Hour Exam 1 Average 76.5% (no scaling) Mechanics Lecture 12, Slide 2

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1 Your Comments This entire prelecture/checkpoint confused me, but especially the double mass but half speed checkpoint. Hey kid in Seat D23, your shoe is untied! Switching reference frames makes this way too hard. i'd rather just solve quadratic equations I feel like this topic should be relatively simple, but I think I need to go over more examples to help clear up some confusions. I'd like to talk about the center of mass velocity affecting the system as a whole, and I want to go over collisions made at angles. I'm also confused because we say the velocity is the same after the collision as before the collision, but the example we did earlier showed the box having a different velocity starting out compared to finishing. It's really amazing how good I feel when I realize the homework says "starts tomorrow" and not "due tomorrow". Can we call elastic collisions "Bouncy Smacks"? I think this would make things very interesting. How in the almighty universe are we supposed to know what direction those balls will end up going in after looking at measly arrows.

2 Hour Exa Average 76.5% (no scaling) Mechanics Lecture 12, Slide 2

3 I did poorly on the first exam, do you recommend dropping the course? No. Don t Panic. Each hour exam is worth 10% of your grade. Even if you got 50% on the exam, it means you have lost only 50 points out of There are 950 other points you haven't lost yet! We have posted a spreadsheet to help you estimate your grade Office Hour Reminder

4 Physics 211 Lecture 13 Today s Concepts: a) Elastic Collisions b) Center-of-Mass Reference Frame

5 In high school I was taught that the main difference between elastic and inelastic collisions is that in inelastic collisions the blocks stick together and in elastic collisions they don't. However, after the last few lectures, I'm starting to think there's a lot more to it. It would be really helpful if we could just make a list of the key differences between the two. Also, please play Simon and Garfunkel! In elastic collisions, E = K + U is the same before & after Mechanics Lecture 12, Slide 5

6 A collision is inelastic if non-conservative forces do work: - Friction - Deformation - Sticking together - Just about anything This is usually the case in real life

7 Review: Center of Mass & Collisions so far F M A net, ext tot cm dptot dt The CM behaves just like a point particle If F net, ext 0 then momentum is conserved P M V tot tot cm If you are in a reference frame moving along with the CM then the total momentum you measure is 0.

8 CheckPoint A box sliding on a frictionless surface collides and sticks to a second identical box which is initially at rest. Compare the initial and final kinetic energies of the system. A) K initial > K final B) K initial K final C) K initial < K final initial final

9 CheckPoint Response A) K initial > K final B) K initial K final C) K initial < K final initial final A) This is not an elastic collision since they stick together and therefore the KE isn't conserved after they collide. B) Because there is no friction, the kinetic energy is conserved in the system.. C) two boxes are moving after the collision.

10 Relationship between Momentum & Kinetic Energy K 1 2 mv 2 1 2m mv p since p mv 2m This is often a handy way to figure out the kinetic energy before and after a collision since p is conserved. initial final K K Initial final 2 p 2M 2 p 2(2 M ) same p

11 Center-of-Mass Frame I would like more examples and elaboration on problems that incorporate center of mass and relative motion. videos v CM m v m v m m P M tot tot..\..\movies\cmel.mpg. mpg..\..\movies\cminel. mpg.mpg In the CM reference frame, v CM 0 In the CM reference frame, P TOT 0

12 Center of Mass Frame & Elastic Collisions The speed of an object is the same before and after an elastic collision if viewed in the CM frame: v * 1,i m 2 v * 2,i m 2 v * 1, f m 2 v * 2, f

13 Example: Using CM Reference Frame A glider of mass 0.2 kg slides on a frictionless track with initial velocity v 1,i 1.5 m/s. It hits a stationary glider of mass m kg. A spring attached to the first glider compresses and relaxes during the collision, but there is no friction (i.e. energy is conserved). What are the final velocities? v 1,i CM v 2,i 0 * m 2 CM * m 2 x v 1, f CM * m 2 v 2, f

