1. Conservation of Momentum Problems Answers
1. A cannonball of mass 4.0 kg is placed in a cannon of mass 650 kg.
After firing, the cannonball is moving to the east at 76 m/s.
Before
76 m/s east
4.0 kg
After
650 kg
mcannon = mc = 650 kg
mball = mb = 4.0 kg
vball = vb = m/s

2. Before
vb = 76 m/s east
mb = 4.0 kg
After
mc = 425 kg
(a) What is the momentum of the cannon-cannonball system
before firing?
p = m v
Before firing, neither the cannon nor the cannonball is moving.
pbefore = 0

3. Before
vb = 76 m/s east
mb = 4.0 kg
After
mc = 650 kg
(b) What is the momentum of the cannonball after firing?
pb = mb vb
= ( 4.0 kg )( + 76 m/s )
pb = kg m/s
OR kg m/s east

4. Before
vb = 76 m/s east
mb = 4.0 kg
After
pb = kg m/s
mc = 650 kg
(c) What is the momentum of the cannon after firing?
pbefore = pafter
pbefore = part (a)
pafter = pball + pcannon
= pb + pc
pb = kg m/s
part (b)

5. Before
vb = 76 m/s east
mb = 4.0 kg
After
pb = kg m/s
mc = 425 kg
(c) What is the momentum of the cannon after firing?
pbefore = pafter
pafter = pb + pc
0 = kg m/s + pc
- 304 kg m/s
- 304 kg m/s
Momentum of the cannon is the same as the momentum of the ball, but in the opposite direction
pc = kg m/s
OR kg m/s west

6. Before
vb = 76 m/s east
vc = ?
mb = 4.0 kg
After
pb = kg m/s
pc = kg m/s
mc = 650 kg
(d) What is the velocity of the cannon after firing?
pc = mc vc mc mc pc
- 304 kg m/s
vc = =
= vc = m/s mc
650 kg
OR m/s west

7. 2. A cannonball of mass 4.0 kg is fired from a cannon of mass
2.0 x 103 kg. After firing, the velocity of the cannonball is 125 m/s
to the north. Calculate the velocity of the cannon after firing.
mb = 4.0 kg mc = 2.0 x 103 kg vb = m/s (north)
pb = mb vb
= ( 4.0 kg )( m/s )
= kg m/s (north)
Momentum of cannon is same as that of the ball, but in the opposite direction
So momentum of cannon = kg m/s (south)
pc = mc vc mc mc pc
- 500 kg m/s
vc = =
= vc = m/s mc
2000 kg
OR m/s south

8. 2. A cannonball of mass 4.0 kg is fired from a cannon of mass
2.0 x 103 kg. After firing, the velocity of the cannonball is 125 m/s
to the north. Calculate the velocity of the cannon after firing.
mb = 4.0 kg mc = 2.0 x 103 kg vb = m/s (north)
pbefore = pafter
pbefore = 0
pafter = pb + pc
0 = pb + pc
pb = mb vb
0 = kg m/s + pc
= ( 4.0 kg )( m/s )
- 500 kg m/s
- 500 kg m/s
= kg m/s
pc = kg m/s
pc = mc vc mc mc pc
- 500 kg m/s
vc = =
= vc = m/s mc
2000 kg
OR m/s south

9. 2. A cannonball of mass 4.0 kg is fired from a cannon of mass
2.0 x 103 kg. After firing, the velocity of the cannonball is 125 m/s
to the north. Calculate the velocity of the cannon after firing.
mb = 4.0 kg mc = 2.0 x 103 kg vb = m/s (north)
pbefore = pafter
0 = pb + pc
pb = mb vb
pc = mc vc
0 = mb vb + mc vc
vc = ?
- mb vb
- mb vb
- mb vb
- mb vb
= mc vc
vc = mc mc mc
- ( 4.0 kg )( +125 m/s ) =
2.0 x 103 kg
vc = m/s

