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 p ,
and is defined as

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
  
  dv d ( mv ) dp
ΣF= 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.

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.
 

(
 dp d 4.8t i − 8.0 j − 8.9t k
2ˆ
) 
=
F =
dt dt
= ( )
9.6t ˆi − 8.9 k N

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Example - Force of a tennis serve

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.
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.

Example: Rifle 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.
Collisions and Impulse

During a collision, objects are

deformed due to the large forces
involved.

Since , we can

write

Integrating,
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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Example -

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 -

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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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.
 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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 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: mA(vA - v´A) (vA + v´A) = mB (v´B - vB) (v´B + vB). 

/ gives: vA + v´A = v´B + vB ,

or 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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 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 θ + m2v2f cos φ.

• After the collision, the momentum in the y-

direction is

m1v1f sin θ − m2v2f sin φ.

• 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
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.
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 → ( mA + mB ) v = mA vA′ + mBvB′ →

mBvB′ ( 5.70 kg ) (10.0 m s )
vA′ =
− =
− = −0.966 m s
mA ( 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:
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

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Example:
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

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 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′ ) → vB′ =

vA − vB = 1.50 m s + vA′
Substitute this relationship into the momentum conservation equation for the collision.

NOTE: SUBSTITUTE 𝑽𝑽’𝑩𝑩 ABOVE INTO THE PLACE OF 𝑽𝑽’𝑩𝑩 BELOW. THE SOLVE FOR 𝑽𝑽’𝑨𝑨 .
AFTER FINDING 𝑽𝑽’𝑨𝑨 SUBSTITUTE 𝑽𝑽’𝑨𝑨 INTO THE EQUATION ABOVE TO FIND 𝑽𝑽’𝑩𝑩

mA vA + mBvB = mA vA′ + mBvB′ → mA vA + mBvB = mA vA′ + mB (1.50 m s + vA′ ) →

mA vA + mB ( vB − 1.50 m s ) ( 0.060 kg )( 4.50 m s ) + ( 0.090 kg )( 3.00 m s − 1.50 m s )
vA′ =
mA + m B 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. 39
 Example:
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:
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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