mass into the kinetic energy equation andcompare the total values before and after. Any lost energy went into thermal energy.Evaluation(l) Some possible sources of error:• The table surface may not be flat.• There may be some frictional or air current forces acting on the pucks.• The air supply line to the pucks may exert some lateral forces on the pucks.• Masses of the pucks may not be identical.• Frequency of the spark timer may not be perfectly uniform.Synthesis(m) This would apply to all of the categories because all of them have the total momentum conserved.(n) The momentum would only be conserved if you considered both pucks.(o) It is wish to avoid dots during the impulse and collision because there is a changing speed during these interactions. Theanalysis of this experiment is not designed to include a puck that is accelerating.(p) It is better to have steel barriers because they will absorb some energy and prevent the cars from bouncing off as much asa rubber barrier would.CHAPTER 5 SELF QUIZ(Pages 267–268)True/False1. F The impulse is the same in magnitude and is in the same direction.2. T3. F You have increased the amount of time the same force is applied.4. F The total momentum before and after is the same.5. T6. T7. F The kinetic energies will be different.8. F The momentum is conserved in the snowball-earth system, even though the collision is completely inelastic.9. T10. F The final kinetic energy can be greater (e.g., an exploding cart system).Copyright © 2003 Nelson Chapter 5 Momentum and Collisions 335

Multiple Choice11. (e)12. (d) Momentum will increase by 2 × 2, and kinetic enegy by 2 × 2 2 .13. (d)14. (d)15. (e)16. (d) Some energy is lost to thermal energy of the collision.17. (b) The momentum is p = 2 mEK so p∝ EK.18. (c) M only stops, while T must continue to act on R to give it momentum in the opposite direction.19. (a) They have both stopped, so the total momentum is zero, but the initial kinetic energy is stored as electric potential.20. (d)⎛1⎞21. (a) mv + (2 m) ⎜ ( − v) ⎟= ( m + 2 m)v′⎝2⎠mv − mvv′ =3mv′ = 022.(d) 0 = mv −( M − m)VmvV =M − mCHAPTER 5 REVIEW(Pages 269–271)Understanding Concepts1. Impulse is the product of the force and the time. A smaller force exerted over a long period of time can impart a largerimpulse than a large force for a short time.2. As the meteor comes to a stop, the kinetic energy is converted into thermal energy which melts the material at the impactsite.3. This observation does not contradict the law of conservation of momentum. The momentum of the earth increases towardthe falling object. The amount of change in the earth’s velocity is so small it is imperceptible, so it appears the law ofconservation is being contradicted, but it is not.4. The change in momentum of the ball that bounces is greater than the putty that sticks to the floor. Assuming both have thesame mass and are dropped from the same height:For the putty,∆ p = m( v −v)2 1= m(0 −v)∆ p = −mv11For the ball∆ p = m( v −v)2 1= m(( −v ) −v)∆ p =−2mv11 12mv5. E K =22 2mv=2m2pEK=2mThe momentum of the two players is the same, but the person with the larger mass will have a smaller kinetic energy. Itwould be better to avoid the faster moving lighter player, p 2 = m 2 v 2 .6. A car crashing into a tree is an example. Momentum is not conserved in the tree-car system because the tree is attached tothe earth. The earth supplies a net external force to the system of objects so that momentum is not conserved.336 Unit 2 Energy and Momentum Copyright © 2003 Nelson

Multiple Choice11. (e)12. (d) Momentum will increase by 2 × 2, and kinetic enegy by 2 × 2 2 .13. (d)14. (d)15. (e)16. (d) Some energy is lost to thermal energy of the collision.17. (b) The momentum is p = 2 mEK so p∝ EK.18. (c) M only stops, while T must continue to act on R to give it momentum in the opposite direction.19. (a) They have both stopped, so the total momentum is zero, but the initial kinetic energy is stored as electric potential.20. (d)⎛1⎞21. (a) mv + (2 m) ⎜ ( − v) ⎟= ( m + 2 m)v′⎝2⎠mv − mvv′ =3mv′ = 022.(d) 0 = mv −( M − m)VmvV =M − mCHAPTER 5 REVIEW(Pages 269–271)Understanding Concepts1. Impulse is the product of the force and the time. A smaller force exerted over a long period of time can impart a largerimpulse than a large force for a short time.2. As the meteor comes to a stop, the kinetic energy is converted into thermal energy which melts the material at the impactsite.3. This observation does not contradict the law of conservation of momentum. The momentum of the earth increases towardthe falling object. The amount of change in the earth’s velocity is so small it is imperceptible, so it appears the law ofconservation is being contradicted, but it is not.4. The change in momentum of the ball that bounces is greater than the putty that sticks to the floor. Assuming both have thesame mass and are dropped from the same height:For the putty,∆ p = m( v −v)2 1= m(0 −v)∆ p = −mv11For the ball∆ p = m( v −v)2 1= m(( −v ) −v)∆ p =−2mv11 12mv5. E K =22 2mv=2m2pEK=2mThe momentum of the two players is the same, but the person with the larger mass will have a smaller kinetic energy. Itwould be better to avoid the faster moving lighter player, p 2 = m 2 v 2 .6. A car crashing into a tree is an example. Momentum is not conserved in the tree-car system because the tree is attached tothe earth. The earth supplies a net external force to the system of objects so that momentum is not conserved.336 Unit 2 Energy and Momentum Copyright © 2003 Nelson

7. Diagram is not to scale.NE2m = 1.3×10 kgv = 8.7 m/s [44 ° E of N]pp= ?= ? p = mv2= (1.3 × 10 kg)(8.7 m/s)p = 1131 kg ⋅ m/s [44 ° E of N]ppppNNEE= pcos 44°= (1131 kg ⋅ m/s)(cos 44 ° )2= 8.1× 10 kg ⋅m/s= psin 44°= (1131 kg ⋅ m/s)(sin 44 ° )2= 7.9× 10 kg ⋅m/sThe northward component of the boat’s momentum is8. Diagram is not to scale.28.1× 10 kg ⋅ m/s . The eastward component is27.9× 10 kg ⋅ m/s .pE= 2.6× 10 kg ⋅m/sm = 1.1×10 kgθ = 22°v = ?34Copyright © 2003 Nelson Chapter 5 Momentum and Collisions 337

pE= mvcos 22°pEv =m cos 22°42.6× 10 kg ⋅m/s=3(1.1× 10 kg)(cos 22 ° )v = 25 m/sThe car is travelling at a speed of 25 m/s.9. Choose west as positive.3m = 1.2×10 kgv1= 53 km/h [W] = 14.72 m/s [W]v2= 0−2∆ t = 55 ms = 5.5×10 sΣ Fx= ?ΣFx∆ t = ∆pmv ( 2 − v1)Σ Fx=∆t3(1.2 × 10 kg)(0 m/s −14.72 m/s)=−25.5×10 s5=− 3.2×10 N [W]5Σ Fx= 3.2×10 N [E]5The average force exerted on the car by the pole is 3.2× 10 N [E] .10. (a) Σ F x= 324 N−3∆ t = 5.1 ms = 5.1×10 sΣFx∆ t = ?−3ΣFx∆ t = (324 N)(5.1×10 s)ΣFx∆ t = 1.7 N ⋅s [fwd]The impulse on the ball is 1.7 N⋅s [fwd].(b) m = 0.059 kgv1= 0v2= ?ΣFx∆ t = ∆pΣFx∆ t = m( v2 −v1)ΣFx∆ t = mv2ΣFx∆tv2=m−3(324 N)(5.1×10 s)=0.059 kgv2= 28 m/s [fwd]The velocity of the ball just as it leaves the racket is 28 m/s [fwd].11. m1= 0.112 kgm2= 0.154 kgv′ 1= 1.38 m/sv′ = ?2338 Unit 2 Energy and Momentum Copyright © 2003 Nelson

p = p′0 = mv1 1′ + m2v2′mv1 1′v′ 2=−m2(0.112 kg)(1.38 m/s)=−0.154 kgv′ 2=−1.00 m/sThe speed of the second car is 1.00 m/s.−2712. m P= 1.67×10 kg−27mα= 6.64×10 kgvP= 1.57 km/svα= 0vP′ =−0.893 km/svα′ = ?mvP P+ mvα α= mv′ P P+ mv′α αmvP P+ 0 = mv′ P P+ mv′α αmvP P− mv′P Pv′ α=mαmP( vP − vP′)=mα−27(1.67 × 10 kg)(1.57 km/s −( −0.893 km/s))=−276.64×10 kgvα′ = 0.620 km/sThe speed of the alpha particle is 0.620 km/s.13. m1= 2.67 kgm2= 5.83 kg2v1= 1.70×10 m/s [toward Jupiter]v1′ = 185 m/s [toward Jupiter]v′ 2= 183 m/s [toward Jupiter]v = ?2Choose the direction toward Jupiter as positive.mv 1 1+ m2v2= mv 1 1′ + mv 2 2′mv 2 2= mv 1 1′ − mv 1 1+mv 2 2′m1( v1′ − v1)+ m2v2′v2=m2m1( v1′ − v1)= + v2′m22(2.67 kg)(185 m/s − 1.70×10 m/s)= + 183 m/s5.83 kg2v2= 1.90×10 m/s2v = 1.90×10 m/s [toward Jupiter]The velocity of the more massive rock is221.90× 10 m/s [toward Jupiter] .Copyright © 2003 Nelson Chapter 5 Momentum and Collisions 339

14. (a) v1i= 6.0 m/sv2i= 0v1f= v2f= 2.0 m/sThe two objects have the same final velocity. Having the same final velocity is an indication of a ‘hit-and-stick’collision. All collisions that have the objects with the same speed (in the same direction) after are completely inelasticcollisions.(b) v1i= 24 m/sv2i= 0v1f=−4.0 m/sv = 14 m/s2f1 2 1 2EKi = mv1 1i+ m2v2i2 21 1= (2.0 kg)(24 m/s) + (4.0 kg)(0 m/s)2 2E = 576 JKi2 2There is some loss of kinetic energy. This is an inelastic collision.(c) v1i = 12 m/sv2i= 0v1f=−4.0 m/sv = 8.0 m/s2f1 2 1 2EKf = mv1 1f+ m2v2f2 21 1= (2.0 kg)( − 4.0 m/s) + (4.0 kg)(14 m/s)2 2E = 408 JKf2 21 2 1 21 2 1 2EKi = mv1 1i+ m2v2iEKf = mv1 1f+ m2v2f2 22 21 2 121 1= (2.0 kg)(12 m/s) + (4.0 kg)(0 m/s)= (2.0 kg)( − 4.0 m/s) + (4.0 kg)(8.0 m/s)2 22 2EKi= 144 JEKf= 144 JThere is no loss of kinetic energy, so this collision is elastic.15. Choose north as positive.2 2Conservation of Energy1 1 1mv = mv′ + m v′2 2 22 1 1 2 1 1 2 2 2mv mv′ mv′2 2 21 1=1 1+2 20.253(1.80) 0.253 ′ 0.232 ′2 2 2= v1 + v23.24 = v′ + 0.917 ′ (Equation 1)2 21 v2Conservation of Momentummv = mv′ + m v′1 1 1 1 2 2(0.253)(1.80) = 0.253v′ + 0.232v′1 21 21.80 = v′ + 0.917v′v′ = 1.80 −0.917 v′(Equation 2)1 2Substitute Equation 2 into Equation 1:3.24 = (1.80 − 0.917 v′ ) + 0.917v′2 22 23.24 = (3.24 − 3.30v′ + 0.841 v′ ) + 0.917v′0 =− 3.30v′ + 1.758 ′2 22 2 222v20 = v′ ( − 3.30 + 1.758 v′)2 2v ′ = 0 or − 3. 30 + 1. 758v′= 02 2Since v′ = 0 is not valid (no change in speed), then21.758v′ = 3.302v′ = 1.88 m/s [N]2340 Unit 2 Energy and Momentum Copyright © 2003 Nelson

Substitute back into Equation 2:v1′ = 1.80 −0.917(1.88)v1′ = 0.08 m/s [N]The velocity of the first cart is 0.08 m/s [N]. The velocity of the second cart is 1.88 m/s [N].16. (a) Diagram not to scale.(b)0 = mvBy−mv0 = vBy−vAyvBy= vAyvBsin 46°= vAsin 33°sin 33°vB= vAsin 46°AAymvA1 = mvAx+ mvBxvA1 = vAx+ vBx5.4 = vAcos33°+ vBcos 46°⎛sin 33°⎞5.4 = vAcos33°+ ⎜ vA⎟cos 46°⎝sin 46°⎠5.4 = 1.365vAv = 3.957 m/s, or 4.0 m/sSubstitute back into equation for vB:sin 33°vB= vAsin 46°sin 33 °= (3.957 m/s)sin 46°vB= 3.0 m/sThe speed of the first puck is 4.0 m/s after the collision. The speed of the second puck is 3.0 m/s after the collision.Copyright © 2003 Nelson Chapter 5 Momentum and Collisions 341

17. θA= 67.8°θ = 30.0°BvBvAmmBA= 3.30= ?0 = p′ Ay− p′By0 = mAvAsin 67.8°− mBvBsin 30.0°mvB Bsin 30.0°= mvA Asin 67.8°mBvAsin 67.8°=mAvBsin 30.0°sin 67.8°=vB(sin 30.0 ° )vAsin 67.8°=(3.30)(sin 30.0 ° )mB= 0.561mAThe ratio of the masses of the particles is 0.561.18. Diagrams not to scale.342 Unit 2 Energy and Momentum Copyright © 2003 Nelson

Conserve momentum in the y direction.p1y + p2y =− p′ 1y + p′2y0 + mv2y =− mv′ 1y + mv′2yv2y =− v1′ y + v2′y=− v′ 1sin 69.2°+ v′2sin 62.8°=− (2.37 m/s)(sin 69.2 ° ) + (2.49 m/s)(sin 62.8 ° )v = 0.00 m/s2 yConserve momentum in the x direction.p1x + p2x = p1′ x − p2′xmv1x + mv2x = mv1′ x−mv2′xv1x + v2x = v′ 1x −v′2xv2x = v′ 1x −v′2x −v1x= v′ 1cos 69.2°− v′2cos 62.8°−v1= (2.37 m/s)(cos69.2 ° ) − (2.49 m/s)(cos62.8 ° ) −2.70 m/s=−3.00 m/s [E]v2x= 3.00 m/s [W]There is no y component for velocity, so the unknown initial velocity is 3.00 m/s [W]219. (a) g = 9.8 m/s∆ y = 0.26 mv = ?Use conservation of mechanical energy for the swing.1 2mg∆ y = mv2v = 2g∆y2= 2(9.8 m/s )(0.26 m)= 2.2574 m/sv = 2.3 m/sThe speed of the ball just before the collision is 2.3 m/s.(b) Conservation of EnergyConservation of Momentum1 2 1 2 1 2mv 1 1 = mv 1 1′ + mv 2 2′mv2 2 21 1= mv1 1′ + m2v2′2 2 2mv 1 1 mv 1 1′ mv 2 2′(0.25)(2.2574)= += 0.25v1′ + 0.21v2′2 2 22.2574 = v1′ + 0.84v2′0.25(2.2574) = 0.25v′ 1+ 0.21v′2v2 21′ = 2.2574 −0.84 v2′(Equation 2)5.096 = v′ + 0.84 v′(Equation 1)1 2Substitute Equation 2 into Equation 1:2 25.096 = (2.2574 − 0.84 v′ 2) + 0.84v′25.096 = (5.096 − 3.792v′ 2+ 0.7056 v′ 2) + 0.84v′220 =− 3.792v′ 2+ 1.5456v′20 = v′ 2( − 3.792 + 1.5456 v′2)v ′ = 0 or − 3.792 + 1. 5456v′= 02 22 2Copyright © 2003 Nelson Chapter 5 Momentum and Collisions 343

