Conservation of Momentum Examples
Example 1
A man of mass 70. kg and a boy of mass 35 kg are standing together on a
smooth ice surface. They push each other apart so that the man moves away
with a speed of 0.30 m . What is the final velocity of the boy?
s
If the system includes both the man and the boy it is then an isolated, closed system and
momentum is conserved.
Initial
Final
Object / Mass / Velocity
Object / Mass / Velocity
Man / 70. kg
kg / 0 m/s
kg / -0.30 m/s
Man / 70. kg
kg / 0 m/s
Boy / 35 kg
Boy / 35 kg
kg / ?
-
0
Momentum
+
-
 21kgm
0
Momentum
s
+
Conservation of Momentum Examples
Example 1
A man of mass 70. kg and a boy of mass 35 kg are standing together on a
smooth ice surface. They push each other apart so that the man moves away
with a speed of 0.30 m . What is the final velocity of the boy?
s
Final
Initial
Object / Mass / Velocity
Object / Mass / Velocity
Man / 70. kg
kg / 0 m/s
kg / -030 m/s
Man / 70. kg
kg / 0 m/s
Boy / 35 kg
Boy / 35 kg
kg / ?
-
0
Momentum
Momentum Conservation Equation:
 
pi  p f


0  pmf  pbf


pbf   pmf
+
-
 21kgm
0
Momentum


mbvbf  mmvmf
mm 

vbf  
vmf
mb
s
+
Conservation of Momentum Examples
Example 1
A man of mass 70. kg and a boy of mass 35 kg are standing together on a
smooth ice surface. They push each other apart so that the man moves away
with a speed of 0.30 m . What is the final velocity of the boy?
s
Final
Initial
Object / Mass / Velocity
Object / Mass / Velocity
Man / 70. kg
kg / 0 m/s
kg / -0.30 m/s
Man / 70. kg
Boy /
kg / 0 m/s
Boy / 35 kg
-
0
Momentum
 21kgm
 21kgm
0.60 m
35 kg
kg / ?
s
-
+
s
s
0
Momentum
Momentum Conservation Equation:
mm 

vbf  
vmf
mb
70. kg

vbf  
 0.30 m
s
35 kg



vbf  0.60 m
s
Notice the boy
has half the
mass so twice
the velocity!
+
Conservation of Momentum Examples
Example 2
A bullet of mass 10. g moves horizontally with a speed of 400. m and embeds
s
itself in a block of wood of mass 0.39 kg that is initially at rest on a frictionless table
as shown below.
a. What is the final velocity of the bullet and the block after the collision?
If the system includes both the bullet and the wood it is then an isolated, closed system
and momentum is conserved.
Initial
Final
Object / Mass / Velocity
Object / Mass / Velocity
Bullet / 0.010 kg
kg / 400. m/s  4.0
kgm
kg / ?
Combo / 0.40 kg
s
kg / 0 m/s
Wood / 0.39 kg
-
0
Momentum
+
-
0
Momentum
+
Conservation of Momentum Examples
Example 2
A bullet of mass 10. g moves horizontally with a speed of 400. m and embeds
s
itself in a block of wood of mass 0.39 kg that is initially at rest on a frictionless table
as shown below.
a. What is the final velocity of the bullet and the block after the collision?
Initial
Final
Object / Mass / Velocity
Object / Mass / Velocity
kgm
Bullet / 0.010 kg
kg / 400. m/s  4.0
s
kg / ?
Combo / 0.40 kg
kg / 0 m/s
Wood / 0.39 kg
-
-
0
+
Momentum
Momentum Conservation Equation:
 
pi  p f


pBi  p f
0
Momentum


mbvbi  mb  mw v f

vf 
mb 
vbi
mb  mw
+
Conservation of Momentum Examples
Example 2
A bullet of mass 10. g moves horizontally with a speed of 400. m and embeds
s
itself in a block of wood of mass 0.39 kg that is initially at rest on a frictionless table
as shown below.
a. What is the final velocity of the bullet and the block after the collision?
Initial
Final
Object / Mass / Velocity
Object / Mass / Velocity
kgm
Bullet / 0.010 kg
kg / 400. m/s  4.0
? .m s
kg / 10
Combo / 0.40 kg
s
 4.0 kgm
s
kg / 0 m/s
Wood / 0.39 kg
-
0
+
Momentum
Momentum Conservation Equation:
mb 

vf 
vbi
mb  mw

vf 
-
0
Momentum

0.010 kg
400. m
s
0.010 kg  0.39 kg

v f  10. m
s
+

Example 2
Conservation of Momentum Examples
A bullet of mass 10. g moves horizontally with a speed of 400. m and embeds
s
itself in a block of wood of mass 0.39 kg that is initially at rest on a frictionless table
as shown below.
b. What are the initial and final kinetic energies of the bullet, wood block system?
1
Ekf  mb  mw v 2f
2
1
Eki  mbvbi2
2

1
Eki  0.010 kg  400. m
s
2
Eki  8.0 102 J

2

1
Ekf  0.010 kg  0.39 kg  10. m
s
2
Ekf  20. J
This is an inelastic collision.
Most of the energy is dissipated and is stored as increased thermal energy in the
bullet/block system.

2