Chapter 9 05

Chapter 9
9.1 Figure 9-36 shows a three particle system. What are (a) the x coordinate and (b) the y
coordinate of the center of mass of the three particle system . (c) What happens to the center of
mass as the mass of the m 3
particle is increased.
We can calculate the center of mass in the x and y directions.
x cm
=
=
=
1
(m 1
+ m 2
+ m 3
) ⋅ (m 1 x 1 + m 2 x 2 + m 3 x 3 )
1
⋅ (3.0kg⋅ 0m + 4.0kg ⋅ 2.0m + 8.0kg ⋅1.0m)
(3.0kg + 4.0kg + 8.0kg)
1
(15kg) ⋅ (16.0kgm)
= 16
15 m
1
y cm
=
(m 1
+ m 2
+ m 3
) ⋅ (m y + m y + m y )
1 1 2 2 3 3
1
=
⋅ (3.0kg⋅ 0m + 4.0kg ⋅1.0m + 8.0kg ⋅ 2.0m)
(3.0kg + 4.0kg + 8.0kg)
1
=
(15kg) ⋅ (20.0kgm)
= 20
15 m = 4 3 m
As the topmost mass grows larger, the cm will move toward that mass.

9.5 In the ammonia (NH3) molecule of Fig 9-39, three hydrogen atoms form an equilateral
triangle, with the center of the triangle at a distance d = 9.40 ×10 −11 m from each hydrogen atom.
The nitrogen atom (N) is at the apex of a pyramid, with the three hydrogen atoms forming the base.
The nitrogen to hydrogen atomic ratio is 13.9 and the nitrogen to hydrogen distance is
L = 10.14 × 10 −11 m . What are (a) x and (b) y coordinates of the molecules center of mass.
(a). By symmetry, we can see that the x coordinate of the center of mass is 0.
(b) To find the y component, we replace the 3 H atoms, with one particle with the mass of the
three H atoms at the cm of the three H. The cm of the 3H is at (0,0) by symmetry.
We now find the y CM by for the N and the 3H particle
y N
= L 2 − d 2 = (10.14 ×10 −11 ) 2 − (9.4 × 10 −11 ) 2 = 3.803 ×10 −11 m
y 3H
= 0.0m
m N
= 13.9m H
y = 3m H ⋅ y 3H + m N ⋅ y N
3m H
+ m N
= 3m H ⋅0 +13.9m H ⋅ 3.803× 10−11 m
3m H
+13.9m H
= 3.128 × 10 −11 m
= 13.9 ⋅ 3.803 ×10 −11 m
3 + 13.9
9.10 Two skaters , one with mass 65 kg and the other with mass 40 kg stand on an ice rink hold a
ole with a length of 10m and negligible mass. Starting from the ends of the pole, the skaters pull
themselves along the pole until they meet. How far will the 40 kg skater move?
Assume that the 65 kg is at 10 and the 40 kg is at 0m. The cm does not move, so both skaters end
up at the cm. The 40 kg skater moves from 0 to the cm.
x cm
=
40 ⋅ 0 + 65⋅10
105
= 6.19m
The 40 kg skater moves 6.19 m.

9.15 A shell is fired from a gun with a muzzle velocity of 20m/s at an angle of 60˚. At the top of
the trajectory, the shell explodes into two equal mass fragments. One fragment, whose speed
immediately after the explosion is zero falls vertically. How far from the gun does the other
fragment land, assuming that the terrain is level and that the air drag is negligible.
Let’s sketch what happens first
Examine how long it took to reach the explosion point.
v fy
= 0
v iy
= 20m / s⋅ sin60 =17.32m / s
v ix
= 20m / s⋅ cos60 =10m / s
a = −g
v fy
= v iy
− gt
t = v iy
g =1.767s
Next we find the distance traveled before the explosion.
x L
= v ix
t =17.67m
The “left” half-shell lands at x L
= 17.67m. If the shell had not exploded, we know that it would
land at x cm
= 2⋅17.67m = 35.34m Since the forces involved in the explosion were entirely internal,
the center of mass of the two shells still lands at exactly this point. Knowing where the cm is
allows us to find where the second half shell will land.