Since v′2= 0 is not valid (no change in speed), then:1.5456v2′ = 3.792v2′ = 2.5 m/s [fwd]The speed of the 0.21-kg ball just after the collision is 2.5 m/s [fwd].20. (a) m1= 0.45 kgm2= 0.79 kgv1= 2.2 m/sv′ 12= ?mv1 1= ( m1+m2)v12′mv1 1v′ 12=m1+m2(0.45 kg)(2.2 m/s)=0.45 kg + 0.79 kgv12′ = 0.80 m/sThe speed of the ball and the box just after the collision is 0.80 m/s.(b) ∆ d = 0.051 mvi= 0.80 m/svf= 0Σ F = ?⎛vi+ vf⎞∆ d = ⎜2⎟∆t⎝ ⎠⎛ 2 ⎞∆ t = ⎜ ⎟∆d⎝vi+ vf⎠⎛ 2 ⎞= ⎜ ⎟ (0.051 m)⎝0.80 m/s + 0 m/s ⎠∆ t = 0.1275 s21.ΣF∆ t = ∆pmv (2− v1)Σ F =∆t(0.79 kg + 0.45 kg)(0 m/s −0.80 m/s)=0.1275 sΣ F = −7.8 NThe magnitude of the friction force acting on the ball and the box is 7.8 N.4m1= 1.9×10 kg4m2= 1.7×10 kg3v1′ = 3.5×10 km/h3v′ = 3.4×10 km/h2θ = 180.0º – 5.1º – 5.9º = 169.0ºv = ?344 Unit 2 Energy and Momentum Copyright © 2003 Nelson

Using the sine law:sin 5.1° sinθ=mv2 2′ ( m1+m2)vmv′2 2sinθv =( m1+ m2)sin5.1°4 3(1.7 × 10 kg)(3.4× 10 km/h)sin169.0°=4 4(1.9 × 10 kg + 1.7× 10 kg)sin 5.1°3v = 3.4×10 km/hThe original speed of the spacecrafts was 3.4× 10 km/h .22. The total momentum before is zero, so the total momentum after must also be zero.Diagram not to scale.3m = 2.0 kgm123123= 3.0 kgm = 4.0 kgv = 1.5 m/s [N]v= 2.5 m/s [E]v = ?2 2 23=1+2p p p2 2 23 3=1 1+2 2( mv) ( mv) ( mv )2 2 2 23 3= (1 1) + (2 2)mv mv mv2 2( mv1 1) + ( m2v2)3=2m3v3=2 2(2.0 kg(1.5 m/s)) + (3.0 kg(2.5 m/s))v = 2.0 m/s(4.0 kg)2mvtanθ=mv−1⎛(2.0 kg)(1.5 m/s) ⎞θ = tan ⎜ ⎟⎝(3.0 kg)(2.5 m/s) ⎠θ = 22°The final velocity of the third piece is 2.0 m/s [22º S of W].1 12 2Copyright © 2003 Nelson Chapter 5 Momentum and Collisions 345

Applying Inquiry Skills23.24.25. (a) The coefficient of restitution is a measure of how quickly an object returns to its shape after it has been compressed.The apparatus shown could help determine the coefficient of restitution because the speed the sphere returns to itsoriginal shape corresponds to the height the ball bounces to.26. (a) The bed sheet spreads the force applied over a greater distance (and therefore time) than if the egg was to strike theground directly. The same impulse is imparted to the egg (i.e. it is brought to rest), so the force required is muchsmaller.(b) This can be used to cushion the landing of a person who was falling from a high position by reducing the average forceneeded to stop them. You could test several heights with a sack of potatoes and plot the results on a graph. For ahuman, you could extrapolate from the graph.Making Connections27. It is better for safety to have telephone poles that collapse upon impact. In this way, they absorb some energy and increasethe time of collision. Both of these reduce the force imparted to the vehicle and occupants in a collision.28. High speed photography and spark timers both show the location of an object at fixed time intervals. High speedphotography gives more thorough information because you can see the state of the objects during each stage of thecollision.29. Many arrester cables are connected to a water squeezer that dampens the motion and converts the kinetic energy of theaircraft to kinetic energy of ejected water.30. The main purpose of the ablation shield is to protect the shuttle during re-entry into the earth’s atmosphere.346 Unit 2 Energy and Momentum Copyright © 2003 Nelson

Extension31. (a) Conservation of Momentum Conservation of Energy1 2 1 2 1 2 1 2mv 1 1+ mv 2 2= mv 1 1′ + m2v2′(but v2=0)mv1 1+ m2v2= mv1 1′ + mv2 2′(but v2=0)2 2 2 2mv 1 1+ 0 = mv 1 1′ + m2v2′2 2 2andmv 1 1+ 0 = mv 1 1′ + m2v2′mv 1 1− mv 1 1′ = m2v2′2 2 2mv 1 1− mv 1 1′ = m2v2′m1( v1 − v1′ ) = m2v2′(Equation 1)2 2 2m ( v − v′ ) = m v′(Equation 2)Divide Equation 2 by Equation 1:2 2 2m1( v1 − v′ 1) m2v′2=m ( v − v′ ) m v′1 1 1 2 2( v1 + v′ 1)( v1 −v′1)= v′2( v − v′)1 11 1 1 2 2v′ = v + v′ (Equation 3) or v′ = v′−v(Equation 4)2 1 1 1 2 1Substitute Equations 3 and Equation 4 back into the original conservation of momentum equation:mv + 0 = mv′ + mv′mv + 0 = mv′ + m v′1 1 1 1 2 2mv = mv′ + m ( v + v′)1 1 1 1 2 1 1mv = mv′ + m v + mv′1 1 1 1 2 1 2 1mv′ + m v′= mv −mv1 1 2 1 1 1 2 1v′ ( m + m ) = ( m −m ) v1 1 2 1 2 1⎛m− m ⎞v′ = ⎜ ⎟v⎝ ⎠1 21 1m1+m21 1 1 1 2 2mv = m( v′ − v)+ mv′1 1 1 2 1 2 2mv = mv′ − mv + mv′1 1 1 2 1 1 2 2mv′ + mv′= 2mv1 2 2 2 1 1v′ ( m + m ) = 2mv2 1 2 1 1⎛ 2m⎞v′ = ⎜ ⎟v⎝ ⎠12 1m1+m232.⎛m1−m2⎞(b) v′ = ⎜ ⎟v⎝m1+m2⎠1 1⎛m−m⎞= ⎜ ⎟v⎝m+m⎠v′ =10⎛m1−m2⎞(c) v′ = ⎜ ⎟v⎝m1+m2⎠11 11= ⎜ ⎟v1m1+ 0⎝v′ = v1 1⎛m− 0 ⎞⎛m1−m2⎞(d) v′ = ⎜ ⎟v⎝m1+m2⎠⎠1 1⎛0− m2= ⎜ ⎟ 10 + m2⎝v′ =−v1 1⎞ v⎠m = 1.0×10 kgmvvPBPB33= 2.0×10 kg= 50 m/s= 0∆ d = ?⎛ 2m ⎞v′ = ⎜ ⎟v⎝ ⎠12 1m1+m2⎛ 2m ⎞= ⎜ ⎟v⎝m+m⎠v′ = v2 1⎛ 2m ⎞v′ = ⎜ ⎟v⎝ ⎠12 1m1+m211= ⎜ ⎟ v1m1+ 0⎝v′ = 2v⎛ 2m2 1⎞⎠⎛ 2m ⎞v′ = ⎜ ⎟v⎝ ⎠12 1m1+m2⎛2m1= ⎜ ⎟v10 + m2⎝2mv′ = v12 1m2⎞⎠Copyright © 2003 Nelson Chapter 5 Momentum and Collisions 347

For the collision:mv + mv = ( m + m ) vP P B B P B PBvvPBPBmv=mP P B BP+ mv+ mB3 3(1.0 × 10 kg)(50 m/s) + (2.0×10 kg)(0)=3 31.0× 10 kg + 2.0×10 kg= 16.67 m/sCalculate the frictional force on the plane and on the barge:For the plane1FK= mg41 (1.0 103 kg)(9.8 m/s2= ×)4F = 2450 N [backward]KAcceleration of the planeΣ F = maΣFa =m−2450 N=31.0×10 kg2a =−2.45 m/sDistance plane will travel during collision2 2vf= vi+ 2a∆d2 2vf− vi∆ d =2a2 2(16.67 m/s) − (50 m/s)=22( −2.45 m/s )∆ d = 453.5 mFor the bargeBy Newton’s third law, F K = 2450 N [forward]Acceleration of the bargeΣ F = maΣFa =m2450 N=32.0×10 kga = 1.225 m/s2Distance barge will travel during collision2 2f iv = v + 2a∆d2 2f− viv∆ d =2a2 2(16.67 m/s) − (0 m/s)=22(1.225 m/s )∆ d = 113.4 mThe required length of the barge is 453.5 – 113.4 = 3.4 × 10 2 m.348 Unit 2 Energy and Momentum Copyright © 2003 Nelson

CHAPTER 6 GRAVITATION AND CELESTIAL MECHANICSReflect on Your Learning(Page 272)1. (a) The force of gravity exerted by Saturn keeps the rings in their orbit around Saturn.(b) The force of gravity exerted by Earth keeps the Hubble Space Telescope in a stable orbit around Earth. The forceexerted by the rockets of the telescope counteracts the small amount of friction of the upper atmosphere.2. (a) The two probes would have the same minimum speed (escape speed) even though the probes have different masses.Students will discover that the escape speed depends on the mass of Earth, not on the mass of the probe, according to2GMEthe equation: v = .r(b) The minimum kinetic energy of the probe of mass 2m would be twice that of the probe of mass m. The minimumkinetic energy of the probe is equal to the gravitational potential energy, which is proportional to the mass of the probe.EK=−EgGMm∴ EK=r3. Since the gravitational potential energy E g is inversely proportional to the distance from the main body, E g will approachzero as the distance increases, according to the equation presented later in the chapter:GMmEg=−r4. Students should be able to answer this question based on their study of the law of universal gravitation in Section 3.3.5. As students have seen on page 147 of the text, a black hole is an extremely dense celestial body. The gravitational field ofa black hole is so strong that nothing, including electromagnetic radiation, can escape from its vicinity. Since light canenter, but not escape, a black hole appears totally black. The surface of the body is called its event horizon because noevent can be observed from outside the surface. At the core of a black hole is a dense centre called a singularity. Thedistance from the centre of the singularity to the event horizon is the Schwartzschild radius.Black holes are thought to form during the course of stellar evolution. When the nuclear fuels are exhausted in thecore of the star, the star collapses. If the mass of the core is greater than a critical value that is almost twice as great as themass of the Sun, the core may collapse into a black hole.Try This Activity: Drawing and Comparing Ellipses(Page 273)(a) The eccentricity (e 1 ) of the first ellipse is less than the eccentricity (e 2 ) of the second ellipse.c15.0 cme1= = = 0.33a115 cmc27.5 cme2= = = 0.60a21.5 cm(b) The Sun is at one focus of the elliptical orbit of each planet. Earth is at one focus of the elliptical orbit of the Moon.Copyright © 2003 Nelson Chapter 6 Gravitation and Celestial Mechanics 349

6.1 GRAVITATIONAL FIELDSPRACTICE(Pages 276–277)Understanding Concepts1. As they are moving at a high speed in their orbits, space vehicles are attracted to Earth by the force of gravity, whichkeeps them in their orbits. Also, the vehicles have small booster rockets to control the location of the orbit and tocounteract low amounts of friction in the upper atmosphere.−11 2 22. (a) G = 6.67× 10 N ⋅ m /kgMmrEMoonMoon24= 5.98×10 kg22= 7.35×10 kg8= 3.84×10 mFGM mE MoonG =2rMoon−11 2 2 24 22(6.67× 10 N ⋅ m /kg )(5.98× 10 kg)(7.35×10 kg)=8 2(3.84×10 m)20FG= 1.99×10 N [toward Earth's centre]The gravitational force exerted on the Moon by Earth is 1.99× 10 20 N [toward Earth’s centre].GM EmMoon(b) FG =2rFGMoon−11 2 2 24 22(6.67× 10 N ⋅ m /kg )(5.98× 10 kg)(7.35×10 kg)=8 2(3.84×10 m)20= 1.99×10 N [toward Moon'scentre]The gravitational force exerted on Earth by the Moon is 1.99× 10 20 N [toward Moon’s centre].GM EGM E3. Let g r represent the gravitational field strength at the desired radius r. Thus, gr = and g = .22( r+rE)rE(a) r = rEggEr 2( rE+ rE)r=GMGM=4rg=4E2EThe magnitude of the gravitational field strength (in terms of g) at 1.0 Earth radius is 4g .(b)r = 3rEg=GMEr 2(3 rE+ rE)GM=16rggr=16E2EThe magnitude of the gravitational field strength (in terms of g) at 3.0 Earth radii is 16g .350 Unit 2 Energy and Momentum Copyright © 2003 Nelson

(c)r = 4.2rEg=GMEr 2(4.2 rE+ rE)GM=27rggr=27E2EgThe magnitude of the gravitational field strength (in terms of g) at 4.2 Earth radii is . 27GMEGME4. Let g P represent the magnitude of the gravitational field strength of the planet. Thus, gP = and g = .22(0.50 rE)rE4.0GMEgP =2rEgP= 4.0gThe planet has a surface gravitational field strength of magnitude 4.0g.5. (a) g = 1.6 N/kg6r = 1.74×10 mGMMoong =2r2grMMoon=G6 2(1.6 N/kg)(1.74×10 m)=−11 2 26.67 × 10 N ⋅m /kg22MMoon= 7.3×10 kgThe mass of the Moon is 7.3 × 10 22 kg.22(b) MMoon= 7.35×10 kgmstudent= 55kg (example mass of student)GMMoonmstudentFG =2r−11 2 2 22(6.67× 10 N⋅ m /kg )(7.35×10 kg)(55kg)=6 2(1.74 × 10 m)FG= 89 NA 55-kg student would have a weight of 89 N on the Moon. (Students can also determine the weight by applying thefact that the ratio of the Moon’s gravitational field strength to Earth’s gravitational field strength is 1.6:9.8.)−96. (a) g = 5.42×10 N/kgm = 1.00kgFG= mg−9= (1.00 kg)(5.42×10 N/kg)−9FG= 5.42×10 NThe magnitude of the gravitational force is 5.42 × 10 −9 N.5(b) m = 8.91×10 kgFG= mg5 −9= (8.91× 10 kg)(5.42×10 N/kg)−3F = 4.83×10 NGThe magnitude of the gravitational force is 4.83 × 10 −3 N.Copyright © 2003 Nelson Chapter 6 Gravitation and Celestial Mechanics 351

Applying Inquiry Skills7. It is an inverse square relationship. The graphs for (a) and (b) are the same except that the vertical axes are labelleddifferently.Making Connections8. (a) If Earth’s density were greater, then its mass would also be greater. Therefore, Earth’s surface gravitational fieldstrength would also be much greater since it is proportional to Earth’s mass.(b) Since the surface gravitational field strength of Earth would be much greater if its density were greater, a human’sweight would also be greater. To support the increased weight, bones would need to be very sturdy (and heavier).Shorter, thicker bones would be more efficient at supporting the larger weight.(c) Answers will vary. Some effects in nature could be shorter maximum heights of plants and more powerful winds.Examples of effects on human activities include different sizes of equipment used in sports (balls, bats, rackets, etc.)and small sizes of some transportation vehicles, especially aircrafts.Section 6.1 Questions(Page 277)Understanding Concepts1. The weight of a space probe decreases as the probe travels away from Earth because the gravitational field strength ofEarth, g E , decreases. However, as it approaches the Moon, the space probe experiences the gravitational field strength ofthe Moon, g M . There is a location where g E and g M are equal, but in opposite directions. At this location, the weight of theprobe is zero, although the mass of the space probe does not change. Since Earth is almost 100 times more massive thanthe Moon, this location is much closer to the Moon.2. (a) Let the subscript s represent the satellite.m = 225kg24ME= 5.98×10 kg6rs= 8.62×10 m6rE= 6.38×10 mr = r + rsEFFEG 2GGM m=r−11 2 2 24(6.67× 10 N⋅ m /kg )(5.98×10 kg)(225kg)=6 6 2(8.62× 10 m + 6.38×10 m)2= 3.99×10 NThe magnitude and direction of the gravitational force on the satellite is 3.99 × 10 2 N [toward Earth’s centre]. (b) ∑ F = ma FGa =m23.99×10 N [toward Earth's centre]=225kg2a = 1.77 m/s [toward Earth's centre]The resulting acceleration of the satellite is 1.77 m/s 2 [toward Earth’s centre].352 Unit 2 Energy and Momentum Copyright © 2003 Nelson

