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

A PowerPoint Presentation by

Paul E. Tippens, Emeritus Professor

Southern Polytechnic State University

A PowerPoint Presentation byA PowerPoint Presentation by

Paul E. Tippens, Emeritus ProfessorPaul E. Tippens, Emeritus Professor

Southern Polytechnic State UniversitySouthern Polytechnic State University

© 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 this Objectives: After completing this module, you should be able to:module, you should be able to:

•• State the law of conservation of momentum State the law of conservation of momentum and apply it to the solution of problems.and apply it to the solution of 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 bodies Predict the velocities of two colliding bodies when given the coefficients of restitution, when given the coefficients of restitution, masses, and initial velocities.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 m2v2

AfterAfter

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 m2u1

u2

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

Example 1.Example 1. A A 22--kgkg mass moving at mass moving at 4 4 m/sm/s collides with a collides with a 11--kgkg mass initially at rest. After mass initially at rest. After the collision, the 2the collision, the 2--kg mass moves at kg mass moves at 1 1 m/sm/s and the 1and the 1--kg mass moves at kg mass moves at 3 3 m/sm/s. What . What energy was lost in the collision?energy was lost in the collision?

ItIt’’s important to draw and label a sketch with s important to draw and label a sketch with appropriate symbols and given information.appropriate symbols and given 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 energy What energy was lost in the collision? Energy is was lost in the collision? Energy is conserved.conserved.

m2

uu22 = 0= 0

m1

uu1 1 = = 4 4 m/sm/s

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

m2

vv22 = = 2 2 m/sm/s

m1

vv1 1 = = 1 1 m/sm/s

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

BEFORE:BEFORE:2 2 21 1 1

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

1 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 JEnergy Loss = 15 J

Impulse and MomentumImpulse and Momentum

A BuA

uB

A BvA vB

B--FF AA tt FF B B tt

Opposite but Equal F t

Ft = mvf – mvo

FB t = -FA t

Impulse = p

mB vB - mB uB = -(mA vA - mA uA )

mA vA + mB vB = mA uA + mB uB mA vA + mB vB = mA uA + mB uBSimplifying:

Conservation of MomentumConservation of Momentum

A BuA

uB

A BvA vB

B--FF AA tt FF B B tt

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

Recall that the total energy is also conserved:

KA0 + KB0 = KAf + KBf + LossKK A0A0 + K+ K B0B0 = = KK AfAf + + KK BfBf + Loss+ Loss

Kinetic Energy: K = Kinetic Energy: K = ½½mvmv22

mA vA + mB vB = mA uA + mB uB mA vA + mB vB = mA uA + mB uB

Example 2:Example 2: A A 22--kgkg block block A A and a and a 11--kgkg block block BB are pushed together against a spring and are pushed together against a spring and tied with a cord. When the cord breaks, tied with a cord. When the cord breaks, the the 11--kgkg block moves to the right at block moves to the right at 8 m/s8 m/s. . What is the velocity of the What is 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 release is zero.zero.

mmAA vvAA + + mmBB vvBB = = mmAA uuAA + + mmBB uuBB0 0

mA vA = - mB vB vA = -mB vB

mA

Example 2 (Continued)Example 2 (Continued)

mmAA vvAA + + mmBB vvBB = = mmAA uuAA + + mmBB uuBB0 0

mA vA = - mB vB vA = -mB vB

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, how Ignoring friction, how much energy was released by the spring?much energy was released by the spring?

A B

2 kg1 kg A B

8 m/s4 m/s

Cons. of E: ½½kxkx22 = = ½½ mmAA vvAA + + ½½mmBB vvBB 2222

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

½½kxkx2 2 = 16= 16 J + 32 J = 48 JJ + 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 CollisionsCompletely Inelastic Collisions

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 6060--kgkg football player stands football player stands on a frictionless lake of ice. He catches a on a frictionless lake of ice. He catches a 22--kgkg football and then moves at football and then moves at 40 cm/s40 cm/s. . What was the initial velocity of the football?What was the initial velocity of the football?

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

AA

BB

mmAA vvAA + + mmBB vvBB = = mmAA uuAA + + mmBB uuBBMomentum:0

(m(mAA + + mmBB )v)vCC = = mmAA uuAA

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

Inelastic collision:

uA = 12.4 m/suuAA = 12.4 m/s= 12.4 m/s

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

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

Conservation of Momentum:Conservation of Momentum:

Conservation of Energy:Conservation of Energy:

( )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

Example 4.Example 4. An An 8787--kgkg skater skater BB collides with a collides with a 2222--kgkg skater skater AA initially at rest on ice. They move together initially at rest on ice. They move together after the collision at after the collision at 2.4 m/s2.4 m/s. Find the velocity of the . Find the velocity of the skater skater BB before the collision.before the collision.

AA BB

uuBB = ?= ?uuAA = 0= 0

Common speed after Common speed after colliding: colliding: 2.4 m/s.2.4 m/s.

