Your Comments I understand the concepts, but the math formulas being spat out like machine gun bullets are really confusing me. Is there a way we can see the formulas worked out by hand? I have a hard time connecting steps when shown on the prelecture/lecture slides I like physics better than chemistry. All they teach you how to do in chemistry is make crystal meth. And that is very bad indeed. It's 4 AM on Friday and I couldn't sleep. What do I do? Do Tuesday's homework, prelecture and checkpoint. Who knew having insomnia could be so productive! Conceptual examples would be nice, since that is what the test consists of. May your mother be with you. Get it? May the F=ma be with you? Oh, please put this on the screen before class!! I overall understand this concept more than the rest. But please go over the Double Mass but Half Speed problem in detail! This all seems to make sense - momentum/elastic/inelastic tie together well. Why isn't the answer to the first question in the pre-lecture elastic? Don't both of the boxes move? I liked the Hitchhiker's Guide themed lecture a few weeks ago, can we have another? Preferably Doctor Who, I know there are plenty of Whovians in this class. You could have a collision problem where the TARDIS collides with a Dalek traveling through the Time Vortex and see where they end up.

Question: “why is there no curve on the test we just took?” Answer: The average was 80% and we only scale an exam up if the average is less than 75%, we never scale it down. The only time there is ever any curving in this class is if we mess up and make something too hard. This is extremely rare. “That test was nothing like the previous midterms. Why did you have to make it harder?” Question: Answer: It wasn’t harder. The average was 80% without any curve.

Physics 211 Lecture 12 Today’s Concepts: a) Elastic Collisions b) Center-of-Mass Reference Frame

Review: Center of Mass & Collisions so far The CM behaves just like a point particle If then momentum is conserved If you are in a reference frame moving along with the CM then the total momentum you measure is 0. What is meant by "velocity of the center of mass of this system is zero"?

”…And can we go over more on how to tell the difference between an inelastic and elastic collision and the kinetic energies related to them.?” “How are elastic and nonelastic collision different??” Inelastic if nonconservatve forces do work: Friction Deformation Sticking together Just about anything

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. initial final A) Kinitial > Kfinal B) Kinitial = Kfinal C) Kinitial < Kfinal

CheckPoint Response A) Kinitial > Kfinal B) Kinitial = Kfinal C) Kinitial < Kfinal A) Since the boxes stuck together, this is not an elastic collision, therefore Kinetic Energy is not maintained. B) Since it sticks, all kinetic energy is conserved. C) This is not an elastic collision, therefore some Ke is lost.

Relationship between Momentum & Kinetic Energy since This is often a handy way to figure out the kinetic energy before and after a collision since p is conserved. same p initial final

In the CM reference frame, PTOT = 0 In the CM reference frame, vCM = 0 Center-of-Mass Frame videos In the CM reference frame, PTOT = 0 In the CM reference frame, vCM = 0 “If two objects collide in opposite directions with the same momentum, DO THEY MOVE AFTER THE COLLISION??? WHAAAAA???”

Center of Mass Frame & Elastic Collisions The speed on an object is the same before and after an elastic collision is viewed in the CM frame: m1 m2 v*1,i v*2,i m1 m2 m2 v*1, f v*2, f m1

Example: Using CM Reference Frame A glider of mass m1 = 0.2 kg slides on a frictionless track with initial velocity v1,i = 1.5 m/s. It hits a stationary glider of mass m2 = 0.8 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? CM m1 m2 v1, f v2, f * x v1,i v2,i = 0 CM CM

Example Four step procedure: Step 1: First figure out the velocity of the CM, vCM. Step 1: 0 in this case

Example Now consider the collision viewed from a frame moving with the CM velocity VCM. m1 m2 v*1,i v*2,i m1 m2 m2 v*1, f v*2, f m1

Example Step 2: Calculate the initial velocities in the CM reference frame (all velocities are in the x direction): v v = v* +vCM v* = v - vCM vCM v* v*1,i = v1,i - vCM = 1.5 m/s - 0.3 m/s = 1.2 m/s v*2,i = v2,i - vCM = 0 m/s - 0.3 m/s = -0.3 m/s v*1,i = 1.2 m/s v*2,i = -0.3 m/s

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 m1 m2 V*1,i V*2,i m1 m2 x v*1, f = - v*1,i = -1.2m/s v*2, f = - v*2,i =.3 m/s m2 V*1, f V*2, f m1

Example Calculate the final velocities back in the lab reference frame: Step 4: v v = v* +VCM vCM v* v1, f = v*1, f + vCM = -1.2 m/s + 0.3 m/s = -0.9 m/s v2, f = v*2, f + vCM = 0.3 m/s + 0.3 m/s = 0.6 m/s v1, f = -0.9 m/s v2, f = 0.6 m/s Four easy steps! No need to solve a quadratic equation!