3.7r = 7.4×10 m24ME= 5.98×10 kgGMEg =2r−11 2 2 24(6.67× 10 N ⋅ m /kg )(5.98×10 kg)=7 2(7.4×10 m)−2g = 7.3×10 mThe magnitude and direction of the gravitational field strength is 7.3 × 10 –2 N/kg [toward Earth’s centre].4. (a) Let the subscript s represent the satellite.gs= 4.5 N/kg6rE= 6.38×10 mg = 9.8 N/kgGMEGMESince gs = and g = :2 2rrsE⎛2GM ⎞⎛E r ⎞⎜ 2r ⎟ ⎝ E ⎠⎝GME⎠2g= ⎜ ⎟⎜ ⎟gg r=2gsrEgr=rEgs6= 6.38×10 mr = 9.42×10 m69.8 N/kg4.5 N/kgThe distance the satellite is above Earth = r – r E= 9.42 × 10 6 m – 6.38 × 10 6 m= 3.0 × 10 6 mThus, the satellite is 3.0 × 10 6 m above Earth’s surface.2(b) m = 6.2×10 kgFG= mg2= (6.2×10 kg)(4.5 N/kg)3FG= 2.8×10 NThe gravitational force on the satellite is 2.8 × 10 3 N.75. rN= 2.48×10 m26MN= 1.03×10 kgGMNg =2r−11 2 2 26(6.67× 10 N⋅ m /kg )(1.03×10 kg)=7 2(2.48×10 m)g = 11.2 N/kgThe magnitude of Neptune’s surface gravitational field strength is 11.2 N/kg. The value given in Table 1 is1.14 × 9.8 N/kg = 11.2 N/kg. Therefore, the values are the same.76. (a) r = 2.5×10 mm = 456kg3v = 3.9 km/s = 3.9×10 m/sCopyright © 2003 Nelson Chapter 6 Gravitation and Celestial Mechanics 353

2vac=r3 2(3.9×10 m/s)=72.5×10 m2ac= 0.61m/sThe acceleration of the satellite is 0.61 m/s 2 [toward Earth’s centre].GM Em(b) FG =2r−11 2 2 24(6.67× 10 N ⋅ m /kg )(456kg)(5.98×10 kg)=7 2(2.5×10 m)2F = 2.9×10 NGThe gravitational force on the satellite is 2.9 × 10 2 N [toward Earth’s centre].7. (a) g = 1.3 N/kg23M = 1.3×10 kgGMg =2rGMr =g= 2.6×10 m3r = 2.6×10 kmTitan’s radius is 2.6 × 10 3 km.(b) m = 0.181kgFG=6−11 2 2 23(6.67× 10 N⋅ m /kg )(1.3×10 kg)1.3N/kg= mg= (0.181kg)(1.3N/kg)FG= 0.24 NThe force of gravity on a 0.181-kg rock on Titan is 0.24 N.8. Let r 2 be the distance from Earth’s centre at which the gravitational field strength has a magnitude, g 2 , of 3.20 N/kg.gE= 9.80 N/kg at Earth's surfaceg = 3.20 N/kg at distance r2 2Using ratio and proportion:g Gm Gm= ÷2 2gE r2 rE2rE=2r222 gErEr2=g22gErEr2=g2 E E22 E2E(9.80 N/kg)( r )=(3.20 N/m)r = 1.75r354 Unit 2 Energy and Momentum Copyright © 2003 Nelson

d = 1.75rE−1.00rEd = 0.75rEThe gravitational field strength is 3.20 N/kg at a distance of 0.75r E above Earth’s surface.Applying Inquiry Skills9. The diagram below shows examples of the required FBDs.Making Connections10. Based on the data in Table 1, some astronomers might argue that Pluto should not be considered to be a planet becausethe gravitational field strength of Pluto is so much smaller than the other planets. On average, Pluto’s gravitational fieldstrength is less than one-tenth the gravitational field strength of the other planets.6.2 ORBITS AND KEPLER’S LAWSPRACTICE(Page 279)Understanding Concepts1. The Moon does not fall into Earth because it travels at a specific speed around Earth. This keeps the Moon at anapproximately constant distance from Earth’s centre called the orbital radius.2. Since the space probe is in a circular orbit, the direction of the gravitational force is perpendicular to the direction of theinstantaneous velocity. Thus, the force of gravity does not do any work on the probe, and there is no change in the kineticenergy (or speed) of the probe.3. Let the subscript s represent the satellite.M E = 5.98 × 10 24 kgr E = 6.38 × 10 6 mr s = 525 km = 5.25 × 10 5 mr = r E + r sGME(a) vs=rs−11 2 2 24(6.67× 10 N ⋅ m / kg )(5.98×10 kg)=6 5(6.38× 10 m) + (5.25×10 m)3v = 7.60×10 m/sThe speed of the satellite is 7.60 × 10 3 m/s.Copyright © 2003 Nelson Chapter 6 Gravitation and Celestial Mechanics 355

d = 1.75rE−1.00rEd = 0.75rEThe gravitational field strength is 3.20 N/kg at a distance of 0.75r E above Earth’s surface.Applying Inquiry Skills9. The diagram below shows examples of the required FBDs.Making Connections10. Based on the data in Table 1, some astronomers might argue that Pluto should not be considered to be a planet becausethe gravitational field strength of Pluto is so much smaller than the other planets. On average, Pluto’s gravitational fieldstrength is less than one-tenth the gravitational field strength of the other planets.6.2 ORBITS AND KEPLER’S LAWSPRACTICE(Page 279)Understanding Concepts1. The Moon does not fall into Earth because it travels at a specific speed around Earth. This keeps the Moon at anapproximately constant distance from Earth’s centre called the orbital radius.2. Since the space probe is in a circular orbit, the direction of the gravitational force is perpendicular to the direction of theinstantaneous velocity. Thus, the force of gravity does not do any work on the probe, and there is no change in the kineticenergy (or speed) of the probe.3. Let the subscript s represent the satellite.M E = 5.98 × 10 24 kgr E = 6.38 × 10 6 mr s = 525 km = 5.25 × 10 5 mr = r E + r sGME(a) vs=rs−11 2 2 24(6.67× 10 N ⋅ m / kg )(5.98×10 kg)=6 5(6.38× 10 m) + (5.25×10 m)3v = 7.60×10 m/sThe speed of the satellite is 7.60 × 10 3 m/s.Copyright © 2003 Nelson Chapter 6 Gravitation and Celestial Mechanics 355

(b)2πrv =T2πrT =v6 52π6.38× 10 m+ 5.25×10 m=37.60×10 m/s3T = 5.71×10 s( )The period of revolution of the satellite is 5.71 × 10 3 s or (5.71× 10 3 ⎛ 1 h ⎞s) ⎜ ⎟ = 1.59 h.⎝3600 s ⎠4. Let the subscript s represent the satellite.M Moon = 7.35 × 10 22 kgr Moon = 1.74 × 10 6 mvss=GMMoonr−11 2 2 22(6.67× 10 N ⋅ m / kg )(7.35×10 kg)=61.74×10 m3v = 1.68×10 m/sThe speed of the satellite is 1.68 × 10 3 m/s.Applying Inquiry Skills5. (a) The speed of a satellite around a central body is inversely proportional to the square root of the radius of the satellite’s1orbit. Thus, vs∝ .rs(b)Making Connections6. Recommended web sites are:www.space.com/spacewatch/space_junk.htmlhttp://earthwatch.unep.net/solidwaste/spacejunk.htmlStudents are likely to encounter the following points about the problem of space junk:• a study done in 1999 estimated 4 million pounds of space junk in low-Earth orbit• objects that are baseball-size and bigger may threaten the safety of astronauts in space; collisions involving even thesmallest of objects may be damaging due to the high speeds of the objects• the U.S. Space Command agency counted the number of objects in space as of June 21, 2000 and found 2671 satellites,90 space probes, and 6096 chunks of debris• some objects re-enter Earth’s atmosphere, but most burn up on re-entry, or land in water or uninhabited land• NASA calculates that if the amount of debris equal to or larger than 1 centimeter exceeds 150 000, it could make spaceflight impossible356 Unit 2 Energy and Momentum Copyright © 2003 Nelson

• as technology improves, more and more satellites and space objects are being launched into space, not only by spaceagencies, but other industries including telecommunications and organizations interested in astronomy• possible solutions include better tracking of space junk, improved methods of bringing down satellites to Earth, andlaunching satellites to sweep up the space junk that is already in orbitPRACTICE(Page 283)Understanding Concepts7. Relative to the rest of the solar system, Earth’s frame of reference is accelerating, so the geocentric model is thenoninertial frame of reference. The heliocentric model is an inertial frame of reference if the solar system is considered tobe isolated. However, it is a noninertial frame with respect to the Milky Way Galaxy and the rest of the universe.8. Tycho Brahe (1546–1601) made precise, comprehensive astronomical measurements of the solar system and more than 700stars. For 20 years, he made countless naked-eye observations using large instruments he made himself. He was able tocollect data for Mercury, Venus, Earth, Mars, and Saturn, because those were planets that he could see. The other planetswere beyond his scope of vision and the telescope was not invented until the early 17th century.9. Using Kepler’s second law, Earth sweeps out equal areas in equal time intervals. Therefore, when Earth is closest to the Sun,it is moving fastest. Conversely, if the orbit is divided into 180° “halves,” the portion closest to the Sun will have a smallerarea, and therefore a shorter time. Since there are three fewer days between September 21 and March 21, Earth must beclosest to the Sun at that time.3r10. The ratio is calculated for each planet in Table 1.2TTable 1ObjectMeanRadius ofOrbit (m)Period ofRevolutionof Orbit (s)r 3 ∝ T 2rC =T2(m 3 /s 2 )Mercury 5.79 × 10 10 7.60 × 10 6 3.36 × 10 18Venus 1.08 × 10 11 1.94 × 10 7 3.35 × 10 18Earth 1.49 × 10 11 3.16 × 10 7 3.31 × 10 18Mars 2.28 × 10 11 5.94 × 10 7 3.36 × 10 18Jupiter 7.78 × 10 11 3.75 × 10 8 3.35 × 10 18Saturn 1.43 × 10 12 9.30 × 10 8 3.38 × 10 18Uranus 2.87 × 10 12 2.65 × 10 9 3.37 × 10 18Neptune 4.50 × 10 12 5.20 × 10 9 3.37 × 10 18Pluto 5.91 × 10 12 7.82 × 10 9 3.38 × 10 18All proportionality constants calculated in Table 1 are within 1.5% of the average value. This verifies Kepler’s third law.11. (a) The average value (in SI base units) of the constant of proportionality in r 3 ∝ T 2 is 3.36 × 10 18 m 3 /s 2 .(b) C S = 3.36 × 10 18 m 3 /s 2 (from (a))GMSCS =24π2CS4πMS=G18 3 2 2(3.36×10 m /s )4π=−11 2 26.67 × 10 N ⋅m / kg30M = 1.99×10 kgSThe mass of the Sun is 1.99 × 10 30 kg.3Copyright © 2003 Nelson Chapter 6 Gravitation and Celestial Mechanics 357

12. (a) r Moon = 3.84 × 10 8 mT Moon = 2.36 × 10 6 sCC3rMoonE=2TMoonE8 3(3.84×10 m)=6 2(2.36×10 s)= 1.02×10 m /s13 3 2Thus, Kepler’s third law constant, C E , is 1.02 × 10 13 m 3 /s 2 for objects orbiting Earth.(b) Let the subscript s represent the satellite.M E = 5.98 × 10 24 kg⎛3600 s ⎞T s = 4.0 h = (4.0 h) ⎜ ⎟⎝ 1 h ⎠ = 1.4 × 104 sSince C3GMErsE 2 24πTs= = , then23 GMETss =2rGM4π3Ers2 2Ts= .−11 2 2 24 4 2(6.67× 10 N ⋅ m / kg )(5.98× 10 kg)(1.4×10 s)3s=2s4πr4π7 4r = 1.3× 10 m or 1.3×10 kmThe satellite must be 1.3 × 10 4 km above the centre of Earth.2πrvs=T72π( 1.26×10 m)=41.4×10 s3v = 5.6×10 m/ssThe speed of the satellite is 5.6 × 10 3 m/s.Applying Inquiry Skills13.In the sample diagram above, the shape of each figure is approximately that of a triangle, so the approximate areas are:11A1 = bh1 1A2 = b2h2221 1= (5.0 mm)(54 mm)= (12 mm)(22 mm)222 22 2A = 1.3×10 mmA = 1.3×10 mm1In the diagram, d 2 > d 1 , but t 2 = t 1 , so v 2 > v 1 .2Making Connections14. (a) Let the subscript C represent the central body around which another body (subscript B) revolves in an orbit of known32 3GM C rB4πrBperiod and average radius. Since = , the equation for the mass of the central body is M2 2C = .24πTGTBB358 Unit 2 Energy and Momentum Copyright © 2003 Nelson

(b) According to the equation derived in (a) above, G is known and if the mass of the star can be estimated, the period ofrevolution of the planet must be found in order to solve for the radius of the orbit. Astronomers have discovered thatthe mass of a star can be estimated by determining its luminosity. (This applies to “main-sequence” stars, by far themajority of stars.) The period of revolution is determined by measuring the period of the wobble of the star as the2GMStar( TPlanet)planet tugs on the star. The radius of the orbit is found using the equation r = 3.24πSection 6.2 Questions(Page 284)Understanding Concepts1. According to Kepler’s first law, a comet travels in an elongated elliptical orbit. Kepler’s second law implies that theportion of the comet’s orbit closet to the Sun will have a much smaller area than the rest of its orbit. Thus, the time spentin this region (where the comet may be visible to observers on Earth) will be far less compared to the total orbital period.2. Kepler’s second law states that Earth sweeps out equal areas in equal time intervals. Therefore, on January 4 when Earthis closest to the Sun, it is moving most rapidly because the distance travelled is greatest for equal time intervals. The Earth ismoving least rapidly on July 5 when it is farthest from the Sun.3. Although the nonrotating frame of reference is placed at the centre of the Sun, the Sun is in orbit around the centre of theMilky Way Galaxy, so it is constantly accelerating. This means it is a noninertial frame of reference within the galaxy.4. r = 4.8 × 10 11 mM S = 1.99 × 10 30 kgGMSCS =24π−11 2 2 30(6.67× 10 N ⋅ m / kg )(1.99×10 kg)=24π18 3 2CS= 3.36×10 m /s3rCS =2T32 rT =CS11 3(4.8×10 m)T =18 3 23.36×10 m /s8T = 1.8×10 sThe orbital period of the asteroid is 1.8 × 10 8 s.5. Let the subscript P represent the unknown planet.3rECS =2TE3rPCS =2TP3 3rPrE=2 2TPTETP= 2TE3 3rPrE=2 2(2 T ) TEE2 34T3E rEP=2TE3= 4rErrP= 1.6rEThe small planet would be 1.6 times farther from the Sun than Earth.Copyright © 2003 Nelson Chapter 6 Gravitation and Celestial Mechanics 359

