00:01
In this question, we have two particles, one particle at rest, and another particle that is a neutron that collides head on with the previous particle.
00:12
We are asked to find what is kinetic energy, what is the fraction of the kinetic energy lost by this neutron after the collision.
00:20
So let's write down the expression that we're trying to find.
00:23
We want to find the change of kinetic energy, that is initial of a, minus, kinetic energy final, of a.
00:37
A divided by the kinetic energy initial of a.
00:43
So that's just a fraction of kinetic energy lost by a.
00:46
But we can simplify this a little bit because we know the risk of kinetic energy, meaning whatever energy is lost by a is equal to the energy gained by b.
00:55
But since b is initially at rest with no kinetic energy, the energy gained by particle b is just the final kinetic energy over kinetic energy initial.
01:10
So this gets us half of mbb prime square over one half an a va square.
01:25
Or we can cancel the one half if we want.
01:27
So from here what we need to know is we want to find the expression that has vb prime in terms of va.
01:35
So we can plug in that expression and va will cancel and we'll have a fractional number.
01:40
So that is great, but we can write down the conservation of kinetic energy and we can write down the conservation of momentum and we can work it out but if you have already read section 7 .4 and 7 .5 you'll find that it's already worked out in example 7 .8.
01:59
So example 7 .8 has this similar configuration where you have two particles one at rest another collides head on with the previous particle and 7 .8 walks through the whole step and it tells you what is the energy and what is the velocity of the two particles after the collision.
02:18
So i think we will just skip that and we'll just grab the increasing that's already derived in the book.
02:25
So in the example 7 .8 we have vb prime equals va prime, va actually, times 2ma over m .a plus mb.
02:41
This is awesome because in this equation we already have vb prime in terms of va, which is what we're looking for.
02:48
So it saves us some work and we can just plug this into mb times vb square becomes va square times 2ma over m .a plus mb square and it plus m a b a square and va square cancels.
03:07
So now this becomes 4 times mamb over m a plus mb square so this is a fraction of kinetic energy loss by particle a we will be using this simplified equation for the different cases we have to deal with so for the first case we have m a equals m b equals 1 .01 u we can plug this in and because m a equals mb we will have the fraction equals um it's four m square both in the numerator and denominator equals one but actually we don't even have to do that for particle a or for case a because we already know from the problems we've already walked through say like problem 28 or from example 7 .7 we know that when two particles have the same mass and they collide they will exchange speed so in this case um particle and particle b will exchange the speed, meaning particle b will be moving at the speed of particle which is v afterwards.
04:25
And particle will be moving at the speed that particle was, which is zero because particle b was at rest.
04:32
That means that if they are of the same mass after the collision, a will be at rest and b will be moving.
04:39
So since a becomes at rest, all of its kinetic energy is lost.
04:43
So we have the fraction equals 1...