Chapter 9B - Conservation of Chapter 9B - Conservation of MomentumMomentum

A PowerPoint Presentation byA PowerPoint Presentation by

Paul E. Tippens, Emeritus Paul E. Tippens, Emeritus ProfessorProfessor

Southern Polytechnic State Southern Polytechnic State UniversityUniversity

A PowerPoint Presentation byA PowerPoint Presentation by

Paul E. Tippens, Emeritus Paul E. Tippens, Emeritus ProfessorProfessor

Southern Polytechnic State Southern Polytechnic State UniversityUniversity© 2007

Momentum is conserved in this rocket launch. The

velocity of the rocket and its

payload is determined by the mass and velocity of the

expelled gases. Photo:

NASA

NASA

Objectives: After completing Objectives: After completing this module, you should be this module, you should be able to:able to:• State the law of conservation of State the law of conservation of

momentum and apply it to the solution of momentum and apply it to the solution of problems.problems.

• Distinguish by definition and example Distinguish by definition and example between elastic and inelastic collisions.between elastic and inelastic collisions.

• Predict the velocities of two colliding Predict the velocities of two colliding bodies when given the coefficients of bodies when given the coefficients of restitution, masses, and initial velocities.restitution, masses, and initial velocities.

A Collision of Two MassesA Collision of Two MassesWhen two masses m1 and m2 collide, we will

use the symbol u to describe velocities before collision.

The symbol v will describe velocities after collision.

BeforeBeforem1

u1m2

u2

m1

v1 m2

v2AfterAfter

A Collision of Two BlocksA Collision of Two Blocks

m1 Bm2

“u”= Before “v” = After

m1

u1m2

u2BeforeBefore

m2v2m1

v1AfterAfter

Collision

Conservation of EnergyConservation of Energy

m1 m2

u1u2

The kinetic energy beforebefore colliding is equal to the kinetic energy afterafter colliding plus the energy lostlost in the collision.

2 2 2 21 1 1 11 1 2 2 1 1 2 22 2 2 2m u m u m v m v Loss 2 2 2 21 1 1 1

1 1 2 2 1 1 2 22 2 2 2m u m u m v m v Loss

Example 1.Example 1. A A 2-kg2-kg mass moving at mass moving at 4 m/s4 m/s collides with a collides with a 1-kg1-kg mass initially at rest. mass initially at rest. After the collision, the 2-kg mass moves After the collision, the 2-kg mass moves at at 1 m/s1 m/s and the 1-kg mass moves at and the 1-kg mass moves at 3 3 m/sm/s. What energy was lost in the . What energy was lost in the collision?collision?

It’s important to draw and label a sketch It’s important to draw and label a sketch with appropriate symbols and given with appropriate symbols and given

information.information.

m2

u2 = 0

m1

u1 = 4 m/s

m1 = 2 kg m1 = 1 kg

BEFOREBEFORE

m2

v2 = 2 m/s

m1

v1 = 1 m/s

m1 = 2 kg m1 = 1 kg

AFTERAFTER

Example 1 (Continued).Example 1 (Continued). What What energy was lost in the collision? energy was lost in the collision? Energy is conserved.Energy is conserved.

m2

uu22 = 0= 0

m1

uu1 1 = = 4 m/s4 m/s

mm1 1 = = 2 kg2 kg mm1 1 = = 1 kg1 kg

m2

vv22 = = 2 m/s2 m/s

m1

vv1 1 = = 1 m/s1 m/s

mm1 1 = = 2 kg2 kg mm1 1 = = 1 kg1 kg

BEFORBEFORE:E:

2 2 21 1 11 1 2 22 2 2 (2 kg)(4 m 0 16 J/s)m u m u

2 2 2 21 1 1 11 1 2 22 2 2 2(2 kg)(1 m/s) (1 kg)(2 m/s) 3 Jm v m v AFTERAFTER

Energy Conservation: K(Before) = K(After) + Loss

Loss = 16 J – 3 J Energy Loss = 15 J

Energy Loss = 15 J

Impulse and MomentumImpulse and Momentum

A BuA

uB

A BvA vB

B--FFAA tt FFB B tt

Opposite but Equal F t

Ft = mvf– mvo

FBt = -FAt

Impulse = p

mBvB - mBuB = -(mAvA - mAuA)

mAvA + mBvB = mAuA + mBuBmAvA + mBvB = mAuA + mBuBSimplifying:

Conservation of Conservation of MomentumMomentum

A BuA

uB

A BvA vB

B--FFAAtt FFB B tt

The total momentum AFTER a collision is equal to the total momentum BEFORE.

Recall that the total energy is also

conserved:

KKA0A0 + K + KB0B0 = K = KAfAf + K + KBfBf + Loss + LossKKA0A0 + K + KB0B0 = K = KAfAf + K + KBfBf + Loss + Loss

Kinetic Energy: K = ½mvKinetic Energy: K = ½mv22

mAvA + mBvB = mAuA + mBuBmAvA + mBvB = mAuA + mBuB

Example 2:Example 2: A A 2-kg2-kg block block A A and a and a 1-kg1-kg block block BB are pushed together against a are pushed together against a spring and tied with a cord. When the spring and tied with a cord. When the cord breaks, the cord breaks, the 1-kg1-kg block moves to block moves to the right at the right at 8 m/s8 m/s. What is the . What is the velocity of the velocity of the 2 kg2 kg block? block?

A B

The initial velocities are The initial velocities are zerozero, so that the total , so that the total

momentum momentum beforebefore release is zero.release is zero.

mmAAvvAA + m + mBBvvBB = m = mAAuuAA + m + mBBuuBB0 0

mAvA = - mBvB vA = - mBvB

mA

Example 2 (Continued)Example 2 (Continued)

mmAAvvAA+ m+ mBBvvBB = m = mAAuuAA + m + mBBuuBB0 0

mAvA = - mBvB vA = - mBvB

mA

A B

2 kg1 kg A B

8 m/svA2

vA = - (1 kg)(8 m/s)

(2 kg)vA = - 4 m/svA = - 4 m/s

Example 2 (Cont.): Example 2 (Cont.): Ignoring friction, Ignoring friction, how much energy was released by the how much energy was released by the spring?spring?

A B

2 kg1 kg A B

8 m/s4 m/s

Cons. of E: ½½kxkx22 = = ½½ mmAAvvAA + ½ + ½mmBBvvBB2222

½½kxkx2 2 = ½= ½(2 kg)(4 m/s)(2 kg)(4 m/s)22 + ½(1 kg)(8 m/s) + ½(1 kg)(8 m/s)22

½½kxkx2 2 = 16= 16 J + 32 J = 48 J J + 32 J = 48 J ½kx2 = 48 J½kx2 = 48 J

Elastic or Inelastic?Elastic or Inelastic?

An elastic collision loses no energy. The deform-ation on collision is fully restored.

In an inelastic collision, energy is lost and the deformation may be permanent. (Click it.)

Completely Inelastic Completely Inelastic CollisionsCollisions

Collisions where two objects stick together and have a common velocity

after impact.

Collisions where two objects stick together and have a common velocity

after impact.

Before After

Example 3:Example 3: A A 60-kg60-kg football player football player stands on a frictionless lake of ice. He stands on a frictionless lake of ice. He catches a catches a 2-kg2-kg football and then football and then moves at moves at 40 cm/s40 cm/s. What was the . What was the initial velocity of the football?initial velocity of the football?

Given: uB= 0; mA= 2 kg; mB= 60 kg; vA= vB= vC vC = 0.4 m/s

AA

BB

mmAAvvAA + m + mBBvvBB = m = mAAuuAA + m + mBBuuBBMomentum:0

(m(mAA + m + mBB)v)vCC = m = mAAuuAA

(2 kg + 60 kg)(0.4 m/s) = (2 kg)uA

Inelastic collision:

uuAA= 12.4 m/s= 12.4 m/s uuAA= 12.4 m/s= 12.4 m/s

Example 3 (Cont.):Example 3 (Cont.): How much How much energy was lost in catching the energy was lost in catching the football?football?

