# Elastic collision - finding ratio between m2 and m1, v2 and v1

1. Nov 23, 2005

### an_mui

A ball of mass m and a ball of unknown mass M approach each other from opposite directions and have the same speed Vo (but oppositely directed velocities). The ball of M is reduced to rest by the impact, while the ball of mass m has a velocity V1'. What are the ratios

a) M / m
b) V1' / Vo

This is what I've done so far. I ended up with 2 answers, but we're only supposed to have one.

Conservation of momentum: mVo + M(-Vo) = mV1' + 0
- divide both sides by mVo, and let x = M / m, y = V1' / Vo
... equation (1'): 1 - x = y
Conservation of kinetic energy: 1/2mVo^2 + 1/2M(-Vo)^2 = 1/2mV1'^2
- divide both sides by 1/2mVo^2
... equation (2'): 1 + x = y^2

equations (1') + (2'): y^2 + y - 2 = 0
(y + 2)(y - 1) = 0
... y = -2 or 1
Case 1:
If y = -2, x = 3

Case 2:
If y = 1, x = 0

2. Nov 23, 2005

### Andrew Mason

You have:

$$1 + x = (1 - x)^2 = 1 - 2x + x^2$$

$$x^2 - 3x = 0$$

So: x = 3 = M/m so M = 3m

AM

3. Nov 24, 2005

### an_mui

thanks for your answer, but that doesn't solve the problem though... because from the above equations i had y = -2, x = 3 and y = 1, x = 0. From your answers x could also be 3 or 0. I need to eliminate one of the answers but i don't know how.

4. Nov 24, 2005

### mezarashi

The two equations you must solve are:

Momentum conservation
$$mv_o - Mv_o = mv_1$$

Energy Conservation
$$\frac{1}{2}mv_o^2 + \frac{1}{2}Mv_o^2 = \frac{1}{2}mv_1^2$$

Express v1 in terms of m, M, and vo from the first equation and substitute it into the second equation. You should be able to solve that without a problem.

Edit: About what you claim about x being zero. This is a trivial solution that is usually present in all types of problems. You can conserve momentum and kinetic energy if you have nothing happening with nothing right? But why bother.

Last edited: Nov 24, 2005
5. Nov 24, 2005

### Andrew Mason

If x = 0, then M=0. That possibility is excluded by your fact situation.

AM