# Conservation of energy or conservation of momentum- which one is applicable in this problem?

There was this question from Kleppner and Kolenkow's Classical Mechanics:

PROBLEM:

And I found its solution from this Link:
It is quite an extended solution and I am not drawing you into the entire working. You can wish to see it if you want. I just want to point out one portion in his solution where the momentum of the system is said to be conserved in the vertical direction which seemed quite illogical to me. It goes as follows:

My point is that gravitational force is acting downwards on the system or in other words an external force is acting on the system along the y-axis. Then how can momentum be conserved along y-axis as the solution says?

My answer does not match that of the solution. But I have used conservation of energy and it is obvious that energy is conserved in a gravitational field since it is conservative.

My equations go as:
At the highest point i.e. at height $H$, $$K.E._{initial} + P.E._{initial} = K.E._{final} + P.E._{final}$$ $$\frac{1}{2}mv_0^2 + mgh = 0 + (M+m)gH$$ and then I solve for $H$.

Which method do you think is correct?

## 1 Answer

My point is that gravitational force is acting downwards on the system or in other words an external force is acting on the system along the y-axis. Then how can momentum be conserved along y-axis as the solution says?

If we model an interaction as taking place over a very short period of time, then we simply say that gravity doesn't change the vertical momentum (much) during that time. For things like collisions, this is a good approximation. This is only true for the short period of interaction, not for the rise from the trampoline (where gravity has a significant effect).

...it is obvious that energy is conserved in a gravitational field since it is conservative.

And if all the interactions were with the gravitational field, you would be correct. But here, the acrobat also interacts with (grabs) the monkey. For the purposes of the problem, you can consider this a completely inelastic collision (two particles meet and join). Momentum is conserved, but KE is not. There must be some loss of energy.