Determine the fraction of kinetic energy lost by a neutron $(m_1 = 1.01 \textrm{ u})$ when it collides head-on and elastically with a target particle at rest which is

VIDEO TRANSCRIPT

This is Giancoli Answers with Mr. Dychko. We have a neutron approaching some other nucleus which is initially at rest and there's going to be an elastic collision and our job is to figure out what fraction of the kinetic energy that this neutron initially has is lost after the collision? So we'll write down our conservation of momentum formula here and then also write down our formula we have because the collision is elastic and then work with these formulas by substituting one into the other until we have an equation that tells us what v 1 prime is the velocity of the neutron after the collision so that we can calculate kinetic energies because the change in kinetic energy of the neutron is gonna be its one-half m 1 times v 1 prime squared minus one-half m 1v 1 squared divided by the total initial kinetic energy which is just that of the neutron since the other particle is initially at rest. So we need to figure out v 1 prime and we'll take equation 2 and solve it for v 2 prime so that we can substitute for that variable in equation 1 to get rid of it and we know that the second nucleus is initially at rest so v 2 is zero so both these terms disappear. So we have v 2 prime is v 1 plus v 1 prime because we can move this term to the left side by adding it to both sides and then switch the sides around. So rewriting equation 1 with this term gone, we have m 1v 1 equals m 1v 1 prime plus m 2 times v 1 plus v 1 prime instead of v 2 prime. And then distribute this m 2 into the brackets here, we have m 2 times v 1 plus m 2 times v 1 prime and just copying these terms here and we wanna solve for v 1 prime so we'll factor out this v 1 prime from these two terms and it's gonna be multiplied by m 1 plus m 2 and then move this term to the left side and it has a common factor v 1 which we factor out and we get m 1 minus m 2 because this is subtracted from both sides so it becomes a negative m 2 there and then switch the sides around so we have the v 1 prime on the left and then divide both sides by m 1 plus m 2 and here we have our formula that works for any elastic collision when the second projectile's initially at rest; the first projectile's velocity after the collision will be its velocity before the collision multiplied by m 1 minus m 2 divided by m 1 plus m 2 and if you examine this equation a little bit, you can figure out the general behavior of this system after the collision. If m 1 equals m 2, this difference would be zero when the masses are equal, it means that the incoming projectile will stop because v 1 prime will be zero and likewise, if m 1 is less than m 2 that means this difference will be negative in which case v 1 prime is gonna be in the opposite direction to its approach velocity and so it rebounds, in other words. And if m 1 is greater than m 2, this is going to be a positive v 1 prime in which case, it keeps on going in the same direction. Now our real question has to do with kinetic energy so let's use this result for v 1 prime in this kinetic energy stuff here. So the fractional change in kinetic energy of the first projectile is this, it's the change in kinetic energy divided by the initial kinetic energy. So that's kinetic energy final minus kinetic energy initial divided by the kinetic energy initial and this one-half m 1 is a common factor everywhere so we can just divide everything by it and so it disappears and then I have substituted for v 1 prime by writing v 1 times m 1 minus m 2 all over m 1 plus m 2 in place of v 1 prime and it's squared from the formula here so let's keep that there and then minus v 1 squared all over v 1 squared. And it turns out that v 1 squared is a common factor for every term here so divide everything by v 1 squared. So that makes this term here, m 1 minus m 2 over m 1 plus m 2 squared and then v 1 squared divided by v 1 squared is 1— careful you don't just make this term disappear entirely— you get a 1 there when you divide v 1 squared by v 1 squared and this v 1 squared divided by v 1 squared is 1 and there's no need to write this whole thing over 1, that's just silly. So let's square the numerator that gives us m 1 squared minus 2m 1m 2 plus m 2 squared— that's the squaring a binomial— and then we'll take this 1 that's here and multiply it by m 1 plus m 2 squared over m 1 plus m 2 squared so we have that common denominator and so we have minus this m 1 squared plus 2m 1m 2 plus m 2 squared where I have squared that to make this here and all of that now gets written over m 1 plus m 2 squared because they have that as a common denominator. And then we'll make everything each of these three terms negative, that's all I did there, and then you can see that m 1 squared minus m 1 squared makes zero, m 2 squared minus m 2 squared makes zero and we are left with negative 2m 1m 2 minus 2m 1m 2 makes negative 4m 1m 2; the negative sign means that the change in kinetic energy is a loss, there's a loss in kinetic energy for this projectile. So since the whole question is what is the fraction of kinetic energy lost, we don't need the negative sign there so we'll just write it without the negative and now we are gonna substitute in all the numbers that we have for each of these parts. So for part 1, we have a neutron colliding with a hydrogen nucleus and they both have a mass of 1.01 atomic mass units. So we have 4 times 1.01 atomic mass units times 1.01 divided by the sum of the masses squared and that gives 1.00. All of the kinetic energy is lost which is another way of saying the neutron comes to rest which we kind of expected anyway based on this formula because we know that if the projectile's are of equal masses, this difference will be zero and v 1 prime will be zero; all of its kinetic energy will be lost, in other words. So part (b), we are plugging in 2.01 atomic mass units for heavy hydrogen and we have 4 times 1.01 times 2.01 divided by 1.01 plus 2.01 atomic mass units and square that denominator and we get 0.890 is the fraction of kinetic energy lost; it mostly slows down significantly, most of its kinetic energy is lost. And for part (c), a neutron colliding with a carbon nucleus, we get 4 times 1.01 times 12.00 divided by the sum of those masses squared 1.01 plus 12.00 and that gives 0.286 is the fraction of kinetic energy lost. And then colliding with a lead nucleus, which is a very heavy one, we get 4 times 1.01 times 208 atomic mass units divided by 1.01 plus 208 all squared in the denominator and we get 0.019 which means almost no kinetic energy is lost and what's happening is that the neutron is bouncing off the lead and the lead is so massive that that it rebounds with practically the same speed that it had when it was approaching. So lead makes a very firm—it's like hitting a wall— it looks like a very firm surface to bounce off of so that kinda makes sense that there is very little loss in kinetic energy. It's velocity is completely turned around because it goes back the other way and there we go.