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M2 collisions

A small smooth sphere A of mass 2kg moves at 4ms^-1 on a smooth horizontal table. It collides directly with a second equal-sized smooth sphere B of mass 3kg, which is moving away from A in the same direction at a speed of 1ms^-1. If the loss of kinetic energy due to the collision is 3J find the speeds and the directions of the two spheres after the collision.

So I set up two equations in terms of momentum and energy.

Let x = velocity of A after collision and y = velocity of B after collision

Then from momentum, 2x+3y=11

And from energy, 2x²+3y²-29=0

Solving these simultaneously obviously gives two combinations of solutions, whereas the answer only has the answer as x (speed of A)=1ms^-1 and y (speed of B)=3ms^-1 with both moving in the same direction as before colliding.

Now I can get these two answers from these equations, but I also get two others and I can't think of any reason to rule these two out. Such a situation also arose on another question. Can anybody clarify this, thanks...

Reply 1

ghostwalker

Original post by fayled

Solving these simultaneously obviously gives two combinations of solutions, whereas the answer only has the answer as x (speed of A)=1ms^-1 and y (speed of B)=3ms^-1 with both moving in the same direction as before colliding.

Now I can get these two answers from these equations, but I also get two others and I can't think of any reason to rule these two out. Such a situation also arose on another question. Can anybody clarify this, thanks...

Your other answers are

x=3.4m/s

and

y=1.4m/s

This would imply that both A and B are going in the same direction as before, and that A is travelling faster than B.

This is impossible, as A would have to leapfrog over B, or somehow pass through B in some way as yet only known to science fiction.

Reply 2

fayled

Original post by ghostwalker

Your other answers are

x=3.4m/s

and

y=1.4m/s

Silly me. Thankyou :smile:

Reply 3

fayled

Original post by ghostwalker

Your other answers are

x=3.4m/s

and

y=1.4m/s

Actually, could you take a look at a previous question please?

A particle of mass m moves in a straight line with velocity v when it explodes into two parts, one of mass m/3 and the other of mass 2m/3 both moving in the same direction as before. If the explosion increases the energy of the system by mu²/4, where u is a positive constant, find the velocities of the particles immediately after the explosion. Give your answers in terms of u and v.

If x is the velocity of the particle of mass m/3 and y is the velocity of the particle of mass 2m/3, then

x=3v-2y
6v²+3u²=2x²+4y²

Solving these simultaneously gives

x=v-u, y=v+u/2
and
x=v+u, y=v-u/2

Now the top combination is given as the correct answer, but I can't see a physical reason why this is the case as there is only one particle beforehand...
Can you understand why or is my working flawed? Thanks.

(edited 11 years ago)

Reply 4

ghostwalker

Original post by fayled

Can you understand why or is my working flawed? Thanks.

Assuming there's no diagram. I can't see any flaw, or error in your working.

The only flimsy thing I can see supporting their answer is that they mention the m/3 mass before the 2m/3 and interpret that as the 2m/3 is further along in the posititive v direction, and hence the m/3 mass must be the slower.