Conservation of Momentum
(in 1D)
SPH4U
Newton’s 3rd Law
FBA   FAB
m a A  mBTheaforce
B A
The force BA
exerts on A
exerts on B
 v A 
 vB 
mA 
   mB 

 t 
 t 
m A v A  mB vB
p A  pB
Newton’s 3rd Law
FBA   FAB
m A a A   mB a B
 v A 
 vB 
mA 
   mB 

 t 
 t 
m A v A  mB vB
p A  pB
Newton’s 3rd Law
FBA   FAB
m A a A   mB a B
 v A 
 vB 
mA 
   mB 

 t 
 t 
that 
the 
time
for
thev
m ANote

v
m
A must be B
B
interaction
the same.
p A  pB
Newton’s 3rd Law
FBA   FAB
m A a A   mB a B
 v A 
 vB 
mA 
   mB 

 t 
 t 
m A v A  mB vB
p A  pB
Newton’s 3rd Law
FBA   FAB
m A a A   mB a B
 v A 
 vB 
mA 
   mB 

 t 
 t 
m A v A  mB vB
p A  pB
The Law of Conservation of
Momentum
During an interaction between two objects
(where there is no external net force), the
change in momentum of the first object is
equal in magnitude but opposite in direction
to the change in momentum of the second.
Conservation and Impulse: Example
A rubber bullet R and a metal bullet M of equal
mass strike a target with the same speed. The
metal bullet comes to rest inside the target
while the rubber bullet bounces back.
Which exerts a greater impulse on the target?
Conservation and Impulse: Example
A rubber bullet R and a metal bullet M of equal
mass strike a target with the same speed. The
metal bullet comes to rest inside the target
while the rubber bullet bounces back.
Which exerts a greater impulse on the target?
R
Conservation and Impulse: Example
A rubber bullet R and a metal bullet M of equal
mass strike a target with the same speed. The
metal bullet comes to rest inside the target
while the rubber bullet bounces back.
Which exerts a greater impulse on the target?
R – changing the momentum from [fwd] to
[back] requires a greater impulse than
changing momentum from [fwd] to zero.
Conservation of Momentum
m A v A  mB vB
m A v A 'v A   mB vB 'vB 
m Av A 'm Av A  mB vB ' mB vB
m Av A ' mB vB '  m Av A  mB vB
Conservation of Momentum
m A v A  mB vB
m A v A 'v A   mB vB 'vB 
m v 'm v  m v ' mB vB
“Primes” denote the velocity
interaction:
v’=vB
2
A A after A
A
B
m Av A ' mB vB '  m Av A  mB vB
Conservation of Momentum
m A v A  mB vB
m A v A 'v A   mB vB 'vB 
m Av A 'm Av A  mB vB ' mB vB
m Av A ' mB vB '  m Av A  mB vB
Conservation of Momentum
m A v A  mB vB
m A v A 'v A   mB vB 'vB 
m Av A 'm Av A  mB vB ' mB vB
m Av A ' mB vB '  m Av A  mB vB
The Law of Conservation of
Momentum (Version 2)
mAvA 'mB vB '  mAvA  mB vB
During an interaction between two objects
(where there is no external net force), the
total momentum of the system after is equal
to the total momentum of the system before.
Conservation: Example 1
A baseball of mass m leaves a pitching machine
off mass M (where M includes the mass of the
ball m) with a speed v.
What is the recoil speed of the machine after
shooting the baseball?
Conservation: Example 1
A baseball of mass m leaves a pitching machine
off mass M (where M includes the mass of the
ball m) with a speed v.
What is the recoil speed of the machine after
shooting the baseball?
Note that total momentum before is zero.
Conservation: Example 1
m v  ( M  m)v M  m  0
( M  m)v M  m   m v
vM  m
 mv

M m
Conservation: Example 1
m v  ( M  m)v M  m  0
( M  m)v M  m   m v
vM  m
 mv

M m
Conservation: Example 1
m v  ( M  m)v M  m  0
( M  m)v M  m   m v
vM  m
 mv

M m
Since the question asks for the (positive) speed,
we would express this as: mv
M m
Conservation: Example 2
A student of mass 75 kg is standing on a
stationary raft of mass 55 kg. The student
then moves toward one end of the raft at a
speed of 2.3 m/s [N] relative to the water.
Neglecting fluid friction, what is the velocity of
the raft relative to the water?
Conservation: Example 2
ms vs ' mr vr '  0
mr vr '   ms vs '
 ms v s '
vr ' 
mr
 (75 kg )(2.3 ms )
vr 
55 kg
vr  3.1 ms
i.e., 3.1 m/s opposite the
direction of motion of the student
Conservation: Example 2
ms vs ' mr vr '  0
mr vr '   ms vs '
 ms v s '
vr ' 
mr
 (75 kg )(2.3 ms )
vr 
55 kg
vr  3.1 ms
And the velocity of the
raft relative to the student
is 2.3 m/s + 3.1 m/s [back]
More Practice
Textbook Questions:
• p. 243 #5, 6, 7
• p. 245 #7
http://www.youtube.com/watch?v=r8E5dUnLmh4