MomentumMomentumMomentum is defined as the product of themass × velocity of an object.

p = m × v

where m is the mass in kg, v is the velocity

in m/s, and the momentum, p, in kg•m/s.

Momentum is a vector quantity meaning that it

has magnitude and direction.

The direction of momentum is that of the velocity

of the object.From Newton’s 2nd Law:

Fnet = ma

a = Δv

Δt

Therefore, Fnet= mΔvΔt

FnetΔt = mΔvFnetΔt is called impulse where Fnet is

measured inN and Δt measured in s.

FnetΔt = N•s Impulse is a vector quantity and its

direction is that of Fnet.

Impulse and MomentumImpulse and MomentumMomentum can be thought of as “inertia inmotion”.

When a car on Thurber’s Avenue crashes into a

guard rail, a large force is exerted on the guard

rail.

The force of impact comes from a change

in velocity or deceleration of the car.

The force of impact comes from a change

in velocity or deceleration of the car.

The force of impact is directly proportional

to the change in velocity of the moving car and to its mass.

Fnet α m × v

To change the momentum of an object, youmust consider the impulse.

Impulse depends on the magnitude of the

force and the time of impact (time the objects are in physical contact).

Hitting a golf ball for distance (teeing off), throwing a baseball or football for distancerequires a “follow through”.

These situations call for the largest change in momentum (mΔv) which is achieved by using a large force for as long as possible (FΔt).

When receiving a punch, it is best to “roll withthe punch”.

By rolling with the punch, you extend the time of impact which minimizes Fnet.

Fnet = mΔv/Δt

The same change in momentum will occur in either case but it is best to maximize

the value of Δt rather then the Fnet.

When throwing a punch, you try to pull your fist

back as quickly as possible.

By pulling back your fist as quickly as possible, it minimizes the Δt which maximizes the Fnet.

Didn’t you use this idea as a youngster jumping

from a high height?

Momentum ProblemsMomentum ProblemsA 0.144 kg baseball with a velocity of 7.5 m/sstrikes a wall perpendicularly and rebounds at8.9 m/s.

m = 0.144 kgv1 = 7.5 m/s v2 = -8.9 m/s

(a) What is the change in momentum of the baseball?

Δmv = mΔv = 0.144 kg × (7.5 m/s – -8.9 m/s) = 2.4 kg•m/s2

(b) If the force of impact acted 0.045 s, what was the average force on the wall?

(c) What average force is exerted on the baseball?

By Newton’s 3rd Law, 53 N.

Fnet = mΔvΔt

2.4 kg•m/s2Fnet = 0.045 s = 53 N

A 26 kg boy is riding on a wagon going at3.0 m/s N. The boy jumps off the back of

awagon in such a way that he has no

velocitywhen he lands on the ground. The mass

of thewagon is 10. kg.

mb = 26 kg mw = 10. kgvb = 3.0 m/s Nvb’ = 0 m/s

(a) What was the boy's change in momentum?

Δpb = mΔv = 26 kg × -3.0 m/s = -78 kg•m/s S

(b) What was the wagon's change in momentum?

Δpsystem = 0, therefore the change in momentum of the wagon is 78 kg•m/s N.

(c) How fast does the wagon go after the boy jumps off?

Δpw = mΔv

Δv = Δpw/m = 78 kg•m/s/10. kg = 7.8 m/s N

vf = vi + Δv = 3.0 m/s + 7.8 m/s = 10.8 m/s N

Conservation of MomentumConservation of MomentumWhen studying the momentum changes incollisions, you must focus on a closed andisolated system.

A system is defined as a collection of objects.

A system is closed if no objects enter or leave the system.

A system is isolated if no external forces are exerted on it.

Consider a game of pool when there is only the

8-ball and the cue ball.

The system is closed because both balls only interact with each other.

The system is isolated because rolling friction (µk = 0) can be ignored.

8

AB

pApB

where mA = mB, vA > vB, and pA > pB

Assume neither ball is spinning (no englishor angular momentum) because that wouldcomplicate matters.

