Unit 7:
Linear Momentum and Collisions
Momentum and its Relation to Force
a vector symbolized by the symbol
, and is defined as
Linear Momentum is a vector symbolized by the symbol
is defined as
p , and
The rate of change of momentum is equal to the net force:
The rate of change of momentum is equal to the net force:
This can be shown using Newton’s second law.
This can be shown using Newton’s second law.
Newton’s Second Law and M o m e n t u m
Newton’s Second Law can be used to relate the momentum of a particle to
the resultant force acting on it
d  mv  dp
dv
F  ma  m


dt
dt
dt
with constant mass.
The time rate of change of the linear momentum of a particle is equal to the
net force acting on the particle.
• This is the form in which Newton presented the Second Law.
• It is a more general form than the one we used previously.
Conservation of Momentum
During a collision, measurements show that the total momentum
does not change:
Conservation of Momentum
For more than two objects,
Or, since the internal forces cancel,
where

 Fext is the sum of all external forces acting on the system.
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Conservation of Momentum
If the net external force is zero, then dP/dt = 0, so ΔP = 0 or
P = constant.
Thus
when the net external force on a system of objects is zero, the
total momentum of the system remains constant.
This is the law of conservation of linear momentum:
Equivalently,
the total momentum of an isolated system remains constant.
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Conservation of Momentum
Momentum conservation works for a rocket as long as we consider the
rocket and its fuel to be one system, and account for the mass loss of the
rocket.
(a) A rocket, containing fuel, at rest in some reference frame. (b) In the same
reference frame, the rocket fires, and gases are expelled at high speed out the r
The total vector momentum, P = pgas + procket, remains zero.
Collisions and Impulse
During a collision, objects are
deformed due to the large forces
involved.
Since
write
Integrating,
, we can
Collisions and Impulse
This quantity is defined as the impulse, J:
The impulse is equal to the change in momentum:
• Impulse is a vector quantity
• Its direction is parallel to the total force
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Impulse and Variable Forces
 The force does not need to be constant.
 The magnitude of the force grows rapidly from
zero to a maximum value.
 The force then decreases to zero.
 Impulse = area under the force-
time curve.
Collisions and Impulse
Since the time of the collision is often very short, we may be able to use the average
force, which would produce the same impulse over the same time interval.
Airbag example
An example of extending the time is an airbag,
• Your collision with the airbag involves a much longer interaction time than if
you were to collide with the steering column,
• This leads to a smaller force
Conservation of Momentum
• Impulse and momentum concepts can
be applied to collisions
• The total momentum just before the
collision is equal to the total momentum
just after the collision
• The total momentum of the system is
conserved
Collisions
• A collision changes the particles’ velocities
• The kinetic energies of the individual particles will also change
• Collisions fall into two categories
• Elastic collisions
•
The system’s kinetic energy is conserved
• Inelastic collisions
•
Some kinetic energy is lost during the collision
• Momentum is conserved in both types of collisions.
More About Energy in Collisions
• Elastic collisions
• Kinetic energy is converted into potential energy and then back into kinetic energy
• So kinetic energy is conserved
• Inelastic collisions
• If the object does not return the kinetic energy to the system after the collision, the
collision is inelastic
• The kinetic energy after the collision is less than the kinetic energy before the collision
Elastic Collision Example
Recognize the Principle
• External forces are zero
• Total momentum is conserved
Sketch the problem
• Shown to the right
• Everything is along the x-axis
Elastic Collision Example, cont.
• Identify the relationships
• Elastic collision, so kinetic energy is conserved
• Equations:
• Solve for the unknowns
Elastic Collisions in 1-D
From conservation of momentum:
mAvA + mBvB = mAv´A + mBv´B