x cm
=
1
2 m ⋅ x + 1 L
2 m⋅ x R
m
x cm
=
1 2 x L + 1 2 x R
x R
= 2x cm
− x L
= 2⋅ 35.34 −17.67
= 53.01m
9.18 A 0.70 kg ball is moving horizontally with a speed of 5.0 m/s when it strikes a vertical wall.
The ball rebounds with a speed of 2.0 m/s. What is the magnitude of the change in the linear
momentum of the ball
Δ p =
p f
− p i
= 0.7kg ⋅ 5m / s− 0.7kg ⋅ (−2 m / s)
= 4.9kgm / s
9.26 A 1.2 kg ball drops vertically onto a floor, hitting with a speed of 25 m/s. It rebounds with an
initial speed of 10 m/s. It rebounds with an initial speed of 10 m/s. (a) What impulse acts on the
ball during the contact? (b) If the ball is in contact with the floor for 0.020 s, what is the magnitude
of the average force on the floor from the ball
Δ p =
p f
− p i
=1.2kg ⋅ (−25m / s) −1.2kg⋅ (10m / s)
= 42kg m / s
F = Δ p
Δ t = 42kg m / s
0.020s
= 2100N
9.35 A 91 kg man lying on a surface of negligible friction shoves a 68 g stone away from himself,
giving it a speed of 4 m/s. What speed does he acquire as a result?
This is a momentum conservation problem. The total momentum is zero (man and stone at rest at
the beginning). The total momentum must remain zero. In the final state

p f
= 0 = p m
+ p s
p m
= − p s
m m
v m
= −m s
v s
v m
= − m s
v
m
s
m
= − 0.068kg
91.0kg ⋅ 4m / s
= 0.002989m / s
9.42 In Fig 9-57, a stationary block explodes into two pieces L and R that slide across a
frictionless floor and then into regions with friction where they stop. Piece L, with a mass of 2.0
kg, encounters a coefficient of kinetic friction µ L
= 0.40 and slides to a stop in a distance
d L
= 0.15m . Piece R encounters a coefficient of kinetic friction µ R
= 0.50and slides to a stop in
distance d R
= 0.25m What was the mass of the block?
Since we know the mass of the L block, we’ll begin with it. The L block stops because friction
does work on it. We can compute the work done
W L
= −F f
d L
= −µ L
N L
d L
= −µ L
m L
gd L
= −0.40⋅ 2kg ⋅9.8m / s 2 ⋅ 0.15m
= −1.176J
Now that we know the work done, we can compute the velocity it had after the explosion. This was
the velocity it had when it hit the friction
K f
− K i
= W = −1.176J
0 − 1 2 m L v L 2 =W
v L
= − 2W 2⋅ (−1.176J)
= −
m L
2kg
=1.08m / s
Now lets consider the right side. As before, we can compute the velocity from the work done.

K f
−K i
=W
0 − 1 2 m R v R 2 =W
− 1 2 m v 2 R R
= −µ R
m R
gd R
v R
= 2µ R
g d R
=1.565m / s
Now we know the velocities. Given that the momenta of the right and left pieces must be equal in
magnitude, we can solve for the unknown mass.
m L
v L
= m R
v R
m R
= v L
v R
m L
m R
=
1.08m / s
1.565m / s ⋅ 2kg
=1.38kg
9.45 A 5.20 g bullet moving at 672 m/s strikes a 700 g wooden block at rest on a frictionless
surface. The bullet emerges traveling in the same direction with its speed reduced to 428 m/s. (a)
What is the resulting speed of the block? (b) What is the speed of the bullet-block center of mass?
This is a conservation of momentum problem.
p i
= m b
v b
+ M B
v B
p i
= 5.2 ×10 −3 kg ⋅ 672m / s + 0.7kg ⋅ 0m / s
= 3.4944 kgm / s
p f
=
p i
p f
= m bv b ′ + M Bv ′ B
5.2 ×10 −3 kg ⋅ 428m / s + 0.7kg ⋅ ′ v B
= 3.4944 kg m / s
= m b
′ v b
+ M B
′ v B
= 5.2 ×10 −3 kg ⋅ 428m / s + 0.7kg ⋅ ′ v B
v ′ B
=
3.4944 kg m / s− 5.2 ×10−3 kg ⋅ 428m / s
0.7kg
=1.812m / s
We can find the CM velocity because it remains unchanged (since the forces are all internal to the
bullet-block system)
v cm
=
m
bv b
+ M B
v B
=
m b
+ M B
3.4944kg m / s
5.2 ×10 −3 kg + 0.7kg = 4.955m / s
9.46 A bullet of mass 10 g strikes a ballistic pendulum of mass 2.0 kg. The center of mass of the
pendulum rises a vertical distance of 12 cm. Assuming that the bullet remains embedded in the
pendulum, calculate the bullet’s initial speed.

