Momentum and Impulse
Physics 6A
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Momentum
any moving object will have both momentum and kinetic energy
We use a lowercase p for momentum – here is the formula


p  mv
Notice that this is a vector, so you will have to break into
components when adding up the momenta of several objects.
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Momentum
any moving object will have both momentum and kinetic energy
We use a lowercase p for momentum – here is the formula


p  mv
Notice that this is a vector, so you will have to break into
components when adding up the momenta of several objects.
Impulse
if the momentum of an object changes, then either its mass or its velocity must have
changed. In most cases, we will consider the mass to be constant, so a change in
momentum will mean the object accelerated (velocity changed).
So there must have been a force applied to it. This gives us a formula:


p  Favg  t
This is just Newton’s 2nd Law – can you see why?
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Momentum
any moving object will have both momentum and kinetic energy
We use a lowercase p for momentum – here is the formula


p  mv
Notice that this is a vector, so you will have to break into
components when adding up the momenta of several objects.
Impulse
if the momentum of an object changes, then either its mass or its velocity must have
changed. In most cases, we will consider the mass to be constant, so a change in
momentum will mean the object accelerated (velocity changed).
So there must have been a force applied to it. This gives us a formula:


p  Favg  t
Solve this for Favg to get
This is just Newton’s 2nd Law – can you see why?




p m  v
Favg 

 ma
t
t
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Typical Impulse Example:
A golf ball (initially at rest) is struck by a club, and the ball is given a
velocity of 50 m/s.
If the mass of the ball is 46 grams and the club is in contact with
the ball for 5 ms, what is the average force applied to the ball by
the club?
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Typical Impulse Example:
A golf ball (initially at rest) is struck by a club, and the ball is given a
velocity of 50 m/s.
If the mass of the ball is 46 grams and the club is in contact with
the ball for 5 ms, what is the average force applied to the ball by
the club?
We can find the impulse (change in momentum) directly in this problem:
Take a minute and compute it for yourself…
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Typical Impulse Example:
A golf ball (initially at rest) is struck by a club, and the ball is given a
velocity of 50 m/s.
If the mass of the ball is 46 grams and the club is in contact with
the ball for 5 ms, what is the average force applied to the ball by
the club?
We can find the impulse (change in momentum) directly in this problem:
The ball starts at rest, so the initial momentum is 0.
After the club hits it, the ball has momentum pfinal=(0.046kg)(50m/s)=2.3kg-m/s
So the impulse is Δp=2.3kg-m/s
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Typical Impulse Example:
A golf ball (initially at rest) is struck by a club, and the ball is given a
velocity of 50 m/s.
If the mass of the ball is 46 grams and the club is in contact with
the ball for 5 ms, what is the average force applied to the ball by
the club?
We can find the impulse (change in momentum) directly in this problem:
The ball starts at rest, so the initial momentum is 0.
After the club hits it, the ball has momentum pfinal=(0.046kg)(50m/s)=2.3kg-m/s
So the impulse is Δp=2.3kg-m/s
Now we can divide by Δt to find Favg=460N
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Notes on Impulse
- Impulses always occur as action-reaction pairs
- Longer impact time means smaller force, and vice-versa
“Real-Life” Examples:
Automobile safety:
Dashboards
Seatbelts and Airbags
Crumple zones
Product packaging
Styrofoam spacers
Egg cartons
Sports
Baseball: Padded catcher’s mitt
Golf: Deformation of ball during brief impact
Boxing gloves: Padding means smaller impact to the hands (and face, etc.)
Cycling: Always wear your helmet!
Sleep technology
Head on a pillow more comfortable than head on a concrete floor.
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Conservation of Momentum
As long as there are no outside forces, the total momentum of a system of objects will not change.
We will use this idea mainly in collision problems (but it is much more fundamental, and always applies)
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Conservation of Momentum
As long as there are no outside forces, the total momentum of a system of objects will not change.
