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?
Best Answer
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.