00:01
This question is about collision of two spheres.
00:10
A sphere 1 has m1 0 .5 kg and the initial velocity is given to be 2 i minus 3 j plus k meters per second and the second sphere has mass 1 .5 kg the initial velocity is given to be minus i plus 2j minus 3k meters per second okay then in part a given that the final velocity of m1 is minus i plus 3j minus it's k we want to find v2 f okay so to do this we'll be using conservation of momentum.
01:27
So it is m1 v1i plus m2 v2 i equals to m1 f plus m2 v2 f so putting in the numbers.
01:49
I'm going to use the column form of vectors to minus 3 1 this is for m1 and for m2 v2 i this is for m1 and for m2 v2 i this is what we have and then m1v1f plus 1 .5 v2f okay so we arrange so evaluate the right -hand side or no the left hand side then bringing in the terms from the right -hand side okay you can see that these two are equal so you get zero which means that v2 f is zero okay and then we'll look at compare kinetic energy so compare kinetic energy before and after collision okay so the kinetic energy before is half mv1 m1 i square plus half and m2 v2 i square and then putting in the numbers, m1 is 0 .5 and calculating v1 i squared will be summing the squares of the components.
04:01
Then we repeat the step for m2 and then you calculate this initial kinetic energy to be 14 juice.
04:19
And then the final kinetic energy is just m1v1 square, half m m2, 1v1 f square so okay so this is 18 .5 shoes okay so we have k i less than k f so which means that collision in this case will be inelastic okay yeah it's not completely it's not perfectly it's not perfectly inelastic but it's just inelastic okay okay so this is this answer for part a and as well as this.
05:23
Then for parts b, so we'll assume that the final velocity of m1 is given to be minus 0 .25i plus 0 .75j minus 2k meters per second then we want to find v2f again so we just we just repeat from this equation so this is the part that needs to be changed so we have 1 .5 v2f is equal to minus 0 .5 1 .5 minus 4 minus 0 .5 this is m1 and then v1 f is given as such okay this is found to be 0 .375 1 .125 minus 3 3 and then after dividing 1 .5 both sides it gets v2 f to be minus 0 .25 i 0 .75 j and minus 2k and minus 2k and this is v1f.
07:19
So the collision is perfectly elastic.
07:33
Because both spheres have the same velocity, final velocity, after the collision.
07:42
So the answer for part b, right? so we move on to part c.
08:17
Okay, now we are given that v1f.
08:23
V1f is minus i plus 3j plus a k meters per second we want to find v2 f again okay so to do this on the conservation of momentum okay so we have 1 .5 v2 f is equal to minus 0 .5, 1 .5, 4, minus 4, minus 0 .5, minus 1 ,3a, is equal to 0 ,0, minus 4, minus 0 .5 a vector.
09:27
Okay? so v2f is equal to 1 over 1 .5 minus 4 plus 0 .5 a k meters per second.
09:46
Then you need another equation because the collision is elastic.
09:53
So we have conservation of kinetic energy...