0

½(2 kg)(12.4 m/s)2 = ½(62 kg)(0.4 m/s)2 + Loss

154 J = 4.96 J + Loss Loss = 149 JLoss = 149 JLoss = 149 JLoss = 149 J

97% of the energy is lost in the collision!!

2 2 21 1 12 2 2 ( ) LossA A B B A B Cm u m u m m v

General: Completely General: Completely InelasticInelastic

Collisions where two objects stick together and have a common velocity vC

after impact.Conservation of Momentum:Conservation of Momentum:

Conservation of Conservation of Energy:Energy:

( )A B c A A B Bm m v m u m u ( )A B c A A B Bm m v m u m u

2 2 21 1 12 2 2 ( )A A B B A B cm u m u m m v Loss 2 2 21 1 1

2 2 2 ( )A A B B A B cm u m u m m v Loss

Example 4.Example 4. An An 87-kg87-kg skater skater BB collides with a collides with a 22-kg22-kg skater skater AA initially at rest on ice. They initially at rest on ice. They move together after the collision at move together after the collision at 2.4 m/s2.4 m/s. . Find the velocity of the skater Find the velocity of the skater BB before the before the collision.collision.

AABB

uuBB = ?= ?uuAA = 0= 0

Common speed Common speed after colliding: after colliding: 2.4 2.4

m/s.m/s.

22 kg22 kg

87 kg87 kg( )A A B B A B Cm u m u m m v ( )A A B B A B Cm u m u m m v

vvBB= v= vA A = v= vCC = = 2.4 m/s2.4 m/s

(87(87 kg)kg)uuBB = (87 kg + 22 kg)(2.4 = (87 kg + 22 kg)(2.4

m/s)m/s)(87 kg)(87 kg)uuBB =262 kg =262 kg

m/sm/s

uB = 3.01 m/s

Example 5:Example 5: A A 50 g50 g bullet strikes a bullet strikes a 1-1-kgkg block, passes all the way block, passes all the way through, then lodges into the through, then lodges into the 2 kg2 kg block. Afterward, the 1 kg block block. Afterward, the 1 kg block moves at moves at 1 m/s1 m/s and the and the 2 kg2 kg block block moves at moves at 2 m/s2 m/s. What was the . What was the entrance velocity of the bullet?entrance velocity of the bullet?

2 kg1 kg1 m/s 2 m/s

1 kg 2 kguA= ?

2 kg1 kg1 m/s 2 m/s

1 kg 2 kgFind entrance velocity of bullet: mA= 0.05 kg; uA= ?

(0.05 kg)uuAA =(1 kg)(1 m/s)+(2.05 kg2.05 kg)(2 m/s)

mAuA + mBuB + mCuC = mBvB + (mA+mC) vAC

Momentum After = Momentum

Before

50 gA C

B

0 0

(0.05 kg) uuAA =(5.1 kg m/s)

uA= 102 m/s uA= 102 m/s

Completely Elastic Completely Elastic CollisionsCollisions

Collisions where two objects collide in such a way that zero energy is lost in the

process.

APPROXIMATIONS!APPROXIMATIONS!

Velocity in Elastic Velocity in Elastic CollisionsCollisions

A B

A B

uBuA

vA vB

1. Zero energy lost.

2. Masses do not change.

3. Momentum conserved.

(Relative v After) = - (Relative v Before)

Equal but opposite impulses (F t) means that:

For elastic collisions: vA - vB = - (uA - uB)vA - vB = - (uA - uB)

Example 6:Example 6: A A 2-kg2-kg ball moving to ball moving to the right at the right at 1 m/s1 m/s strikes a strikes a 4-kg4-kg ball ball moving left at moving left at 3 m/s3 m/s. What are the . What are the velocities after impact, assuming velocities after impact, assuming complete elasticity?complete elasticity?