10. 3. A steel ball of mass 4.0 kg and moving at 7.5 m/s collides with a
stationary brass ball of mass 2.0 kg. After the collision, the brass
ball moves forward at 10.0 m/s. Find the velocity of the steel ball
after the collision.
v1b = m/s
m2 = 2.0 kg
Before
m1 = 4.0 kg
After
v1a = ?
v2a = m/s
pbefore = pafter
pbefore = p1b + p2b
p2b = 0
pbefore = p1b
= m1 v1b
= ( 4.0 kg )( m/s )
pbefore = kg m/s

11. 3. A steel ball of mass 4.0 kg and moving at 7.5 m/s collides with a
stationary brass ball of mass 2.0 kg. After the collision, the brass
ball moves forward at 10.0 m/s. Find the velocity of the steel ball
after the collision.
v1b = m/s
m2 = 2.0 kg
Before
m1 = 4.0 kg
After
v1a = ?
v2a = m/s
pbefore = pafter
= kg m/s
pafter = p1a + p2a
= m1 v1a + m2 v2a
= ( 4.0 kg ) v1a
= ( 2.0 kg )( m/s )
= kg m/s

12. 3. A steel ball of mass 4.0 kg and moving at 7.5 m/s collides with a
stationary brass ball of mass 2.0 kg. After the collision, the brass
ball moves forward at 10.0 m/s. Find the velocity of the steel ball
after the collision.
v1b = m/s
m2 = 2.0 kg
Before
m1 = 4.0 kg
After
v1a = ?
v2a = m/s
pbefore = pafter
= kg m/s
pafter = p1a + p2a
= m1 v1a + m2 v2a
+ 30 kg m/s = ( 4.0 kg ) v1a kg m/s
- 20 kg m/s kg m/s
+ 10 kg m/s = ( 4.0 kg ) v1a
4.0 kg kg
v1a = m/s

13. steel ball of mass 0.150 kg. After the collision, the billiard ball
4. A billiard ball of mass kg collides elastically with a stationary
steel ball of mass kg. After the collision, the billiard ball
continues forward with a velocity of 1.5 m/s, while the steel ball is
propelled forward at 9.0 m/s. Find the speed of the billiard ball
before the collision.
m1 = kg
m2 = kg
Before
v1b = ?
v2a = m/s
After
v1a = m/s
pbefore = pafter
p1b + p2b = p1a + p2a
p2b = 0
m1 v1b = m1 v1a + m2 v2a m1 m1

14. m1 v1a + m2 v2a v1b = m1 (0.225 kg)(1.5 m/s) + (0.150 kg)(9.0 m/s) =
v2a = m/s
v1a = m/s
m1 v1a + m2 v2a
v1b = m1
(0.225 kg)(1.5 m/s) + (0.150 kg)(9.0 m/s) =
0.225 kg
v1b = m/s

15. 8. A train car of mass 1800 kg is carrying a metric ton (1000 kg) of coal.
It moves down the track at 3.6 m/s and collides inelastically with a
stationary car of mass 3500 kg. Find the velocity of the cars after the
collision.
Before
2800 kg
3500 kg
v1b = m/s
va = ?
After
pbefore = pafter
M = kg kg
= kg
p1b + p2b = pa
m1 v1b = M va M M
m1 v1b
( 2800 kg )( 3.6 m/s )
va = =
= va = 1.6 m/s M
6300 kg

16. 6. A block of wood of mass 1.5 kg is hanging from a tree by a string.
A bullet of mass 15.0 g is fired at 180 m/s into the block. Find
the speed at which the bullet/block system moves forward after the
bullet is embedded into the block.
m1 = 15.0 g
va = ?
Before
After
1.5 kg
v1b = m/s
m1 = g
= kg
M = 1.5 kg kg
= kg
pbefore = pafter
p1b + p2b = pa
m1 v1b = M va M M
m1 v1b
( kg )( 180 m/s )
va = =
= va = 1.8 m/s M
1.515 kg