Because the collision is assumed elastic, KE is also conserved:
½ mAv²A + ½mBv²B = ½mAv´²A + ½mBv´²B .
From the first equation:
mA(vA - v´A) = mB (v´B - vB).
From the second equation:
mA(v²A - v´²A) = mB (v´²B - v²B)
Last equation:
/ gives:
or
mA(vA - v´A) (vA + v´A) = mB (v´B - vB) (v´B + vB). 
vA + v´A = v´B + vB ,
vA - vB = - (v´A - v´B)
Inelastic Collisions in 1-D
• In many collisions, kinetic energy is not conserved
• The KE after the collision is smaller than the KE before the collision
• These collisions are called inelastic
• The total energy of the universe is still conserved
• The “lost” kinetic energy goes into other forms of energy
• Momentum is conserved
• Momentum gives the following equation:
• Leaves two unknowns
Completely Inelastic Collisions
• In a completely inelastic
collision, the objects stick
together.
• They will have the same
velocity after the collision.
• Therefore, there is only one
unknown and the equation can
be solved.
Kinetic Energy in Inelastic Collisions
• Although kinetic energy is not conserved, total energy is conserved
• The kinetic energy is converted into other forms of energy
• These could include
•
Heat
•
Sound
•
Elastic potential energy
Inelastic Collisions - Example
Ballistic pendulum.
The ballistic pendulum is a device used to
measure the speed of a projectile, such as a
bullet. The projectile, of mass m, is fired into a
large block of mass M, which is suspended like a
pendulum. As a result of the collision, the
pendulum and projectile together swing up to a
maximum height h. Determine the relationship
between the initial horizontal speed of the
projectile, v, and the maximum height h.
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Example – Ballistic Pendulum
(a) Diagram of a ballistic pendulum. Notice that v1A is the velocity of the projectile
immediately before the collision and vB is the velocity of the projectile’s lock system
immediately after the perfectly inelastic collision. (b) Multiflash photograph of a ballistic
pendulum used in the laboratory.
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Collisions in Two or Three Dimensions
Conservation of energy and momentum can also be used to analyze collisions
in two or three dimensions, but unless the situation is very simple, the math
quickly becomes unwieldy.
Here, a moving object collides with an object initially at rest.
Knowing the masses and initial velocities is not enough; we need
to know the angles as well in order to find the final velocities.
Two-Dimensional Collision, example
• Particle 1 is moving at velocity
v1i
and
particle 2 is at rest.
• In the x-direction, the initial momentum is
m1v1i.
• In the y-direction, the initial momentum is 0.
Two-Dimensional Collision, example cont.
• After the collision, the momentum in the x-
direction is
m1v1f cos q + m2v2f cos f
• After the collision, the momentum in the y-
direction is
m1v1f sin q - m2v2f sin f
• The negative sign is due to the component of
the velocity being downward.
• If the collision is elastic, apply the kinetic energy
equation.
Problem Solving
 Recognize the principle
 The momentum of the system is conserved when the external forces are zero
 Sketch the problem
 Make a sketch of the system. Show the coordinate axes
 Show the initial and final velocities of the particles in the system
• Identify the relationships
• Write the conservation of momentum equation for the system
• Is the kinetic energy conserved?
•
If KE is conserved, then the collision is elastic
•
•
Write the kinetic energy equation for both particles
If KE is not conserved, then the collision is inelastic
•
Use the conservation of momentum equation
• Solve for the unknown(s)
• Check
• Consider what your answer means. Check that the answer makes sense
Example
Two objects are known to have
the same momentum. Do these
1) yes
two objects necessarily have the
2) no
same kinetic energy?
Example
Two objects are known to have
the same momentum. Do these
1) yes
two objects necessarily have the
2) no
same kinetic energy?
If object #1 has mass m and speed v and object #2 has
mass
1
2
m and speed 2v, they will both have the same
momentum. However, because KE =
1
2
mv2, we see
that object #2 has twice the kinetic energy of object #1,
due to the fact that the velocity is squared.
Example: Force of a tennis ser ve
For a top player, a tennis ball may leave the racket on the serve
with a speed of 55 m/s. If the ball has a mass of 0.060 kg and is
in contact with the racket for about 4 ms (4 x 10-3 s), estimate
the average force on the ball.
Solution:
Solution:
Fav = Δp/Δt = (mv2 - mv1 )/Δt
Fav = Δp/Δt = (mv2 - mv1 )/Δt
= 0.06 (55 – 0)/0.004  800 N.
= 0.06 (55 – 0)/0.004  800 N.
This is larger force, larger than the weight of a 70-kg person,
This is larger force, larger than the weight of a
which would require a force mg = (70)(9.8)  700N.
60-kg person, which would require a force mg =
(60)(9.8)  600N.
Example
The momentum of a particle in SI units, is given by
p = 4.8t² i – 8.0 j – 8.9t k. What is the force as a function of time?
Solution:
The force is the derivative of the momentum with respect to time.
F
dp
dt


