Momentum conservation on 2 buggies Question

Attempt 
I tried taking a system in which there was visibly no external force and applied momentum conservation. But my answer came wrong. Where did I go wrong? Please help. 

I am in high school so I dont know much advanced concepts.
 A: Your initial condition is correct. The buggy without the man would have $p = M \cdot v_1$ and the man $p = m \cdot v_1$. Now for the second buggy the same with $v_2$. For the first buggy to completely stop moving after the jump, its momentum would need to be $0$ but due to conservation of momentum it is also equal to the sum of the momentum of the second man and the first buggy before the jump: $0 = m \cdot v_2 + M \cdot v_1$. The second buggy now moves at speed $v$ so: $p_{f,\ buggy\ 2} = (m + M) \cdot v = m \cdot v_1 + M \cdot v_2$ also due to conservation of momentum. We have 2 equations and 2 variables so this is solvable!
Isolate $v_1$ from the first equation: $v_1 = \frac{-m\cdot v_2}{M}$. We fill it in in the second equation: $(m+M) \cdot v = m \cdot \frac{-m\cdot v_2}{M} + M \cdot v_2$ from which we get: $v_2\cdot (\frac{-m^2}{M} + M) = (m+M) \cdot v$ or $v_2 = \frac{(m+M) \cdot v}{\frac{-m^2}{M} + M} = \frac{(m+M) \cdot v\cdot M}{M^2-m^2} = \frac{v \cdot M}{M-m}$. Filling into the formula for $v_1$: $v_1 = \frac{-m\cdot v_2}{M} = \frac{-v\cdot m}{M-m}$, the minus sign indicates them moving in opposite directions. Assuming the men weigh less than a buggy, the first buggy moves slower than the second buggy in the beginning. You can also see that $v < v_2$ which is also expected.