We will use this idea mainly in collision problems (but it is much more fundamental, and always applies)
Collisions (2 main categories)
Elastic Collisions – KE is conserved (no dissipative forces)
A good approximation for things like billiard balls
Your textbook has shortcut formulas for head-on elastic collisions (page 266).
Use these formulas if you want to avoid some possibly difficult algebra
Inelastic Collisions – some KE is lost due to dissipative forces during impact
Special Case: When the objects stick together, the collision is perfectly inelastic.
This will be the easiest type of problem to solve, because there is only 1 object after
the collision.
Momentum is conserved in all collisions.
Use conservation of momentum first in all collision problems.
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Example: A 100kg man and a 50kg boy are at a father-son ice skating jamboree.
The man skates toward his son at a speed of 5 m/s, and picks the boy up and puts
him on his shoulders. If their final speed is 3 m/s, find the initial velocity of the boy.
vboy=??
v=3m/s
v=5m/s
BEFORE
AFTER
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Example: A 100kg man and a 50kg boy are at a father-son ice skating jamboree.
The man skates toward his son at a speed of 5 m/s, and picks the boy up and puts
him on his shoulders. If their final speed is 3 m/s, find the initial velocity of the boy.
vboy=??
v=3m/s
v=5m/s
BEFORE
AFTER
This is a perfectly inelastic collision (they stick together). So we only need conservation of momentum.
Write down the initial and final momentum for the system, then set them equal.
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Example: A 100kg man and a 50kg boy are at a father-son ice skating jamboree.
The man skates toward his son at a speed of 5 m/s, and picks the boy up and puts
him on his shoulders. If their final speed is 3 m/s, find the initial velocity of the boy.
vboy=??
v=3m/s
v=5m/s
BEFORE
AFTER
This is a perfectly inelastic collision (they stick together). So we only need conservation of momentum.
Write down the initial and final momentum for the system, then set them equal.
pf  pi
(100kg  50kg)(3 ms )  (100kg)(5 ms )  (50kg)( v boy )
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Example: A 100kg man and a 50kg boy are at a father-son ice skating jamboree.
The man skates toward his son at a speed of 5 m/s, and picks the boy up and puts
him on his shoulders. If their final speed is 3 m/s, find the initial velocity of the boy.
vboy=??
v=3m/s
v=5m/s
BEFORE
AFTER
This is a perfectly inelastic collision (they stick together). So we only need conservation of momentum.
Write down the initial and final momentum for the system, then set them equal.
pf  pi
(100kg  50kg)(3 ms )  (100kg)(5 ms )  (50kg)( vboy )
vboy  1ms
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Example: A 100kg man and a 50kg boy are at a father-son ice skating jamboree.
The man skates toward his son at a speed of 5 m/s, and picks the boy up and puts
him on his shoulders. If their final speed is 3 m/s, find the initial velocity of the boy.
vboy=??
v=3m/s
v=5m/s
BEFORE
AFTER
This is a perfectly inelastic collision (they stick together). So we only need conservation of momentum.
Write down the initial and final momentum for the system, then set them equal.
pf  pi
(100kg  50kg)(3 ms )  (100kg)(5 ms )  (50kg)( vboy )
vboy  1ms
Why is the answer negative? What does that mean?
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Example: A 100kg man and a 50kg boy are at a father-son ice skating jamboree.
The man skates toward his son at a speed of 5 m/s, and picks the boy up and puts
him on his shoulders. If their final speed is 3 m/s, find the initial velocity of the boy.
vboy=-1m/s
v=3m/s
v=5m/s
BEFORE
AFTER
This is a perfectly inelastic collision (they stick together). So we only need conservation of momentum.
Write down the initial and final momentum for the system, then set them equal.
pf  pi
(100kg  50kg)(3 ms )  (100kg)(5 ms )  (50kg)( vboy )
vboy  1ms
The boy is initially moving toward his father.
