1. Conservation of Momentum:
The total vector momentum of an isolated system remains constant.
pbefore = pafter

2. Conservation of Momentum Examples Answers
1. A cannonball of mass 5.0 kg is placed in a cannon of mass 425 kg.
After firing, the cannonball is moving to the east at 85 m/s.
Before
85 m/s east
5.0 kg
After
425 kg
mcannon = mc = 425 kg
mball = mb = 5.0 kg
vball = vb = m/s

3. Before
vb = 85 m/s east
mb = 5.0 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

4. Before
vb = 85 m/s east
mb = 5.0 kg
After
mc = 425 kg
(b) What is the momentum of the cannonball after firing?
pb = mb vb
= ( 5.0 kg )( + 85 m/s )
pb = kg m/s
OR kg m/s east

5. Before
vb = 85 m/s east
mb = 5.0 kg
After
pb = kg m/s
mc = 425 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)

6. Before
vb = 85 m/s east
mb = 5.0 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
- 425 kg m/s
- 425 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

7. Before
vb = 85 m/s east
vc = ?
mb = 5.0 kg
After
pb = kg m/s
pc = kg m/s
mc = 425 kg
(d) What is the velocity of the cannon after firing?
pc = mc vc mc mc pc
- 425 kg m/s
vc = =
= vc = m/s mc
425 kg
OR 1.0 m/s west

8. 2. A gun of mass 3. 0 kg fired a bullet of mass 10
2. A gun of mass 3.0 kg fired a bullet of mass 10.0 g with a velocity of
625 m/s. Determine the recoil velocity of the gun.
mb = g = kg mg = 3.0 kg
vb = m/s vg = ?
pb = mb vb
= ( kg )( +625 m/s )
= kg m/s
pg = mg vg
= ( 3.0 kg ) vg
pbefore = pafter
0 = pb + pg
0 = kg m/s + ( 3.0 kg ) vg
-6.25 kg m/s kg m/s
-6.25 kg m/s = ( 3.0 kg ) vg
3.0 kg kg
vg = m/s

9. 2. A gun of mass 3. 0 kg fired a bullet of mass 10
2. A gun of mass 3.0 kg fired a bullet of mass 10.0 g with a velocity of
625 m/s. Determine the recoil velocity of the gun.
mb = g = kg mg = 3.0 kg
vb = m/s vg = ?
pbefore = pafter
0 = pb + pg
pb = mb vb
pg = mg vg
0 = mb vb + mg vg
vg = ?
- mb vb
- mb vb
- mb vb
- mb vb
= mg vg
vg = mg mg mg
- ( kg )( m/s ) =
3.0 kg
vg = m/s

10. 3. A bowling ball of mass 7.0 kg moving at 4.8 m/s collides with a
stationary pin of mass 4.4 kg. After the collision, the pin flies forward
at 5.9 m/s. Find the velocity of the bowling ball after the collision.
m2 = 4.4 kg
m1 = 7.0 kg
Before
v1b = m/s
After
v1a = ?
v2a = m/s
pbefore = pafter
pbefore = p1b + p2b
p2b = 0
= p1b
= m1 v1b
= ( 7.0 kg )( m/s )
pbefore = kg m/s

11. 3. A bowling ball of mass 7.0 kg moving at 4.8 m/s collides with a
stationary pin of mass 4.4 kg. After the collision, the pin flies forward
at 5.9 m/s. Find the velocity of the bowling ball after the collision.
m2 = 4.4 kg
m1 = 7.0 kg
Before
v1b = m/s
After
v1a = ?
v2a = m/s
pbefore = pafter = kg m/s
p2a = m2 v2a
= ( 4.4 kg )( m/s )
pafter = p1a + p2a
= kg m/s
kg m/s = p1a kg m/s
kg m/s
kg m/s
kg m/s = p1a

