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Collisions and Conservation of MomentumPowerPoint Presentation

Collisions and Conservation of Momentum

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u1

u2

m1

m2

v2

v1

m2

m1

A Collision of Two MassesWhen 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

A Collision of Two BlocksCollision

u2

u1

m1

m2

Conservation of EnergyThe 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

-FADt

B

vA

vB

B

A

Impulse and MomentumImpulse = Dp

FDt = mvf– mvo

Opposite but Equal F Dt

FBDt = -FADt

mBvB - mBuB = -(mAvA - mAuA)

mAvA + mBvB = mAuA + mBuB

Simplifying:

uB

uA

B

A

-FADt

FB Dt

B

vA

vB

B

A

Conservation of MomentumThe 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

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= -

Example 2 (Continued)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.)

Before

Completely Inelastic CollisionsCollisions where two objects stick together and have a common velocity after impact.

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

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?

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 Collisions1. 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:

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

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

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

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

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.

vB1=0

5 kg

7.5 kg

A

B

vB

vA

A

B

Example 8.(b) Elastic collision: Find vA2 and vB2Conservation 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 . . .

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 & vBSolve 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:

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