A small block of mass m is moving on a horizontal table surface at initial speed v0. It then moves smoothly onto a sloped big block of mass M. The big block can also move on the table surface. Assume that everything moves without friction
Find the speed v of the small block after it leaves the slope.
Conservation of Momentum: mv0 = (M+m)v → v=(mv0)/(M+m)
Conservation of Mechanical Energy: fairly obvious
The height that it rises to is h = (1/2g)(Mv0^2/M+m), I derived this from conservation of momentum and conservation of mechanical energy
The Attempt at a Solution
I set up the conservation of momentum: (M + m)v = -mv1 + Mv2 Is this correct?
Now I ask, how should I set up the conservation of mechanical energy equation? Should I solve for one variable in terms of the other? The answer should be in terms of v0.
The answer is v1 = ((M-m)/(M+m))v0