Confused about elasticity and collisions

Question

I was solving the following problem and the explanation to it confused me.

There are two objects with mass $m$ and $M$, respectively. The object with mass $m$ has a velocity of $\sqrt{2gl}$ and collides with the other object that is initially at rest. If the collision is elastic, what is the velocity of the object with mass $M$ right after the collision? Friction is negligible.

My understanding is, for these kind of collision problems momentum is conserved so,

$$mv_{m} +Mv_{M} =mv'_{m}+Mv'_{M}$$

or in other words,

$$mv_m=mv'_{m}+Mv'_{M}$$

Since the initial velocity is 0 for the object with mass $M$.

In addition, elastic collision means that kinetic energy is conserved and intuitively, I am imagining that the smaller object (say, $m$) will "bounce" to the opposite direction as its initial velocity while the bigger object starts moving in the direction that is as same as the initial velocity of $m$, just a little slower.

However, the answer explanation tells me that right after the collision the momentum is preserved, so

$$mv_m=(m+M)v'_{m+M}$$.

I thought that this equation represents the situation where the collision is completely INELASTIC and the objects stick to each other.

Does it have to do with the wording "right after the collision" ?

Or, is the explanation not making sense ?

Answer 1

There are two objects with mass $m$ and $M$, respectively. The object with mass $m$ has a velocity of $\sqrt{2gl}$ and collides with the other object that is initially at rest. If the collision is elastic, what is the velocity of the object with mass $M$ right after the collision? Friction is negligible.

This is an old question but it may still interest someone: the velocity of mass M, according to the formula will be: $$v'_M = \frac{\sqrt{8gl}}{m+M}$$