6. M E = 5.98 × 10 24 kgr E = 6.38 × 10 24 mr s = 2r EGM E4πGME4π3(2 rE)=2 2T38rE=2 2T =T32πGM2 3rEE2 6 332 π (6.38×10 m)=−11 2 2 24(6.67× 10 N ⋅ m / kg )(5.98×10 kg)4 ⎛ 1 h ⎞= 1.43× 10 s ⎜ ⎟⎝3600s⎠T = 4.0 hThe period of revolution is 4.0 h.7. Let the subscript P represent Phobos, the subscript D represent Deimos, and the subscript M represent Mars.⎛T D = 30 h 18 min = 30 h 3600 s ⎞⎜ ⎟⎝ 1 h ⎠ + 18 min ⎛ 60 s ⎞⎜ ⎟⎝1 min ⎠ = 1.09 × 105 sr D = 2.3 × 10 4 km⎛T P = 7 h 39 min = 7 h 3600 s ⎞⎜ ⎟⎝ 1 h ⎠ + 39 min ⎛ 60 s ⎞⎜ ⎟⎝1 min ⎠ = 2.75 × 104 sCC3rDM =2TDM7 3(2.3×10 m)=(1.09 × 10 s)12= 1.02×10 s5 2Substitute the value of C M into the equation for Phobos:3rPCM =2TP3 2r = C TP M P312 4 2P = (1.02× 10 s)(2.75×10 s)6P= 9.2×10 mrrPhobos is 9.2 × 10 6 m from the centre of Mars.Applying Inquiry Skills8.GMr=2 2(N ⋅m / kg )(kg)m2 2 2(kg ⋅ m/s )(m / kg )(kg)=m3 2m/s=m2 2= m/s= m/sTherefore, the SI base units are metres per second.360 Unit 2 Energy and Momentum Copyright © 2003 Nelson

9. Since r 3 = CT 2 (from Kepler’s third law), the line on the graph is straight and the slope is Kepler’s third law constant forthe Sun.Making Connections10. (a) Galileo was born near Pisa, on February 15, 1564. In 1609, after learning that a telescope had been invented inHolland, he built his own telescope of 20 times magnification. The strength of this magnification allowed Galileo to seemountains and craters on the Moon, and to discover the four largest satellites of Jupiter.Tycho’s work was done between 1581 and 1601, during which he made numerous naked-eye observations withlarge instruments. Kepler began analyzing Tycho’s data in 1601. In 1609, Kepler published his first two laws. Hepublished his third law in 1619.(b) Relating Kepler’s third law and the law of universal gravitation, we find that the mass of Jupiter is given by2 24πrMMJ = , where r2M and T M are the radius of the orbit and period of revolution, respectively, of any moon aroundGTMJupiter. Thus, Galileo would need to know the values of r M and T M for at least one moon, as well as the universalgravitation constant, G.(c) Calculating the mass of Jupiter was not possible until the value of G was determined, which was not possible untilKepler’s third law was formed. (As mentioned in Section 3.3, it was Cavendish who first determined that value in1798.)6.3 GRAVITATIONAL POTENTIAL ENERGY IN GENERALPRACTICE(Pages 287–288)Understanding Concepts1. M E = 5.98 × 10 24 kgm = 0.0123M Er = 3.84 × 10 5 kmGMEmEg=−r2GME0.0123=−r2−11 2 2 24( 6.67× 10 N ⋅ m /kg )( 5.98×10 kg) ( 0.0123)=−83.84×10 m28Eg=− 7.64×10 JThe gravitational potential energy of the Earth-Moon system is –7.64 × 10 28 J.2. (a) m = 1.0 kgr E = 6.38 × 10 6 mr = 1.0 × 10 2 km = 1.0 × 10 5 mCopyright © 2003 Nelson Chapter 6 Gravitation and Celestial Mechanics 361

2GMEr =2c−11 2 2 242(6.67 × 10 N ⋅ m /kg )(5.98×10 kg)=8 2(3.00×10 m/s)−3= 8.86×10 mr = 8.86mmThe theoretical Schwartzschild radius of the black hole is 8.86 mm.1(b) The very small radius of the black hole implies that it is extremely dense, according to ρ ∝ . Thus,3r3ρ rE=3ρEr6 3(6.38×10 m)=−3 3(8.86×10 m)ρ= 3.73×10ρE26Therefore, the black hole is 3.73 × 10 26 times more dense than Earth.Making Connections12. The escape energy of an object from the Moon is much less than the escape energy of the same object from Earth. (In fact,calculations show that the ratio of the escape energy from Earth is more than 22 times as great as that from the Moon.)Thus, less fuel is required to send the spacecraft from the Moon to Earth than from Earth to the Moon.CHAPTER 6 LAB ACTIVITIESLab Exercise 6.3.1: Graphical Analysis of Energies(Page 295)Procedure1. Changing the numerical values in the given data to megametres and gigajoules is suggested just to make data entry in asoftware program easier. The results of the graphing and the analysis are not affected by this suggestion. The data are:r (Mm)E g (GJ)1.6 3.2 4.8 6.4 8.0 9.6 1.12 1.28 1.44 1.60−3.0 −1.5 −1.0 −7.5 −6.0 −5.0 −4.3 −3.8 −3.3 −3.02 – 5. The four lines required on the graph are shown below.372 Unit 2 Energy and Momentum Copyright © 2003 Nelson

Analysis(a) To analyze the energy data of a spacecraft-moon system, graphs of gravitational potential energy, binding energy, andkinetic energy are useful. All graphs should be as a function of the distance from the centre of the moon, r.Evaluation(b) Answers will vary. An advantage of a spreadsheet program is that it performs calculations and plots relationships ongraphs very quickly. However, a disadvantage is that learning how to use the program for a specific application can bemore time consuming.Synthesis(c) Energy considerations are important in planning space missions in order to determine the amount of fuel required tosafely launch and transport the vehicles to and from the destination. The amount of fuel affects the total cost of anymission.CHAPTER 6 SUMMARYMake a Summary(Page 297)Details on the diagram will vary. The figure below shows the start of the diagram. Urge students to add enough detail to helpthem understand and remember as much as possible about the chapter.CHAPTER 6 SELF QUIZ(Pages 298–299)True/False1. T2. T3. TCopyright © 2003 Nelson Chapter 6 Gravitation and Celestial Mechanics 373

Analysis(a) To analyze the energy data of a spacecraft-moon system, graphs of gravitational potential energy, binding energy, andkinetic energy are useful. All graphs should be as a function of the distance from the centre of the moon, r.Evaluation(b) Answers will vary. An advantage of a spreadsheet program is that it performs calculations and plots relationships ongraphs very quickly. However, a disadvantage is that learning how to use the program for a specific application can bemore time consuming.Synthesis(c) Energy considerations are important in planning space missions in order to determine the amount of fuel required tosafely launch and transport the vehicles to and from the destination. The amount of fuel affects the total cost of anymission.CHAPTER 6 SUMMARYMake a Summary(Page 297)Details on the diagram will vary. The figure below shows the start of the diagram. Urge students to add enough detail to helpthem understand and remember as much as possible about the chapter.CHAPTER 6 SELF QUIZ(Pages 298–299)True/False1. T2. T3. TCopyright © 2003 Nelson Chapter 6 Gravitation and Celestial Mechanics 373

Analysis(a) To analyze the energy data of a spacecraft-moon system, graphs of gravitational potential energy, binding energy, andkinetic energy are useful. All graphs should be as a function of the distance from the centre of the moon, r.Evaluation(b) Answers will vary. An advantage of a spreadsheet program is that it performs calculations and plots relationships ongraphs very quickly. However, a disadvantage is that learning how to use the program for a specific application can bemore time consuming.Synthesis(c) Energy considerations are important in planning space missions in order to determine the amount of fuel required tosafely launch and transport the vehicles to and from the destination. The amount of fuel affects the total cost of anymission.CHAPTER 6 SUMMARYMake a Summary(Page 297)Details on the diagram will vary. The figure below shows the start of the diagram. Urge students to add enough detail to helpthem understand and remember as much as possible about the chapter.CHAPTER 6 SELF QUIZ(Pages 298–299)True/False1. T2. T3. TCopyright © 2003 Nelson Chapter 6 Gravitation and Celestial Mechanics 373

4. T5. F “Evidence” is to the “Analysis” as Tycho Brahe’s work was to Kepler’s work.6. F In Figure 1 where the path distances d 1 and d 2 are equal, the speeds along those path segments are not equal because thearea swept out by these paths, and thus the time intervals, are different according to Kepler’s second law.7. F Kepler’s third law constant for Earth is the same for the Moon as for all satellites in orbit around Earth; the constantdepends on Earth’s mass.8. F The gravitational potential energy of the Earth-Moon system is inversely proportional to the distance between thecentres of the two bodies.9. T10. F Your escape energy and binding energy are equal because you are at rest, and therefore you have no kinetic energy.Multiple Choice11. (a) g ∝ m12. (d) g ∝ 1 r13. (c) v ∝ m114. (d) v ∝r15. (a) area ∝ ∆t16. (c) r ∝ 3 T2 or r ∝17. (a) r 3 ∝ T 218. (d) E K ∝ 1 r23T19. (e) E g ∝ – 1 rGM E20. (c) Kepler’s third law allows us to determine Earth’s mass according to the equation CE = .24π1GM GM21. (c) Since g ∝ , then r22 = 3r 1 . Using the equations g1 = and g22 = :2rrr22. (d) r 1 = 4r 2gggv⎛GM⎞⎜ 23 ⎟ ⎛ ⎞ 1= = ⎜ ⎟ =⎛GM⎞ ⎝3r⎠ 922 ⎝ r1⎠ r11 1⎜ 2r ⎟1⎝1= g92 1GM⎠GMS121= and24rS2r2⎛ GM ⎞Ev⎜4r⎟12=⎝ ⎠v2 ⎛ GME ⎞⎜r ⎟⎝ 2 ⎠r2=4r2v11=v24v = 0.5v1 2v=13 1374 Unit 2 Energy and Momentum Copyright © 2003 Nelson

23. (c) Let the subscript 1 represent the mass of the Sun at its current value and the subscript 2 represent the mass of the Sun athalf its current value. Since M S2 = 1 2 M S1, C S2 = 1 2 C S1. Thus,33rrCS1 = and C2 S2= .2TT⎛3r ⎞⎜ 2 ⎟C TS1 1=⎝ ⎠C3S2 ⎛ r ⎞⎜ 2 ⎟⎝T2⎠2T2= 22T1T2= 2T124. (c) A satellite in geosynchronous orbit has a period of revolution of 24 h.25. (a) The speed of the comet increases as it comes closer to the Sun. Position A is closet to the Sun, and therefore has thegreatest speed. Position C is furthest from the Sun, and so has the slowest speed. Positions B and D are equidistantfrom the Sun and are between positions A and C. Thus, v A > v B = v D > v C .26. (e) M P = M Er P = 1 4 r Ev2GM2GM8GM= =EE EE= andPr1ErrEE⎛ 8GM⎞Ev⎜r ⎟PE=⎝ ⎠vE⎛ 2GME ⎞⎜r ⎟⎝ E ⎠vP= 4vvEP= 2vEv4CHAPTER 6 REVIEW(Pages 300–301)Understanding Concepts1. The escape energy (and thus the escape speed) from the Sun is much greater than that from Earth, so the rocket given thespeed needed to escape from Earth would not have enough speed to escape from the solar system. Space vehicles sent toexplore distant planets have a much lower binding energy by the time they reach those distant locations, and could acquireenough energy to escape from the solar system by taking advantage of the force of gravity of the distant planet.2. Since Earth rotates eastward, an eastward orientation of the rocket as it is being launched means that the rocket alreadyhas a component of the required velocity before blasting off. This means that less energy will be needed to launch therocket eastward than would be required to launch it westward in order to achieve the same speed.3. g U = 1.0 N/kgM U = 8.80× 10 25 kgr U = 2.56× 10 7 mCopyright © 2003 Nelson Chapter 6 Gravitation and Celestial Mechanics 375

23. (c) Let the subscript 1 represent the mass of the Sun at its current value and the subscript 2 represent the mass of the Sun athalf its current value. Since M S2 = 1 2 M S1, C S2 = 1 2 C S1. Thus,33rrCS1 = and C2 S2= .2TT⎛3r ⎞⎜ 2 ⎟C TS1 1=⎝ ⎠C3S2 ⎛ r ⎞⎜ 2 ⎟⎝T2⎠2T2= 22T1T2= 2T124. (c) A satellite in geosynchronous orbit has a period of revolution of 24 h.25. (a) The speed of the comet increases as it comes closer to the Sun. Position A is closet to the Sun, and therefore has thegreatest speed. Position C is furthest from the Sun, and so has the slowest speed. Positions B and D are equidistantfrom the Sun and are between positions A and C. Thus, v A > v B = v D > v C .26. (e) M P = M Er P = 1 4 r Ev2GM2GM8GM= =EE EE= andPr1ErrEE⎛ 8GM⎞Ev⎜r ⎟PE=⎝ ⎠vE⎛ 2GME ⎞⎜r ⎟⎝ E ⎠vP= 4vvEP= 2vEv4CHAPTER 6 REVIEW(Pages 300–301)Understanding Concepts1. The escape energy (and thus the escape speed) from the Sun is much greater than that from Earth, so the rocket given thespeed needed to escape from Earth would not have enough speed to escape from the solar system. Space vehicles sent toexplore distant planets have a much lower binding energy by the time they reach those distant locations, and could acquireenough energy to escape from the solar system by taking advantage of the force of gravity of the distant planet.2. Since Earth rotates eastward, an eastward orientation of the rocket as it is being launched means that the rocket alreadyhas a component of the required velocity before blasting off. This means that less energy will be needed to launch therocket eastward than would be required to launch it westward in order to achieve the same speed.3. g U = 1.0 N/kgM U = 8.80× 10 25 kgr U = 2.56× 10 7 mCopyright © 2003 Nelson Chapter 6 Gravitation and Celestial Mechanics 375

gGMUU 2( r+rU)r=GMU= −rUgU−11 2 2 25(6.67× 10 N ⋅ m /kg )(8.80×10 kg)7= − 2.56×10 m1.0 N/kg7= 5.1×10 m4r = 5.1×10 kmUranus has a gravitational field strength of 1.0 N/kg at an elevation of 5.1 × 10 4 km above the surface.4. M = 1.48 × 10 23 kgr = 5.55 × 10 3 km = 5.55 × 10 6 mGMg =2r−11 2 2 23(6.67× 10 N ⋅ m /kg )(1.48×10 kg)=6 2(5.55×10 m)g = 0.318 N/kgThe magnitude of Ganymede’s gravitational field strength at a point in space 5.55 × 10 3 km from its centre is 0.318 N/kg.5. Use the spacecraft-to-Earth line as the reference for the coordinate system.M E = 5.98 × 10 24 kgM Moon = 7.35 × 10 22 kgr E-spacecraft = 3.07 × 10 8 mr Moon-spacecraft = 2.30 × 10 8 m gT = gE + gMoongT, x= gEg = gT, y MoonggEE 2EGM=r−11 2 2 24(6.67× 10 N ⋅ m /kg )(5.98×10 kg)=8 2(3.07×10 m)−3= 4.23×10 N/kggMoonMoon 2gMGM=r−11 2 2 22(6.67× 10 N ⋅ m /kg )(7.35×10 kg)=8 2(2.30×10 m)−5= 9.26×10 N/kgg g g2 2T=T, x+T, yT−3 −5( 4.23 10 N/kg) ( 9.26 10 N/kg)= × + ×−3g = 4.23×10 N/kg2 2376 Unit 2 Energy and Momentum Copyright © 2003 Nelson

gtanθ=gT, yT, x⎛ g1 T, y⎞−θ = tan⎜g ⎟⎝ T, x ⎠θ = 1.26°The total gravitational field strength (magnitude and direction) of the Earth-Moon-spacecraft system is 4.23 × 10 −3 N/kg[1.26° from the spacecraft-to-Earth line].6. r E = 6.38 × 10 6 md M = 0.38d Eg E = 9.8 N/kgg M = 0.38g ESince Mercury’s diameter is 0.38 times that of Earth’s, Mercury’s radius is also 0.38 times that of Earth. Therefore,r M = 0.38 r E .GMMgM =2rMGM MgM =20.38r( )Substituting g M = 0.38 g E :0.38g=GM0.38rMMEME 2( E )MM=( 0.38g)( 0.38r)EGE26( 0.38 )(9.80 N/kg) ((0.38)(6.38×10 m) )=6.67 × 10 N ⋅m /kg23= 3.3×10 kgTherefore, Mercury’s mass is 3.3 × 10 23 kg.7. v = 7.15 × 10 3 m/sM E = 5.98 × 10 24 kgr E = 6.38 × 10 6 mGME(a) v =rGMEr =2v−11 2 2 246.67× 10 N ⋅ m /kg 5.98×10 kg=237.15×10 m/s( )( )( )r = 7.80×10 m6−11 2 2In terms of Earth’s radius, the satellite’s distance from Earth’s centre is6r 7.80×10 m=6rE6.38×10 mr = 1.22 rEThus, the satellite is 1.22 r E from Earth’s centre.2Copyright © 2003 Nelson Chapter 6 Gravitation and Celestial Mechanics 377

