Physics Answers

QUESTION ONE
A car is moving at constant speed along the hard sand on a beach.
It moves into some very soft sand.
Discuss what would happen to the car.
In your answer include:
The forces acting on the car
The size of the forces
The effect on the car’s motion
QUESTION ONE ANSWER
For Achievement your answer should talk about several (at least two or
three) points such as:
The forces acting are the friction force and the engine force.
There is also the weight force and the support force
Engine force is acting forwards and friction force acts backwards
The forces are balanced / the same size on the hard sand (as there is a
constant speed)
F net is zero on the hard sand
There is more friction on the soft sand
The friction force is bigger than the engine force on the soft sand
The forces are not balanced on the soft sand (so the car slows down)
F net is not zero on the soft sand
When the forces are balanced there is a constant speed
When the forces are not balanced the car changes speed
Merit would be to explain two or three points:
The forces acting are the friction force (acting backwards) and the engine force
acting forwards. There is also the cars weight force (acting downwards) and a
support force acting upwards.
On the hard sand these forces are the same size (as they cancel out to give F net
= 0; = no acceleration = constant speed)
On the soft sand the tyres would sink in and friction would increase
As the friction force is now bigger than the engine force, the forces are
unbalanced and there is a net force acting backwards which will cause the car to
slow down (decelerate or negative acceleration)
As the car slows down the frictional force would decrease until a new equilibrium
/ balance is reached between the engine force and the friction force and the car
will continue with a new (reduced) speed.
Excellence would be to give all the points above:

speed (m s -
1 )
25
20
15
10
5
0
QUESTION TWO
A family drives their 800 kg car to a holiday house to enjoy
their summer holiday. Part of their journey is represented by
the speed-time graph shown below:
(a) Calculate the acceleration acting during the first 20
seconds.
(b)
Section A
0 20 40 60 80 100 120
Calculate the Work done during the first 20 seconds
time (s)
QUESTION TWO ANSWERS
Part a) and b)
Achievement is correct calculation of acceleration
OR calculation of Force (using incorrect value for a)
OR correct calculation of distance travelled. Units must be given.
a = gradient of line = change in v / t = 25 / 20 = 1.25 m s -2
Merit is correct calculation of acceleration
AND correct calculation of Force (using incorrect value for a)
OR correct calculation of distance travelled. Units must be given.
W = Fd
F = ma = 800 x 1.25 = 1000 N
d = area under line for first 20 seconds = ½ x base x height
d = 0.5 x 20 x 25 = 250 m
Excellence is correct answer for Work with all working shown and units
W = Fd = 1000 x 250 = 250 000 J
Part C
Achievement would be to give two or three of the points given below:
The car has a constant velocity (is not accelerating)
The total force acting on the car is zero
The forces on the car are balanced
Work = Change in energy
E k = ½ mv 2 = ½ x 800 x 25 2 = 250 000 J
(c)
Explain (in a Physics sense) why the graph shows that
there was no Work done between time = 20 seconds
and time = 60 seconds. Include in your answer:
The motion of the car during that time and what
that tells us about the forces involved
What work is
At least one energy calculation
Merit would be to give at least one of the points listed below
Because the car has a constant velocity it is not accelerating (a = 0) and so the
total Force acting on the car is 0 (F = ma = m x 0 = 0)
Work = Fd and as F = 0 then Work = 0 x distance = 0 J
Work = Change in energy and the energy is not changing between t = 20 s and t
= 60 s because the car has the same (Kinetic) energy over that time as velocity is
not changing (E k = ½ mv 2 )
Excellence would be to give at least two of the points required for Merit

