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# Collisions and Conservation of Momentum PowerPoint PPT Presentation

Collisions and Conservation of Momentum. u 1. u 2. m 1. m 2. v 2. v 1. m 2. m 1. A Collision of Two Masses. When two masses m 1 and m 2 collide, we will use the symbol u to describe velocities before collision. Before. The symbol v will describe velocities after collision.

Collisions and Conservation of Momentum

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## Collisions and Conservation of Momentum

u1

u2

m1

m2

v2

v1

m2

m1

### A Collision of Two Masses

When two masses m1 and m2 collide, we will use the symbol uto describe velocities before collision.

Before

The symbol v will describe velocities after collision.

After

u1

u2

Before

m1

m2

“u”= Before

“v” = After

m1

m2

B

v1

After

v2

m2

m1

Collision

u2

u1

m1

m2

### Conservation of Energy

The kinetic energy before colliding is equal to the kinetic energy after colliding plus the energy lost in the collision.

u2 = 0

u1 = 4 m/s

v1 = 1 m/s

v2 = 2 m/s

m1

m1

m2

m2

m1 = 2 kg

m1 = 1 kg

m1 = 2 kg

m1 = 1 kg

AFTER

BEFORE

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

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

u2= 0

u1 = 4 m/s

v1 = 1 m/s

v2 = 2 m/s

m1

m1

m2

m2

m1 = 2 kg

m1 = 1 kg

m1 = 2 kg

m1 = 1 kg

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

BEFORE:

AFTER

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

Loss = 16 J – 3 J

Energy Loss = 15 J

uB

uA

B

A

FB Dt

B

vA

vB

B

A

### Impulse and Momentum

Impulse = Dp

FDt = mvf– mvo

Opposite but Equal F Dt

mBvB - mBuB = -(mAvA - mAuA)

mAvA + mBvB = mAuA + mBuB

Simplifying:

uB

uA

B

A

FB Dt

B

vA

vB

B

A

### Conservation of Momentum

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

mAvA + mBvB = mAuA + mBuB

Recall that the total energy is also conserved:

Kinetic Energy: K = ½mv2

KA0 + KB0 = KAf + KBf + Loss

A

B

0

0

mBvB

mA

vA= -

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

The initial velocities are zero, so that the total momentum before release is zero.

mAvA + mBvB = mAuA + mBuB

mAvA = - mBvB

vA2

2 kg

8 m/s

1 kg

A

B

mAvA+ mBvB = mAuA + mBuB

A

B

mAvA = - mBvB

0

0

(1 kg)(8 m/s)

(2 kg)

vA = -

mBvB

mA

vA= -

vA = - 4 m/s

### 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.)

After

Before

### Completely Inelastic Collisions

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

0

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

A

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

B

mAvA + mBvB = mAuA + mBuB

Momentum:

(mA + mB)vC = mAuA

Inelastic collision:

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

uA= 12.4 m/s

0

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

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 J

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

### General: Completely Inelastic

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

Conservation of Momentum:

Conservation of Energy:

Common speed after colliding: 2.4 m/s.

uA= 0

uB= ?

87 kg

B

A

22 kg

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

vB= vA = vC = 2.4 m/s

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

(87 kg)uB =262 kg m/s

uB = 3.01 m/s

uA= ?

2 kg

1 kg

1 m/s

2 m/s

1 kg

2 kg

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

C

A

B

2 kg

1 kg

1 m/s

2 m/s

Momentum After = Momentum Before

1 kg

2 kg

0

0

50 g

Find entrance velocity of bullet: mA= 0.05 kg; uA= ?

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

(0.05 kg)uA=(1 kg)(1 m/s)+(2.05 kg)(2 m/s)

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

uA= 102 m/s

### Completely Elastic Collisions

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

APPROXIMATIONS!

uB

uA

A

B

vA

vB

A

B

### Velocity in Elastic Collisions

1. Zero energy lost.

2. Masses do not change.

3. Momentum conserved.

Equal but opposite impulses (F Dt) means that:

(Relative Dv After) = - (Relative DvBefore)

vA - vB = - (uA - uB)

For elastic collisions:

A

B

1 m/s

3 m/s

vA

vB

A

B

1 kg

2 kg

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

vA - vB = - (uA - uB)

vA - vB = uB - uA

vA - vB= (-3 m/s) - (1 m/s)

From conservation of energy (relative v):

vA - vB = - 4 m/s

3 m/s

1 m/s

A

B

vA

vB

1 kg

2 kg

A

B

### Example 6 (Continued)

Energy: vA - vB =- 4 m/s

Momentum also conserved:

mAvA + mBvB = mAuA + mBuB

(1 kg)vA+(2 kg)vB=(1 kg)(1 m/s)+(2 kg)(-3 m/s)

vA + 2vB = -5 m/s

Two independent equations to solve:

vA - vB = - 4 m/s

3 m/s

1 m/s

vA + 2vB = -5 m/s

A

B

vA - vB = - 4 m/s

vA

vB

1 kg

2 kg

A

B

### Example 6 (Continued)

Subtract:

0 + 3vB2 = - 1 m/s

vB = - 0.333 m/s

vA2 - (-0.333 m/s) = - 4 m/s

Substitution:

vA= -3.67 m/s

vA - vB = - 4 m/s

B

A

uB = 0

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

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

0.150vA+ (80 m/s) =(107 m/s)

0.150vA =27.2 m/s)

vA= 181 m/s

2 m/s

uB=0

5 kg

7.5 kg

A

B

vC

Common vC after

A

B

### Example 8a: Inelastic collision: Find vC.

After hit: vB= vA= vC

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

12.5 vC =10 m/s

vC = 0.800 m/s

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

2 m/s

vB1=0

5 kg

7.5 kg

A

B

vB

vA

A

B

### Example 8.(b) Elastic collision: Find vA2 and vB2

Conservation of Momentum:

(5kg)(2 m/s) = (5 kg)vA2 + (7.5 kg) vB

5 vA + 7.5 vB= 10 m/s

For Elastic Collisions:

Continued . . .

2 m/s

vB=0

x (-5)

5 kg

7.5 kg

5 vA + 7.5 v B= 10 m/s

A

B

vB

vA

B

A

### Example 8b (Cont). Elastic collision: Find vA & vB

Solve simultaneously:

5 vA + 7.5 vB= 10 m/s

-5 vA+ 5 vB= +10 m/s

vA- 1.60 m/s = -2 m/s

vA = -0.400 m/s

12.5 vB = 20 m/s

vB = 1.60 m/s

### General: Completely Elastic

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

Conservation of Momentum:

Conservation of Energy:

For elastic only:

### Summary of Formulas:

Conservation of Momentum:

Conservation of Energy: