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Puck $A$ (mass $m_{A} )$ is moving on a frictionless, horizontal air table in the $+x$ -direction

with velocity $\vec{v}_{A 1}$ and makes an elastic, headon collision with puck $B\left(\text { mass } m_{B}\right),$ which is initially at rest. After the collision, both pucks are moving along the $x$ -axis. (a) Calculate the velocity of the center of mass of the two-puck system before the collision. (b) Consider a coordinate system whose origin is at the center of mass and moves with it. Is this an inertial reference frame? (c) What are the initial velocities $\vec{u}_{\Delta 1}$ and $\vec{u}_{B 1}$ of the two pucks in this center-of-mass reference frame? What is the total momentum in this frame? (d) Use conservation of momentum and energy, applied in the center-of-mass reference frame, to relate the final momentum of each puck to its initial momentum and thus the final velocity of each puck to its initial velocity. Your results should show that a one-dimensional, elastic collision has a very simple description in center-of-mass coordinates. (e) Let $m_{A}=0.400 \mathrm{kg}$ ,

$m_{B}=0.200 \mathrm{kg},$ and $v_{A 1}=6.00 \mathrm{m} / \mathrm{s}$ . Find the center-of-mass velocities $\vec{u}_{A 1}$ and $\vec{u}_{B 1},$ apply the simple result found in part (d),

and transform back to velocities in a stationary frame to find the final velocities of the pucks. Does your result agree with Eqs. $(8.24)$ and $(8.25) ?$

a. $v_{\mathrm{sm} x}=\frac{m_{A} v_{A \mathrm{I}}}{m_{A}+m_{B}}$

b. The center of mass moves with constant speed so this coordinate system is an inertial frame.

c. 0

d. $p_{B 2 x}=-p_{B 1 x}$

e. $8.00 \mathrm{m} / \mathrm{s}$

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Rutgers, The State University of New Jersey

University of Michigan - Ann Arbor

Numerade Educator

Hope College

{'transcript': "problem. 8.21. So we have some bucks moving around frictionless Lee on an air table. Hey, is moving to the right with some speed that we don't know where's Bee, which is initially address, hear their masses and then sometime later they will collide. And then after that okay is moving to the left to the speed of 0.12 meters per second. And he is moving to the right of the speed of 65 meters per cent. And so the two things we want to find or what was the initial speed of a sort of a big mystery here, like, you know, very interesting to know that. And then what is the change in the kinetic energy? Because it will turn out This vision is not the last. It's partially elastic. So let's start with party. We know from the conservation of momentum that the initial momentum, which since is the only thing moving will be just the massive eh times. It's eager to you like to know, And now this is going to be equal to the sum of the Momenta over here. That's a day because that has not changed. Piece of a final. It should be a negative number because it's moving to the lab. Plus, I meant to be now, in order to solve this, it's pretty easy. You just divide both sides by a and so we have the final speed of A. This is going to be negative. Plus the ratio of the masses, the final speed. And so now we just put all of these things in since so bye here. But it's numbers. Together you get its initial speed 0.7 i zero now to find a change in Connecticut, all we have to do this computer be a final and initial that again because now we know how fast everything was moving in both situations. So the final Connecticut take this one outside. You say that you did it massive, eh? It's finalised, which is naked recall, but they're squaring it. Explain. My kinetic energy had to begin with you. And so then put in all of these numbers we get that change in Medic energy is negative, which they asked, zero are naked because this is nothing going on in this problem. Externally add energy to the system. And so the Connecticut you can either be in only either be conserved or if the deformation of the pucks is not perfectly elastic. So that will turn into, you know, very slight temperature rise and the pots. And this comes out to be a fairly small number. 0.2 the jewels. Some energy was lost, but not a lot. If you'd like to compute what the percentage changes, you divide the final kinetic energy by the initial what percentage of the energy was concerned? But we don't need to worry too much about that. Here. We could do that yourself. Curious would be a good good exercise for you, perhaps, But this is the second thing we were asked if I This was the first thing I asked if I So we've solved the mystery of B, and we've also determined that this was not an elastic"}