A B

A B

3 3 m/sm/s

1 m/s1 m/s

vvAA vvBB1 kg1 kg 2 kg2 kg

vvAA - v - vBB = - (u = - (uAA - u - uBB))

vvA A - v- vBB = u = uBB - u - uAA

vvAA - v - vBB = (-3 m/s) - (1 m/s)

From conservation of energy (relative v):

vA - vB = - 4 m/s vA - vB = - 4 m/s

Example 6 (Continued)Example 6 (Continued)

A B

A B

3 3 m/sm/s1 m/s1 m/s

vvAA vvBB1 kg1 kg 2 kg2 kg

mmAAvvAA + m + mBBvvBB = m = mAAuuAA + m + mBBuuBB

Energy: Energy: vvAA - v - vBB = = - 4 m/s- 4 m/s

(1 kg)vvAA+(2 kg)vvBB=(1 kg)(1 m/s)+(2 kg)(-3 m/s)

vvAA + 2v + 2vBB = -5 m/s

Momentum also conserved:

vvAA - v - vBB = = - 4 m/s- 4 m/s

Two independent equations to

solve:

Example 6 (Continued)Example 6 (Continued)

A B

A B

3 3 m/sm/s1 m/s1 m/s

vvAA vvBB1 kg1 kg 2 kg2 kg

vA + 2vB = -5 m/s

vvAA - v - vBB = = - 4 m/s- 4 m/s

Subtract:0 + 3vvB2B2 = - = - 1 m/s1 m/s

vB = - 0.333 m/svB = - 0.333 m/s

Substitution:

vvAA - v - vBB = = - 4 m/s- 4 m/s

vvA2A2 - - (-0.333 m/s)(-0.333 m/s) = = - 4 m/s- 4 m/s

vA= -3.67 m/svA= -3.67 m/s

Example 7.Example 7. A A 0.150 kg0.150 kg bullet is fired at bullet is fired at 715 m/s715 m/s into a into a 2-kg2-kg wooden block at rest. The velocity of wooden block at rest. The velocity of block afterward is block afterward is 40 m/s40 m/s. The bullet passes through . The bullet passes through the block and emerges with what velocity?the block and emerges with what velocity?

A A B B A A B Bm v m v m u m u A A B B A A B Bm v m v m u m u BB

AA

uuB B = 0= 0

(0.150 kg)(0.150 kg)vvAA+ + (2 kg)(40 m/s) =(2 kg)(40 m/s) = (0.150 kg)(715 (0.150 kg)(715

m/s)m/s)0.1500.150vvAA+ + (80 m/s) =(80 m/s) = (107 (107

m/s)m/s)0.1500.150vvAA = = 27.2 27.2

m/s)m/s)

27.2 m/s

0.150Av

vA = 181 m/s

Example 8a: Example 8a: Inelastic collision: Find Inelastic collision: Find vvCC..

AA BB

5 kg5 kg 7.5 kg7.5 kg

uuBB=0=02 m/s2 m/s

AA BB

CommoCommon n vvCC afterafter

vvCC

( )A A B B A B Cm u m u m m v ( )A A B B A B Cm u m u m m v

After hit: After hit: vvBB= v= vAA= v= vCC

(5(5 kg)(2 m/s) = (5 kg + 7.5 kg)kg)(2 m/s) = (5 kg + 7.5 kg)vvCC

12.5 12.5 vvCC =10 m/s =10 m/s

vC = 0.800 m/svC = 0.800 m/s

In an completely inelastic collision, the two balls stick together and move as one

after colliding.

In an completely inelastic collision, the two balls stick together and move as one

after colliding.