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Example: A golf ball (45 grams) rolls toward a billiard ball (135 grams). The initial
speed of the golf ball is 10 m/s, and the billiard ball is initially at rest. Find the final
velocity of each ball. Assume the collision is elastic.
10 m/s
BEFORE
V=??
V=??
AFTER
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Example: A golf ball (45 grams) rolls toward a billiard ball (135 grams). The initial
speed of the golf ball is 10 m/s, and the billiard ball is initially at rest. Find the final
velocity of each ball. Assume the collision is elastic.
V=??
V=??
10 m/s
AFTER
BEFORE
Since the collision is elastic we can use the formulas on page 266 of your book:
v A,f  v A,i 
mA  mB
mA  mB
vB,f  vA,i 
2mA
mA  mB
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Example: A golf ball (45 grams) rolls toward a billiard ball (135 grams). The initial
speed of the golf ball is 10 m/s, and the billiard ball is initially at rest. Find the final
velocity of each ball. Assume the collision is elastic.
b
a
Vb=??
Va=??
10 m/s
b
a
AFTER
BEFORE
Since the collision is elastic we can use the formulas on page 266 of your book:
2mA
mA  mB
m  mB
v A,f  v A,i  A
mA  mB
vB,f  v A,i 
45g  135g
v A,f  10 m 
s 45g  135g
vB,f  10 m 
v A,f  5 m
s
s
2(45g)
45g  135g
vB,f  5 m
s
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Example: A golf ball (45 grams) rolls toward a billiard ball (135 grams). The initial
speed of the golf ball is 10 m/s, and the billiard ball is initially at rest. Find the final
velocity of each ball. Assume the collision is elastic.
b
a
Vb=??
Va=??
10 m/s
b
a
AFTER
BEFORE
Since the collision is elastic we can use the formulas on page 266 of your book:
2mA
mA  mB
m  mB
v A,f  v A,i  A
mA  mB
vB,f  v A,i 
45g  135g
v A,f  10 m 
s 45g  135g
vB,f  10 m 
v A,f  5 m
s
s
2(45g)
45g  135g
vB,f  5 m
s
Important note: The relative speed of the balls is the same before and
after the collision. This will always hold true for head-on elastic collisions.
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Example: A golf ball (45 grams) rolls toward a billiard ball (135 grams). The initial
speed of the golf ball is 10 m/s, and the 8-ball is moving toward the golf ball at 5 m/s.
Find the final velocity of each ball. Assume the collision is elastic.
a
Vb=??
Va=??
10 m/s
b
a
b
5 m/s
BEFORE
AFTER
Now let’s see what happens if BOTH objects are moving before the collision.
The shortcut formulas will only work if one of the objects is at rest, so we have
to switch our point of view before we can plug in the numbers.
We need to subtract the velocity of one of the objects from both initial
velocities, then use the formula, then add back the velocity to get the final
answer.
In this case, let’s say the 8-Ball is at rest. That gives a new initial velocity for
the golf ball of 10 m/s – (-5 m/s) = 15 m/s.
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Example: A golf ball (45 grams) rolls toward a billiard ball (135 grams). The initial
speed of the golf ball is 10 m/s, and the 8-ball is moving toward the golf ball at 5 m/s.
Find the final velocity of each ball. Assume the collision is elastic.
a
Vb=??
Va=??
10 m/s
b
b
a
5 m/s
BEFORE
AFTER
2mA
mA  mB
m  mB
v A,f  v A,i A
mA  mB
vB,f  v A,i 
45g  135g
v A,f  15 m 
s 45g  135g
vB,f  15 m 
v A,f  7.5 m
s
v A,f  7.5 m  (5 m)  12.5 m
s
s
s
s
2(45g)
45g  135g
vB,f  7.5 m
s
vB,f  7.5 m  (5 m)  2.5 m
s
s
s
Notice that the relative speed of the balls is still the same before and after
the collision. This will always hold true for head-on elastic collsions.
