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A 10.0 -g marble slides to the left with a velocity of magnitude 0.400 $\mathrm{m} / \mathrm{s}$ on the

frictionless, horizontal surface of an icy New York sidewalk. and has a head-on, elastic collision with a larger $30.0-\mathrm{g}$ marblesliding to the right with a velocity of magnitude 0.200 $\mathrm{m} / \mathrm{s}$ (Fig. 8.38$) .$ (a) Find the velocity of each marble (magnitude and direction) after the collision. (Since the collision is head-on, all the motion is along a line. (b) Calculate the change in momentum (that is, the momentum after the collision minus the momentum before the collision) for each marble. Compare the values you get for each marble. (c) Calculate the change in kinetic energy (that is, the kinetic energy after the collision minus the kinetic energy before the collision) for each marble. Compare the values you get for each marble.

(a) $-0.100 \mathrm{m} / \mathrm{s}$ ; $0.500 \mathrm{m} / \mathrm{s}$

(b) $+9\mathrm{g} \cdot \mathrm{m} / \mathrm{s}$

(c) $+4.5 \times 10^{-4} \mathrm{J}$

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in this question of 30 g. Marble is moving to the right with the velocity off 0.2 m per second, while a 10 g marble is moving to the left with the velocity off 0.4 m per second. The first item off this question, we have to calculate what is the velocity off marbles? Numbers one and two after the collision, provided that the collision is elastic. For that, we have to use two equations. One is this equation, which is the equation for the relative velocities after and elastic collision when the two bodies are moving on. The other equation is the equation that comes from using the principle of conservation of momentum. So by using the principle of conservation of momentum, we have the following the net momentum before the collision is equals to the net momentum. After the collision before the collision, we have two marbles moving. Let me choose my reference frame to be such that everything that is pointing to the right, it's positive and everything that is pointing to the left is negative. Then, in this reference frame before the collision, the momentum was given by the some off the momentum off marble number one on the momentum off marble number two. The momentum off marble number one is given by M one times If you won the momentum off marble number two is given by M two times we chew on This is equals to the net momentum after the collision after the collision, both marbles still moving. So we have a one view on prime plus M two times V two Prime, where the primate velocities are the velocities after the collision. Now, by plugging in the values that the problem gives us, we got the following m one is a cost authority V one is plus 0.2 in our reference frame. Now we have minus because V Chu will be negative so minus M. Chu which is the cost of 10 g times 0.4 and this is equals two and one, which is 30 times view one prime plus M. Chu, which is 10 times V chew prime. Then, by working on the left hand side, we got the following 30 times View one prime plus 10 times V two prime. Is he, of course, to chew. Note that we have only one equation, but two variables. So we need another equation to be able to solve. This problem on this other equation comes from this one, which is the equation for the relative velocities. So let me call this equation. We had just arrived. Equation number one and let's go in the right equation number two. So equation number two thousands. That V two prime minus V one prime is equal to minus V Chew, which is minus 0.4 minus view one, which is plus 0.2. This leads us The following conclusion V two prime minus. View one prime is the cost to minus minus 0.6. Therefore V two prime minus. View one prime is equals to 0.6. Let me call this equation number two. So now we have two equations and two variables. Let me organize my board so I can continue the solution. Okay. Now, using equation number two, we can say the following According to equation number two V two, prime is equals to 0.6 plus view on prime. Now, let us substitute this result in equation number one. By doing that, we get the following 30 times view one prime plus 10 times v Chew Prime which is 0.6 plus view one prime is a cost to chew Then Now we have Onley one variable so we can solve this equation for a view on prime Let's do it So here we have 30 times view on prime plus six plus 10 times view one prime being equals To choose to reform, we have 40 times view one prime being equals to choose minus six So 40 times with you on prime is minus four. The reform view one prime is minus four divided by 40 which results in minus 0.1 m per second. The reform after the collision the big marble is moving to the left with a velocity off 0.1 m per second. Then to calculate the velocity off marble number two after the collision, all you have to do is substitute. This result entered