In an elastic collision between 2 objects of mass and show that the speeds of the approach is equal to the speed of the seperation . The symbols represent speeds not velocities.
I did this question twice and I got the result u2+u1=v1+v2. On the third attempt I got the result that the question gives since I realised that this question is worded very
shittly(adverb of the adjective "shit"?). Anyways, what's happening is that initially both objects are travelling in the same direction, but once they collide, they move apart. I deduced this from the equations that were given for speed of approach and seperation. At first, I assumed that they move towards eachother and then move apart, but in that case the two equations given for speed of speration and approach would be different, and that's the shitty wording i was refering to earlier.
Take the initial direction to be positive. Hence the conservation of momentum implies:
The negative sign comes from the assumption that object 2 is the one that rebounds back while object 1 continues going in the positive direction. This is because the expression u2-u1 implies that u2>u1 since this is a speed(always positive) Hence object 2 must be comming from behind, catching up to object 1 in order to make the collision. Because object 2 is behind, it is the one that rebounds back.
Kinetic Energy:
momentum equation implies that:
By doing a similair thing to the kinetic energy equation, u can get the result:
By equating the two expressions for
and factorising using difference of squares, then cancelling out factors etc. you will arrive at the result