(b) altitude = r−rE= 1.22rE−rEaltitude = 0.22 rEThe satellite has an altitude of 0.22 r E .8. v Tethys = 1.1 × 10 4 m/sM S = 5.67 × 10 26 kgGM(a) v = SrGMSr =2v−11 2 2 266.67× 10 N ⋅ m /kg 5.67×10 kg=421.1×10 m/s( )( )( )8 5r = 3.1× 10 m, or 3.1×10 kmThe orbital radius of Tethys is 3.1 × 10 5 km.2πr(b) T =v8( )2π 3.1×10 m=41.1×10 m/s5 ⎛ 1 h ⎞⎛ 1 d ⎞= (1.8 × 10 s) ⎜ ⎟⎜ ⎟⎝3600 s ⎠⎝24 h ⎠T = 2.1 dThe orbital period of Tethys is 2.1 d.9. M S = 1.99 × 10 30 kgT V = 1.94 × 10 7 mCrSS 23GM=4π= C2 SV23 TV GMSr =2T4π2T3 V GMSr =24π7 2 −11 2 2 30(1.94× 10 s) (6.67× 10 N ⋅ m /kg )(1.99×10 kg)= 324π11r = 1.08×10 mThe average Sun-Venus distance is 1.08 × 10 11 m.10. M E = 5.98 × 10 24 kgr E = 6.38 × 10 6 mv = 9.00 km/s = 9.00 × 10 3 m/sm R = 4.60 kg378 Unit 2 Energy and Momentum Copyright © 2003 Nelson

(a) r′ = ?Applying the law of conservation of energy:Eg + EK = E ′g+ E ′KGM EmR 1 2 GM EmR− + mRv= − + 0rE2r′GM E 1 2 GM E− + v =−rE2 r′GMEGME 1 2= − vr′rE2GMEr′ =GME 1 2− vrE2−11 2 2 24( 6.67× 10 N ⋅ m /s )( 5.98×10 kg)=−11 2 2 24( 6.67 × 10 N ⋅ m /s )( 5.98×10 kg)1 3−69.00 × 10 m/s6.38×10 m27r′ = 1.85×10 m( )Let the altitude be A.A= r′−rE7 6= 1.85× 10 m − 6.38×10 m7A = 1.21×10 mThe altitude above Earth’s surface is 1.21 × 10 7 m or 1.21 × 10 4 km.(b) At the altitude found in (a), the gravitational potential energy is negative, the kinetic energy is zero (because the speedis zero), and the binding energy, E B ′, is the extra energy needed to give the rocket a total energy of zero.E ′ + E ′ + E ′ = 0B g KE ′B=−E′g⎛ GMEm=− ⎜ −⎝ r′GMEmR=r′B=R−11 2 2 24( 6.67× 10 N ⋅ m /s )( 5.98×10 kg)( 4.60 kg)7E ′ = 9.92×10 JThe binding energy is 9.92 × 10 7 J.11. M T = 1.35 × 10 23 kg (mass of Titan)r T = 2.58 × 10 3 km = 2.58 × 10 6 mm R = 2.34 × 10 3 kg (mass of rocket)(a) v esc = ? (escape speed)2GMTvesc=rvescT⎞⎟⎠71.85×10 m−11 2 2 23( × ⋅ )( × )2 6.67 10 N m /s 1.35 10 kg=62.58×10 m3= 2.64×10 m/sThe escape speed from Titan is 2.64 × 10 3 m/s.2Copyright © 2003 Nelson Chapter 6 Gravitation and Celestial Mechanics 379

(b) E esc = ? (escape speed)At the surface of Titan, the rocket is at rest, so its kinetic energy is zero. Thus, its total energy is E g and the escapeenergy is the extra energy needed to give the rocket a total energy of zero.Eg+ Eesc= 0Eesc=−Eg⎛ GMTmR⎞=−⎜−⎟⎝ rT⎠GMTmR=rEesc=T−11 2 2 23 3( 6.67× 10 N ⋅ m /s )( 1.35× 10 kg)( 2.34×10 kg)9= 8.17×10 J62.58×10 mThe escape energy is 8.17 × 10 9 J. This value can also be found by using the escape speed of the rocket in the equation1E ( ) 2K = mR vesc.212. M E = 5.98 × 10 24 kgr E = 6.38 × 10 6 mm R = 1.00 × 10 4 kgr R = 1.00 × 10 10 m(a) E g = ?GM EmREg=−rR−11 2 2 24 4( 6.67× 10 N ⋅ m /s )( 5.98× 10 kg)( 1.00×10 kg)=−101.00×10 m8E =− 3.99×10 JgThe gravitational potential energy is −3.99 × 10 8 J.(b) Since the total energy, E K + E g , must be at least zero, the kinetic energy needed to escape is +3.99 × 10 8 J.(c) v esc = ?2GMEvesc=rvescR−11 2 2 24( 6.67× 10 N ⋅ m /s )( 5.98×10 kg)=101.00×10 m2= 2.82×10 m/sThe escape speed from this position is 2.82 × 10 2 m/s. The escape speed can also be found by applying the escapeenergy found in (b) to the equation involving the kinetic energy, vesc=2EKm, where E K = E esc .13. M S = 1.99 × 10 30 kgM E = 5.98 × 10 24 kgr E = 1.49 × 10 11 mGMSMEEg=−rE−11 2 2 30 24N ⋅ m /kg × 10 kg 5.98×10 kg)111.49×10 m( 6.67× 10=−)( 1.99 )(E33=− 5.33×10 JgThe gravitational potential energy of the Sun-Earth system is −5.33 × 10 33 J.380 Unit 2 Energy and Momentum Copyright © 2003 Nelson

14. (a) M M = 3.28 × 10 23 kgr M = 2.44 × 10 6 m2GMv =rMM−11 2 2 23( × ⋅ )( × )2 6.67 10 N m /kg 3.28 10 kg=62.44×10 m3v = 4.23×10 m/s, or 4.23 km/sThe escape speed from Mercury is 4.23 km/s.(b) M Moon = 7.35 × 10 22 kgr Moon = 1.74 × 10 6 m2GMMoonv =rMoon−11 2 2 22( × ⋅ )( × )2 6.67 10 N m /kg 7.35 10 kg=61.74×10 m3v = 2.37×10 m/s, or 2.37 km/sThe escape speed from Earth’s Moon is 2.37 km/s.15. (a) M star = 3.4 × 10 30 kg41.7×10 m 3r star == 8.5 × 10 m22GMstarv =rstar−11 2 2 30( × ⋅ )( × )2 6.67 10 N m /kg 3.4 10 kg=38.5×10 m8v = 2.3×10 m/sThe escape speed from a neutron star is 2.3 × 10 8 m/s.(b) c = 3.00 × 10 8 m/s8v 2.3×10 m/s=8c 3.00×10 m/sv= 0.77c⎛v⎞Thus, the percentage equals ⎜ × 100% = 77%c⎟.⎝ ⎠The escape speed from a neutron star is 77% the speed of light.75.06×10 m 716. (a) r == 2.53 × 10 m2v = 24 km/s = 2.4 × 10 4 m/s2GMv =r2rvM =2G=7 4( 2.53× 10 m)( 2.4×10 m)−11 2 226.67 ( × 10 Nm/kg ⋅ )26M = 1.1×10 kgThe planet’s mass is 1.1 × 10 26 kg.2Copyright © 2003 Nelson Chapter 6 Gravitation and Celestial Mechanics 381

(b) According to the data in Appendix C, the planet is Neptune. (There is a 6% difference in mass from Appendix C.)17. M S = 1.99 × 10 30 kgM P = 1.67 × 10 −27 kgr = 1.4 × 10 9 m (initial position)v = 3.5 × 10 5 m/s (initial speed)(a) r′ = 2.8 × 10 9 m (final position)v′ = ? (final speed)Applying the law of conservation of energy:E + E = E ′ + E ′g K g KGMSmP 1 2 GMSmP1− + mvP= − + mPvr 2 r′2GMS1 2 GMS 1 2− + v = − + ( v′)r 2 r′22 ⎛ 1 1⎞2( v′ ) = 2GMS⎜ − ⎟+v⎝r′r⎠⎛ 1 1⎞v′ = 2GMS⎜ − ⎟+v⎝r′r⎠The proton’s speed is 1.7 × 10 5 m/s.(b) v esc ′ = ? (escape speed at the final position)vvescesc′ =2GMr′S( ′)22−11 2 2 30 ⎛ 1 1 ⎞5( )( ) ⎜⎟ ( ) 2= 2 6.67 × 10 N ⋅ m /s 1.99 × 10 kg − + 3.5×10 m/s8 9⎝ 2.8× 10 m 1.4 × 10 m ⎠5v′ = 1.7×10 m/s−11 2 2 302(6.67× 10 N ⋅ m /kg )(1.99×10 kg)=92.8×10 m′5= 3.1×10 m/sThe escape speed is 3.1 × 10 5 m/s at the location indicated; this is greater than the speed found in (a), so the proton willnot escape.18. When light strikes a piece of black paper, a small portion of the light is reflected. However, when light strikes a blackhole, the light is absorbed, making the black hole even blacker than black paper.19. m = 1.1 × 10 11 M SM S = 1.99 × 10 30 kg8c = 3.00×10 m/s2GMr =2c−11 2 2 11 302(6.67 × 10 N ⋅ m /kg )(1.1× 10 (1.99×10 kg))=8 2(3.00×10 m/s)14r = 3.2×10 mThe Schwartzschild radius of the black hole is 3.2 × 10 14 m.Applying Inquiry Skills20. Table 1 provides the missing answers concerning some of the moons of Uranus.3r(a) Kepler’s third law constant for Uranus (C U ) can be calculated using the ratio2T .(b) The average of the C U values of the calculations in (a) is 1.48 × 10 14 m 3 /s 2 .382 Unit 2 Energy and Momentum Copyright © 2003 Nelson

(c) M U = 8.80 × 10 25 kgGMUCU =24π−11 2 2 25(6.67× 10 N ⋅ m / kg )(8.80×10 kg)=24π14 3 2CU= 1.49×10 m /sThe values agree (0.7% difference).3r3 2(d) Table completed using equations T = and r = CUT.CU(e) Students who speculate that only the larger moons can be observed using Earth-based telescopes are right. Thus, onlythe larger moons were discovered hundreds of years ago. Students who research the physical data of the moons thatorbit Uranus will find that Titania and Oberon have diameters greater than 1500 km, whereas all the other moons listedare less than 100 km in diameter.Table 1 Data of Several Moons of the Planet Uranus for Question 20Moon Discovery r average (km) T (Earth days) C U (m 3 /s 2 )Ophelia Voyager 2 (1986) 5.38 × 10 4 0.375 1.48 × 10 14Desdemona Voyager 2 (1986) 6.27 × 10 4 0.475 1.46 × 10 14Juliet Voyager 2 (1986) 6.44 × 10 4 0.492 1.48 × 10 14Portia Voyager 2 (1986) 6.61 × 10 4 0.512 1.48 × 10 14Rosalind Voyager 2 (1986) 6.99 × 10 4 0.556 1.48 × 10 14Belinda Voyager 2 (1986) 7.52 × 10 4 0.621 1.48 × 10 14Titania Herschel (1787) 4.36 × 10 5 8.66 1.48 × 10 14Oberon Herschel (1787) 5.85 × 10 5 13.46 1.48 × 10 1421. (a) Some students may think the problem makes sense. However, many students will realize that the (theoretical) radius ofan orbit that has a period of 65 min would be less than Earth’s radius. (Students may recall that the typical orbitalperiod of a satellite in low-altitude orbit is about 80 min. For example, see question 22 on page 168 of the text.)(b) M E = 5.98 × 10 24 kgT = 65 min = (65 min)(60 s/min) = 3.90 × 10 3 sr = ?3r GME=2 2T 4π3 ⎛GME ⎞ 2r = ⎜ T2 ⎟⎝ 4π⎠2GM3 ETr =24π3−11 2 2 24 3( 6.67 × 10 N ⋅ m /s )( 5.98× 10 kg)( 3.90×10 s)=24π6r = 5.36×10 mThe theoretical radius of the orbit is 5.36 × 10 6 m.(c) Earth’s radius (6.38 × 10 6 m) is larger than the theoretical radius found in (b), so the calculated orbit cannot exist.(d) The skill of analyzing a situation is valuable in order to reduce the chances of wasting time on calculations that don’tmake sense and to increase the chances of being able to estimate whether or not a solution to a problem is logical.2Copyright © 2003 Nelson Chapter 6 Gravitation and Celestial Mechanics 383

22. (a) The rocket’s mass can be calculated from the gravitational potential energy at rest (on Earth’s surface at r E ). From thegraph E g = –10 × 10 10 J = –1.0× 10 11 J.−GM EmEg=r−rEgm = GM E6 11− ( 6.38× 10 m)( − 1.0×10 J)=−11 2 2 246.67× 10 N ⋅ m /kg 5.98×10 kg( )( )m = 1.6×10 kg3The rocket’s mass is 1.6 × 10 3 kg.(b) The escape energy can be determined using the value of gravitational potential energy at rest (1.0 × 10 11 J).(c) The launch speed of the rocket can be calculated using the value of the initial kinetic energy E K (on Earth’s surface atr E ). From the graph E K = –12 × 10 10 J = –1.2× 10 11 J.1 2EK= mv22EKv =m112(1.2 × 10 J)=31.6×10 kg4v = 1.2×10 m/sThe launch speed is 1.2 × 10 4 m/s.(d) Extrapolating from the graph, the kinetic energy E K , approaches 2.0 × 10 10 J as the distance approaches infinity, whereE g would approach zero. This can be approximated: at 5r E , the kinetic energy is 4.0 × 10 10 J and E g is –2.0 × 10 10 J.1 2EK= mv22EKv =m102(2.0×10 J)=31.6×10 kg3v = 5.0×10 m/sThe speed is 5.0 × 10 3 m/s.Making Connections23. (a) Turning the high-speed craft around would require a fairly large amount of energy, so mission control decided to havethe craft continue on toward the Moon. The idea was to take advantage of the Moon’s gravity to act as a sort of slingshotto help the craft accelerate in turning around and begin its return journey at the highest speed possible.(b) One major risk was the chance that there would not be enough electrical power available to guide the craft around theMoon at the most crucial times.Extension24. Let L represent the large planet and S represent the small planet.r L = 2r SD L = D S (densities)384 Unit 2 Energy and Momentum Copyright © 2003 Nelson