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Example: This one will be 2-dimensional. Two hockey players are initially skating as
shown in the figure. They collide, and stick together. Find their final velocity
(magnitude and direction).
vfinal=?
40°
Player B
Player A
Mass 80kg
Mass 90kg
v=4 m/s
v=6 m/s
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Example: This one will be 2-dimensional. Two hockey players are initially skating as
shown in the figure. They collide, and stick together. Find their final velocity
(magnitude and direction).
This is a perfectly inelastic collision, so we should use conservation of
momentum. We need to have separate formulas for x- and for y-directions
vfinal=?
40°
Player B
Player A
Mass 80kg
Mass 90kg
v=4 m/s
v=6 m/s
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Example: This one will be 2-dimensional. Two hockey players are initially skating as
shown in the figure. They collide, and stick together. Find their final velocity
(magnitude and direction).
This is a perfectly inelastic collision, so we should use conservation of
momentum. We need to have separate formulas for x- and for y-directions
vfinal=?
We should start by writing down the
components of each player’s initial velocity:
40°
Player B
Player A
Mass 80kg
Mass 90kg
v=4 m/s
v=6 m/s
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Example: This one will be 2-dimensional. Two hockey players are initially skating as
shown in the figure. They collide, and stick together. Find their final velocity
(magnitude and direction).
This is a perfectly inelastic collision, so we should use conservation of
momentum. We need to have separate formulas for x- and for y-directions
vfinal=?
Components of each player’s initial velocity:
v a,x  (6 m )  cos(40 )  4.6 m
s
s
v a,y  (6 m )  sin( 40 )  3.9 m
s
s
vb,x  0
vb,y  4 m
40°
s
Player B
Player A
Mass 80kg
Mass 90kg
v=4 m/s
v=6 m/s
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Example: This one will be 2-dimensional. Two hockey players are initially skating as
shown in the figure. They collide, and stick together. Find their final velocity
(magnitude and direction).
This is a perfectly inelastic collision, so we should use conservation of
momentum. We need to have separate formulas for x- and for y-directions
vfinal=?
Components of each player’s initial velocity:
v a,x  (6 m )  cos(40 )  4.6 m
s
s
v a,y  (6 m )  sin( 40 )  3.9 m
s
s
vb,x  0
vb,y  4 m
40°
s
Player B
Next we can write down the formula for
conservation of momentum in each direction:
Player A
Mass 80kg
Mass 90kg
v=4 m/s
v=6 m/s
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Example: This one will be 2-dimensional. Two hockey players are initially skating as
shown in the figure. They collide, and stick together. Find their final velocity
(magnitude and direction).
This is a perfectly inelastic collision, so we should use conservation of
momentum. We need to have separate formulas for x- and for y-directions
vfinal=?
Components of each player’s initial velocity:
v a,x  (6 m )  cos(40 )  4.6 m
s
s
v a,y  (6 m )  sin( 40 )  3.9 m
s
s
vb,x  0
vb,y  4 m
40°
s
Player B
Conservation of momentum in each direction:
Player A
Mass 80kg
Mass 90kg
v=4 m/s
v=6 m/s
x  direction
(90kg)  ( 4.6 m )  (80kg)  (0 m )  (170kg)  v f ,x
s
s
Total mass = 90kg + 80kg
y  direction
(90kg)  (3.9 m )  (80kg)  ( 4 m )  (170kg)  v f ,y
s
s
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Example: This one will be 2-dimensional. Two hockey players are initially skating as
shown in the figure. They collide, and stick together. Find their final velocity
(magnitude and direction).
This is a perfectly inelastic collision, so we should use conservation of
momentum. We need to have separate formulas for x- and for y-directions
vfinal=?
