3
$\begingroup$

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?

$\endgroup$
2
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ Though an elastic collision between a monkey and an acrobat would be hilarious. $\endgroup$ – Neil Mar 8 '16 at 7:48
  • $\begingroup$ Subtle point about why momentum is conserved! $\endgroup$ – buzaku Mar 8 '16 at 9:22
  • $\begingroup$ 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? $\endgroup$ – buzaku Mar 8 '16 at 14:47

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.