QUESTION THREE
While a 1200 kg car is moving at 20 m s -1 , a dog runs across the road in
front of it. When the driver applies the brakes, the car stops in a
distance of 25 metres.
a) Calculate the average force acting on the car when the
driver applies the brakes. Your answer should include:
Any energy transformations that take place
Any assumptions you have made
All calculations with working
an appropriate unit with your answer.
b) Use your answer from a) to calculate the time it takes for the car to
stop
c) If the car had been travelling uphill would it have taken a greater or
lesser distance to stop? Discuss your answer in terms of:
The forces acting on the car
Any energy transformations that take place
QUESTION THREE ANSWERS
Achievement would be to give two of the points given below:
Kinetic energy is changed into heat energy (Work done by friction)
Assuming all Kinetic energy is changed into Heat (Work done by friction)
Work done to stop car = Kinetic energy of car
E k = ½ mv 2 = ½ x 1200 x 20 2 = 240 000 J (Working must be shown)
(Or Merit calculation done correctly with incorrect E k value)
Merit is all points for Achieved as well as force calculation done
W = Fd = 240 000 so F = 240 000 / d = 240 000 / 25 = 9600 N
Part b)
Achievement is one calculation done correctly
F = ma so a = F / m = 9600 / 1200 = 8 m s -2 (Or carried error using your answer
from a))
a = change in v / t so t = change in v / a = 20 m s -1 / 8 = 2.5 s (Or carried error if
the calculation of a is incorrect)
Merit is both calculations done correctly
Part c)
Achievement would be to give two of the points given below:
The car would take less distance to stop
The forces slowing the car would be bigger so the car would stop quicker (in less
distance) OR The weight force of the car is partly acting backwards slowing the
car
Kinetic energy is being turned into heat and sound AND Gravitational Potential
Energy
Merit would be to give at least one of the points listed below
Because the weight force is (partly) acting backwards this means that the total
force stopping the car will be bigger than just the force being applied by the
brakes
On the flat all the E k of the car would have to be turned into heat and sound by
the brakes. But as the car is going uphill some of the E k is being turned into E p so
there is less to be transformed by the brakes so the car stops sooner.
Excellence is to extend the points given for Merit
If F (back) is bigger; than a = F/m is bigger and v will decrease to 0 faster.
If some E k is turned into Ep then W (done by brakes) will be less and since w =
Fd then d = W / F so smaller W = smaller stopping distance

QUESTION FOUR
A student cycles to school each day. He must ride over the steep hill
between his home and school. The combined mass of the bike and
student is 70kg.
He has a new speedometer and records his speed at various times as in
the table.
Time (s) 0 120 240 360 480 540 600
Speed (ms -1 ) 0 2.5 2.5 1.2 1.2 9.0 0
QUESTION FOUR ANSWERS
For Achievement you require Part (i) OR part of (ii) correct (a)
a = change in speed = 2.5 = 0.02 ms -2
Change in time 120
OR
W=Fd - correct calculation of force or distance (units not needed) or carry
through error from part (i) in F=ma
This data is shown on the grid below.
Speed (m s -1 )
10
9
8
7
6
5
4
3
2
1
0
0 100 200 300 400 500 600 700
Time (s)
(i) Calculate the acceleration of the student on his bike in the
first 120 seconds.
For Merit you require part (i) correct (a = 0.02 ms -2 ) AND part of (ii) with
correct force or distance with unit (m)
F = ma
OR
= 70 x 0.02 = 1.4 N (allow 1.46 if a unrounded)
Distance = area under graph or d = v (av) t
= 120 x 2.5
2 = 150m
For Excellence you require ALL answers correct with units (e)
W=Fd
F= ma = 70 x 0.02 = 1.4 N
d = area under graph or d = v (av) t
= 120 x 2.5
2 = 150m
(ii)
Calculate the work done by the cyclist in this time.
Additional Question over page
W = Fd = 1.4 x 150 = 210J

(iii)
Compare the forces as the student on his bike goes up the hill
(between 240 and 360 s) with the forces as he goes down the
hill (between 480 and 540 s). In your answer:
Explain which requires more effort and why.
Use appropriate force calculations in your
comparisons.
For Achievement you require ONE correct statement (a)
More effort is needed going uphill as he must overcome gravity / weight force (a)
OR
Net force is much higher downhill as he is accelerating with assistance from
gravity (a) OR
Calculation of forces ie.
Uphill F = ma = 70 x 1.3/120 =0.76N
Downhill F = ma = 70 x 7.8/60= 9.1N (a)
For Merit you need to correctly compares forces OR effort with explanation
(m)
More effort is needed going uphill as he must overcome gravity / weight force is
added, whereas going downhill gravity / weight is in the direction of travel so aids
the thrust from the cyclist (m)
OR
Net force is much higher downhill as he is accelerating due to assistance of
gravity.
Calculation of force downhill
F=ma = 70 x 7.8/60 = 9.1 N
Uphill F=ma = 70 x 1.3/120 = 0.76 N ie. lower / decelerating as cyclist’s thrust
force is acted against by the weight force he must overcome (m)
For Excellence you need to have a comparison of forces and effort both
uphill and downhill (e)
Going uphill requires more effort because the cyclist must overcome his 700N
weight force, ie. he is lifting himself and his bike uphill as well as moving along.
This causes deceleration, and a net force of only 0.76 N, ie. it is difficult to
maintain his speed due to the extra effort in lifting the bike against gravity. When
travelling downhill, the bike will roll naturally as its weight force acts in a
downward direction, carrying the bike and rider down the hill. Any extra effort the
rider adds as thrust by peddling effectively adds to this, ie. the bike accelerates
rapidly (from 1.2 m/s to 9 m/s) with a net force of 9.1 N

