Here we will look at an example of subatomic elastic...

Here we will look at an example of subatomic elastic collisions. High-speed neutrons are produced in a nuclear reactor during nuclear fission processes. Before a neutron can trigger additional fissions, it has to be slowed down by collisions with nuclei of a material called the moderator. In some reactors, the moderator consists of carbon in the form of graphite. The masses of nuclei and subatomic particles are measured in units called atomic mass units, abbreviated u, where 1u = 1.66Ã—10^(-27) kg. Suppose a neutron (mass 1.0 u) traveling at 2.6Ã—10^7 m/s makes an elastic head-on collision with a carbon nucleus (mass 12 u) that is initially at rest. What are the velocities after the collision? If the neutron's kinetic energy is reduced to 9/16 of its initial value in a single collision, what is the mass of the moderator nucleus? Express your answer in atomic mass units as an integer.

Transcript

00:01 Here we're looking at two subatomic particles with masses of 1 u and 12 u and initial speeds, and they collide elastically, and we want to know what are their speeds after the collision.

00:22 So in order to do that, we're going to need to apply both conservation of energy and conservation of momentum.

00:31 In order to find those two unknowns.

00:35 So when it comes to conservation of energy, oh, sorry about that.

00:44 When it comes to conservation of energy, i'm going to skip a few steps for two objects that are colliding elastically.

00:54 Conservation of energy can always be simplified down to velocity, of one initial minus velocity of the second item initial is equal to minus velocity one final minus velocity two final or speed to final we know that this is going to be zero and we need both conservation of energy and conservation of momentum because we need to be able to solve for two unknowns, v1 final and v2 final.

01:39 So i'm going to go ahead and solve this for v2 final so that i can then plug that in in the future to other things.

01:50 So if we solve for v2 final, you get that v2 final is equal to v1 initial plus b1 final.

02:06 Now, we want to apply conservation of momentum.

02:13 And conservation of momentum will help us to solve.

02:15 So initially, again, only object one is moving.

02:20 And so the only momentum we have initially is going to be, sorry, my touchscreen is being a little bit unresponsive.

02:29 So we'll have m1b1 initial.

02:36 That's our initial momentum because only object one is moving.

02:40 That has to be equal to the final momentum, which will be the momentum of object 1 plus the momentum of object 2.

02:48 So we'll have m1 v1 final plus m2 v2 final.

02:58 But of course we know that v2 final is v1 initial plus v1 final, so we can go ahead and make that substitution, which is going to allow us to solve for v1 final.

03:14 So if you make that substitution, we'll have that, sorry, well, if you make that substitution, you'll have that m1v1 initial is equal to m1v1 final plus m2 v1 initial, plus m2 v1 final.

03:49 And again, we want to solve for v1 final.

03:52 So we'll have m1 plus m2 v1 final, if we gather those together.

04:05 And then on the other side, we'll be left with m1 v1 initial minus m2 v1 initial.

04:14 And so we're almost there.

04:15 We can go ahead now and solve for v2.

04:17 Final, or sorry, the one final, the one final is just going to be m1 v1 initial minus m2, v1 initial, all divided by m1 plus m2.

04:43 And we can go ahead and plug in all of the values that we were given in the problem, making sure to convert from atomic mass units to kilograms, and you would get a velocity of negative 2 .2 times 10 to the 7 meters per second for object 1, which we can then plug in back at the top to get a velocity of about 4 times 10 to the 6 meters per second for.

05:18 Object 2...

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