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A stationary object with mass $m_{B}$ is struck head-on by an object with mass $m_{A}$ that is moving initially at speed $v_{0}$ , (a) If the collision is elastic, what percentage of the original energy does each object have after the collision? (b) What does your answer in part (a) give for the special cases $(1) m_{A}=m_{B}$ and $(i i) m_{A}=5 m_{B} ?(c)$ For what values, if any, of the mass ratio $m_{A} / m_{B}$ is the original kinetic energy shared equally by the two objects after the collision?

(a) $\frac{4 m_{A} m_{B}}{\left(m_{A}+m_{B}\right)^{2}}$

(b) $\frac{K_{A 2}}{K_{1}}=\frac{4}{9}$ and $\frac{K_{m 2}}{K_{1}}=\frac{5}{9}$

(c) $\frac{K_{n 2}}{K_{1}}=\frac{1}{2}$

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Rutgers, The State University of New Jersey

Numerade Educator

University of Washington

Hope College

{'transcript': "So it's exercise. We have a perfect elastic collusion between two particles. So particle A and particle be so In the beginning, Particle A is moving in the horizontal access with initial velocity V zero and the park will be is stead still on? After that, the particle a collides head on with particle B and both particles changes Velocities of particle begin a velocity V b Q and particle way gained a A velocity of V h. Okay, so in exercise A let me write it here. We want to find what fraction off the initial kinetic energy that I'll call. Okay, zero goes for particle A and what fraction goes to particle be okay, So we have We want to find expressions for this and this. It was so starting with the fraction with the expression from the fraction off cannot energy that goes to park away. So we have that the initial kinetic energy is just a massive particle. A times the initial velocity squared over two and then the final kinetic energy of particle A is it with you the massive Parco a V H two squared or Richard? Um so this fraction, uh, begin becomes V H two squared over 30 squared. Okay, for the more we also have that leszek collusions have, uh, a relation between the initial and final velocity off this first particle here. So I can write that they the final velocity V two is equal to, um, a minus M b over a blues M b these times be sorry, v zero. Then you should also forgo it. Okay, This is a equation. Uh, that works for elastic collisions on Lee on, and it is written in Equation 8.24 in the test book. Okay, so we just use this and to be astute in our expression, so we have that. Okay, He's zero. So, So big V A squared. We have I am a minus m b over. I am a plus m b squared. Okay. The V zero squared cancels out with zero from the dominator. So this is the first expression we wanted to find. The second expression is for the amount off the fraction of energy that goes to particle Be What? So, um, we can argue that since we're working with a plastic collusion, we have conservation off kinetic energy such that uh um que a plus k b has to be equal to zero, which also implies that if I divide the conservation off kinetic energy by K zero in both sides, we have that the some off the fraction that goes to particle A with the some off the fraction, with the fraction that goes to particle be has to be equal to one. Okay, this is go thesis school from this come from conservation of energy. So conservation off kinetic energy. Okay, okay. So in this case, I can isolate the fraction k B zero to be one minus k over K zero. So I only pick one minus the expression I found previously. So one miners m a minus m b over, um a plus m b Everything squared. So performing this objection, we find that the fraction that goes to, uh, K B is equal to and May waas MBI squared. So this is going to be a really big term. So here, this one becomes m a plus m b squared. So I'm going to expand this here inside off this parenting, so am re squared. Plus m b squared was too, um, may and be miners and may minus them B squared. I'll also expand this so we have and may squared less M b squared minus shoe and may M b Okay, so because of this minus sign here and may squared cancels with Amy squared and B squared also cancels with and B squared. And we can sum up these two terms such that we have that the fraction of energy the fraction of the initial energy that goes to particle B is going to be equal to 4 a.m. a. M B over m A plus m b squared. So this is the final answer. All questions. A now question. Be asked us to evaluate the americal value off this fractions. Okay, que zero and K B zero for some cases. So the first case is when the mass off particle way is equal to the mass off particle. Be so we can go back to the case expression. And we have that when m A is equal to m B, this term become zero. OK, zero goes to zero, so que que zero is equal to zero. Okay. Which means that the the particle a well, we lose all its initial velocity will be standing still after the collision and from the conservation off kinetic energy we have that baby zero will be obviously equal to one. And the second case is when, um sorry. Um, A is equal to five times and be okay. So this time I'll rewrite the expression again. Here. Eso we have that This is going to be I am a, uh sorry. Um, so this would be six and sorry, uh, for M b over six. M b squared. Okay, because I said be stewed this by five MB. So five and B minus and be becomes for M B and five and B plus and B six and b. Okay, so from this, we can cancel all today's with this, and we have to over three. So two squared over three squared is equal to 4/9. So the first answer and now sorry, this is not over. KB is over zero. So now we can once again use conservation off, uh, kinetic energy and but one minus for over nine. And find that the amount of energy that will go to particles to park O B is going to be 5/9. Okay, Uh, now, questions see asked us to find the values off the mass ratio and may end over and be such that the kinetic energy will be equal, equally shared between the two particles. So questions see Okay, we have We want to find this fraction in the case where K B over zero is equal to K A over zero. And we know from conservation off kind of energy that this fraction will be will have to be equals. You wanna have? Okay, so it just pick the expression from K P over K zero that we found on question A and, uh, an equal to the expression we found for K A over K zero. So we pick this expression here, they'll call one and put this equals to this expression to So we have expression. One equals true expression to this leads to, um m m a sorry for em, eh? And be over. Um um, A plus and B squared has to be equal to m A minus m b over a may plus, um, be both terms squared. So notice that we can cancel out the denominators and expanding this term here on the right hand side, we have four m A and be equal to M a squared plus M b squared minus two. Um, a m B. Okay, so if we divide, uh, the both sides of the equation by m. B squared, we'll have that. So remember that we want to find this fraction here. I'm sorry. So we want to find this fraction here. So dividing both sides of the equation by and B squared we have that, uh, So I m a over m b squared plus one minus. So passing for, um a m b to the other side of the equation minus six. Uh um a over m B. This is And this is it for 20 So notice that this is a credit credit IQ formula on which my variable is m A over M b. So solving the quadratic formula, we have that the truth possible? Uh, answers are six plus minus the square root off 32 over to, and this leads to the two possible fractions. We can have so five point 83 and the second one to be 0.172 on this concludes the exercise"}