Energy conservation in superball collision

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1. Dec 16, 2016

Alettix

1. The problem statement, all variables and given/known data
There are two elastic "superballs" of mass M and m placed on top of eachother with a smal distance. The lighter ball of mass m is on top of the bigger ball of mass M. The balls are released from a height h and have velocity u when they hit the ground. Prove that the top ball will have velocity $v = \frac{3M-m}{M+m} \cdot u$ when it leaves the bigger ball.

2. Relevant equations
momentum conservation $p_{before} = p_{after}$
energy conservation $E_{before} = E_{after}$
zero momentum frame velocity: $v_{ZMF} = \frac{p_{tot}}{m_tot}$

3. The attempt at a solution
I didn't find it problematic to prove the requried result (see attached solution). What I find problematic is the result itself. I must have made some very silly misstake in my reasoning, but the end result I get is that the speed of the top ball after all collisions is $u + 2v_{ZMF}$ and of the bottom ball $u$. However, before the two balls collide, they both had velocity $u$. Clearly, energy conservation cannot be satisfied!

Can somebody please point out my probably very stupid misstake to me? Thank you! :)

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2. Dec 16, 2016

BvU

Perhaps the big ball loses some speed when colliding with the smaller ball ? How did you find this $u$ for the bottom ball ?

3. Dec 16, 2016

Alettix

Oh wait, I really messed it up! The speed of the big ball should be $u - 2v_{ZMF}$, shouldn't it?

4. Dec 16, 2016

Buffu

$E_{ki} = 1/2mu^2 + 1/2Mu^2$
$E_{kf} = 1/2m(u^2 + 2(2v_{ZMF})u + (2v_{ZMF})^2) \\+ 1/2M(u^2 + 2(2v_{ZMF})u - (2v_{ZMF})^2)$
$E_{kf} = (1/2M+1/2m)u^2 + m(2v_{ZMF})u + 2m(v_{ZMF})^2 + M(2v_{ZMF})u - 2M(v_{ZMF})^2$
$E_{kf} = (1/2M+1/2m)u^2 + 2(v_{ZMF})^2(m - M) + (M + m)(2v_{ZMF})u$
$E_{ki} \ne E_{kf}$
I think I did this correctly.

5. Dec 16, 2016

Alettix

Sooo, $u - 2v_{ZMF}$ is not right either?

6. Dec 16, 2016

Buffu

I think using momentum conservation will give right answer.

7. Dec 16, 2016

Alettix

As far as I am concerned ZMF is meant to make it quicker and easier, so I would like to use that method. But yes, momentum conservation should always give the right answer.

8. Dec 16, 2016

Buffu

No that's not what I meant.
I mean that we can get right velocity for big ball after collision using momentum conservation.

$mu + Mu = m(u + 2v_{ZMF}) + M\color{red}{V}$

Last edited: Dec 16, 2016
9. Dec 16, 2016

haruspex

Yes, and with that I think you will find the energy is conserved in the lab frame.

10. Dec 19, 2016

Alettix

Yes. Thank you for helping me get it right! :)

11. Dec 19, 2016

Buffu

That looks correct but I did not get it.

Initial velocity in lab frame :-
Big mass - u
Small mass - u

Final velocity in lab frame :-
Big mass - $u - 2v_{ZMF}$
Small mass - $u + 2v_{ZMF}$
As per OP,

That is not going to get conserved. Either I am getting final velocity of small mass wrong or the big mass.
Can you help me please ?

12. Dec 19, 2016

haruspex

Are you sure, or are you just guessing?

13. Dec 19, 2016

Buffu

I checked if it is going to conserve in post number 4.
Can you please check my calculations ?

14. Dec 19, 2016

haruspex

You have some signs wrong in the second equation.