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A neutron with mass $m$ makes a head-on, elastic collision with a nucleus of mass $M$ . which is initially at rest. (a) Show that if the neutron's initial kinetic energy is $K_{\infty}$ the kinetic energy that it loses during the collision is 4$m M K_{0} /(M+m)^{2} .$ (b) For what value of $M$ does the incident neutron lose the most energy? (c) When $M$ has the value calculated in part (b), what is the speed of

the neutron after the collision?

(a) $\frac{4 K_{0} m M}{(M+m)^{2}}$

(b) $M=m$

(c) $v_{A 2}=0$

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Rutgers, The State University of New Jersey

University of Michigan - Ann Arbor

Simon Fraser University

Hope College

{'transcript': "in six er size. We have a new trim off mass small m, and this neutral initially has a velocity V one and kinetic energy ke zero, and it's going to collide head on with, uh, nuclear nuclear's off mass capital M. That is initially at rest. So after the collision, the nucleus gain a velocity V B one that points to the same direction as the initial velocity off the neutron and the neutron gain a velocity v a to that points to the negative direction. Okay, so in question. Hey, we want to find, um, What is the laws on the nutrients? Kinetic energy after the collision. So right here the los off the neutrals, kinetic energy they'll take a and capital a small and for neutral. So as the kinetic energy in the beginning, minus the kinetic energy in the end, which is equal to, um hmm v a to squared over to Okay. Furthermore, I can rewrite zero in terms off the one. So we have, um, over to times V a one squared minus v a to squared. Okay, so something you didn't mention Is that the collision off the new strong with the nucleus is an elastic collision. So let me write it here. Last tick collision. So in a elastic collision, we have conservation off linear momentum. And we also have conservation off kinetic energy. And by this, we can use uh, And because of this off these two conservations, we have the follow relations so we can relate the initial velocity off the neutral with its final velocity as follows. So we have v u two is equal. True? Um, minus capital. Um, over, um, plus capital, um, times we a one. Okay, so this is equation, um eight, uh, points 24 from the book. Okay, so this equation is just for, uh, elastic collisions. So we can substitute this expression in here, and we have that The variation on the nucleus kinetic energy is going to be, um um, be a one squared over two times one minus, um, minus capital. Um, over, um less skeptical. Uh um um, everything squared. Okay, So, uh, so we can make this super traction here inside off the square brackets. And we have, um v a one squared over two. Sorry, We can already rewrite thes first term as K zero because we want to find the expression in terms off zero. So okay. Zero, um, am plus am squared mines. I am minus, um, squared over blood. Um, squared. Okay. So we can expand these squares terms here and here. So if we expanded, it will appear some terms that we can cancel out. So let me do this real quick. So we have for the first squares term have m squared plus m squared. Plus two. Um um and this is minus AM squared plus m squared minus two. Um um, this over m plus m squared. Okay. Eso noticed that because of this negative sign this M capital m squared consoles with this capital m squared this m squared cancels with this and this to sum up so that we find that the variation on the neutrons kinetic energy is equal to four que zero and small m over m plus m squared. So this is the answer off the first exercise. No question. Be as the should find. What is the mass off the nucleus that maximizes the loss on the kinetic energy? So, question be, we want to find what is the value off capital M that, uh, extra my XYZ Delta Cayenne. Okay, so to do so, two extra mayes a function so we can notice that Delta Kate and is a function off capital m so we can solve this exercise by, uh, deriving Delta K and over em and picking the point on which the derivative is equal to zero. So the extreme point will be here. Okay, So Okay, so let's do so. So we have that de delta km d m is equal to So I will put four k zero. I am in evidence. So for que zero small m times. So the derivative off the capital m here, So this becomes just one over m plus m is squared, plus the derivative off one over m plus small. I am everything squared so deserve a tive is just minus two AM over. And plus, I am Cube. Okay, So if you don't know how to calculate this last derivative, I just use it The chain room, the chain role. Sorry, on which we have that, um, the derivative of one over and plus, um, squared over em over D m. Yeah. Is equal to the derivative off all over. Um m plus m squared in, um, the variable m plus, um, times derivative off the verbal M plus M over em. Okay, so this first derivative here is just, um is just to minus two over m plus, um, cube. Okay. And deserve a tive is just one. Okay. Okay, So we have that. This has to be equal to zero in order to find the mass. That extra Mayes is the loss on the nutrients energy. So, for this should be close to zero. We know that none of these terms we can put this terms on the other side of the equation, so they vanish so we can perform the subtraction here. And we have something like this, so let me rewrite it. Here we have. I am plus M minus 2 p.m. Over M plus, um, cube. Okay. This has to be equal to zero. So this is just disobey attraction here so we can pass, uh, determine the denominator to the other side. So it vanishes, and we have that the mass off the nucleus has should be equal to the mass off the neutron. Okay, no question. See, as the ship calculate what is going to be the final velocity off the neutrons. So what is going to be V two if, um is equal to a small M? If the mass off the nucleus is equal to the mass off the neutral so we can recall once again equation 8.2 24. Thanks. This where we have that V H two is going to be equal to, uh, small M minus capital M over small m plus m This times the initial velocity okay, so indicates where capital M zero swim. This term here goes to zero, and we have that The final velocity is going to be equal to zero, and this is the answer off the exercise."}