1. Meh, tried to make a physics/unofficial joke. I got nothing. If a Physics 211 student collides with another Physics 211 student on Unofficial, will the collision be elastic or inelastic? Can we please go over the kinetic energy of the system part? THE THIRD PART IS CONFUSING.. K-REFRENCE FRAME AND BLABLABLA. ARE THERE ANY EXAMPLES? Everything was fine until the energy of a system of particles and thats where I was lost. Could you go over the Pushing a block problem in class? can you go over the bowling ball and ping pong ball example? i didnt quite grasp the concept please go over bouncing ball Everything. This unit makes me want to cry. I get really overwhelmed with all the symbols and subscripts and asterisks* in the prelecture SERIOUSLY...HOW DO NEUTRINOS TRAVEL FASTER THAN THE SPEED OF LIGHT? I love Impulse! I like today's prelecture. WHY DIDN'T WE LEARN ABOUT SAME RELATIVE SPEEDS BEFORE!!! THIS MAKES ELASTIC PROBLEMS SO MUCH EASIER. I think you should make an announcement about how you can rate problems.. Mechanics Lecture 13, Slide 1 Your comments:

2. Physics 211 Lecture 13 Today’s Concepts: a) More on Elastic Collisions b) Average Force during Collisions Mechanics Lecture 13, Slide 2

3. In the last slides of the prelecture, it talked about KE in terms of different reference frames and in one of the reference frames the KE  0 why? Throw Something Mechanics Lecture 13, Slide 3

4. In CM frame, the speed of an object before an elastic collision is the same as the speed of the object after.  v * 1,i  v * 2,i  But the difference of two vectors is the same in any reference frame. Rate of approach before an elastic collision is the same as the rate of separation afterward, in any reference frame!  v * 1, f  v * 2, f   v * 1, f  v * 2, f  v * 1, f  v * 2, f  So the magnitude of the difference of the two velocities is the same before and after the collision. More on Elastic Collisions m1m1 m2m2 V * 1,i V * 2,i m2m2 V * 1, f V * 2, f m1m1 Just Remember This Mechanics Lecture 13, Slide 4

5. Clicker Question Consider the two elastic collisions shown below. In 1, a golf ball moving with speed V hits a stationary bowling ball head on. In 2, a bowling ball moving with the same speed V hits a stationary golf ball. In which case does the golf ball have the greater speed after the collision? A) 1 B) 2 C) same V 1 V 2 Mechanics Lecture 13, Slide 5

6. Clicker Question A small ball is placed above a much bigger ball, and both are dropped together from a height H above the floor. Assume all collisions are elastic. What height do the balls bounce back to? A H Before BC After Mechanics Lecture 13, Slide 6

7. Explanation For an elastic collision, the rate of approach before is the same as the rate of separation afterward: v v m M Rate of approach  2V Rate of separation  2V v m M 3v3v v v v Mechanics Lecture 13, Slide 7

8. A block slides to the right with speed V on a frictionless floor and collides with a bigger block which is initially at rest. After the collision the speed of both blocks is V/3 in opposite directions. Is the collision elastic? A) Yes B) No CheckPoint Mechanics Lecture 13, Slide 8 V V/3 Before Collision After Collision m M M m

9. F t titi tftf tt Impulse Forces during Collisions Mechanics Lecture 13, Slide 9 F ave

10. Clicker Question The change in momentum of block B is: A) Twice the change in momentum of block A B) The same as the change in momentum of block A C) Half the change in momentum of block A F B F A Two blocks, B having twice the mass of A, are initially at rest on frictionless air tracks. You now apply the same constant force to both blocks for exactly one second. air track Mechanics Lecture 13, Slide 10

11. Clicker Question Two boxes, one having twice the mass of the other, are initially at rest on a horizontal frictionless surface. A force F acts on the lighter box and a force 2F acts on the heavier box. Both forces act for exactly one second. Which box ends up with the biggest momentum? A) Bigger box B) Smaller box C) same F 2F2F M 2M2M Mechanics Lecture 13, Slide 11

12. CheckPoint A) Four times as long. B) Twice as long. C) The same length. D) Half as long. E) A quarter as long. Lets try it again ! A constant force acts for a time  t on a block that is initially at rest on a frictionless surface, resulting in a final velocity V. Suppose the experiment is repeated on a block with twice the mass using a force that’s half as big. For how long would the force have to act to result in the same final velocity? Mechanics Lecture 13, Slide 12 F

13. The experiment is repeated on a block with twice the mass using a force that’s half as big. For how long would the force have to act to result in the same final velocity? A) Four times as long. B) Twice as long. C) The same length. A) Ft=mv, if v is the same, m is doubled, F is cut in half, so t has no choice but to be four times as big B) The change in momentum is simply the force times the time of collision. Since the force is halved, the time it would take to reach the same momentum must be doubled. C) Ft=mv, the 2 and the one half cancel Mechanics Lecture 13, Slide 13 F

14. Case 1 Case 2 Identical balls are dropped from the same initial height and bounce back to half the initial height. In Case 1 the ball bounces off a cement floor and in Case 2 the ball bounces off a piece of stretchy rubber. In which case is the average force acting on the ball during the collision the biggest? A) Case 1 B) Case 2 Lets try it again ! CheckPoint Mechanics Lecture 13, Slide 14

15. Case 1 Case 2 In which case is the average force acting on the ball during the collision the biggest? A) Case 1 B) Case 2 A) The time of contact is less so the force must be greater. B) Ball is on the ground for a greater amount of time so more force acts on it. Mechanics Lecture 13, Slide 15 CheckPoint Responses

16. HW Problem with demo Mechanics Lecture 13, Slide 16

17. pfpf pipi pfpf -p i P  pf piP  pf pi |  P X | = 2mv cos(  ) cos  |  P Y | = 0   mv Another way to look at it : Mechanics Lecture 13, Slide 17

18. | F AVE | = |  P | /  t = 2mv cos(  ) /  t Mechanics Lecture 13, Slide 18

19. pfpf pipi |  P | = | p f – p i | = m | v f – v i |  t =  P / F AVE pfpf -p i  P = p f - p i F AVE =  P /  t Another way to look at it: Mechanics Lecture 13, Slide 19

20. H h vivi vfvf = reading on scale Scale time between bounces Demo – similar concept Mechanics Lecture 13, Slide 20