this equation. By doing that, we get the following so it's substituting that equation back into equation. Number two leads us to V two. Prime equals to 0.6 minus 0.1. Then V two prime is equals to 0.5 m per second. So after the collision, marble number two is moving to the right with a velocity off 0.5 m per second. And this is the answer to the first item off this question. Now, before moving into the second item, let me organize my board in the second item off this question, we have to calculate the variations in the momentum's off marbles number one and marbles number two on. Then compare. So let us begin by marble. Number one The variation in the momentum off marble number one is given by the momentum off that marble after the collision, which is m one times V one prime minus the momentum off the marble before the collision and one V one. So, as you can see, we can factor the mass. So we have m one times V one prime minus V one. By plugging in the values that we got, we have the following. Everyone is 30 grands view one prime is minus 0.1 and then we have minus V one, which is 0.2. These results in Delta P one equals to 30 times, minus 0.3 on these results in a variation off minus 9 g times meters per second. This is a variation of the momentum off marble number one. So let me write it here minus 9 g meters per second. Now let's do the same for marble number two for marble number two, we have the following variation in the momentum. Delta Bi Chu is because of the momentum after m two V two prime minus the momentum before m two v chew. Then we can factor m two so that we got em two times V two prime minus V two. Now plug in the values that the problem gives us. We got 10 times V two prime, which is 0.5 minus the velocity number two before the collision which was minus 0.4. This results in 10 times plus 0.9 because here we've got a plus sign on this results in a variation the momentum that is 9 g times meters per second. So as you can see the variation, the momentum off marble number one is because to minus the variation the momentum off marble number two. This is expected because momentum is conserved. So any variation of the momentum off marble number one must be compensated by some variation in the momentum off marble number two Until this precisely. What's happening here now, Before going to the final item, let me erase these calculations that I have done for item B. Okay for items, See, we have to evaluate the changes in the kinetic energies off marbles number one and number two. The variation in the kinetic energy off marble number one is given by the kinetic energy after the collision and one times view one prime squared, divided by two minus the kinetic energy before the collision. And one times we want squared, divided by two plugging in the results that we have, we get the following Delta K one is equal to the mass off marble number one, which is 30 g. Here. I want to convert 2 kg so that we get our answer in jewels. So let's plug in a tanto ministry to convert from grams to kilograms. So this is divided by true and animal to apply by view one prime which is minus 0.1 and then square it for the kinetic energy. After the collision, we have 30 times stand to ministry divided by two times view one squared view one waas zero point shoes. So we have 0.2 squared. This gives us a variation. The kinetic energy off minus 0.0 45 which we can write as minus 4.5 times 10 to minus four Jun's This is a variation the kinetic energy off marble number one. Now let us do the same for marble number two. The variation in the kinetic energy off marble number two is given by the kinetic energy off marble number two After the collision m two V two prime squared divided by two minus the kinetic energy off marble chew before the collision Plugging in the values that we have we got the following m two is 10 times stand to ministry again. I'm converting from grams to kilograms. That's why I'm plugging. Attend to ministry here times V two prime squared So 0.5 squared and this is divided by two. Now we subtract m Chu, which is 10 times stand toe ministry times v chew squared before the collision, virtue was minus 0.4 squared. Now we divide by two These results in a variation, the kinetic energy off marble number two that is given by 0.0 45 Jews, which we can write as 4.5 times, stand to minus four jewels. So the variation in the kinetic energy off marble number two is 4.5 times 10 to minus four Jews. As you can see, the variation in the kinetic energy off marble number two is you cause to minus the variation in the kinetic energy off marble number one Again, this is to be expected because we're talking about an elastic collision on in that kind of collision, the kinetic energy is conserved. So any change in the kinetic energy off marble number one must be compensated by a change in the kinetic energy off marble number two.

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