Thus, using V for volume, the ratio of the masses is:mS DSVS=mL DLVLVS=VL⎛4⎞ 3⎜ ⎟πrS3=⎝ ⎠⎛ 4 ⎞ 3⎜ ⎟πrL⎝3⎠3⎛ r ⎞S= ⎜ 3r ⎟⎝ L ⎠3mS⎛1⎞= ⎜ ⎟m ⎝2⎠LThe centripetal acceleration of the satellite is caused by the force of gravity in each case. Thus, using magnitudes:2 2mv 4πmrΣ F = mac = =2r T2GMm 4πmr=2 2r T3r GM=2 2T 4π2 2T 4π=3r GM2⎛2T ⎞L⎛ 4π⎞⎜ 3 ⎟ ⎜ ⎟⎝ rL⎠ GmL=⎝ ⎠2 2⎛T⎞ ⎛S4π⎞⎜ 3 ⎟ ⎜ ⎟⎝ r GmS ⎠ ⎝ S ⎠2 3TLrSmS× =3 2r mLTSL2 3TLrLmS= ×2 3TSrSmL32 2⎛rL⎞⎛ mS⎞TL = TS ⎜ 3⎟⎜r ⎟ ⎟⎝ S ⎠⎝mL⎠T2 2LTS2 2L= TS3 3⎛2⎞ ⎛1⎞= ⎜ ⎟ ⎜ ⎟⎝1⎠ ⎝2⎠TTL= TSThe shortest possible period is 40 min.25. Since the radius of the path is 2.0 × 10 11 m, the distance between the stars is 2(2.0 × 10 11 m) = 4.0 × 10 11 m,M S = 3.0 × 10 30 kg (mass of each star). The only force acting on each star is the force of gravity of the other star, whichcauses the circular motion of one star around the other.Copyright © 2003 Nelson Chapter 6 Gravitation and Celestial Mechanics 385

Σ F = maGM Mrc2S SMSv=2( 2r)GMS24rGM4rGM4rGM4rTSSS2=222⎛2πr ⎞= ⎜ ⎟⎝ T ⎠2 24πr=2T2 316πr=GMS2 316πrGMS2 11316π( 2.0×10 m)−11 2 2 30( 6.67 × 10 N ⋅ m /s )( 3.0×10 kg)T = 7.9×10 sThe period of one complete cycle is 7.9 × 10 7 s.vr= vT ==726. E is the amount of energy per unit area, and that area is proportional toplanet, so:1E ∝2rkE =r212r, where r is the distance from the Sun to theFrom Kepler’s third law:3r= C2T3 2r = CT where C is Kepler’s third law constant for the Sunr2 42 3 3= C TSubstituting into the first equation:kE =2 4C3T3−2 −43 3E = kC TE = (constant) TThus, E is proportional to−43−4T3. (Solving for the “constant” is unnecessary.)386 Unit 2 Energy and Momentum Copyright © 2003 Nelson

UNIT 2 PERFORMANCE TASKSAFETY IN TRANSPORTATION AND SPORTS(Pages 302–303)The three options in this task relate most directly to the principles studied in Chapters 4 and 5. Thus, a good time to discuss theoptions with the students is at the start of Chapter 5. Share with the students the method of assessment you intend to use withthe task.The first two options are research-oriented. Option 1 involves a sport, activity, or piece of equipment of the student’s ownchoice. Option 2 is somewhat more complex because the students are required to identify the social issues involved beforethey research and analyze them. Option 3 is a hands-on option that is an alternative to the typical egg-drop device.Option 1: Protective Equipment in Sports and Recreational ActivitiesIn this option, some students will analyze the safety equipment used in a particular sport or activity, such as hockey orbicycling. Other students will analyze the physics principles involved in the use of a particular piece of equipment, such as asafety helmet, used in several sports and activities. In both cases, students should relate what they discover to the principlespresented in Chapters 4 and 5, especially collisions in which energy and momentum play an important role.Students will likely find that the Internet is the best resource for this task.Option 2: Vehicle Safety FeaturesVehicle safety is an important issue in our society, and students will benefit from analyzing the issue and relating it to thephysics principles in Chapters 4 and 5. Energy, momentum, and collisions are the topics most closely related to vehicle safety.Students can refer to Appendix A3, text pages 765–766, for suggestions about decision making and defining, researching,and analyzing issues. Students can find information about vehicle safety in reference books, in automobile magazines, and onthe Internet.Option 3: The Bouncing EggThis option applies many of the principles of Chapters 4 and 5, especially the law of conservation of energy, Hooke’s law, andthe physics of collisions. The task can be started in conjunction with Section 4.5.As in the case with the model roller coaster (Unit 1 Performance Task, Option 1), it is wise to develop the criteria used toevaluate the success of the device before the students begin their task. Some of the quantitative criteria might be:• height of the drop• number of clear bounces• ability of the device to remain within a horizontal circle of predetermined diameter during its bounces• a maximum allowable mass• maximum allowable outside dimensions (e.g., the device must be able to fit inside a 4-L bucket)Some of the qualitative criteria might be:• originality/creativity• obvious protection of the egg• wise use of materials• aesthetic appealStudents can refer to Appendix A4, text page 767, for a summary of technological problem solving.AnalysisAs in the Unit 1 Performance Task, the answers to the Analysis questions depend on various factors, particularly the choice ofoptions. With any of the three options, students are urged to create their own questions and answer them, a feature that helps tomake the task more open-ended.EvaluationHere the students evaluate their own task. Answers will depend on the option chosen.Copyright © 2003 Nelson Unit 2 Performance Task 387

UNIT 2 SELF QUIZ(Pages 304–306)True/False1. F One joule is one kilogram metre squared per second squared.2. T3. F The thermal energy will be the same if the size of the force of kinetic friction is the same.4. F The impulse is the change in momentum.5. T6. F The only type of collision in which momentum is not conserved is one with a net external force.7. F The gravitational field strength is inversely proportional to the square of the distance.8. F The work done on the satellite by Earth is zero.9. F The geocentric model has Earth at the centre of the universe.10. F The Sun is located at one focus of a planets orbit.11. F The speed in orbit depends on the location in the elliptical orbit.12. F A black hole has an extremely strong gravitational field.13. T14. F No form of electromagnetic radiation can escape.Multiple Choice15. (c) Assuming a mass of 75 kg, and a height of 1.8 m:W =∆Eg= mg∆y2= (75 kg)(9.8 m/s )(1.8 m)3W = 1.3×10 JThe power of 10 is 10 3 .16. (c) Σ F = ma = 0F + F sinφ− mgcos β = 0NA17. (a) Σ F = ma = 0F cosφ−mgsin β − F = 0AyxF = mgcosβ −FsinφNKFKF =AA+ mgsinβcosφ388 Unit 2 Energy and Momentum Copyright © 2003 Nelson

18. (d) W = ( Fcos θ ) ∆ dW = F A cos φ(L)hSince sin β =LhL =sin βFh ATherefore,cos φW = .sin β19. (e) The force is always perpendicular to the motion.1 m 220. (c) E ⎛ ⎞K = ⎜m+⎟v2⎝5 ⎠2= (0.5)(1.2 mv )2E = 0.6mvK1 ⎛ m m⎞2E′ K= ⎜m+ + ⎟(0.80 v)2⎝5 5 ⎠2= (0.5)(1.4 m)(0.64 v )2= 0.448mv2= 0.75(0.6 mv )EK′ = 0.75EK21. (e) All three stones have the same initial total energy, and they all lose the same amount of gravitational energy. Therefore,they will all have the same increase in kinetic energy when they reach the water (even though they will reach the waterat different times).22. (d)23. (c) To conserve momentum, the same mass must have the same velocity if all of the energy was given to the billiard ballthat was stationary.24. (a) Escape speed is defined as the speed needed to escape from the surface. The current kinetic energies of the rockets arenot relevant.25. (c) The initial kinetic and gravitational energies are the same, giving a total energy of zero.26. (a) Escape energy is defined as the energy needed to escape from the surface. The current kinetic energies of the rocketsare not relevant.127. (d) The gravitational field is proportional to .2r328. (d) r ∝ MCompletion29. (a) Galileo Galilei(b) Johannes Kepler(c) James Prescott Joule(d) Tycho Brahe(e) Robert Hooke(f) Karl Schwartzschild30. (a) work(b) the force constant of a spring(c) impulse(d) force(e) thermal energy(f) the mass of Earth31. completely inelastic; equals; completely inelastic collision32. zero33. a singularity; Schwartzschild radiusCopyright © 2003 Nelson Unit 2 Self Quiz 389

Matching34. (a) A(b) E35.impulselaw of conservation of momentumkinetic energythermal energyelastic potential energyescape speedgravitational potential energyKepler’s third-law constantfrequency of a mass spring system in SHM(e)(g)(h)(j)(k)(d)(b)(a)(m)390 Unit 2 Energy and Momentum Copyright © 2003 Nelson

UNIT 2 REVIEW(Pages 307–311)Understanding Concepts1. One situation is a person carrying a book at a constant height across a level floor. A satellite in circular orbit or pushingon a brick wall are two other examples.2. The work done is transformed into another form of energy. One example is pushing a crate across a level floor. The workdone is transformed into thermal energy through friction.3. Momentum and energy can be related as follows:2mvEK=22 2mv=2m2pEK=2mp = 2mEKIf both the baseball and the shot have the same kinetic energy, the shot will have a larger momentum because it has thelarger mass.4. If both objects have the same kinetic energy, they both require the same amount of work done on them to stop them. If thefrictional force is the same and the work done is the same, the distance the frictional force acts is the same. Therefore, thetwo toboggans will have the same stopping distance.5. During the collision, the initial kinetic energy of the carts is stored as elastic potential energy in the spring.6. Several common devices that can store elastic potential energy are an elastic band, a bow for shooting arrows, golf balls,and a bungee cord.7. Bumpers made of springs are impractical because the kinetic energy of the collision would be stored in the spring, andthen converted back into kinetic energy again. This last stage could project the car into oncoming traffic or concretebarriers and cause further collisions.8. One possibility is:9. m = 1.5 10 3 kg∆d = 0.5 mF = 3.5 10 5 Nh = ?Copyright © 2003 Nelson Unit 2 Review 391

Using conservation of energy:ET= E′Tmg∆ y = F∆dmg( h + 0.50 m) = F∆dF∆d( h + 0.50 m) =mgF∆dh = −0.50 mmg5(3.5×10 N)(0.50 m)= −0.50 m3(1.5×10 kg)(9.8 N/kg)h = 11 mThe pile driver must start from a height of 11 m above the pile.10. m = 10.0 kg(a) d = b = 2.00 mW = ?The work done is equal to the area under the graph, therefore:1W = bh21= (2.00 m)(10 N)21W = 1.0×10 JThe amount of work done is 1.0 10 1 J.(b) d = b = 3.00 mE K = ?The change in E K will be equal to the work done, therefore:EK= W1= bh + lw21= (2.00 m)(10 N) + (10 N)(3.00 m − 2.00 m)21E = 2.0×10 JKThe block’s kinetic energy is 2.0 10 1 J, or 20 J.(c) d = 3.00 mE K = 20 Jv = ?1 2EK= mv22EKv =m2(20 J)=(10.0 kg)v = 2.0 m/s [W]The velocity at the 3.00-m mark is 2.0 m/s [W].392 Unit 2 Energy and Momentum Copyright © 2003 Nelson

11. W = E K = 1.25 10 4 Jv = 50.0 m/s(a) m = ?(b) ∆d = 5.00 mF = ?1 2EK= mv22EKm =2v42(1.25×10 J)=2(50.0 m/s)m = 10.0 kgW = F∆dWF = ∆ d41.25×10 J=5.00 m3F = 2.50×10 N [E]A constant force of 2.50 10 3 N [E] would give the object the same final velocity.12. m = 5.0 kgE K = 5.0 10 2 Jp = ?2mvEK=22 2mv=2m2pEK=2mp = 2mE= 2(5.0 kg)(5.0×10 J)p = 71 kg ⋅m/sThe momentum of the sled is 71 kg⋅m/s.13. x 1 = 0.10 mF = 5.0 Nm = 4.5 kgv = 2.0 m/sx 2 = ?First we must find the force constant of the spring:F = kxFk =x5.0 N=0.10 mk = 50 N/mK2Copyright © 2003 Nelson Unit 2 Review 393

Using the law of conservation of energy:E = E′TT1 2 1 2mv = kx2 2mvx =k(4.5 kg)(2.0 m/s)=50 N/mx = 0.60 mThe maximum compression of the spring will be 0.60 m.14. x = 2.00 cm = 0.0200 m(a) E e = ?1 2Ee= kx21 (50.0 N/m)(0.0200 m)2=2−2E = 1.00×10 JeThe elastic potential energy is 1.00 10 –2 J.(b) x i = 0.0200 mx f = 6.00 cm = 0.0600 m∆E e = ?1 2 1 2∆ Ee = kxf − kxi2 21 (2 2= kxf−xi)21= (50.0 N/m) 0.0600 m − 0.0200 m2−2∆ Ee= 8.00×10 JThe change in elastic potential energy is 8.00 10 –2 J.(c) m = 4.00 kgv = ?E = E′TT1 1kx = mv2 22kxv =m2 2222 2(( ) ( ) )2(50.0 N/m)(0.0600 m)=0.400 kgv = 0.671 m/sThe cart leaves the spring at a speed of 0.671 m/s.15. When a small object bounces off a larger stationary object, the change in momentum is greater than if the object sticks toit. If the collision takes the same amount of time, the force applied will therefore have to be much larger (twice as much).When riot police use rubber bullets, the rubber bullet exerts a large force on the person or object it hits.16. When the net force consists of only one force, impulse is a very useful tool in analyzing the changes in motion producedby that force.394 Unit 2 Energy and Momentum Copyright © 2003 Nelson

17. m = 78 kg∆t = 1 min = 60 sv 1 = 24 m/sv 2 = 0 m/sF = ?ΣF∆ t = ∆pmv ( 2 − v1)F =∆t78 kg(0 m/s − 24 m/s)=60 sF =−31 NThe magnitude of the force is 31 N.18. Let the subscript D represent the dog and W represent the wagon.m D = 9.5 kgv D = 1.2 m/sv W = –3.0 m/sm W = ?Using the conservation of momentum:p = p′0 = mDv′ D+mWv′W−mDvD′mW=v′W−(9.5 kg)(1.2 m/s)=−3.0 m/smW= 3.8 kgThe mass of the wagon is 3.8 kg.19. m 1 = 2.4 kgv 1 = 1.5 m/s [W]m 2 = 3.6 kg(a) E T = ?E = EETTK1 2= mv 1 121= (2.4 kg)(1.5 m/s)2= 2.7 JThe total energy of the system before the system is 2.7 J.(b) v´ = ?p = p′mv + 0 = mv′ + mv′1 1 1 1 2 2At minimum separation, the two velocities of the carts will be the same, therefore:mv1 1= ( m1+m2)v′mv 1 1v′ =m + m1 2(2.4 kg)(1.5 m/s)=2.4 kg + 3.6 kgv′ = 0.60 m/s [W]The velocity of each cart will be 0.60 m/s [W].2Copyright © 2003 Nelson Unit 2 Review 395

