Why is relative velocity taken with respect to final velocity of affected object?

Question

I am practicing a conservation of momentum question:

Two identical buggies move one after the other due to inertia (without friction) with the same velocity $v_0$. A man of mass m rides the rear buggy. At a certain moment the man jumps into the front buggy with a velocity u relative to his buggy. Knowing that the mass of each buggy is equal to M, find the velocities with which the buggies will move after that.

In the solution for this question mentioned in my book,

Conservation of momentum in rear buggy: $(M+m)v_0 = m[u +v_r] + Mv_r$,

where velocity of man with respect to ground is $u+v_r$ and final velocity of rear buggy is $v_r$.

My question is, why is velocity of man with respect to ground given as $u+v_r$? Shouldn't it be $u + v_0$? Isn't relative velocity here supposed to be taken from before the man starts moving?

Why have they considered the relative velocity of man with respect to the final velocity of buggy? (I don't think their answer is incorrect, though, a few other online sources detail a similar working).

Answer 1

The question states that the man's velocity is measured with respect to the rear buggy, not with respect to the ground. When the man jumps forward, the rear buggy moves backwards relative to him - the buggy does not remain static at the moment he jumps. If you take $u$ to be with respect to the ground, you will find that the man is moving away from the rear buggy at a speed greater than $u$, contradicting the premise that the jumping man is moving at speed $u$ relative to the rear buggy.

You're suggesting that it's acceptable to compare the velocity of the man after he jumps to the velocity of the buggy before he jumps, but that doesn't represent one single point in time. At the moment the man has relative forward velocity, the buggy has relative backward velocity. There is no time at which the man is moving forward where the buggy is still moving at $v_0$.