Lecture 13: Collisions, Impulse, & Reference Frames

1. Lecture 13: Collisions, Impulse, & Reference Frames Today’s Concepts: a) More on Elastic Collisions b) Average Force during Collisions

2. Stuff you asked about: I did not understand the equation of kinetic energy showed at the second last of the pre lecture..too much frame has to be considered Stiil confuse to lab frme and centre of mass frame BLA BLA.... need to be clarified Impulse. I am hopelessly confused. I have a hard time identifying elastic collisions and understanding how they work. The whole part about velocity of objects approaching each other is the same as the velocity at which they separate was a little confusing, but other than that everything was pretty straight forward. I hope there are some good demos for Impulse. the part where the guy did the thing and then said the thingy about the things... please explain about blocks hitting the ground and the forces during this collision. it's so confusing. I found the kinetic energy equation in the pre-lecture to be slightly confusing.. I think it would be pretty epic if for Halloween this year Bruce had an Ewok costume - robes, spear and all.

3. More on Elastic Collisions In CMframe, the speed of an object before an elastic collision is the same as the speed of the object after. m2 v*1,i v*2,i m1 |v*1i-v*2i| = |-(v*1f-v*2f)| = |v*1f-v*2f | = |-v*1f+ v*2f | So the magnitude of the difference of the two velocities is the same before and after the collision. But the difference of two vectors is the same in any reference frame. m2 m1 Just Remember This v*2, f v*1, f Rate of approach before an elastic collision is the same as the rate of separation afterward, in any reference frame!

4. ACT: Golf Ball and Bowling Ball Consider the two elastic collisions shown below. In 1, a golf ball moving with speed Vhits a stationary bowling ball head on. In 2, a bowling ball moving with the same speed Vhits a stationary golf ball. In which case does the golf ball have the greater speed after the collision? A) 1B)2C)same Both cases: |v1i – v2i | = V = |v1f – v2f | Case 1: v2f~ 0 v1f = –V Case 2:v1f~ Vv2f ~ 2V M >> m: Vbb ~ constant V V 1 2

5. ACT: Little Ball on Big Ball A small ball is placed above a much bigger ball, and both are dropped together from a height Habove the floor. Assume all collisions are elastic. To what height do the balls bounce back? Before After H A B C

6. Explanation For an elastic collision, the rate of approach before is the same as the rate of separation afterward: 3V m V m V V V V M V M Rate of separation=2V Rate of approach=2V

7. CheckPoint: Another 2-Box Collision A block slides to the right with speed Von 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 About 60% of you got this correct V M Before Collision m V/3 V/3 M m After Collision

8. Playing Baseball In professional baseball, a good hitter can swing a bat so that the end of the bat travels at a maximum speed of about 40 mph. A good pitcher can throw a 95 mph fastball. Estimate the maximum possible distance (in feet) that the ball can travel. 1 mile = 5280 feet

9. Forces during Collisions F Favg t Δt Impulse ti tf

10. CheckPoint: Bouncing Ball Identical balls are dropped from the same initial height and bounce back to half the initial height. In Case 1the 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 Case 1 Case 2

11. CheckPoint Results: Bouncing Ball In which case is the average force acting on the ball during the collision the biggest? A) Case 1 B) Case 2 Case 2 Case 1 A) Equal height means that the impulse for both balls must be the same. Therefore the ball that hit the cement must have a greater force since it is applied for a shorter period of time. B) since the collision time is longer in case two, it would result in a larger average force

12. ACT: After 1 second, the speed of block B is: A) Twice the speed of block A B) The same as the speed of block A C) Half the speed of block A F A Two identical 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. The same in both cases! B F air track air track

13. ACT: Two boxes, one having twice the mass of the other, are initially at rest on a horizontal frictionless surface. A force Facts on the lighter box and a force 2Facts on the heavier box. Both forces act for exactly one second.Which box ends up with the largest speed? A) Bigger boxB) Smaller boxC) same 2F F 2M M

14. CheckPoint: Pushing a Block A) Four times as long. B) Twice as long. C) The same length. D) Half as long. E) A quarter as long. A constant force acts for a time Δton 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? F 50% of you got this right.

15. CheckPoint: Results 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? F A) Four times as long. B) Twice as long. C) The same length. A) Since the mass is doubled, the block's momentum change is doubled also. But the force is only as half, thus the time needed has to be increased four times.

16. Homework Problem Review

17. ΔP =pf-pi pi θ Another way to look at it: -pi pf θ pf mv |ΔPX | = 2mv cosθ |ΔPY| = 0

18. |Favg | = |ΔP| /Δt = 2mv cosθ /Δt

19. pi Another way to look at it: pf ΔP = pf-pi pf -pi Favg = ΔP/Δt |ΔP| = |pf–pi| = m|vf–vi| Δt = ΔP/Favg