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

[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

I don't know how we're supposed to determine which is the correct case. Please help!

 

[Andrew Mason]

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
I don't know how we're supposed to determine which is the correct case. Please help!

You have:

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

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

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

AM

 

[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.

 

[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.

 

[Andrew Mason]

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.

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

AM