I don't understand the purpose of switching reference frames I don't understand the purpose of switching reference frames. Isn't it enough just to say that momentum and KE are conserved in elastic collisions? Yes “I get why switching to center of mass makes things easier. But at the same time, I feel like quadratic equations would be easier than trying to keep track of switching and switching back” Try it ! “Can we just play a few rounds of pool for this week's lab?” 2D demo

Summary: m1 m2 v1, f v2, f x v1,i v2,i = 0

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 Before Collision After Collision m M We will now move on to consider a very interesting class of problems – that of collisions – and we will start with a very specific example to guide our discussion. Consider a box of mass m1 sliding along a horizontal frictionless floor in the positive x direction with an initial velocity vi. Suppose that there is a second box of mass m2 which is initially at rest, and that the first box collides with the second one. After the collision the boxes stick together and move with the same final velocity vf. Our job is to find a relationship between the initial and final velocity. In this problem the system we are interested in is made up of the two boxes. Since the floor is horizontal and frictionless, the total external force on the system in the x direction is zero which means that total momentum of the two boxes in the x direction will be the same before and after the collision. At this point you might be wondering about the forces that will act between the boxes during the actual collision itself – wont these forces change the momentum of the system? [show middle picture] The answer is no – they will not – and the reason is simple: The forces between the boxes are not from an external source since both boxes are part of the system. In other words, the force by box 1 on box 2 will change the momentum of box 2 and the force by box 2 on box 1 will change the momentum of box 1, but the total momentum of the two boxes combined will not change since these forces are equal and opposite by Newton's third law. For this reason we never actually have to worry about what happens during the instant when the boxes collide – we can just focus on the total momentum before and after [fade out the middle picture]. In this example the initial momentum of the system in the x direction is due entirely to box 1 [show Pinitial = m1vi]. The final momentum of the system is due to both boxes [show Px,final=(m1+m2)vf]. Since the initial and final of the system has to be the same, we can solve for the final velocity in terms of the initial. We see that the final velocity of the combined boxes is lower than the initial velocity of box 1 by itself. This makes because the mass of the moving stuff is bigger after the collision so the velocity after the collision has to be smaller in order to keep the momentum the same. 19

CheckPoint A) m > M B) M = m C) M > m M m A) m is has greater momentem, enough to push block M B) if m was greater than M, it would keep rolling, and if m was less than M, it would bounce back. C) If m>M then it wouldnt have stopped moving. Before Collision After Collision m M We will now move on to consider a very interesting class of problems – that of collisions – and we will start with a very specific example to guide our discussion. Consider a box of mass m1 sliding along a horizontal frictionless floor in the positive x direction with an initial velocity vi. Suppose that there is a second box of mass m2 which is initially at rest, and that the first box collides with the second one. After the collision the boxes stick together and move with the same final velocity vf. Our job is to find a relationship between the initial and final velocity. In this problem the system we are interested in is made up of the two boxes. Since the floor is horizontal and frictionless, the total external force on the system in the x direction is zero which means that total momentum of the two boxes in the x direction will be the same before and after the collision. At this point you might be wondering about the forces that will act between the boxes during the actual collision itself – wont these forces change the momentum of the system? [show middle picture] The answer is no – they will not – and the reason is simple: The forces between the boxes are not from an external source since both boxes are part of the system. In other words, the force by box 1 on box 2 will change the momentum of box 2 and the force by box 2 on box 1 will change the momentum of box 1, but the total momentum of the two boxes combined will not change since these forces are equal and opposite by Newton's third law. For this reason we never actually have to worry about what happens during the instant when the boxes collide – we can just focus on the total momentum before and after [fade out the middle picture]. In this example the initial momentum of the system in the x direction is due entirely to box 1 [show Pinitial = m1vi]. The final momentum of the system is due to both boxes [show Px,final=(m1+m2)vf]. Since the initial and final of the system has to be the same, we can solve for the final velocity in terms of the initial. We see that the final velocity of the combined boxes is lower than the initial velocity of box 1 by itself. This makes because the mass of the moving stuff is bigger after the collision so the velocity after the collision has to be smaller in order to keep the momentum the same. 20

Newton’s Cradle

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 m 2m 2v v Before Collision This is the CM frame

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 m 2m 2v v + Before Collision

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 + This is the CM frame Before Collision m 2m 2v v B) The objects just change direction with the same speed in the cm reference frame. C) The total momentum of the system before and after the collison is zero. D) Initial Velocity and Final Velocity are the same in elastic motion.

do a problem were we have to go between the enter of mass reference frame and the labs reference frame

vcm is the same = final speed B C Do same steps for car 2 vcm is the same = final speed Compare 1/2 mv 2 before and after for both cases