Solving an acrobat problem in 3 methods

A circus acrobat of mass $M$ leaps straight up with initial velocity $v_{0}$ from a trampoline. As he rises up, he takes a trained monkey of mass $m$ off a perch at a height $h$ above the trampoline. What is the maximum height attained by the pair?

Center of Mass

Initially the center of mass(CM) is at $y_{0} = \dfrac{mh}{M+m}$. The only force on the CM is gravity and its initial velocity is $\dfrac{Mv_{0}}{M+m}$. So it is going to reach its highest point traveling a distance of $\dfrac{M^{2}v_{0}^{2}}{2(M+m)^{2}g}$ from its initial position $y_{0}$. This leaves the pair at a final distance of $\dfrac{mh}{M+m} + \dfrac{M^{2}v_{0}^{2}}{2(M+m)^{2}g}$ above the ground

Conservation of Energy

Initial Energy of the pair, $E_{0} = \frac{1}{2}Mv_{0}^{2} + mgh$ Final Energy of the pair, $E_{1} = (M+m)gH$, where $H$ is the maximum height above the ground that the pair reaches.

By conservation of energy, $E_{0} = E_{1} \implies (M+m)gH = \frac{1}{2}MV_{0}^{2} + mgh \implies H = \dfrac{Mv_{0}^{2} + 2mgh}{2(M+m)g}$

Conservation of momentum

At a height of $h$, the velocity of M is $v = \sqrt{v_{0}^{2} - 2gh}$. Just after picking up the monkey, the velocity of the pair, $v^{'}$ is given by $Mv = (M+m)v^{'}$ (Why should the momentum be conserved for the acrobat-monkey system?), they will go on to a height of $\dfrac{v^{'2}}{2g}$ from here.

The three approaches don't seem to tally. Where am I wrong?

• Commented Mar 8, 2016 at 6:56

The two first methods assume that the collision between the monkey and the acrobat is elastic, which is clearly not the case.

The third method uses simple kinematic equations. More importantly, the calculation gets past the collision part of the process using conservation of momentum, which does hold for this collision.

• Though an elastic collision between a monkey and an acrobat would be hilarious.
– Neil
Commented Mar 8, 2016 at 7:48
• Subtle point about why momentum is conserved! Commented Mar 8, 2016 at 9:22
• Does the conservation of momentum here serve as a best model of approximation given the short interaction time between the objects? Or am I missing a subtlety on the validity of momentum conservation itself? Commented Mar 8, 2016 at 14:47

Momentum Conservation can be applied (approximately) along the $$y$$-axis!

The reason is not so obvious. We can never conserve momentum along $$y$$-axis because gravity is always there to provide a force in the $$y$$-axis of our system (jumper+monkey); hence momentum conservation is invalid.

But we know something about gravity, that it does not play a significant role in very short intervals of time, and does not change the momentum of the system appreciably in that small period. But do note that the change is the momentum is very small, but it's not equal to zero!

So with this reasoning, one can apply momentum conservation provided that the time interval is super small.

Correct application of Newton's second law leads to the correct answer. Here's the full solution