Physics 203 College Physics I Fall 2012

1. Physics 203College Physics IFall 2012 S. A. Yost Chapter 7 Part 2 Momentum and Collisions

2. Announcements • Read Ch. 7, but you can skip sections 7.7 and 7.9. • A problem set HW07A was due today. • A problem set HW07B on sections 4 – 6 (collisions) is due Thursday. • Next class: Rotational Motion, Ch. 10. • Read sec. 1 – 5 for next time. • Homework set HW08 on sections 1 – 6 will be due next Thursday. Sections 7 – 8 will be combined with chapter 9 in a later set. Skip section 9.

3. Exam 2 • The average was 36%, high 67.5%.

4. Momentum and Impulse • Momentum is defined to be • p = m v • The change in momentum is the impulse. • Dp= Favg t= J • This is a consequence of Newton’s 2nd Law. • In an isolated system of objects, the total momentum of the objects is always conserved. • This is a consequence of Newton’s 3rd Law. → → → → →

5. 1D Collisions • In a one dimensional collision, two masses m1and m2 approach with velocities v1and v2, and then collide. Their velocities v1’and v2’ are measured after the collision. • m1 v1 + m2 v2 = m1 v1’ + m2 v2’ • Dp1 + Dp2 = 0 m1 m2

6. Elasticity of a Collision • Not all collisions are equally “bouncy”. • This is quantified by elasticity. • In a perfectly elastic collision, the relative speeds of the colliding objects is the same, though the direction may be different.

7. v v’ Inelastic Collision • One possibility is that the red ball can come in with velocity v, stick to the blue ball, and both can travel off with final velocity v’. m m

8. Inelastic Collision This case is called an inelastic collision. In this case, the relative velocity of the masses is zero after the collision, since the balls stick together. m1 m2 v’

9. Example • For example, a railroad car of mass M1= 1200 kg could travel at v = 2.5 m/s and strike a second railroad car of mass M2 = 2800 kg. If the cars collide and hitch together, how fast do they move together? 2.5 m/s 1200 kg 2800 kg

10. Example • Initial momentum: • pi = M1v = 1200 kg x 2.5 m/s = 3000 kg m/s. • Final momentum: pf = (M1 + M2) v’ • v’ = 3000 kg m/s / 4000 kg = 0.75 m/s. 2.5 m/s 1200 kg 2800 kg

11. Energy in a Collision • How much energy is converted to internal energy (vibrations, heat, etc.) in this collision? • Ei = ½ M1 v2 = ½ (1200 kg)(2.5 m/s)2 = 3750 J. • Ef = ½ (M1+M2)v’2 = ½ (4000 kg)(0.75 m/s)2 • = 1125 J. • Ei – Ef = 2625 J.

12. Question • Which of the inelastic collision(s) shown bring(s) the car at left to a halt? Choose D if all of them, E if none of them. A B C

13. Question • In which collision does the car on the left experience the greatest force? Choose D if they are all equal, E if more information is needed. A B C

14. Question • In collision C, which car experiences the greater force? • A = left, B = right, C = equal. A B C

15. Question • Which of the labeled cars has the greatest change in energy, comparing all of the cases together? Pick A, B, C, D, or E. (One of these is correct.) A B D C E

16. Elastic Collision • An elastic collision conserves energy. • In one dimension, a collision is elastic if the relative speeds are the same, • v2’ – v1’ = v1– v2. but notice the sign reversal m2 m1 m2 m1 v1 v2 v1’ v2’

17. Bat and Ball Example • A bat is swung at a ball pitched at 35 m/s, hitting the ball back on a line-drive at 56 m/s. • What was the speed of the bat before and after the collision if the bat has mass M = 850 g? • Assume an elastic collision. v’ = 56 m/s v = - 35 m/s m = 145 g M = 850 g

18. Bat and Ball Example • Let v1, v1’ be the speed of the bat before and after the collision. • Momentum conservation : • m1v1+ m2v2= m1v1’ + m2v2’ • Elastic collision: v1– v2= v2’ – v1’ v2’ = 56 m/s v2= - 35 m/s m 2= 145 g m1= 850 g

19. Bat and Ball Example • m1v1+ m2v2= m1 v1’ + m2v2’ • v1– v2= v2’ – v1’ • 2m1v1+ (m2 – m1) v = (m1 + m2) v’ • 2m1v1= (m1 + m2) v’ + (m1 – m2) v • 1.7 v1= 0.995(56 m/s) + 0.705(–35 m/s) • = 31.0 m/s • v1= 18 m/s. m1m1m1 m1 v1’ = v2’ + v2– v1= 3 m/s.

20. Force Due to a Stream of Mass This mini-UZI fires 16 rounds per second, withm = 8geach, into a wall at velocity v0 =360 m/s. What is the average force on the wall? Think about the impulse resulting in the change of momentum when the bullets are stopped by the wall.

21. Force Due to a Stream of Bullets The wall stops the bullets, applying a force to the bullets of Fwb = dp/dt. In time t = 1/16 s, a mass m = 8 ghits the wall. The momentum lost is Dp = -m v0. The impulse is Fwb = Dp/t = -mv0/t = - 46 N Fwb

22. F = 46 N Force Due to a Stream of Bullets The stream of bullets could be used as a propulsion mechanism for the gun… this is exactly the mechanism of rocket propulsion. The wall plays no role in this; the gun pushes against the bullets.

23. Question Two people on roller blades throw a ball back and forth. After a couple of throws, they are (ignore friction) A. standing where they were initially. B. standing farther away from each other. C. standing closer together. D. moving away from each other. E. moving toward each other.

24. Recoil • Suppose a box of mass m = 10 kg is thrown out of a small boat of mass M = 100 kg with a speed relative to the boat of 15 m/s. m M What is the speed of the boat after the box is thrown?

25. Recoil • Momentum conservation implies • MV + mv = 0. • We were given the speed of the box relative to the boat: v m V M v0 = 15 m/s = v – V

26. Recoil • MV + mv = 0. • v = v0 + V. • MV + m(v0+ V) = 0. • Solving for V gives • -m v0 -(10 kg) (15 m/s) • M + m 110 kg v m V M V = = = -1.4 m/s

27. Center of Mass • The center of mass a set of objects is the average position of their mass. For two objects in 1D: • x1 m1 + x2 m2 • m1 + m2 x cm = x1 x2 CM m2 m1 xcm

28. rcm= vcm= Center of Mass • In more dimensions you can use vectors to locate the CM: • m1r1 + m2r2 +m3r3 • m1 + m2 + m3 • It moves with velocity m2 m1 r2 r1 m2 CM r3 • m1v1 + m2v2 +m3v3 • m1 + m2 + m3 P = M vcm

29. Motion of the Center of Mass • If an external force F acts on an extended object or collection of objects of mass M, the acceleration of the CM is given by • F= Macm. • You can apply Newton’s 2nd Law as if it were a particle located at the CM, as far as the collective motion is concerned. • This says nothing about the relative motion, rotation, etc., about the CM. That comes up in chapter 8.

30. Motion of Extended Objects • The motion of extended objects or collections of particles is such that the CM obeys Newton’s 2nd Law.

31. Motion of the Center of Mass • The CM of a wrench sliding on a frictionless table will move in a straight line because there is no external force. In this sense, the wrench may be though of as a particle located at the CM. cm motion

32. Motion of the Center of Mass • For example, if a hammer is thrown, its CM follows a parabolic trajectory under the influence of gravity, as a point object would.

33. Motion of the Center of Mass • For example, if a hammer is thrown, its CM follows a parabolic trajectory under the influence of gravity, as a point object would.