This is both a momentum conservation problem and an energy conservation problem. The energy
conservation part is the motion of the pendulum. The total energy of the pendulum at the top of the
swing is equal to the total energy of the pendulum after it has been struck by a bullet.
E f
= (M + m)g y
E i
= 1 2
(M + m)v
2
E i
= E f
1
2 (M + m)v 2 = (M + m)g y
v =
2g y
= 2⋅ 9.8 ⋅ 0.12
=1.534 m / s
Now that we know the velocity of the bullet-pendulum immediately after the bullet strikes the
pendulum, we can use momentum conservation to find the bullet’s velocity
p i
= p f
mv b
= (M + m)v
(M + m)
v b
= v
m
(2.0kg + 0.01kg)
= ⋅1.534m / s
0.01kg
= 308.3m / s
9.50 In Fig. 9-60, a 10 g bullet moving directly upward at 1000m/s strikes nd passes through the
center of mass of a 5.00 kg block initially at res. The bullet emerges from he block moving directly
upward at 400 m/s. Tow what maximum height does the block then rise above its initial position.
As in the ballistic pendulum case, this problem uses both momentum conservation and energy
conservation. We begin with momentum conservation in the collision.
p i
= m b
v b
+ M B
v B
p i
=10.0 ×10 −3 kg ⋅1000m / s + 5.0 kg⋅ 0m / s
=10 kgm / s
p f
=
p i
10 ×10 −3 kg ⋅ 400m / s + 2.0kg ⋅ v ′ B
=10.0 kg m / s
p f
= m bv b ′ + M Bv ′ B
= m b
′ v b
+ M B
′ v B
=10 ×10 −3 kg ⋅ 400m / s + 2.0 kg⋅ ′ v B
v ′ B
= 10.0 kg m / s−10.0 ×10−3 kg ⋅ 400m / s
2.0kg
= 3m / s

Now we know the velocity of the block after the collision. We can use conservation of energy to
find the maximum height.
1
2 Mv2 = Mgh
h = v 2 (3m / s)2
=
2g 2 ⋅ 9.8m / s 2
= 0.459m
9.60 A steel ball of mass 0.5 kg is fastened to a cord that is 70 cm long an fixed at the far end. The
ball is then released when the cord is horizontal (Fig. 9-65). At the bottom of its path, the ball
strikes a 2.5 kg steel block initially at rest on a frictionless surface. The collision is elastic. Find
(a) the speed of the ball and (b) the speed of the block, both just after the collision.
We begin by finding the initial velocity of the ball using energy conservation.
1
2 mv 2 = mgh
v 2 = 2gh
v b
= 2gh = 2 ⋅ 9.8⋅ 0.7m
= 3.7m / s
Now consider the collision. Both momentum and energy are conserved in this elastic collision.
mv b
= m v ′ b
+ M v ′ B
1
2 mv 2 b
= 1 2 m v ′
2 b
+ 1 2 M v ′
2
B
We have two equations and two unknowns. We solve the momentum equation for one of the
unknowns and then plug into the energy equation.

m v ′ b
= mv b
− M v ′ B
v ′ b
= v b
− M v
m ′ B
1
2 mv 2 b
= 1 2 m v ′
2 b
+ 1 2 M v ′
2
B
1
2 mv 2 b
= 1 2 m(v − M b
m v ′ B )2 + 1 2 M v ′ B
1
2 mv 2 b
= 1 2 mv 2 b
+ 1 2
M 2
m v ′
2 B
− Mv b
′
0 = M 2
v
m ′
2 2
B
− 2Mv b
v ′ B
+ M v ′ B
0 = M m v ′
2 2
B
− 2v b
v ′ B
+ v ′ B
⎛ ⎛ M
= v ′ B ⎜
⎝ m +1 ⎞ ⎞
⎜ ⎟ v ′
⎠
B
− 2v b ⎟
⎝
⎠
⎛ ⎛ M
0 = ⎜
⎝ m +1 ⎞ ⎞
⎜ ⎟ v ′
⎠
B
− 2v b ⎟
⎝
⎠
2v
v ′ B
= b
2⋅ 3.7m / s
=
⎛ M
⎜
⎝ m +1 ⎞ ⎛ 2.5kg
⎟
⎠
⎜
⎝ 0.5kg +1
⎞
⎟
⎠
=1.23m / s
2
v B
+ 1 2 M ′
Notice that if the mass of the ball and the mass of the block are the same, the ball will stop and the
block will take all of the velocity of the ball. This is exactly what we saw in the elastic equal mass
collision in lab. We finally find the ball’s velocity after the collision, by substituting back in.
v ′ b
= v b
− M v
m ′ B
= 3.7m / s − 2.5
0.5 ⋅1.23m / s
= −2.45m / s
v 2
B