QUESTION FIVE
New car models are crash tested under laboratory conditions to
ensure that safety standards are met. A 1500 kg car is accelerated
to a speed of 30 ms -1 from stationary in 6 seconds, before hitting a
brick wall without braking.
(a) Calculate the force the causes the car to accelerate.
Modern cars are built with crumple zones, this means that the cars
are designed to fold up (crumple) as they hit an object. This results
in a lot more damage to the car, but reduces injury to the
passengers by increasing the TIME the car takes to stop completely
during a crash.
(b)
Crumple zone
Discuss the energy and forces involved as the car drives
towards the wall, then crashes into it. In your answer:
Describe the energy transformations that take
place
Calculate the energy the car has the instant
before it hits the wall
Explain how the crumple zone works to reduce the
force on impact and reduce injury
QUESTION FIVE ANSWERS
(a) Achievement requires correct acceleration calculation (a)
a = 30/6 = 5 ms -2
Merit requires correct force calculation from acceleration (m)
F = ma = 1500 x 5
= 7500 N
(b) Achievement requires ONE correct statement (a)
Eg:
Kinetic energy heat energy (as friction) + sound (a)
(sound not required)
OR
Kinetic energy calculated
Energy E k = 1/2mv 2 = ½ x 1500 x 30 2 = 675000 J (675kJ)
Merit requires Energy transformation AND kinetic energy calculation OR
explanation of crumple zone (m) Eg:
Kinetic energy heat energy (as friction) + sound AND
Energy E k = 1/2mv 2 = ½ x 1500 x 30 2 = 675000 J (675kJ) (m)
OR
Crumple zone increases stopping time, ie. deceleration is less which decreases
the forces (F = ma) (m)
Excellence needs discussion energy transformation, with kinetic energy or
force linked to explanation of crumple zone (e) Eg:
The car has a large amount of kinetic energy, ie.
E k = 1/2mv 2 = ½ x 1500 x 30 2
= 675000 J (675kJ) just before it hits the wall.
All of this energy must be transformed as the car stops, mostly into heat through
friction, and some sound. As the car hits the wall, it decelerates rapidly, giving a
very large force. A crumple zone allows the car to stop more slowly, ie. it
decreases the deceleration by allowing the energy to be releases as heat over a
longer time. Using F = ma, if a is smaller due to longer stopping time, then force
must be less. If a passenger’s body decelerates more slowly with less force, less
damage/injury will result. A car with no crumple zone will stop completely more
quickly, ie. large deceleration so a larger force and greater damage. (e)

