What is the maximum height reached by two masses after an elastic collision?

[haruspex]

Ah, ok.
I calculate the ratio of the heights, h₂/h₁ = (2m₁/(m₁-m₂))². That agrees with your final answer. It follows that for m₁ to climb higher than m₂ it needs to have at most 1/3 of the mass of m₂.
Consider the case where the masses are the same. m₁ would stop dead and m₂ would climb to height h.

 

[Lolagoeslala]

Ah, ok.
I calculate the ratio of the heights, h₂/h₁ = (2m₁/(m₁-m₂))². That agrees with your final answer. It follows that for m₁ to climb higher than m₂ it needs to have at most 1/3 of the mass of m₂.
Consider the case where the masses are the same. m₁ would stop dead and m₂ would climb to height h.

but...
i don't get it .. if u divide 16 kg by 3 .. you get 5.333 kg
and mine is 6 kg ?

 

[haruspex]

but...
i don't get it .. if u divide 16 kg by 3 .. you get 5.333 kg
and mine is 6 kg ?

Quite so. 6k is greater than one third of 16k, so the 16k ball will climb higher. For the light ball to climb higher it must be less than one third the mass of the other.

 

[Lolagoeslala]

Quite so. 6k is greater than one third of 16k, so the 16k ball will climb higher. For the light ball to climb higher it must be less than one third the mass of the other.

Seriously? :O But i have always been though that the smaller mass will go higher... why would the bigger mass travel further?

 

[haruspex]

Seriously? :O But i have always been though that the smaller mass will go higher... why would the bigger mass travel further?

The scenario is not symmetric. As I suggested, consider two equal masses. The first (initiating) mass will stop dead and transfer all its KE to the other. Now make the first mass a bit smaller. It will bounce back some, but still most of the energy will be transferred. It won't achieve its 'fair' share of energy (i.e. in proportion to mass) until it's down to one third the mass of the other. And to get equal energy it would have to be smaller still.