Example 8.Example 8. (b) Elastic collision: Find (b) Elastic collision: Find vvA2A2 and and vvB2B2

AA BB

5 kg5 kg 7.5 kg7.5 kg

vvB1B1=0=02 m/s2 m/s

A A A A B Bm v m v m v A A A A B Bm v m v m v

Conservation of Conservation of Momentum:Momentum:

(5(5 kg)(2 m/s) = (5 kg)kg)(2 m/s) = (5 kg)vvA2A2 + (7.5 kg) + (7.5 kg)

vvBB

AA BB

vvAAvvBB

5 vA + 7.5 vB = 10 m/s

( )A B A Bv v u u

For Elastic For Elastic Collisions:Collisions:

2 m/sA Bv v 2 m/sA Bv v

Continued . . . Continued . . .

Example 8b (Cont).Example 8b (Cont). Elastic collision: Find Elastic collision: Find vvAA & & vvBB

AA BB

5 kg5 kg 7.5 7.5 kgkg

vvBB =0=02 m/s2 m/s

AABB

vvAAvvBB

Solve Solve simultaneously:simultaneously:

5 vA + 7.5 v B = 10 m/s

2 m/sA Bv v 2 m/sA Bv v

5 5 vvAA + 7.5 + 7.5 vvBB = 10 m/s= 10 m/s

-5 -5 vvAA + 5 + 5 vvBB = +10 m/s= +10 m/s

x (-5)x (-5)

12.5 12.5 vvBB = 20 m/s = 20 m/s

20 m/s1.60 m/s

12.5 Bv

vvAA - 1.60 m/s = -2 m/s- 1.60 m/s = -2 m/s

vA = -0.400 m/svA = -0.400 m/s

vB = 1.60 m/svB = 1.60 m/s

General: Completely General: Completely ElasticElastic

Collisions where zero energy is lost during a collision (an ideal case).

Conservation of Momentum:Conservation of Momentum:

Conservation of Conservation of Energy:Energy:

2 2 2 21 1 1 12 2 2 2

A A B B A A B B

A B B A

m u m u m v m v Loss

v v u u

2 2 2 21 1 1 12 2 2 2

A A B B A A B B

A B B A

m u m u m v m v Loss

v v u u

A A B B A A B Bm v m v m u m u A A B B A A B Bm v m v m u m u

Example 9:Example 9: A A 50 g50 g bullet lodges into a bullet lodges into a 2-kg2-kg block of clay hung by a string. The block of clay hung by a string. The bullet and clay rise together to a height bullet and clay rise together to a height of of 12 cm12 cm. What was the velocity of the . What was the velocity of the 50-g50-g mass just before entering? mass just before entering?

uuAA

BA

B

A 12 cm

The ballistic pendulum!The ballistic pendulum!

Example (Continued):Example (Continued):

B

A 12 cm

50 g

uA

2.05 kg

2 kg

Collision and Momentum:

mAuA+0= (mA+mB)vC

(0.05 kg)uA = (2.05

kg)vC

To find vA we need vC .After collision, energyenergy is conserved for

masses.

vC = 2ghvC = 2gh212 ( ) ( )A B C A Bm m v m m gh

Example (Continued):Example (Continued):

B

A 12 cm

50 g

uA

2.05 kg

2 kgmAuA+0= (mA+mB)vC

(0.05 kg)uA = (2.05 kg)(1.53 m/s)

vC = 2gh = 2(9.8)(0.12)

After Collision: vC = 1.53 m/s

uA = 62.9 m/suA = 62.9 m/s

Momentum Also Conserved:

Summary of Formulas:Summary of Formulas:Conservation of Momentum:Conservation of Momentum:

Conservation of Conservation of Energy:Energy:

2 2 2 21 1 1 12 2 2 2A A B B A A B Bm u m u m v m v Loss 2 2 2 21 1 1 1

2 2 2 2A A B B A A B Bm u m u m v m v Loss

A A B B A A B Bm v m v m u m u A A B B A A B Bm v m v m u m u

For elastic only:For elastic only: A B B Av v u u A B B Av v u u

CONCLUSION: Chapter 9BCONCLUSION: Chapter 9BConservation of Conservation of

MomentumMomentum