QUESTION SIX: IN THE GYM
Two men were lifting weights in the gym. They were lifting a bar
with weights that had a total mass of 80 kg.
(a) State the total weight they were lifting and explain the
difference between mass and weight.
The two men were bench pressing the weights (pushing the
weights upwards from their chest as far as their arms could extend).
One man’s arms were 0.65 m long and the other man’s were 0.60
m long. Both men took exactly 2.5 seconds to fully extend their
arms.
The man with the longer arms said that he had to do more
work than the man with the shorter arms but the man with the
shorter arms was more powerful.
(b)
Discuss the statement (in bold above).
In your answer, you should:
calculate the work done by each man
calculate the power exerted by each man
compare the results with the statement in bold and explain
your findings
QUESTION SIX ANSWERS: IN THE GYM
a) Calculation of weight: F w = mg = 80 x 10 = 800 N
Explanation of difference:
Mass is the amount of material / matter in an object.
Weight is the gravitational force on an object (not the amount of gravity).
b) Calculation of work:
Long arms W = F x d = 800 x 0.65 = 520 J
Short armsW = F x d = 800 x 0.60 = 480 J
Calculation of power:
Long arms P = W/t = 520 / 2.5 = 208 W
Short arms P = W/t = 480 / 2.5 = 192 W
OR statement that both took the same time to lift weights therefore the man that
did the most work also exerted the most power.
Comparison with statement:
Man with longer arms did the most work as he exerted the same force over a
greater distance. However, man with the longer arms was also exerting more
power as he did the greater amount of work in the same time.
For Achievement your answer should have two of:
in (a), calculates the weight correctly OR makes an accurate statement
about mass or weight
in (b), one calculation is undertaken by selecting and substituting into the
correct formula and solving it
in (b), long armed man identified as either doing the most work or
as exerting the most power
For Merit your answer should have both of:
in (a), calculates the weight correctly AND distinguishes between mass
and weight
in (b), correctly calculates work for both men AND calculates power
correctly for one man OR makes a statement comparing work over the
same time with power.
For Excellence your answer should have both of:
in (b), correctly calculates work for both men AND calculates power
correctly for both men OR makes a statement comparing work over the
same time with power for both men.
in (b), compares results with statement and explains either the difference
in work Or difference in power

QUESTION SEVEN: PRESSURE
Hayley and Sarah were running across the football field. Hayley has a
mass of 54 kg and Sarah has a mass of 60 kg.
(a) Sarah was running with a forward force of 65 N and a friction force
of 50N. Calculate the acceleration of Sarah.
Hayley was wearing football boots and Sarah was wearing
running shoes. Hayley noticed that when they both stood still,
she left marks in the field but Sarah did not.
QUESTION SEVEN ANSWERS: PRESSURE
a) Calculation of acceleration:
F net = m x a so a = F net / m = 15 / 60 = 0.25 ms -2
b) Calculation of pressure:
Hayley
P = F/A Force is F w = mg = 54 x 10 = 540N
Area is that of studs A = 0.0001 x 12 = 0.0012 m 2
P = 540 / 0.0012 = 450 000 Pa
Sarah
P = F/A Force is F w = mg = 60 x 10 = 600N
Area of both shoes A = (0.10 x 0.27)x2 = 0.054 m 2
P = 600 / 0.054 = 11 111 Pa
The bottoms of their shoes are shown below:
Each stud
has a surface
area of
Hayley’s
football boots
Sarah’s running
shoes
Length
= 0.27 m
Discussion of physics principles:
1. Sarah’s running shoes have a much larger area than the studs on
Hayley’s boots.
2. P = F/A so as A increases the pressure must get smaller.
3. Even though Sarah is heavier, her weight is distributed over a
larger area while Hayley’s is applied over a small area.
Hayley puts more pressure on the ground causing the studs to ‘sink in’
and leave marks.
For Achievement your answer should have two of:
in (a), net force is calculated OR formula correctly arranged but
incorrect force value substituted
in (b), calculates the area or weight force for one person correctly
in (b), states that Hayley leaves marks as her boots exert more
pressure
0.0001 m 2 Average width = 0.10 m
(b) Discuss the physics principles that caused Hayley to
leave marks while Sarah, who was heavier, did not.
Include calculations in your answer.
For Merit your answer should have two of:
in (a), net force is calculated AND correct acceleration
calculated including unit
in (b), shows an understanding of how the two formula should
be applied to each person but fails to give correct pressures or
units
in (b), shows an understanding of the physics principles
involved by stating that smaller area of studs would exert more
pressure than area of running shoes
For Excellence your answer should have everything