(c) ∆E K = ?∆ E = E′−EK KT KT⎛1 2 1 2⎞ ⎛12⎞= ⎜ mv1 1′ + mv1 2′⎟−⎜ mv1 1⎟⎝2 2 ⎠ ⎝2⎠⎛1 2 1 2⎞ ⎛12⎞= ⎜ (2.4 kg)(0.60 m/s) + (3.6 kg)(0.60 m/s) ⎟−⎜ (2.4 kg)(1.5 m/s) ⎟⎝2 2 ⎠ ⎝2⎠∆ EK= −1.6 JThe change in total kinetic energy of the system is –1.6 J.(d) x = 12 cm = 0.12 mk = ?1 2Ee= kx22Eek =2x2(1.62 J)=2( 0.12 m)2k = 2.2×10 N/mThe force constant of the spring is 2.2 10 2 N/m.20. m 1 = 1.2 kgm 2 = 4.8 kgk = 2.4 10 3 N/mv 2´ = 2.0 m/sx = ?First we must find the speed of the 1.2-kg trolley:p = p′0 = mv 1 1′ + mv 2 2′−mv2 2′v′ 1=m1−(4.8 kg)(2.0 m/s)=(1.2 kg)v′ =−8.0 m/s1We can now calculate the force constant:Ee = EK1+EK21 2 1 2 1 2kx = m1v 1′ + m2v2′2 2 22 2mv1 1′ + m2v2′x =k(1.2 kg)( − 8.0 m/s) + (4.8 kg)(2.0 m/s)=32.4×10 N/mx = 0.20 mThe force constant of the spring is 0.20 m.21. Choose right as positive.m 1 = 15 kgv 1 = –6.0 m/sm 2 = 25 kgv 2 = 3.0 m/sv 2´ = ?2 2396 Unit 2 Energy and Momentum Copyright © 2003 Nelson

(a) v 1´ = –0.30 m/sp = p′mv1 1+ m2v2= mv1 1′ + mv2 2′mv1 1+ m2v2−mv1 1′v2′ =mmv2 2+ m1( v1−v′1)=m2m1( v1 − v1′)= v2+m215 kg( −6.0 m/s −( −0.30 m/s))= 3.0 m/s +25 kgv2′ =−0.42 m/s [right], or 0.42 m/s [left]The velocity of the 25-kg object would be 0.42 m/s [left].(b) v 1´ = 0.45 m/sp = p′mv 1 1+ m2v2= mv 1 1′ + mv 2 2′mv 1 1+ m2v2−mv1 1′v2′ =m2mv2 2+ m1( v1−v′1)=m2m1( v1 − v′1)= v2+m215 kg( −6.0 m/s −0.45 m/s)= 3.0 m/s +25 kgv′ 2=−0.87 m/s [right], or 0.87 m/s [left]The velocity of the 25-kg object would be 0.87 m/s [left].(c) Since the objects stick together, their final velocities will be equal to each other, therefore:p = p′mv1 1+ m2v2= mv1 1′ + mv2 2′mv 1 1+ m2v2= ( m1+m2)v′mv 1 1+m2v2v′ =m1+m2(15 kg)( − 6.0 m/s) + (25 kg)(3.0 m/s)=15 kg + 25 kgv′ =−0.38 m/s [right], or 0.38 m/s [left]The velocity of the 25-kg object would be 0.38 m/s [left].22. m 1 = 1.4 10 4 kgv 2´ = 2.5 10 4 m/sv 1´ = 2.0 10 1 m/sFirst we must calculate the mass of the sled and the rocket combined:p = p′0 = mv 1 1′ + m2v2′−mv1 1′m2=v′24− (1.4 × 10 kg)(50 m/s)=4− 2.5×10 m/sm = 28 kg22The total mass required is 28 kg. At 10 kg/s, it will take 28 ÷ 10 = 2.8 s to burn that much fuel.Copyright © 2003 Nelson Unit 2 Review 397

23. m 1 = 2.3 kgv 1 = 18 m/s [E]v 2 = 19 m/s [W]v´ = 3.1 m/sm 2 = ?Choose east as positive. Using the conservation of momentum:p = p′mv + mv = mv′ + mv′1 1 2 2 1 1 2 2v′ = v′ = v′, therefore:Since it is a completely inelastic collision 1 2mv2 2− mv2′ = mv1′−mv1 1m2( v2 − v′ ) = m1( v′−v1)m1( v′ − v1)m2=( v2− v′)(2.3 kg)(3.1 m/s −18 m/s)=−19 m/s −3.1 m/sm2= 1.6 kgThe mass of the second bird is 1.6 kg.24. v1= vv2= 0vv1′ =−5m = ?2p = p′mv1 1+ mv2 2= mv′ 1 1+mv′2 2mv1 1+ 0 = mv′ 1 1+mv′2 2mv1 1− mv′ 1 1=mv′2 2m1( v1 − v′ 1)= m2v′2mv′= (Equation 1)2 2m1v1−v ′1ET= E′1 1 1 1mv mv mv′ mv′2 2 2 2mv mv′ m v′2 2 2 21 1+ 2 2= 1 1+2 2T2 2 21 1+ 0 =1 1+2 2mv mv′ m v′2 2 21 1− 1 1=2 2m v v′ m v′2 2 21( 1 − 1) = 2 22mv 2 21 =2 21− ′1mv vmv′2′2 2m1=( v + v′ )( v −v′)1 1 1 1(Equation 2)398 Unit 2 Energy and Momentum Copyright © 2003 Nelson

Set Equations 1 and 2 equal to each other:2mv′ 2 2mv′2 2=v1− v1′ ( v1+ v1′ )( v1−v1′)v2′1 =v1+v′1v′ 2= v1+v′1⎛ v ⎞= v + ⎜ − ⎟⎝ 5 ⎠4v′ 2= v (Equation 3)5Substitute Equation 3 into Equation 1:mv 2 2′m1= v − v ′1 1⎛4⎞m⎜v⎟5=⎝ ⎠⎛ v ⎞v − ⎜ − ⎟⎝ 5 ⎠⎛4mv⎞⎜ ⎟5=⎝ ⎠⎛ 6v⎞⎜ ⎟⎝ 5 ⎠2m1= m3The mass of the second nucleus is 2 3 m .25. The total momentum before and after must be the same. The two initial momentum vectors will be the components of thefinal momentum vector. The dotted line on the diagram represents the direction “after.” A completely inelastic collisionmeans they will stick together.m 2 = 2.0 10 3 kgv 2= 2.4 10 1 m/s [E]m 1 = 3.6 10 3 kgv 1= 1.0 10 1 m/s [S]v = ?Copyright © 2003 Nelson Unit 2 Review 399

First we must calculate the initial momentum of each vehicle. For the truck: p = m vp1 1 1143 1= (3.6× 10 kg)(1.0×10 m/s)= 3.6× 10 kg ⋅m/sFor the car:pp2 2 22= m v43 1= (2.0× 10 kg)(2.4×10 m/s)= 4.8× 10 kg ⋅m/sThe momentum of the car and truck coupled together after the collision: 2 2p′ = p + p12 1 24 2 4 2p= (3.6× 10 kg ⋅ m/s) + (4.8× 10 kg ⋅m/s)6.0 10 kg m/s412′ = × ⋅To calculate the final velocity of the car and truck coupled together:p12 ′ = m12 v12′p′12v′ 12=m1246.0× 10 kg ⋅m/s=3 33.6× 10 kg + 2.0×10 kg= 10.714 m/sv′ = 11 m/s12p1tanθ= p2⎛1p1⎞−θ = tan⎜ p ⎟⎝ 2 ⎠4−1⎛3.6× 10 kg ⋅m/s⎞= tan ⎜ 4 ⎟⎝4.8× 10 kg ⋅m/s⎠θ = 37°The final speed of the cars is 11 m/s [37º S of E].26. (a) m 1 = 2.3 10 4 kgv 1 = 15 m/s [51° S of W]m 2 = 1.2 104 kgv´ = 11 m/s [35° S of W]400 Unit 2 Energy and Momentum Copyright © 2003 Nelson

90°= 35°+ θ + 39°θ = 16°Units have been omitted until the final step for clarity. Using the cosine law:2 2 2p = p + p′ −2p p′cosθ2 1 12 1 12m v m v m v′ 2( mv )( m v′)cosθ2 2 2 2 2 22 2 = 1 1 + 12 12 − 1 1 12 12vv′ 2( )( ′ )cosθ2 2 2 2m1 v1 + m12v12 − mv1 1m12v122=2m22=4 2 2 4 4 2 2 4 4 4(2.3× 10 ) (15) + (2.3× 10 + 1.2× 10 ) (11) − 2(2.3× 10 )(15)(2.3 × 10 + 1.2× 10 )(11) cos16°= 9.0871 m/s= 9.1 m/ssinφsinθ=p p1 2(1.2 10)× 4 2−1 ⎛ p1sinθ⎞φ = sin ⎜ ⎟⎝ p2⎠−1 ⎛mv1 1sinθ⎞= sin ⎜ ⎟⎝ mv2 2 ⎠4−1⎛(2.3× 10 kg)(15 m/s)sin16°⎞= sin ⎜ 4⎟⎝ (1.2×10 kg)(9.0871 m/s) ⎠φ = 61°Since 61º – 35º = 26º, the initial velocity of the second truck was 9.1 m/s [26º N of W].(b) % lost = ?ET− E′T% lost = × 100%ET⎛1 2 1 2⎞ ⎛1 2 1 2⎞⎜ mv 1 1+ m2v2⎟− ⎜ mv 1 1′ + mv 2 2′⎟2 2 2 2=⎝ ⎠ ⎝ ⎠× 100%1 2 1 2mv1 1+m2v22 2mv mv mv′ mv′= × 100%2 2 2 21 1+ 2 2− 1 1 − 2 22 2mv1 1+m2v2mv mv′ m v m v′= × 100%2 2 2 21 1− 1 1 + 2 2−2 22 2mv1 1+m2v2m ( v v′ ) m ( v v′)=2 2 2 21 1−1+2 2−22 2mv 1 1+ mv 2 2(( ) ( ) ) ( ) ( )( )4 2 2 42 2(2.3× 10 kg) 15 m/s − 11 m/s + (1.2× 10 kg) 9.0871 m/s − 11 m/s= × 100%4 2 4 2(2.3× 10 kg)(15 m/s) + (1.2 × 10 kg)(9.0871 m/s)% lost = 31%The percentage of the initial kinetic energy lost is 31%.Copyright © 2003 Nelson Unit 2 Review 401

27. θ = 180º – 25º = 155ºm 1 = 82 kgv 1 = 8.3 m/s [N]m 2 = 95 kgv 2 = 6.7 m/s [25° W of N]m 3 = 85 kgv´ = ?′ = + −mv mv mv 2( mv)( mv)cosθ2 2 2p p1 p2 2p1p2cosθ2 2 2 2 2 2′ =1 1+2 2−1 1 2 22 2 2 2mv1 1+ mv2 2− mv1 1mv2 2v′ =22( )( )cosθm2 2 2 2(82 kg) (8.3 m/s) + (95 kg) (6.7 m/s) − 2(82 kg)(8.3 m/s)(95 kg)(6.7 m/s)cos155°=2(82 kg + 95 kg + 85 kg)v′ = 4.9 m/ssinφsinθ=p2p′−1 ⎛ p2sinθ⎞φ = sin ⎜ ⎟⎝ p′⎠−1 ⎛mv2 2sinθ⎞= sin ⎜mv⎟⎝ ′ ⎠−1⎛ (95 kg)(6.7 m/s)sin155°⎞= sin ⎜ ⎟⎝(85 kg + 95 kg + 82 kg)(4.9 m/s) ⎠φ = 12°The final velocity of the combined players is 4.9 m/s [12º W of N]28. We are most likely to see a comet when it is moving at its fastest speed. When a comet is within the solar system, it is atits closest approach to the sun which means most of the comet’s energy is in the form of kinetic energy.402 Unit 2 Energy and Momentum Copyright © 2003 Nelson

29. The farther an orbiting vehicle is from Earth, the slower its speed. To increase the radius of orbit from a particular point,the satellite would first need an increase in speed to increase its total energy. As it moved to a higher orbit, this additionalkinetic energy would be converted into gravitational potential energy. At the higher orbit, the speed of the vehicle wouldbe less than at the lower orbit.GMv =r30. The speed required to escape from the surface of Earth is always 11.2 km/s. If a space craft is not on the surface of Earth,it would not need that much speed. The term escape speed is defined as the launch speed necessary to just escape Earth’sgravitational pull.31. G = 6.67 10 –11 N·m 2 /kg 2M = 5.98 10 24 kgr E = 6.38 10 6 mg = ?First we must calculate the radius of the satellite’s orbit:2r2 = rE+ 6.55×10 km6 5= 6.38× 10 m + 6.55×10 m6r = 7.035×10 m2To calculate the gravitational acceleration:GMg =2r−11 2 2 24(6.67× 10 N ⋅ m /kg )(5.98×10 kg)=6 2(7.035×10 m)2g = 8.06 m/sThe magnitude of the gravitational acceleration is 8.06 m/s 2 .32. Let the subscript X represent the unknown planet and E represent Earth.m X = 0.25M Er X = 0.60r Eg X = ?GMEgE =2rEGM XgX =2rXG(0.25 ME)=2(0.60 rE)0.25 ⎛GM⎞E= 20.36 ⎜r ⎟⎝ E ⎠0.25= ( gE)0.36gX= 0.69gEThe surface gravitational acceleration of the unknown planet is 0.69g E .Copyright © 2003 Nelson Unit 2 Review 403

33. G = 6.67 10 –11 N·m 2 /kg 2M S = 1.99 10 30 kgr S = 6.96 10 8 mr E = 1.49 10 11 mg = ?GMg =2r−11 2 2 30(6.67× 10 N ⋅ m /kg )(1.99×10 kg)=8 11 2(6.96× 10 m + 1.49×10 m)−3g = 5.92×10 N/kg [toward the Sun]The gravitational field of the Sun at the position of Earth is 5.92 10 –3 N/kg [toward the Sun].34. Let the subscript M represent Mars, J represent Jupiter, and S represent the Sun.(a) r M = 2.28 10 11 mr J = 7.78 10 11 mM S = 1.99 10 30 kgT = ?The radius of orbit will be:rM+ rJr =211 112.28× 10 m + 7.78×10 m=211r = 5.03×10 mTo calculate the period:3GMS r=2 24πT4πrT =GM2 3S2 11 34 π (5.03×10 m)=−11 2 2 30(6.67× 10 N ⋅ m /kg )(1.99×10 kg)8T = 1.9456×10 s8 1 hr 1 d 1 a1.9456× 10 s× × × = 6.16 a3600 s 24 h 365.26 dThe period of the asteroid’s orbit in Earth years is 6.16 a.(b) v = ?GMv =r−11 2 2 30(6.67× 10 N ⋅ m /kg )(1.99×10 kg)=115.03×10 m4v = 1.62×10 m/sThe speed of the asteroid is 1.62 10 4 m/s.404 Unit 2 Energy and Momentum Copyright © 2003 Nelson

60 min 60 s35. T = 24 h = 24 h × × = 8.64 10 4 s1 h 1 min(a) M N = 1.03 10 26 kgC = ?GMC =24π−11 2 2 26(6.67× 10 N ⋅ m /kg )(1.03×10 kg)=24π14 3 2C = 1.74×10 m /sKepler’s third-law constant is 1.74 10 14 m 3 /s 2 .(b) r = ?3rC =T2r = CT3 2314 3 2 4 2= (1.74× 10 m /s )(8.64×10 s)8r = 1.09×10 mThe satellite must be 1.09 10 8 m from Neptune to maintain its circular orbit.(c) 1.09 × 10 8 m – 2.48 × 10 7 78.4×10 mm == 8.42 × 10 4 km1000The altitude of the orbit is 8.42 × 10 4 km.24 h 60 min 60 s36. T = 1.77 d = 1.77 d× × × = 1.529 10 5 s1 d 1 h 1 minr = 4.22 10 8 mM J = ?3J r=2 2GM4πT2 34πrMJ=2GT2 8 34 π (4.22×10 m)=(6.67× 10 N ⋅ m /kg )(1.529×10 s)27M = 1.90×10 kgJ−11 2 2 5 2The mass of Jupiter is 1.90 10 27 kg.37. (a) m = 5.32 10 –26 kgv = ?ET= 0EK+ Eg= 0EK=−Eg1 2 ⎛ GMm ⎞mv =− ⎜ − ⎟2 ⎝ r ⎠2GMv =r−11 2 2 232(6.67× 10 N ⋅ m /kg )(3.28×10 kg)=62.44×10 m3v = 4.23×10 m/sIn order to escape, the minimum speed of the molecule must have been 4.23 10 3 m/s.Copyright © 2003 Nelson Unit 2 Review 405

