2. Physics 221 Chapter 9

3. On a Collision Course... Linear Momentum = mass x velocity p = mv Conservation of Momentum: In a collision, momentum BEFORE the collision equals the momentum AFTER the collision

4. Problem 1... The Odd Couple I A model railroad car (mass = 100 g) 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

5. Solution 1... The Odd Couple I Momentum BEFORE = Momentum AFTER 100 x 4 +0 = 200 x v v = 2 m/s

6. Problem 2... The Odd Couple II Is the K.E. BEFORE = K.E. AFTER? In other words, is the K.E. conserved in this collision?

7. Solution 2... The Odd Couple II K.E. BEFORE the collision = (1/2)(0.1)(16) +0 = 0.8J K.E. AFTER the collision = (1/2)(0.2)(4) +0 = 0.4J Half of the K.E. has mysteriously disappeared! BUT this does not mean that ENERGY is not conserved! K.E. was transformed (converted) into other forms of energy.

8. Problem 3... Big Bang! A shell explodes into two unequal fragments. One fragment has twice the mass of the other. The smaller fragment moves in the NE direction with a speed of 6 m/s. The velocity (speed and direction) of the other fragment is most nearly A. 3 m/s SW B. 3 m/s NE C. 4 m/s NE D. 4 m/s SW

9. Solution 3... Big Bang! M x 6 = 2 M x V V = 3 m/s in the SW direction.

10. 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.

11. Problem 4... Elastic I 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

12. Solution 4... Elastic I 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)

13. Problem5... Elastic II Two hockey pucks collide elastically on ice. P2 has twice the mass of P1 and is at rest and P1 strikes it “head-on” with a speed of 3 m/s. Calculate their velocities after the collision.

14. Solution 5... Elastic II P1 moves backward at 1 m/s and P2 moves forward at 2 m/s. HINT: 1. Take 5 sheets of scrap paper ( Trust me!!!) 2. Write the equation for momentum conservation 3. Write the equation for K.E. conservation 4. Solve two equations for two unknowns.

15. Tell me more about momentum! 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 then : mv f = mv i Another interesting observation: If F is NOT zero then momentum WILL change and change in momentum equals F x t. This is called IMPULSE

16. Problem 6... 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

17. Solution 6... 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

18. Problem 7... 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? 0 20 50 100 180 120

19. Solution 7... See-Saw 30 x 180 = what x 120 what = 45 or 95 cm. mark 0 20 50 100 180 120

20. Problem 8... 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. ? 100 300 0 20 50 80 100?

21. Solution 8... Center of Mass (100)( x -20) = (300)(80 - x ) x = (100) (20) + (300)( 80) / (100 + 300) x = 65 0 20 50 80 100 100 300 x

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

23. Problem 9... more C. M. Include the mass of the meter stick (60 g). Where is the new C.M. ? 100 300 0 20 50 80 100?

24. Solution 9... more C. M. C.M. = M1 * X1 + M2 * X2 + M3 * X3 M1 + M2 + M3 C.M. = (100)(20) + (60)(50) +(300)(80) 100 + 60 + 300 C.M. = 63

25. Problem 10... The Hole A circular hole of radius R/2 is drilled in a disc of mass M and radius R as shown in the figure. The center of mass is closest to A. (R/6, 0) B. (R/4, 0) C. (R/3, 0) D. (R/2, 0) (R, 0)

26. Circular plate with hole If the mass of the missing piece is 100 then the mass of the remaining piece is 300 for a total of 400. In other words, if M is the mass of the original disc (no hole) then the mass of the disc with the hole is 3M/4 and the mass of the circular piece removed is M/4. NOTE: Non-Linear thinking! Disc with HALF the radius has only a quarter of the mass!

27. Solution 10 A model for the “hole” problem (100)(R/2) = (300)( x ) x = R/6 100300 x