QUESTION EIGHT: CONSERVATION OF ENERGY
Josh is skateboarding on the local half-pipe. The top of the half-pipe
is 2 m above the bottom of the half-pipe. Josh and his board have a
total mass of 70 kg.
Point B
2 m
Point A
(a) Calculate the gravitational potential energy of Josh at the top
of the half-pipe.
(b) Use your answer from (a) above to calculate the speed Josh
would be expected to be travelling at the bottom of the half-pipe.
(c) Describe the energy changes involved in skating from point
A to point B and discuss whether it is possible for Josh to skate
from point A to point B without putting in any effort of his own.
QUESTION EIGHT ANSWERS: CONSERVATION OF ENERGY
a) Calculation of gravitational potential energy:
E P = mgh = 70 x 10 x 2 = 1400 J
b) Calculation of speed:
E K = 2
1 mv
2
Assume all gravitational energy would be converted to kinetic energy
1400 = 2
1 x 70 v
2
so v 2 = 40 so v = 6.3 m s -1
c) Description of energy changes:
At point A he has all gravitational potential energy and no kinetic energy.
At the bottom of the ramp he has all kinetic energy and no gravitational
potential energy (the gravitational potential energy has been changed to
kinetic).
At point B all the kinetic has been transferred back into gravitational potential
energy.
Discussion of energy conservation:
Between A and B some of the kinetic energy has been converted into heat
due to work done by friction. This loss of energy means that Josh would not
be able to reach all the way to point B.
For Achievement your answer should have two of:
in (a), uses the correct formula in an attempt to calculate gravitational
potential energy OR gives correct value but no unit
in (b), recognises that gravitational energy will be converted to kinetic
energy OR selects the correct formula and attempts the calculation by
substituting all values except v
in (c), identifies the type of energy at each of the three points (A, bottom of
ramp, and B) OR recognises that energy is being lost therefore Josh won’t
reach point B
For Merit your answer should have two of:
in (a), calculates correct gravitational potential energy and gives unit
in (b), selects the correct formula and substitutes correctly (using value
from a) and gives unit
in (c), accurately describes all of the mechanical energy changes taking
place from point A to point B OR recognises that energy is being lost as
heat therefore Josh won’t reach point B
For Excellence your answer should have everything

QUESTION NINE: MOTION-TIME GRAPHS
A passenger training for a harbour swim boards a ferry that travels
from wharf A across the harbour to wharf B.
The ferry leaves wharf A and slowly accelerates. It takes 150 s to
travel the first 500 m. The ferry then travels the next 2000 m at a
constant speed. This section of the journey takes 300 s. The ferry
decelerates over the final 500 m, taking a further 150 s.
(a)
Sketch a distance-time graph for the ferry’s journey.
(b) (i) Calculate the average speed of the ferry during the
journey.
The same passenger then dives off wharf B and swims across the
harbour to the far shore as shown in the speed-time graph below.
Speed
(ms -1 )
0.8
0
Section
A
Speed-time graph for swim across the
harbour
Section
B
400 4000
Section
C
(b) (ii) Calculate the distance the person swam while her
speed was changing.
Time (s)
(c) Discuss whether the forces on the swimmer are balanced or
unbalanced during each section of the swim and how this results in
the types of motion produced.
4500
QUESTION NINE ANSWER: MOTION-TIME GRAPHS
a) Plotting graph correctly and drawing of correct graph shapes:
3000
2500
tra
vell
ed
(m)
500
b) (i) Calculation
0
of average
1 2
speed:
3 4
v = d/t
5 6
= 3000/600 = 5 m s -1
(ii) Calculation of distance:
0
Only
0
applies
0
to
0
section e 0
A and
0
C
Area of section A: 0.5 x
0
400
0
x 0.8
0
= 160
0
m (sec 0 0
Area of section C: 0.5 x 500 x 0.8 = 200 m onds
Total distance = total area 160 + 200 ) = 360 m
c) Explanation of motion:
Section A: Forces are unbalanced; there is a net force in the forward direction causing the
swimmer to accelerate.
Section B: Forces are balanced; there is zero net force, therefore swimmer moving at
constant speed.
Section C: Forces are unbalanced; there is a net force in the backward direction causing
the swimmer to decelerate.
For Achievement your answer should have two of:
in (a), completes the graph accurately showing correct graph shapes for at least two
sections (with the y axis labelled correctly).
in (b), attempts to calculate average speed (E.g. writes the formula and substitutes
values OR finds average speed but does not give the unit). AND recognises the distance
as the area under the speed-time graph (E.g. attempt made at finding an area of any of
the three sections).
in (c), shows understanding that section B represents constant speed and A and C
represent changing speed.
In (c) correctly states whether forces are balanced or unbalanced for each section.
For Merit your answer should have two of:
in (a), draws accurate graph with appropriate shapes for all three sections
in (b), calculates average speed correctly (with unit) AND calculates the area under at
least one section of the graph and states the distance.
in (c), states whether the forces are balanced or unbalanced for two sections of the
graph AND links them to the motion. (balanced = constant speed, unbalanced =
acceleration)
For Excellence your answer should have everything