+

Momentum and CollisionsChapter 8

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Momentum and Impulse

+What do you think?

Imagine an automobile collision in which an older model car from the 1960s collides with a car at rest while traveling at 15 mph. Now imagine the same collision with a 2007 model car. In both cases, the car and passengers are stopped abruptly.List the features in the newer car that are

designed to protect the passenger and the features designed to minimize damage to the car.

How are these features similar?

+What do you think?

What are some common uses of the term momentum?Write a sentence or two using the

term momentum.

Do any of the examples provided reference the velocity of an object?

Do any of the examples reference the mass of an object?

+Momentum

Momentum (p) is proportional to both mass and velocity.

A vector quantity, so direction matters!!Things moving right are positiveTings moving left are negative

SI Units: kg • m/s

+Example

A 2250kg pick-up truck has a velocity of 25m/s to the east. What is the momentum of the truck?

m= 2250kg v=25m/s east (+)

P= mv

P= (2250)(25)

P= 56,250 kg • m/s

+Momentum and Newton’s 2nd Law

Prove that the two equations shown below are equivalent.F = ma and F = p/t

Newton actually wrote his 2nd Law as F = p/t.Force depends on how rapidly the

momentum changes.

+Impulse and Momentum

The quantity Ft is called impulse.SI units: N•m or kg•m/s

Impulse equals change in momentum.Another version of Newton’s 2nd LawChanges in momentum depend on both

the force and the amount of time over which the force is applied.

+Changing momentum

Greater changes in momentum (p) require more force (F) or more time (t) .

A loaded truck requires more time to stop. Greater p for truck with

more mass Same stopping force

+Movie

file:///Volumes/Physics_mac/inquiry_ppts/files/ch06/70598.html

Put Advanced CD in to work

+Example

A 1400kg car moving westward with a velocity of 15m/s collides with a utility pole and is brought to rest in 0.3sec. Find the force exerted on the car during collision.

Given:

m=1400kg t= 0.30sec F=??

vi= 15m/s west (-) vf= 0m/s

+Example

Ft= mvf- mvi

F=70,000N to the east

f imv mvF

t

f imv mvF

t

(1400)(0) (1400)( 15)

0.30F

+Example

A 2240kg car traveling west slows down uniformly from 20m/s to 5m/s. how long does it take the car to slow down if the force on the car is 8410N to the east? How far does the car travel during the time it slows down?

+Example

Given:

m= 2240kg

vi= 20m/s west (-)

vf= 5m/s west (-)

F= 8410N east (+)

t= ?? d=??

t= 4 sec

f imv mvt

F

(2240)( 5) (2240)( 20)

8410t

+Example

d= ½ (vi + vf)t

d= ½ (-20-5) 4

d= -50 m/s or 50 m/s west

+Stopping Time

Ft = p = mv

When stopping, p is the same for rapid or gradual stops.

Increasing the time (t) decreases the force (F).What examples demonstrate this relationship?

Air bags, padded dashboards, trampolines, etc

Decreasing the time (t) increases the force (F).What examples demonstrate this relationship?

Hammers and baseball bats are made of hard material to reduce the time of impact.

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Conservation of Momentum

+What do you think?

• Two skaters have equal mass and are at rest. They are pushing away from each other as shown.• Compare the forces on the two girls.• Compare their velocities after the push.

• How would your answers change if the girl on the right had a greater mass than her friend?

• How would your answers change if the girl on the right was moving toward her friend before they started pushing apart?

+Momentum During Collisions

The change in momentum for one object is equal and opposite to the change in momentum for the other object.

Total momentum is neither gained not lost during collisions.

+Conservation of Momentum

Total momentum remains constant during collisions.

The momentum lost by one object equals the momentum gained by the other object.

Conservation of momentum simplifies problem solving.

+Example

A 76kg boater, initially at rest, in a stationary 45kg boat steps out of the boat onto the dock. If the boater moves out of the boat with a velocity of 2.5m/s to the right, what is the final velocity of the boat?

+Given

m1v1i + m2v2i= m1v1f + m2v2f

(76)(0)+(45)(0)=(76)(2.5)+(45)(v2f)

0= 190 + 45v2f

-190 = 45v2f

v2f= -4.2 m/s to the right but notice the negative

v2f= 4.2 m/s to the left

m1= 76kg

v1i= 0m/s

v1f= 2.5 m/s

m2= 45kg

v2i= 0m/s

v2f= ??

+Now what do you think?

• Two skaters have equal mass and are at rest. They are pushing away from each other as shown.• Compare the forces on the two girls.• Compare their velocities after the push.

• How would your answers change if the girl on the right had a greater mass than her friend?

• How would your answers change if the girl on the right was moving toward her friend before they started pushing apart?

+

Elastic and Inelastic Collisions

+What do you think?

• Collisions are sometimes described as elastic or inelastic. To the right is a list of colliding objects. Rank them from most elastic to most inelastic.

• What factors did you consider when ranking these collisions?

1. A baseball and a bat

2. A baseball and a glove

3. Two football players

4. Two billiard balls

5. Two balls of modeling clay

6. Two hard rubber toy balls

7. An automobile collision

+Collisions

Perfectly inelastic collisions A collision where 2 objects stick

together after colliding

Perfectly inelastic collisions are analyzed in terms of momentumDue to the fact that the objects

become one object after the collision

+Perfectly Inelastic Collisions

Two objects collide and stick together. Two football players A meteorite striking the earth

Momentum is conserved.

