Classical mechanics energy and momentum

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  • #1
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Homework Statement



I try to solve the problem 3.5 in Kleppner Kolenkow ”An introduction to Mechanics” using various approaches but end up with wrong answers. The problem is:
3.5 A circus acrobat of mass M leaps straight up with initial velocity v0 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.


Homework Equations




The Attempt at a Solution



At first I use the fact that the center of mass is at hm/(M+m). The velocity of the center of mass is v1 = Mv0/(M+m).
Since they reach a maximum height v2/2g + x0 the height attained is
xmax = v12/2g + hm/(M+m) = M2*v02/(M+m)2/2g + hm/(M+m)
The first term is right here. The second is not.

Secondly i use energy conservation:
Wbefore = mgh + Mv02/2
Wafter = (M+m)gxmax

This gives:
xmax = hm/(M+m)+Mv02/2(M+m)g

The right answer is v02M2/(M+m)2/2g+h(1- M2/(M+m)2)
I know how to get to the right answer, but what I need is an explanation of what laws of physics i dont obey.

Homework Statement





Homework Equations





The Attempt at a Solution

 

Answers and Replies

  • #2
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For the 2nd approach, it's obvious: mechanical energy is not conserved. When the man picks up the monkey, that's a "collision."

It's more subtle in the 1st approach. Hint: What's the assumption for the equation: x - xo = v2/2g? Further hint: the original one is actually
x - xo = v2/2a with the assumption that a = const.
 
  • #3
97
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Thanks Hikaru!

I think I see. Below h you can show that dv/dt = -gM/(M+m), Whereas above h the acceleration is simply g. I think I will leave the center of mass approach. It is too complicated!

Instead I can divide it into three steps where in the first step the mass is M and in the last step the mass is M+m Then I can use the equations for constant acceleration.

In the energy case I can divide the problem into three steps where mechanical energy is conserved in the first and last step. In the second step I use momentum conservation of course.
 

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