Components of each player’s initial velocity:
v a,x  (6 m )  cos(40 )  4.6 m
s
s
v a,y  (6 m )  sin( 40 )  3.9 m
s
s
vb,x  0
vb,y  4 m
40°
s
Player B
Conservation of momentum in each direction:
Player A
Mass 80kg
Mass 90kg
v=4 m/s
v=6 m/s
x  direction
(90kg)  ( 4.6 m )  (80kg)  (0 m )  (170kg)  v f ,x
s
 v f ,x  2 . 4 m
s
s
y  direction
(90kg)  (3.9 m )  (80kg)  ( 4 m )  (170kg)  v f ,y
s
 v f ,y  3 . 9 m
s
s
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Example: This one will be 2-dimensional. Two hockey players are initially skating as
shown in the figure. They collide, and stick together. Find their final velocity
(magnitude and direction).
This is a perfectly inelastic collision, so we should use conservation of
momentum. We need to have separate formulas for x- and for y-directions
vfinal=?
Components of each player’s initial velocity:
v a,x  (6 m )  cos(40 )  4.6 m
s
s
v a,y  (6 m )  sin( 40 )  3.9 m
s
s
vb,x  0
vb,y  4 m
40°
s
Player B
Conservation of momentum in each direction:
Mass 80kg
Mass 90kg
v=4 m/s
v=6 m/s
x  direction
(90kg)  ( 4.6 m )  (80kg)  (0 m )  (170kg)  v f ,x
s
 v f ,x  2 . 4 m
s
Player A
s
Combine the components to find the
magnitude and direction of the final velocity:
y  direction
(90kg)  (3.9 m )  (80kg)  ( 4 m )  (170kg)  v f ,y
s
 v f ,y  3 . 9 m
s
s
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Assistance Services at UCSB
Example: This one will be 2-dimensional. Two hockey players are initially skating as
shown in the figure. They collide, and stick together. Find their final velocity
(magnitude and direction).
This is a perfectly inelastic collision, so we should use conservation of
momentum. We need to have separate formulas for x- and for y-directions
vfinal=?
Components of each player’s initial velocity:
v a,x  (6 m )  cos(40 )  4.6 m
s
s
v a,y  (6 m )  sin( 40 )  3.9 m
s
s
vb,x  0
vb,y  4 m
40°
s
Player B
Conservation of momentum in each direction:
Mass 80kg
Mass 90kg
v=4 m/s
v=6 m/s
x  direction
(90kg)  ( 4.6 m )  (80kg)  (0 m )  (170kg)  v f ,x
s
 v f ,x  2 . 4 m
s
Player A
s
Combine the components to find the
magnitude and direction of the final velocity:
v f  (2.4 m )2  (3.9 m )2  4.6 m
s
y  direction
s
s
(90kg)  (3.9 m )  (80kg)  ( 4 m )  (170kg)  v f ,y
s
 v f ,y  3 . 9 m
s
s
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Assistance Services at UCSB
Example: This one will be 2-dimensional. Two hockey players are initially skating as
shown in the figure. They collide, and stick together. Find their final velocity
(magnitude and direction).
This is a perfectly inelastic collision, so we should use conservation of
momentum. We need to have separate formulas for x- and for y-directions
vfinal=?
Components of each player’s initial velocity:
v a,x  (6 m )  cos(40 )  4.6 m
s
s
v a,y  (6 m )  sin( 40 )  3.9 m
s
s
vb,x  0
vb,y  4 m
40°
s
Player B
Conservation of momentum in each direction:
Mass 80kg
Mass 90kg
v=4 m/s
v=6 m/s
x  direction
(90kg)  ( 4.6 m )  (80kg)  (0 m )  (170kg)  v f ,x
s
 v f ,x  2 . 4 m
s
Player A
s
Combine the components to find the
magnitude and direction of the final velocity:
v f  (2.4 m )2  (3.9 m )2  4.6 m
s
y  direction
(90kg)  (3.9 m )  (80kg)  ( 4 m )  (170kg)  v f ,y
s
 v f ,y  3 . 9 m
s
s
tan( ) 
s
s
3.9
   58
2.4
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