00:01
So in this exercise we have a perfect elastic collusion between two particles, so particle a and particle b.
00:10
So in the beginning, particle a is moving in the horizontal axis with initial velocity 0 and the particle b is stead still.
00:22
And after that, the particle a collides head on with particle b and both particles changes velocity, so particle b gain a velocity vb2, and particle a gain a velocity of v82.
00:41
Okay, so in exercise a, let me write it here.
00:47
We want to find what fraction of the initial kinetic energy that i'll call k -0 goes for particle a and what fraction goes to particle b.
00:58
Okay, so we have, we want to find the expressions for this and this.
01:09
So starting with the fraction, with the expression from the fraction of kinetic energy that goes to particle a.
01:19
So we have that the initial kinetic energy is just the mass of particle a times the initial velocity squared over 2.
01:30
And then the final kinetic energy of particle a is equal to the mass of particle a.
01:36
V -a -2 squared over 2.
01:43
So this fraction becomes v -a -2 squared over v0 squared.
01:59
Furthermore, we also have that elastic collusions have a relation between the initial and final velocity of this first particle here.
02:17
So i can write that the final velocity v2 is equal to m, a, minus m, b over m .a plus mb, this times b, oh sorry, v zero.
02:40
The initial velocity of particle a.
02:44
Okay, this is a, equation that works for elastic collisions only, and it is written in equation 8 .24 in the test book.
03:01
Okay? so we just use this and substitute in our expression, so we have that kk0, so pick a va squared, we have m .a.
03:16
Minus mb over m a plus m b squared okay the v0 squared cancels out with the v0 from the denominator so this is the first expression we wanted to find the second expression is for the amount of the fraction of energy that goes to particle b.
03:50
So we can argue that since we are working with a elastic collusion, we have conservation of kinetic energy such that k -a plus kb has to be equal to k -0, which also implies that if i divide the conservation of kinetic energy, by k0 in both sides, we have that the sum of the fraction that goes to particle a with the sum of the fraction with the fraction that goes to particle b has to be equal to 1.
04:40
Okay, this is goal from, this comes from conservation of energy.
04:48
Conservation of kinetic energy.
04:55
Okay, so in this case, i can isolate the fraction kb, k0 to be 1 minus k -a over k -0.
05:07
So i only pick one minus the expression i found previously.
05:13
So 1 -m -a -minus m -b over m -a plus m -b.
05:23
Everything squared.
05:25
So performing this objection, we find that the fraction that goes to kb is equal to m a plus m b 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 of this parenthesis so m a squared plus m b squared plus 2 mb mb minus m a minus mb squared i'll also expand this so we have m a squared plus mb squared minus two m a mb okay so because of this minus sign here m a squared cancels with m a squared mb squared also cancels with mb 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 m .a, m .b over m .a plus mb squared.
06:56
So this is the final answer of question a.
07:03
Now question b asks us to evaluate the numerical value of these fractions, so ka, k0 and kb, k0 for some cases.
07:16
So the first case is when the mass of particle a is equal to the mass of particle b.
07:23
So we can go back to the ka's expression, and we have that when m .a is equal to mb, this term becomes zero...