(b) v = ?1ET= Eg21EK + Eg = Eg21EK=− Eg21 2 1⎛GMm ⎞mv =− ⎜ − ⎟2 2⎝r ⎠GMv =r−11 2 2 23(6.67× 10 N ⋅ m /kg )(3.28×10 kg)=62(2.44×10 m)3v = 2.12×10 m/sThe speed of the molecule in an orbit around Mercury would be 2.12 10 3 m/s.(c) Let the subscript M represent Mercury and O represent the orbit.v = ?ET= ET′1EK + EgL = EgO21 2 ⎛ GMm ⎞ 1⎛GMm ⎞mv + ⎜− ⎟= ⎜−⎟2 ⎝ rM⎠ 2⎝ rO⎠1 2 GM GMv = −2 rM2rO⎛ 2 1 ⎞v = GM ⎜ − ⎟⎝ rM rO ⎠⎛ 2 1 ⎞= GM ⎜ − ⎟⎝ rM2 rM ⎠3GM=2rM−11 2 2 233(6.67× 10 N ⋅ m /kg )(3.28×10 kg)=62(2.44×10 m)3v = 3.67×10 m/sThe molecule would need to launch from the surface at a speed of 3.67 10 3 m/s.(d) binding energy = ?1binding energy =− ET=− Eg21 ⎛ GMm ⎞=− ⎜ − ⎟2 ⎝ r ⎠GMm=2r−11 2 2 23 −26(6.67× 10 N ⋅ m /kg )(3.28× 10 kg)(5.32×10 kg)=62(2.44×10 m)−19binding energy = 2.39×10 JIn this orbit, the molecule’s binding energy would be 2.39 10 –19 J.406 Unit 2 Energy and Momentum Copyright © 2003 Nelson

38. (a) Let the subscript J represent Jupiter, and B represent the black hole.M J = 1.90 10 27 kgM B = 85M J = 85(1.90 10 27 kg) = 1.615 10 29 kgr = ?For a black hole, the escape speed is the speed of light, c.ET= 0EK+ Eg= 0EK=−Eg1 2 ⎛ GMm ⎞mv =− ⎜ − ⎟2 ⎝ r ⎠2GMr =2v−11 2 2 292(6.67 × 10 N ⋅ m /kg )(1.615×10 kg)=8 2(3.00×10 m/s)2r = 2.4×10 mThe Schwartzschild radius of the black hole is 2.4 10 2 m.(b) For a black hole, the escape speed is the speed of light, c = 3.00 × 10 8 m/s.Applying Inquiry Skills39.40.Copyright © 2003 Nelson Unit 2 Review 407

41. m = 0.55 kgt = 0.020 s(a) k = ?From the diagram, there are 14 dots one cycle, therefore:T = 14(0.020 s)T = 0.28 sTo calculate the force constant:T = 2π2mkm T=2k 4π24πmk =2T4π0.55 kg=2( )2( 0.28 s)k = 2.8×10 N/m2The force constant of the springs is 2.8 10 2 N/m.(b) The results would be the same whether the paper was pulled quickly or slowly. The only value needed from the paperis how many dots there are in each cycle. This can be determined at almost any speed. The only restrictions on speedare that it must be slow enough to get at least one full cycle on the paper, and fast enough that the dots don’t overlapone another.(c) The most likely source of error would be the friction between the paper and the puck.(d) Safety considerations include exercising caution when using an electrical spark timer, handling the vibrating springcoils carefully so as not to be pinched, and being careful not to receive a paper cut from the sliding paper.42. (a) With no mass hanging from the spring, mark that point as zero. Suspend one weight from the accelerometer and markthat point as 1g. Suspend a second weight and mark the bottom of the spring as 2g. Measure the distances between 0gand 1g, and between 1g and 2g. They should be the same. Measure out the average of the two values beyond 2g andmark 3g and 4g.(b) Some possibilities are:• spinning in a circle with the accelerometer held horizontally• landing after jumping off a stool while holding the accelerometer• quickly lifting the accelerometer into the air(c) Some possible answers are:• the bottom of a roller coaster ride• during a large swing• during a spin ride (if held horizontal)43. m = 0.88 kgk = 36 N/mv = 0.22 m/s∆d = 2.5 mx = 3.6 cm = 0.036 m(a) E e = ?E K = ?To calculate the elastic potential energy:1 2Ee= kx21= (36 N/m)(0.036 m)2= 0.02333 J−2E = 0.023 J, or 2.3×10 Je2408 Unit 2 Energy and Momentum Copyright © 2003 Nelson

To calculate the kinetic energy:1 2EK= mv21 (0.88 kg)(0.22 m/s)2=2= 0.02130 J−2EK= 0.021 J, or 2.1×10 JThe elastic potential energy was 0.023 J and the kinetic energy was 0.021 J. The discrepancy in values was due toenergy lost in the form of sound and thermal energy.(b) F K = ?ET= E′T∆ EK= Eth1 2− mv = F ∆dK0 2FK2FK2mv=− ∆ d=−(0.88 kg)(0.22 m/s)2(2.5 m)−3=− 8.5×10 NThe average kinetic friction is –8.5 10 –3 N.(c) Possible sources of error are:• a desk or track that is not levelled properly• the cart is not initially in contact with the wall• incorrect release of the trigger2Making Connections44. (a) The jarring forces on your legs can be reduced when jogging by keeping your knees bent and allowing them to flexwith each step.(b) By reducing the jarring, you reduce the forces applied internally in your legs. This helps prevent painful damage tobones and tissue, such as cartilage, in the knee.45. (a) An air bag is used to reduce injury in two primary ways. Because it is a bag, the force that stops a passenger in the caris spread out over the entire surface of the body, not just a small area such as the points the hands contact on thesteering wheel. This reduces the pressure on the person’s body. The second way the airbag reduces injury is by slowingthe person down over a longer period of time. This increase in stopping time significantly reduces the total forcerequired to bring the passenger to rest.(b) Other safety devices that used to help are:• crumple zones in a car• snug fit and padding of any type of helmet• design of protective equipment (such as shin pads)46. r = 2.7 10 20 mT = 2.0 10 8 8 365.26 d 24 h 3600 sa = 2.0× 10 a × × × = 6.312 10 15 s1 a 1 d 1 h(a) M = ?3GM r=2 24πT2 34πrM =GT22 20 34 π (2.7×10 m)=−11 2 2 15 2(6.67× 10 N ⋅ m /kg )(6.312×10 s)41M = 2.9×10 kgThe total mass of the stars at the hub of our galaxy is 2.9 10 41 kg.Copyright © 2003 Nelson Unit 2 Review 409

(b)412.9×10 kg11= 1.5×10302.0×10 kgThe approximate number of stars that are the size of our Sun is 1.5 10 11 .Extension47. m 1 = 2.5 kgv 1 = 2.3 m/sm 2 = 2.0 kgv 1´ = ?v 2´ = ?Choose right as positive.Using the conservation of energy (units are omitted):1 2 1 2 1 2mv1 1 = mv1 1′ + m2v2′2 2 22 2 2mv 1 1 = mv 1 1′ + mv 2 2′2 2 22.5(2.3) = 2.5v1′ + 2.0v2′2 25.29 = v′ + 0.8 v′(Equation 1)1 2Using the conservation of momentum (units are omitted):mv 1 1= mv 1 1′ + mv 2 2′(2.5)(2.3) = 2.5v1′ + 2.0v2′2.3 = v1′ + 0.8v2′v′ = 2.3 −0.8 v′(Equation 2)1 2Substitute Equation 2 into Equation 1:2 25.29 = (2.3− 0.8 v2′ ) + 0.8v2′5.29 = (5.29 − 3.68v2′ + 0.64 v2′ ) + 0.8v2′20 =− 3.68v′ 2+ 1.44v′20 = v′ 2( − 3.68 + 1.44 v′2)v ′ = 0, or − 3. 68 + 1. 44v′= 02 22 2Sincev′ 2= 0 is not valid because it represents no change in speed, we use:− 368 . + 144 . v2′= 01.44v2′ = 3.68v2′ = 2.5556v′ = 2.6 m/s [right]2Substitute back into Equation 2:v1′ = 2.3 −0.8(2.5556)v1′ = 0.26 m/s [right]The velocity of ball 1 is 0.26 m/s [right], and the velocity of ball 2 is 2.6 m/s [right].48. Let the subscript b represent the bullet and w represent the block of wood.m b = 4.0 g = 4.0 10 –3 kgv b = 5.0 10 2 m/s [N]v′b= 1.0 10 2 m/s [N]m w = 2.0 kg∆d = 0.40 m410 Unit 2 Energy and Momentum Copyright © 2003 Nelson

(a)v′ w= ?Choose north as positive.p = p′mv b b+ 0 = mv b b′ + mwvw′mv b b−mv b b′v′ w=mwmb( vb − v′b)=mw−3 2 2(4.0× 10 kg)(5.0× 10 m/s − 1.0×10 m/s)=2.0 kgv′ = 0.80 m/s [N]wThe wooden block moves at a velocity of 0.80 m/s [N] just after the bullet exits.(b) E K = ?1 2EK= mv21 (2.0 kg)(0.80 m/s)2=2E = 0.64 JKThe maximum kinetic energy of the block is 0.64 J.(c) F K = ?ET= ET′∆ E = EKth1 2− mv = F ∆dK0 2FK22mv=− ∆ d(2.0 kg)(0.80 m/s)=−2(0.40 m)FK=−1.6 N [N], or 1.6 N [S]The average frictional force stopping the block is 1.6 N [S].(d) ∆E K = ?∆ EK = EK2 −EK11 2 1 2= mv2 − mv12 21 2 2= mv ( 2 −v1)21 3 2222(4.0 10−= × kg) (( 1.0 × 10 m/s ) − ( 5.0 × 10 m/s )2)2∆ EK= − 4.8×10 JThe decrease in kinetic energy of the bullet is –4.8 10 2 J.(e) The collision between the bullet and the block is not elastic, so most of the energy is lost to thermal energy.49. Let the subscript P represent the plane, and B represent the barge.m P = 1.0 Mg = 1.0 10 3 kgm B = 2.0 Mg = 2.0 10 3 kgv P = 5.0 10 1 m/s = 50 m/sv B = 0.0 m/s2Copyright © 2003 Nelson Unit 2 Review 411

For the landing of the plane:mv P P + mv B B = ( mP + mB)vPBmv P P + mv B BvPB=mP+ mB3(1.0 × 10 kg)(50 m/s) + 0=3 31.0× 10 kg + 2.0×10 kgv = 16.67 m/sTo calculate the frictional force on the plane, and on the barge:1FK= mg41 (1.0 103= × kg)(9.8 N/kg)4F = 2450 N [backward]By Newton’s third law, F K = –2450 N [forward]To calculate the acceleration of the plane:Σ F = mPaPΣFaP=mP−2450 N=31.0×10 kg2a =−2.45 m/sTo calculate the acceleration of the barge:Σ F = mBaBΣFaB=mB2450 N=32.0×10 kg2a = 1.225 m/sKPBTo calculate the distance the plane will travel during the landing:2 2vf = vi + 2aP∆d2 2vf− vi∆ d =2aP2 2(16.67 m/s) − (50 m/s)=22( −2.45 m/s )∆ d = 453.5 mPBTo calculate the distance the barge will travel during the landing:2 2vf = vi + 2aB∆d2 2vf− vi∆ d =2aB2 2(16.67 m/s) − (0 m/s)=22(1.225 m/s )∆ d = 113.4 mTherefore, the required length of the barge is 453.5 m – 113.4 m = 3.4 × 10 2 m.412 Unit 2 Energy and Momentum Copyright © 2003 Nelson

50. Let the subscript C represent the chair.(a) m C = ?k = ?For the empty chair:T = 2π2mkm T=2k 4π24πmk =2T24 π ( mC+ 0)=k =For masses on the chair:T = 2π( 0.901 s)24πmC2( 0.901 s)2mkm T=2k 4π24πmk =2T24 π ( mC+ m)k =2TTo calculate the mass of the chair:2 24π mC4 π ( mC+ m)=2 2( 0.901 s)T2T mC= m2 C+ m( 0.901 s)2T mC− m2 C= m( 0.901 s)⎛ 2T ⎞− 1m2 C= m⎜( 0.901 s)⎟⎝⎠mmC=⎛ 2T ⎞−1⎜2( 0.901 s)⎟⎝⎠14.1 kg=⎛2( 1.25 s)⎞−1⎜2( 0.901 s)⎟⎝⎠m = 15.2 kg2CCopyright © 2003 Nelson Unit 2 Review 413

Similarly,23.9 kg 33.8 kg 45.0 kgmC = = 15.4 kg mC = = 15.4 kg m2 2 C= = 15.3 kg2⎛ ( 1.44 s)⎞ ⎛ ( 1.61 s)⎞ ⎛ ( 1.79 s)⎞−1 −1 −1⎜2 2 2( 0.901s)⎟ ⎜( 0.901s)⎟ ⎜( 0.901s)⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠56.1 kg67.1 kgmC = = 15.4 kg m2C= = 15.3 kg2⎛ ( 1.94s)⎞⎛ ( 2.09 s)⎞−1−1⎜2( 0.901s)⎟⎜2⎝⎠( 0.901s)⎟⎝⎠To calculate the average value for the mass of the chair:15.2 kg + 15.4 kg + 15.4 kg + 15.3 kg + 15.4 kg + 15.3 kgmC=6m = 15.3 kgTo calculate the spring constant:CT = 2π2mkm T=2k 4π24πmCk =2T24 π (15.3 kg)=( 0.901 s)k = 744 N/mThe mass of the chair is 15.3 kg, and the spring constant is 744 N/m.(b) f = ?Assuming a mass of 94 kg, we first calculate the period:mT = 2πk94 kg + 15.3 kg= 2π744 N/mT = 2.4 sSince frequency is inversely proportional to period:1f =T1=2.4 sf = 0.42 HzThe frequency is 0.42 Hz.2414 Unit 2 Energy and Momentum Copyright © 2003 Nelson

(c) T 1 = 2.01 sT 2 = 1.98 s∆m = ?T = 2π2mkm T=2k 4π2kTm =24π∆ m = m2 −m12 2kT2 kT1= −2 24π4πk 2 2= ( T2 2−T1)4π744 N/m 2 2=2 (( 1.98 s ) −( 2.01 s ) )4π∆ m =−2.3 kgThe astronaut loses 2.3 kg51. (a) Venus has the most circular orbit because its eccentricity is closest to zero. Pluto has the most elongated orbit becauseits eccentricity is the highest.(b)(c)(d) The typical elliptical orbits shown for most planets do not represent the true shape of the orbit. Most texts exaggeratethe length of the orbit for effect. Even the most elongated orbits have the Sun close to the centre of the ellipse.52. Let the subscript X represent the planet and E represent Earth.1m X =E10 M1r X = E2 rF gE = 6.1 10 2 NF gX = ?Copyright © 2003 Nelson Unit 2 Review 415

Half of the diameter also means half of the radius, therefore:F = mggEFm = ggEEEAlso,gggEE 2rEXX 2rXXGM=GM=⎛ 1 ⎞G⎜ME⎟10=⎝ ⎠2⎛1⎞⎜ rE⎟⎝2⎠110 ⎛GM⎞E= 1 ⎜ 2r ⎟⎝ E ⎠44= ( gE)10= 0.4gESubstituting the value of m of the value of g X :F = mgFgXgXX⎛FgE⎞= ⎜ ⎟(0.4 gE)⎝ gE⎠= 0.4FgE= 0.4(6.1×10 N)2= 2.4×10 NThe astronaut weighs 2.4 10 2 N on the new planet.2416 Unit 2 Energy and Momentum Copyright © 2003 Nelson