12. 3. A bowling ball of mass 7.0 kg moving at 4.8 m/s collides with a
stationary pin of mass 4.4 kg. After the collision, the pin flies forward
at 5.9 m/s. Find the velocity of the bowling ball after the collision.
m2 = 4.4 kg
m1 = 7.0 kg
Before
v1b = m/s
After
v1a = ?
v2a = m/s
p1a = m1 v1a
p1a = kg m/s m1 m1
p1a
kg m/s
v1a = =
= v1a = m/s m1
7.0 kg

13. m2 = 4.4 kg
m1 = 7.0 kg
Before
v1b = m/s
After
v1a = ?
v2a = m/s
pbefore = pafter
p1b + p2b = p1a + p2a
p2b = 0
m1 v1b = m1 v1a + m2 v2a

14. m2 = 4.4 kg
m1 = 7.0 kg
Before
v1b = m/s
After
v1a = ?
v2a = m/s
pbefore = pafter
p1b + p2b = p1a + p2a
p2b = 0
m1 v1b = m1 v1a + m2 v2a
- m2 v2a
- m2 v2a
m1 v1b - m2 v2a = m1 v1a m1 m1
m1 v1b - m2 v2a
v1a = m1

15. m2 = 4.4 kg
m1 = 7.0 kg
Before
v1b = m/s
After
v1a = ?
v2a = m/s
m1 v1b - m2 v2a
v1a = m1
( 7.0 kg )( 4.8 m/s ) – ( 4.4 kg )( 5.9 m/s ) =
7.0 kg
v1a = m/s

16. 4. A golf ball of mass 0.045 kg moving at 28.0 m/s collides with a
stationary baseball of mass kg. The baseball moves forward at
12.9 m/s. Find the velocity of the golf ball after the collision.
m2 = kg
m1 = kg
Before
v1b = m/s
v2a = m/s
After
v1a = ?
pbefore = p1b + 0
p2a = m2 v2a
= kg m/s
= kg m/s
pbefore = pafter = p1a + p2a
p1a = kg m/s
p1a = m1 v1a
v1a = m/s

17. pbefore = pafter p1b + p2b = p1a + p2a p2b = 0
m2 = kg
m1 = kg
v1b = + 28 m/s
v2a = m/s
v1a = ?
pbefore = pafter
p1b + p2b = p1a + p2a
p2b = 0
m1 v1b = m1 v1a + m2 v2a
- m2 v2a
- m2 v2a
m1 v1b - m2 v2a = m1 v1a
m1 v1b - m2 v2a
v1a = m1 m1 m1
(0.045 kg)(28 m/s) – (0.150 kg)(12.9 m/s)
= v1a = m/s
v1a =
0.045 kg

18. 5. A train car of mass 2000 kg and moving at 2.5 m/s collides with a
stationary car of mass 1200 kg. The cars couple together and move
down the track with velocity v. Find v.
Before
2000 kg
1200 kg
v1b = m/s
After
va = ?
pbefore = pafter
M = kg kg
= kg
p1b + p2b = pa
m1 v1b = M va M M
m1 v1b
( 2000 kg )( 2.5 m/s )
va = =
= va = 1.6 m/s M
3200 kg

19. 6. A running back has a mass of 105 kg and is moving in a straight line at
9.4 m/s. He is hit head-on by a linebacker that has a mass of 115 kg
and is moving in the opposite direction at 6.4 m/s. After the inelastic
collision, what is the velocity of the running back-linebacker system?
v1b = m/s
m1 = 105 kg
m2 = 115 kg
v2b = m/s
M = 105 kg kg
= 220 kg
pbefore = pafter
p1b + p2b = pa
m1 v1b + m2 v2b = M va M M

20. v1b = m/s
m1 = 105 kg
m2 = 115 kg
v2b = m/s
M = 105 kg kg
= 220 kg
m1 v1b + m2 v2b
va = M
( 105 kg )( m/s ) + ( 115 kg )( m/s ) =
220 kg
va = m/s