14 Four step procedure: Example Step 1: First figure out the velocity of the CM, v CM. 1 v m v m v m1 vcm v1, i m m v CM 1 1, i 2 2, i m1 m2 CM 1 2.2kg v.2 kg.8kg 1, i 0 in this case

15 Example Now consider the collision viewed from a frame moving with the CM velocity V CM. v * 1,i m 2 v * 2,i m 2 v * 1, f m 2 v * 2, f

16 Example Step 2: Calculate the initial velocities in the CM reference frame (all velocities are in the x direction): v box,lab v box,cm v cm,lab v v * v - v CM v v * v CM v CM v * v * 1,i v 1,i - v CM 1.5 m/s m/s 1.2 m/s v * 2,i v 2,i - v CM 0 m/s m/s -0.3 m/s v * 1,i 1.2 m/s v * 2,i -0.3 m/s

17 Example Step 3: Use the fact that the speed of each block is the same before and after the collision in the CM frame. v * 1, f -v * 1,i v * 2, f -v * 2,i V * 1,i m 2 V * 2,i m 2 x v * 1, f - v * 1,i -1.2m/s v * 2, f - v * 2,i.3 m/s V * 1, f m 2 V * 2, f

18 Example Step 4: Calculate the final velocities back in the lab reference frame: v v * v CM v v CM v * v 1, f v * 1, f v CM -1.2 m/s 0.3 m/s -0.9 m/s v 2, f v * 2, f v CM 0.3 m/s 0.3 m/s 0.6 m/s v 1, f -0.9 m/s v 2, f 0.6 m/s Four easy steps! No need to solve a quadratic equation!

19 Summary: DEMO v 1,i m 2 v 2,i 0 m 2 x v 1, f m 2 v 2, f

20 4-step summary 1. Find velocity of center of mass of the system 2. Transfom the velocity of each object to the center of mass using v 1,i * = v 1,i v cm and v 2,i * = v 2,i v cm 3. Reverse the velocities of each object in the center of mass because the speed of each block is the same before and after the collision in the CM frame: v 1,f * = - v 1,i * v 2,f * = - v 2,i * 4. Transform the velocities back to the lab frame using v 1,f = v 1,f * + v cm

21 CheckPoint A green block of mass m slides to the right on a frictionless floor and collides elastically with a red block of mass M which is initially at rest. After the collision the green block is at rest and the red block is moving to the right. How does M compare to m? A) m > M B) M m C) M > m m M Before Collision m M After Collision

22 CheckPoint A) m > M B) M m C) M > m A) m sets M into motion B) if m > M, then both blocks would move to the right. If m < M, m would move to the left. C) The second block must be larger because it makes the first block stop. m M Before Collision m M After Collision

23 CheckPoint Two blocks on a horizontal frictionless track head toward each other as shown. One block has twice the mass and half the velocity of the other. The velocity of the center of mass of this system before the collision is A) Toward the left B) Toward the right C) zero 2v m v 2m Before Collision This is the CM frame

24 CheckPoint Two blocks on a horizontal frictionless track head toward each other as shown. One block has twice the mass and half the velocity of the other. Suppose the blocks collide elastically. Picking the positive direction to the right, what is the velocity of the bigger block after the collision takes place? A) 2v B) v C) 0 D) -v E) -2v 2v v + m 2m Before Collision

25 CheckPoint Picking the positive direction to the right, what is the velocity of the bigger block after the collision takes place? A) 2v B) v C) 0 D) -v E) -2v 2v m v 2m Before Collision + This is the CM frame B) for cm calculation, v2i = v2f, so the just swap directions because of the collision. What was -v before is +v now. C) If the velocity of the center of mass is 0, neither block can be moving otherwise the Vcm would be nonzero. D) In center of mass frame, initial velocity equals final velocity.