22 kg22 kg

87 kg87 kg

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

vvBB = = vvAA = = vvCC = = 2.4 m/s2.4 m/s

(87(87 kg)kg)uuBB = (87 kg + 22 kg)(2.4 m/s)= (87 kg + 22 kg)(2.4 m/s)

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

uB = 3.01 m/s

Example 5:Example 5: A A 50 g50 g bullet strikes a bullet strikes a 11--kgkg block, passes all the way through, then block, passes all the way through, then lodges into the lodges into the 2 kg2 kg block. Afterward, block. Afterward, the 1 kg block moves at the 1 kg block moves at 1 m/s1 m/s and the and the 2 2 kgkg block moves at block 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)

mA uA + mB uB + mC uC = mB vB + (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 CollisionsCompletely Elastic Collisions

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

APPROXIMATIONS!APPROXIMATIONS!

Velocity in Elastic CollisionsVelocity in Elastic Collisions

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 22--kgkg ball moving to the ball moving to the right at right at 1 m/s1 m/s strikes a strikes a 44--kgkg ball moving ball moving left at left at 3 m/s3 m/s. What are the velocities . What are the velocities after impact, assuming complete after impact, assuming complete elasticity?elasticity?

A B

A B

3 m/s3 m/s1 m/s1 m/s

vvAA vvBB1 kg1 kg 2 kg2 kg

vvAA -- vvBB = = -- ((uuAA -- uuBB ))

vvAA -- vvBB = = uuBB -- uuAA

vvAA -- vvBB = (-3 m/s) - (1 m/s)

From conservation of energy (relative v):

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

Example 6 (Continued)Example 6 (Continued)

A B

A B

3 m/s3 m/s1 m/s1 m/s

vvAA vvBB1 kg1 kg 2 kg2 kg

mmAA vvAA + + mmBB vvBB = = mmAA uuAA + + mmBB uuBB

Energy: Energy: vvAA -- vvBB == -- 4 m/s4 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 -- vvBB = = -- 4 m/s4 m/s

Two independent equations to solve:

Example 6 (Continued)Example 6 (Continued)

A B

A B

3 m/s3 m/s1 m/s1 m/s

vvAA vvBB1 kg1 kg 2 kg2 kg

vA + 2vB = -5 m/s

vvAA -- vvBB = = -- 4 m/s4 m/s

Subtract: 0 + 3vv B2B2 = = -- 1 m/s1 m/s

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

Substitution:

vvAA -- vvBB = = -- 4 m/s4 m/s

vv A2A2 -- ((--0.333 m/s)0.333 m/s) = = -- 4 m/s4 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 22-- kgkg wooden block at rest. The velocity of block afterward is wooden block at rest. The velocity of block afterward is 40 m/s40 m/s. The bullet passes through the block and emerges . The bullet passes through the block and emerges with what velocity?with what velocity?

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

uuBB = 0= 0

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

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

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

0.150Av

vA = 181 m/s

Example 8a: Example 8a: Inelastic collision: Find Inelastic collision: Find vv CC ..

AA BB

5 kg5 kg 7.5 kg7.5 kguuBB =0=02 m/s2 m/s

AA BB

Common Common vvCC afterafter

vvCC

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

After hit: After hit: vvBB = = vvAA = = vvCC

(5(5 kg)(2 m/s) = (5 kg + 7.5 kg)(2 m/s) = (5 kg + 7.5 kg)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 vand vB2B2

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

Conservation of Momentum:Conservation of 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 Collisions:For Elastic Collisions:

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 kg7.5 kgvvBB =0=02 m/s2 m/s

AABB

vvAA vvBB

Solve simultaneously:Solve simultaneously:

5 vA + 7.5 v B = 10 m/s

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 (x (--5)5)

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

20 m/s 1.60 m/s12.5 Bv

vvAA -- 1.60 m/s = 1.60 m/s = --2 m/s2 m/s

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

vB = 1.60 m/svB = 1.60 m/s

General: Completely ElasticGeneral: Completely ElasticCollisions where zero energy is lost during a collision (an ideal case).

Conservation of Momentum:Conservation of Momentum:

Conservation of Energy:Conservation of Energy:2 2 2 21 1 1 1

2 2 2 2

A A B B A A B B

A B B A

m u m u m v m v Lossv v u 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 22--kgkg block of clay hung by a string. The bullet and block of clay hung by a string. The bullet and clay rise together to a height of clay rise together to a height of 12 cm12 cm. What . What was the velocity of the was the velocity of the 5050--gg mass just before mass just before entering?entering?

uuAA

BA

B

A 12 cm

The ballistic pendulum!The ballistic pendulum!

Example (Continued):Example (Continued):

BA 12 cm

50 g

uA

2.05 kg

2 kg

Collision and Momentum:

mA uA +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):

BA 12 cm

50 g

uA

2.05 kg

2 kgmA uA +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 Energy:Conservation of 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

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

CONCLUSION: Chapter 9BCONCLUSION: Chapter 9B Conservation of MomentumConservation of Momentum