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 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/s B. 2 m/s C. 3 m/s D. 4 m/s

Solution The Odd Couple I Momentum BEFORE = Momentum AFTER 100 x 4 +0 = 200 x v v = 2 m/s

Example The Odd Couple II Is the K.E. BEFORE = K.E. AFTER? In other words, is the K.E. conserved in this collision?

Solution 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 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/s B. P1 and P2 move forward at 1.5 m/s C. P1 and P2 move in opposite directions at 1.5 m/s D. P1 moves forward at 1 m/s and P2 moves forward at 2 m/s

Solution Hockey Puck A. P1 stops and P2 moves forward at 3 m/s. Please verify that: 1. Momentum is conserved 2. K.E. is also conserved (elastic)

Equations for Head-on Collision Momentum is conserved(Always!) m 1 v 1 + m 2 v 2 = m 1 v 1 '+ m 2 v 2 ' K.E. is conserved (Elastic) 1/2 m 1 (v 1 ) 2 + 1/2 m 2 (v 2 ) 2 = 1/2 m 1 (v 1 ') 2 + 1/2 m 2 (v 2 ') 2 Momentum and K.E. BOTH conserved (Elastic) v 1 - v 2 = v 2 '- v 1 '

Example 7.3 … one more time! Work out Example 7.3 using: v 1 - v 2 = v 2 '- v 1 ' 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 = ma So F = m(v f - v i ) / t If F = 0 mv f = mv i If F is  0 mv f - mv i = F x t Change in momentum = F x t = IMPULSE

Example 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/s B. 75 m/s C. 95 m/s D. 135 m/s

Solution Impulse to ride the tiger Impulse = F  t = 90 x 0.04 = 3.6 Ns Impulse = change in momentum 3.6 = mv = x v v = 75 m/s

Example Teeter-Totter 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 See-Saw 30 x 180 = what x 120 what = 45 or 95 cm. mark

Example Center of Mass ? 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 Center of Mass (100)( x -20) = (300)(80 - x) x = (100) (20) + (300)( 80) / ( ) x = x

A formula for C.M. C.M. = M1 * X1 + M2 * X2 M1 + M2 0 X1 50 X2 100 M 1 M 2 C.M..

That’s all folks!