d 4.8t 2 ˆi - 8.0 j - 8.9t k
dt

 9.6t ˆi - 8.9 k  N
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Example: Final Exam Fall 2015
A particle moving in the x direction feels a force of F(t) = 3t + 4 newtons
where t is in seconds. What is the momentum change between t = 0 and
t = 2 s?
a.
-2 Ns
b.
0 Ns
c.
6 Ns
d.
12 Ns
e.
14 Ns
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Example: Final Exam Fall 2015
A particle moving in the x direction feels a force of F(t) = 3t + 4
newtons where t is in seconds. What is the momentum change
between t = 0 and t = 2 s?
a.
-2 Ns
b.
0 Ns
c.
6 Ns
d.
12 Ns
e. 14 Ns
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Example: Rif le Recoil
Calculate the recoil velocity of a 5.0-kg
rifle that shoots a 0.020-kg bullet at a
speed of 620 m/s.
Solution:
momentum before = momentum after
mB vB + mR vR = mB v’B + mR v’R
0
+ 0
= mB v’B + mR v’R
v’R = - mB v’B / mR = - (0.020 kg)(620 m/s)/(5.0 kg) = - 2.5 m/s.
Example
An open cart rolls along a
frictionless track while it is
raining. As it rolls, what
happens to the speed of the
cart as the rain collects in it?
(Assume that the rain falls
vertically into the box.)
1) speeds up
2) maintains constant speed
3) slows down
4) stops immediately
Example
An open cart rolls along a
frictionless track while it is
raining. As it rolls, what
happens to the speed of the
cart as the rain collects in it?
(Assume that the rain falls
vertically into the box.)
Because the rain falls in vertically, it
adds no momentum to the box, thus
the box’s momentum is conserved.
However, because the mass of the
box slowly increases with the added
rain, its velocity has to decrease.
1) speeds up
2) maintains constant speed
3) slows down
4) stops immediately
Example
It is said that in ancient times a rich man with a bag of gold coins was stranded on
the surface of a frozen lake. Because the ice was frictionless, he could not push
himself to shore and froze to death. What could he have done to save himself had
he not been so miserly?
Response:
He could have thrown the coins in the direction opposite the shore he was trying to
reach. Since the lake is frictionless, momentum would be conserved and he would
“recoil” from the throw with a momentum equal in magnitude and opposite in
direction to the coins. Since his mass is greater than the mass of the coins, his
speed would be less than the speed of the coins, but, since there is no friction, he
would maintain this small speed until he hit the shore.
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Example
A child in a boat throws a 5.70-kg package out horizontally with a speed of
10 m/s. Calculate the velocity of the boat immediately after, assuming it was
initially at rest. The mass of the child is 24.0 kg and that of the boat is 35.0
kg.
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Example, cont.
The throwing of the package is a momentum-conserving action, if the water
resistance is ignored. Let A represent the boat and child together, and let B represent
the package. Choose the direction that the package is thrown as the positive
direction. Apply conservation of momentum, with the initial velocity of both objects
being 0.
pinitial  pfinal
vA  -
mBvB
mA
 mA + mB  v  mA vA + mBvB 
 5.70 kg  10.0 m s 
 -0.966 m s
 24.0 kg + 35.0 kg 

The boat and child move in the opposite direction as the thrown package, as
indicated by the negative velocity.
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Example: Final Exam Fall 2015
Automobile air bags are effective because they
a. increase impulse time
b. decrease the impulse
c. decrease the momentum change
d. decrease the time that the impulse acts
e. none of the above
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Example: Final Exam Fall 2015
Automobile air bags are effective because they
a.increase impulse time
b. decrease the impulse
c. decrease the momentum change
d. decrease the time that the impulse acts
e. none of the above
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Example
A golf ball of mass 0.045 kg is hit off the tee at a speed of 45 m/s. The golf club
was in contact with the ball for 3.5 x 10-3 s. Find
(a) the impulse imparted to the golf ball, and
(b) The average force exerted on the ball by the golf club.
SOLUTION: (a) The impulse is the change in momentum. The direction of travel
of the struck ball is the positive direction.
J = Δp = mΔv = (4.5 x 10-2 kg)(45 m/s – 0) = 2.0 kg m/s
in the forward direction.
(b) The average force is the impulse divided by the interaction time.
Fav = Δp/Δt = (2.0 kg m/s)/(3.5 x 10-3 s) = 580 N
In the forward direction.
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Example
When a bullet is fired
from a gun, the bullet
and the gun have equal
and opposite momenta.
If this is true, then why
is the bullet deadly
(whereas it is safe to
hold the gun while it is
fired)?
1) it is much sharper than the gun
2) it is smaller and can penetrate your body
3) it has more kinetic energy than the gun
4) it goes a longer distance and gains speed
5) it has more momentum than the gun
Example
When a bullet is fired
from a gun, the bullet
and the gun have equal
and opposite momenta.
If this is true, then why
is the bullet deadly
(whereas it is safe to
hold the gun while it is
fired)?
1) it is much sharper than the gun
2) it is smaller and can penetrate your body
3) it has more kinetic energy than the gun
4) it goes a longer distance and gains speed
5) it has more momentum than the gun
Even though it is true that the magnitudes of the momenta
of the gun and the bullet are equal, the bullet is less
massive and so it has a much higher velocity. Because KE
is related to v2, the bullet has considerably more KE and
therefore can do more damage on impact.
Example 6
If all three collisions below are
1) I
totally inelastic, which one(s)
2) II
will cause the most damage
3) III
(in terms of lost energy)?
4) II and III
5) all three
Example 6
If all three collisions below are
1) I
totally inelastic, which one(s)
2) II
will cause the most damage
3) III
(in terms of lost energy)?
4) II and III
5) all three
The car on the left loses
the same KE in all three
cases, but in case III, the
car on the right loses the
most KE because KE =
1
2
2 mv and the car in case III
has the largest velocity.
Example 7
You are lying in bed and you want to
shut your bedroom door. You have a
superball and a blob of clay (both with
the same mass) sitting next to you.
Which one would be more effective
to throw at your door to close it?
1) the superball
2) the blob of clay
3) it doesn’t matter—they
will be equally effective
4) you are just too lazy to
throw anything
Example 7
You are lying in bed and you want to
shut your bedroom door. You have a
superball and a blob of clay (both with
the same mass) sitting next to you.
Which one would be more effective
to throw at your door to close it?
1) the superball
2) the blob of clay
3) it doesn’t matter—they
will be equally effective
4) you are just too lazy to
throw anything
The superball bounces off the door with almost no loss of
speed, so its Dp (and that of the door) is 2mv.
The clay sticks to the door and continues to move along with
it, so its Dp is less than that of the superball, and therefore it
imparts less Dp to the door.
Example
A 0.060-kg tennis ball, moving with a speed of 4.50 m/s, has a head-on collision
with a 0.090-kg ball initially moving in the same direction at a speed of 3.00 m/s.
Assuming a perfectly elastic collision, determine the speed and direction of each
ball after the collision.
Solution:
Let A represent the 0.060-kg tennis ball, and let B represent the 0.090-kg ball.
The initial direction of the balls is the positive direction.
We have vA = 4.50 m/s and vB = 3.00 m/s.
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Example, cont.
vA - vB  -  vA - vB   vB  1.50 m s + vA
Substitute this relationship into the momentum conservation equation for the
collision.
mA vA + mBvB  mA vA + mBvB  mA vA + mBvB  mA vA + mB 1.50 m s + vA  
vA 
mA vA + mB  vB - 1.50 m s 
mA + m B

 0.060 kg  4.50 m s  +  0.090 kg  3.00 m s - 1.50 m s 
0.150 kg
 2.7 m s
vB  1.50 m s + vA  4.2 m s
Both balls move in the direction of the tennis ball’s initial motion.
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Example: Final Exam Spring 2016
A 1.0 kg model car has an initial velocity of 5 m/s in the +x-direction. It
collides with a model truck, 6.0 kg, which is moving in the -x-direction.
After the collision, the two vehicles stick together and move as one with
4.0 m/s in the -x-direction. Determine the speed of the model truck before
the collision.
a.
4 m/s
b.
5 m/s
c.
2 m/s
d.
8 m/s
e.
6 m/s
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Example: Final Exam Spring 2016
A 1.0 kg model car has an initial velocity of 5 m/s in the +x-direction. It
collides with a model truck, 6.0 kg, which is moving in the -x-direction.
After the collision, the two vehicles stick together and move as one with
4.0 m/s in the -x-direction. Determine the speed of the model truck before
the collision.
a.
4 m/s
b.
5 m/s
c.
2 m/s
d.
8 m/s
e. 6 m/s
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