How can you find the final velocity of a ball hitten by a train? [closed]

Question

I want to clarify that I've already seen this question asked here , but I'm interested in how you'd solve it with the conservation of momentum rather with the relativity mechanism of finding the velocity with respect to the train and then adding the velocity of the train which I already saw here Velocity of the ball after hitting a train.

I have tried doing it for myself but end up getting nowhere, assuming the velocity of the train remains unchanged after collision and no energy is loss in the colision, and the ball is in rest before the collision: $$p_{before}=p_{after}$$ $$Mv=Mv+mv'\Rightarrow v'=0$$ being $M$, $v$ and $v'$ the mass of the train, the velocity of the train and the velocity of the ball after collision.

Answer 1

The trick is to not plug in infinite masses and other extremal values too early. Sure, the momentum of the train will not change by any detectable amount, but it must lose some of it, otherwise it couldn't give any to the ball. So, the massive train reduces its speed by the tiniest amount, but that tiny amount translates to a large velocity increase of the ball, because it has much much less mass. So do everything exact, and only at the last possible moment let $m/M\rightarrow 0$.

In your example, we would use the conservation of momentum $$Mu_\mathrm{T} = Mv_\mathrm{T} + mv_\mathrm{B}$$ and the conservation of energy $$Mu_\mathrm{T}^2 = Mv_\mathrm{T}^2 + mv_\mathrm{B}^2$$ where $u$ is the velocity before, and $v$ after the collision, and T and B stand for train and ball, respectively.

I won't show how to solve these explicitly, this is part of any good text book. (It would also make a good exercise. ) The solution is $$v_\mathrm{B} = \frac{2}{\frac{m}{M} + 1}u_\mathrm{T} \approx 2u_\mathrm{T}\\ v_\mathrm{T} = \frac{1-\frac{m}{M}}{1+\frac{m}{M}}u_\mathrm{T}\approx u_\mathrm{T} $$