I just don't understand should I take u relative to the plane or relative to the ground.
I tried to solve it like this:
$$p_{final}=m_{0}v-m(u-v)-M(u-v)$$
$$p_{initial}=m_{0}v$$
$$\Delta p=-m(u-v)-M(u-v)$$
##m_0## is mass of the plane.
$$F=\Delta p$$
$$F=-m(u-v)-M(u-v)=(m+M)(v-u)$$...
But in the task there is written biggest part of energy, not bigger. So there is only one ratio of masses with which ##m_1##'s KE is biggest, right? And I have to maximize for ##m_1## energy, because it's the top one.
Yes, I made some mistakes in previous solution.
Expression for energy gain should be written for the first ball, not the second.
$$E=\frac{m_{1}v_{1}^{2}}{m_{1}v_{0}^{2}+m_{2}v_{0}^{2}}$$
solving ##v_1## from (2)
$$v_{1}=\frac{v_{0}(3m_{2}-_{}m_{1})}{m_{1}+m_{2}}$$
replacing ##v_1^2## gives...
Ok, so then ##-2v_0=v_2-v_1##
replacing ##v_1## in (2)
$$m_{2}{v}_{0}=m_{2}{v}_{2}+m_{1}{v}_{2}+3m_{1}{v}_{0} $$
$${v}_{2}=\frac{v_{0}(m_{2}-3m_{1})}{m_{1}+m_{2}}$$
Energy converted = first ball energy / total energy
$$E=\frac{m_{2}v_{2}^{2}}{m_{1}v_{0}^{2}+m_{2}v_{0}^{2}}$$...
So then (2) should be
$$m_{2}{v}_{0}-m_{1}{v}_{0}=m_{2}{v}_{2}+m_{1}{v}_{1}$$
then I get ##2v_0=v_1-v_2##, which gives me same result if I square it. I think somehow I should find maximum velocity for ##m_1## after first collision with ##m_2##. Then it would get maximum kinetic energy.
With given information in the problem I can't decide which direction balls will go after collision. I assumed they will go in opposite directions. I know that is not a full solution, but I can't come up with anything else.
From conservation of energy we have...