At the instant of impact.

8

FAFB

8

After impact.

vA = 0vB > 0

The Conservation of Momentum states that in aclosed and isolated system, the momentumbefore the collision equals the momentum afterthe collision.

Δp = 0 pi = pf

Momentum may be redistributed in a collision but Δpsystem = 0.

Internal forces are forces between objects within the system.

External forces are exerted by objects outside a system.

Momentum is conserved only when internal forces are present.

Conservation of Momentum Conservation of Momentum ProblemsProblems

A 7450 kg truck traveling at 4.9 m/s 090°collides with a 1575 kg car traveling at 21

m/s240°. After colliding, the truck and car

movetogether. What is their velocity after thecollision?

m1 = 7450 kg m2 = 1575 kgv1 = 4.9 m/s 090° v2 = 21 m/s 240°

.

m1v1

m 2v 2

θ1

θ2mRvR

θ1 = 30°θ2 = ?

mRvR = ((m1v1)2 + (m2v2)2 - 2×m1v1×m2v2cosθ)½

mRvR = ((7450 kg × 4.9 m/s)2 + (1575 kg × 21 m/s)2

- 2 × 7450 kg × 4.9 m/s × 1575 kg × 21 m/s × cos30°)½

.mRvR = 1.8 x 104 kg•m/svR = 2.0 m/s

sinθ1 mRvR

= sinθ2 m2v2

θ2 =67°

sin30° = sinθ2 1575 kg × 21 m/s1.8 x 104 kg•m/s

vR = 2.0 m/s, 157°

Two angry hockey players push each other.The mass of the first hockey player is 70. kgand the second has a mass of 50. kg. If the firsthockey player has a velocity of 6.2 m/s N, howfar apart are they after 3.0 s? Assume friction isnegligible.

m1 = 70.kg m2 = 50. kgv1 = 6.2 m/s N v2 = ?Δt = 3.0 s µ = 0

Δp = 0pi = pf

m1v1 + m2v2 = m1v1’ + m2v2’

0 = m1v1’ + m2v2’

0 = 70. kg × 6.2 m/s + 50. kg × v2’

V2’ = -8.7 m/s = 8.7 m/s S

ΔxT = Δx1 + Δx2 = v1 × Δt1 + v2 × Δt2

ΔxT = 6.2 m/s × 3.0 s + 8.7 m/s × 3.0 s

ΔxT = 45 m

Conservation of Kinetic EnergyConservation of Kinetic EnergyKinetic energy is the energy due to the

motionof an object.

KE = ½mv2

where m is the mass in kg, v is the velocity in m/s, and KE is the kinetic energy in J (joules).

Kinetic energy is a scalar quantity.

The Law of Conservation of Energy states thatwithin a closed and isolated system, energy canchange form but the total amount of energy isconstant.

ΔE = 0

When two objects collide, the force each objectexerts on the other slightly change the shape ofthe object.

KE PE

Momentum And Energy Momentum And Energy TogetherTogether

These two conservation laws, momentum andenergy, are both needed to give a “completepicture” of what happens in any collision.

The conservation of momentum gives the final momentum after the collision.

What it does not tell us is the distribution between mass and velocity.

The conservation of kinetic energy tells us how

the distribution between mass and velocity by considering the elastic properties of the

collidingobjects.

The conservation of energy puts an upper limit of the velocity of the impacted

object and how close you get to this upper limit depends on the elasticity.

Remember the three types of collisions:

Elastic Collisions

After the collision, the objects separate.

Both momentum and kinetic energy are

conserved.

Perfectly Inelastic Collisions

After the collision, the objects stick together and share a common

velocity. Only momentum is conserved.

Inelastic Collisions

The objects deform during the collision so

the total kinetic energy decreases but the

objects separate after the collision.

Only momentum is conserved.

A 600. mg popcorn kernel feeling the heat is

moving around at 7.4 cm/s before it pops and

breaks into two pieces of equal mass. If one

piece comes to an abrupt halt, determine the

change in kinetic energy.

m1 = 600. mg m1’ = 300. mg v1’ = v1’v1 = 7.4 cm/s m2’ = 300. mg v2’ = 0

Δp = 0pi = pf

m1v1 = m1v1’ + m2v2’

600. mg × 7.4 m/s = 300. mg × v1’

v1’ = 14.8 m/s

ΔKE = 0KEi = KEf

½m1v12 = ½m1v1’2 + ½m2v2’2

½ × (600. mg× 1 g103 mg

× ×1 kg103 g

(7.4 cms × 1 m

102 cm)2 = 1.6 x 10-6 J

(14.8 cms × 1 m

102 cm )2 = 3.3 x 10-6 J

½ × (300. mg× 1 g103 mg

× ×1 kg103 g

ΔKE = KEf – KEi = 3.3 x 10-6 J - 1.6 x 10-6 J

ΔKE = KEf – KEi = 1.6 x 10-6 J

A Toyota with a mass of 575 kg moving at15 m/s collides into the rear of a Ford pick-upwhich has a mass of 1575 kg and is moving

at5.0 m/s. The two vehicles lock together andcontinue sliding forward.

(a) What is their final velocity?

m1 = 575 kg m2 = 1575 kg v1 = 15 m/s v2 = 5.0 m/s v1’ = ? v2’ = ?

Δp = 0pi = pf

m1v1 + m2v2 = (m1+m2)v’

575 kg × 15 m/s + 1575 kg × 5.0 m/s = (575 kg + 1575 kg)

× v’

v’ = 7.7 m/s

(b) How much kinetic energy was “lost” in the collision?

ΔKE = 0KEi = KEf

½m1v12 + ½m2v2

2 = ½(m1 + m2)v’

½ × 575 kg × (15 m/s)2 + ½ × 1575 kg × (5.0 m/s)2 = ½ × (575 kg + 1575

kg) × (7.7 m/s)2

84000 J ≠ 64000 J

ΔKE = 84000 J – 64000 J = 2.0 x 104 J

Therefore, 2.0 x 104 J were converted into heatand sound which represent a wasted form of energy.

A 15 g bullet is shot into a wooden block whichhas a mass of 5085 g. The block is at rest on ahorizontal surface. After the bullet becomesimbedded in the block, the two move at 1.0 m/s.Determine the velocity of the bullet beforestriking the block.

mb = 15 g mw = 5085 gvb = ? vw = 0vb’ = 1.0 m/s vw’ 1.0 m/s

Wrap Up QuestionsWrap Up QuestionsIf two particles have equal kinetic

energies, aretheir momenta always equal?

No, the momenta would only be equal if their

masses were equal.

Does a large force always produce a largerimpulse than a smaller force? Justify youranswer.

The impulse, FnetΔt, is dependent on both thenet force and the time interval the net forceacts.

The statement would be true only if the timeintervals were equal.

The cue ball collides with the 8-ball which isinitially at rest. Is it possible for both the cue ball and 8-ball to be at rest immediately after the collision?

The conservation of momentum (Δp = 0)prohibits this from happening. If the system(the cue ball and the 8-ball) had momentumbefore the collision, then there has to be thesame momentum before the collision.

Would it be possible for one ball to be at rest

after the collision?

Yes, whenever two equal masses experience a

head-on elastic collision, the stationary ball will

have all of the momentum after the collision.

Is momentum conserved when a basketballbounces off the floor?

Yes, if the system is considered to be both the

basketball and the earth. However, due to the

mass of the earth, its momentum would beundetectable.

Too understand why the earth must beincluded, take a look at the next slide.

Assume the ball moving downward to have a

positive velocity, then the ball moving upward

would have a negative velocity. m1v1 -m1v1’

Δp = 0pi = pf

m1v1 = - m1v2

How can v1 = - v2?

Can a collision occur such that all of the kinetic

energy is lost?

Yes, provided that equal masses areapproaching each other with the same

speeds,such that the momentum before the collision

isequal to zero.

If the collision is perfectly inelastic, then thekinetic energy is converted into elastic

potentialenergy which deforms the objects.