Masses combine.

+Perfectly Inelastic Collisions

v1i is + (m1 is moving to the right)

V2i is – (m2 is moving to the left)

+Example

A 1850kg luxury sedan stopped at a traffic light is struck from the rear by a compact car with a mass of 975kg. Te 2 cars become entangled as a result of the collision. If the compact car was moving at a velocity of 22m/s to the north before the collision, what is the velocity of the mass after the accident?

+Example

m1v1i + m2v2i= (m1+ m2) vf

vf= 7.59 m/s north

m1= 1850kg

v1i= 0m/s

m2= 975kg

v2i= 22m/s north (+)

vf= ??

(1850)(0) (975)(22)

1850 975fv

+Classroom Practice Problem

Gerard is a quarterback and Tyler is a defensive lineman. Gerard’s mass is 75.0 kg and he is at rest. Tyler has a mass of 112 kg, and he is moving at 8.25 m/s when he tackles Gerard by holding on while they fly through the air. With what speed will the two players move together after the collision?

Answer: 4.94 m/s

+Classroom Practice Problems

An 2.0 x 105 kg train car moving east at 21 m/s collides with a 4.0 x 105 kg fully-loaded train car initially at rest. The two cars stick together. Find the velocity of the two cars after the collision.Answer: 7.0 m/s to the east

+Inelastic Collisions

Kinetic energy is less after the collision.It is converted into other forms of

energy.Internal energy - the temperature is

increased.Sound energy - the air is forced to

vibrate.

Some kinetic energy may remain after the collision, or it may all be lost.

+Example

2 clay balls collide in a perfect inelastic collision. The 1st ball has a mass of 0.500kg and an initial velocity of 4.0m/s to the right. The 2nd ball has a mass of 0.250 kg. and an initial velocity of 3.0m/s to the left. What is the decrease in kinetic energy during the collision?

+Example

m1= 0.500 kg

v1= 4m/s right (+)

m2= 0.250kg

v2= 3m/s left (-)

KE=

KE= KEf – KEi

KEi= ½ m1v1i + m2v2i

KEf= ½ (m1+m2)vf2

+Example

m1v1i + m2v2i= (m1+ m2) vf

vf= 1.67 m/s

KEi= ½ (0.5)(42) + ½ (0.25)(-32)= 5.125J

KEf= ½ (0.5 + 0.25) 1.672 = 1.05J

KE= 1.05 – 5.125 = -4.08J Means that KE is lost

1 1 2 2

1 2

(0.5)(4) (0.25)( 3)

( ) (0.5 0.25)i im v m v

vfm m

+Elastic Collisions Objects collide and return to their original

shape.

Kinetic energy remains the same after the collision.

Perfectly elastic collisions satisfy both conservation laws shown below.

+Elastic Collisions

Two billiard balls collide head on, as shown. Which of the following possible final velocities satisfies the law of conservation of momentum? vf,A = 2.0 m/s, vf,B = 2.0 m/s

vf,A = 0 m/s, vf,B = 4.0 m/s

vf,A = 1.5 m/s, vf,B= 2.5 m/s

Answer: all three

m = 0.35 kg m = 0.35 kg

v = 4.0 m/sv= 0 m/s

+Elastic Collisions

Two billiard balls collide head on, as shown. Which of the following possible final velocities satisfies the law of conservation of kinetic energy? vf,A = 2.0 m/s, vf,B = 2.0 m/s

vf,A = 0 m/s, vf,B = 4.0 m/s

vf,A = 1.5 m/s, vf,B= 2.5 m/s

Answer: only vf,A = 0 m/s, vf,B = 4.0 m/s

m = 0.35 kg m = 0.35 kg

v = 4.0 m/sv= 0 m/s

+Example

A 0.015 kg marble moving to the right at 0.225 m/s makes an elastic head on collision with a 0.030 kg shooter marble moving to the left at 0.180m/s. After the collision, the smaller marble moves to the left at 0.315m/s. Assume neither marble rotates at all and both marbles are on a frictionless surface. What is the velocity of the 0.030kg marble after the collision?

+Example

m1v1i + m2v2i= m1v1f + m2v2f

(0.015)(0.225) + (0.030)(-0.18) = (0.015)(-0.315) + (0.030)(vf)

0.003375 + -0.0054 = -0.004725 + 0.030 vf

-0.002025= - 0.004725 + 0.030 vf

0.0027 = 0.030vf

vf= 0.09 m/s right

m1= 0.015kg

v1i= 0.225m/s right (+)

v1f= 0.315m/s left (-)

m2= 0.030kg

v2i= 0.18m/s left (-)

vf= ??

+

Click below to watch the Visual Concept.

Types of Collisions

file:///Volumes/Physics_mac/inquiry_ppts/files/ch06/70600.html

Put Advanced CD in to work

+Types of Collisions

+Now what do you think?

• To the right is a list of colliding objects. Rank them from most elastic to most inelastic.

• What factors did you consider when ranking these collisions?

1. A baseball and a bat

2. A baseball and a glove

3. Two football players

4. Two billiard balls

5. Two balls of modeling clay

6. Two hard rubber toy balls

7. An automobile collision