Physics 121 7. Linear Momentum 7.1 Momentum and its relation to Force 7.2 Conservation of Momentum...

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Transcript of Physics 121 7. Linear Momentum 7.1 Momentum and its relation to Force 7.2 Conservation of Momentum...

Physics 121

7. Linear Momentum

7.1 Momentum and its relation to Force

7.2 Conservation of Momentum

7.3 Collisions and Impulse

7.4 Conservation of Energy and Momentum

7.5 Elastic Collisions

7.6 Inelastic Collisions

7.7 Solving Collision Problems

7.8 Center of Mass

Linear Momentum

Linear Momentum = mass x velocity

p = mv

Conservation of Momentum

Law of Conservation of Momentum

In a collision, momentum BEFORE the collision equals the momentum AFTER the

collision

Example 7.1 . . . The Odd Couple I

A model railroad car (mass = 100 g), moving at 4 m/s, collides and locks onto a similar stationary car. The coupled cars move as a unit on a straight and frictionless track. The speed of the moving cars is most nearly

A. 1 m/sB. 2 m/sC. 3 m/sD. 4 m/s

Solution 7.1 . . . The Odd Couple I

Momentum BEFORE = Momentum AFTER100 x 4 +0 = 200 x vv = 2 m/s

Example 7.2 . . . The Odd Couple II

Is the K.E. BEFORE = K.E. AFTER?

In other words, is the K.E. conserved in this collision?

Solution 7.2 . . . The Odd Couple II

K.E. BEFORE the collision = (1/2)(0.1)(16) +0 = 0.8 J

K.E. AFTER the collision = (1/2)(0.2)(4) +0 = 0.4 J

Half of the K.E. has mysteriously disappeared!

Note: However, this does not mean that ENERGY is not conserved! K.E. was transformed (converted) into other forms of energy.

Close collisions of the second kind ...

There are TWO types of collisions

I. Momentum conserved and K.E. also conserved. This type is called an ELASTIC collision. Elastic collisions are “non-sticky” as in billiard balls or steel ball-bearings.

II. Momentum conserved BUT K.E. NOT conserved. This type is called an INELASTIC collision. Inelastic collisions are “sticky” as in coupled railroad cars and putty.

Example 7.3 . . . Hockey Puck

Two hockey pucks collide elastically on ice. P2 is at rest and P1 strikes it “head-on” with a speed of 3 m/s.

A. P1 stops and and P2 moves forward at 3 m/sB. P1 and P2 move forward at 1.5 m/sC. P1 and P2 move in opposite directions at 1.5 m/sD. P1 moves forward at 1 m/s and P2 moves forward at 2 m/s

Solution 7.3 . . . Hockey Puck

A. P1 stops and P2 moves forward at 3 m/s.

Please verify that:1. Momentum is conserved2. K.E. is also conserved (elastic)

Equations for Head-on Collision

Momentum is conserved(Always!)m1 v1 + m2 v2 = m1 v1'+ m2 v2'

K.E. is conserved (Elastic)1/2 m1 (v1 )2 + 1/2 m2 (v2 )2 = 1/2 m1 (v1')2 + 1/2 m2(v2')2

Momentum and K.E. BOTH conserved (Elastic)v1 - v2 = v2'- v1'

Example 7.3 … one more time!

Work out Example 7.3 using:

v1 - v2 = v2'- v1'

and see how effortless it is to arrive at the correct answer!

Impulse

O.K. good boys and girls. Here is everything you always wanted to know about momentum but were afraid to ask!

In the beginning there was F = maSo F = m(vf - vi) / t

If F = 0 mvf = mvi

If F is 0 mvf - mvi = F x t Change in momentum = F x t = IMPULSE

Example 7.4 . . . Tiger in the woods

A golf ball has a mass of 48 g. The force exerted by the club vs. time is a sharp spike that peaks at 180 N for 0.01 s and then drops back to zero as the ball leaves the club head at high speed. It is estimated that the average force is 90 N over a short time (t) of 0.04 s. The speed of the ball is

A. 35 m/sB. 75 m/sC. 95 m/sD. 135 m/s

Solution7.4 . . . Impulse to ride the tiger

Impulse = F t = 90 x 0.04 = 3.6 Ns

Impulse = change in momentum

3.6 = mv -0

3.6 = 0.048 x v

v = 75 m/s

Example 7.5 . . . Teeter-Totter

0 20 50 100

180 120

FB weighs 180 pounds and sits at the 20 cm. mark. Where should SP (120 pounds) sit in order to balance the teeter-totter at the playground?

Solution 7.5 . . . See-Saw

30 x 180 = what x 120

what = 45 or 95 cm. mark

0 20 50 100

180 120

Example 7.6 . . . Center of Mass

100300

0 20 50 80 100?

The C.M. is a weighted average position of a distribution of masses where the system can

be balanced.

Where is the C.M.?

Solution 7.6 . . . Center of Mass

(100)( x -20) = (300)(80 - x)

x = (100) (20) + (300)( 80) / (100 + 300)

x = 65

0 20 50 80 100

100300

x

A formula for C.M.

C.M. = M1 * X1 + M2 * X2

M1 + M2

0 X1 50 X2 100

M 1M 2

C.M..

That’s all folks!