## Homework Statement

A circus acrobat of mass M leaps straight up with an initial velocity of vo from a trampoline. As he rises, he takes a trained monkey of mass m off a perch of height h above the trampoline. What is the maximum height attained by the pair in terms of M, m, Vo, h, and g?

## The Attempt at a Solution

I used energy conservation upto the point where the acrobat reaches the perch and found out his imediate velocity at the perch to be v=square root(Vo^2 - 2gh) . The problem lies after this.

I looked up the solution to this problem and at this point conservation of linear momentum has been used as follows

Mv0 = (M + m)v' , where v' is the common velocity of the acrobat and the monkey immediately after the acrobat has taken the monkey off the perch.

The question I want to ask is, how can the momentum be conserved, when there IS a net external force on the syetm of (acrobat + monkey)? The net external force is (M + m)g downwards, so can we apply the conservation of linear momentum?

You are correct, the impulse of gravity cannot, in general, be ignored.

Let us consider the impulse-momentum theorem, as it applies to a collision that lasts some length of time, $$\tau[tex] [tex]\Delta \vec P = \int _\tau \vec F dt$$ (I beg your pardon for the informal notation)

For a constant force this reduces to:

$$\Delta \vec P = \vec F \cdot \tau$$

Now, we know the initial momentum, $$P_i = Mv_0$$ and the final momentum, as we've defined it, after the monkey and acrobat are moving together is $$P_f = (M+m)v'$$ and we know that the collision lasted some time, $$\tau$$

$$P_f-P_i =-(M+m)g\tau$$

But since we know the collision is nearly instantaneous, that is to say, that $$\tau$$ is a very very short time interval, then it holds, approximately $$P_f-P_i\approx 0$$

There was an external force acting on the system, but you could overlook that fact since the total momentum of the system hardly changed due to its presence.

You find conservation of momentum where there is either no external force, or that any external impulse for the process in question is negligible compared with the momenta involved.

Good question, and one that seems to be slightly overlooked at times. :)

Another approach would be to use a free-falling accelerated reference frame (In which there is no net external force) and eventually transform back to the original, inertial reference frame, but that would run into the same problem of knowing how to transform the velocities back, since the difference in relative velocities of the two reference frames would depend on how long the collision took.

Doc Al
Mentor
The question I want to ask is, how can the momentum be conserved, when there IS a net external force on the syetm of (acrobat + monkey)? The net external force is (M + m)g downwards, so can we apply the conservation of linear momentum?
You're only applying conservation of momentum during the brief time that he grabs the monkey: just before versus just after. Too brief a time to worry about gravity. For longer times--like as they rise together after the grab--you know that momentum is not conserved.

I used energy conservation upto the point where the acrobat reaches the perch
Use energy conservation up to end.

At start point energy is equal to acrobat's kinetic energy plus monkey's potential energy.

At the end it is equal to both's potential energy at max. height

Doc Al
Mentor
Use energy conservation up to end.

At start point energy is equal to acrobat's kinetic energy plus monkey's potential energy.

At the end it is equal to both's potential energy at max. height
Careful: You cannot assume that energy is conserved as the monkey is grabbed.

Careful: You cannot assume that energy is conserved as the monkey is grabbed.

In fact, you cannot assume that, as you would end up with a contradiction.
Applying conservation of momentum:

$$Mv_0 = (M+m)v'$$

Applying conservation of mechanical energy:

$$\tfrac{1}{2}Mv_0^2=\tfrac{1}{2}(M+m)v'^2$$

The first implies: $$v'=\frac{M}{M+m}v_0$$
While the latter implies:

$$v'=\sqrt{\frac{M}{M+m}}v_0$$

An obvious contradiction. When then, can we apply the energy condition and still have it be consistent with momentum conservation? Our mistake is in assuming that they both continue moving in the same direction. Allowing them to move in opposite directions makes the elastic collision possible.

By forcing both momentum conservation (Always true) and conservation of mechanical energy (Only sometimes true), we have in fact hammered the collision into a position where it is both perfectly elastic and perfectly inelastic at the same time.

Careful: You cannot assume that energy is conserved as the monkey is grabbed.
You are right. I imagine, that "monkey grabbing" is not the same as "inelastic collision with monkey", so energy will be conserved.

If we have to use collision point of view, it's necessary to use momentum conservation of course.

deltaP is equal to F*t. We know initial and final momentum. We know F, so we can calculate the time of lifting.

Then by using equation for distance in uniformly variable motion we can calculate maximum height.

Right?

Doc Al
Mentor
You are right. I imagine, that "monkey grabbing" is not the same as "inelastic collision with monkey", so energy will be conserved.
The "monkey grabbing" is a perfectly inelastic collision!

The "monkey grabbing" is a perfectly inelastic collision!
But... acrobat loves monkey, isn't he?

In collisions mechanical energy is not conserve, because of deformation energy, heat energy etc. When acrobate grabbing monkey he minimalize all that cases, and grab animal carefully. Dispersal of energy can be omit, I suppose Whatever, as RoyalCat said, momentum is allways conserve... and it is allways right method.

Doc Al
Mentor
In collisions mechanical energy is not conserve, because of deformation energy, heat energy etc. When acrobate grabbing monkey he minimalize all that cases, and grab animal carefully. Dispersal of energy can be omit, I suppose Sounds to me like you're saying that if he grabs the monkey carefully enough, he can do so without transforming kinetic energy into random internal energy? If so, that's incorrect. As long as he grabs the monkey and they end up traveling together, translational KE will be 'lost'.

Sounds to me like you're saying that if he grabs the monkey carefully enough, he can do so without transforming kinetic energy into random internal energy? If so, that's incorrect. As long as he grabs the monkey and they end up traveling together, translational KE will be 'lost'.
OK. now I read what I wrote... horrible ! You are right, it isn't possible not to dispose energy.

I do not delete my posts only for next generations shame exemplar.

The only excuse that can be scorching heat in July.

@ Royal cat and Doc al

Thanks a ton.....infact I figured out the answer to this half an hour after posting this but still wanted to confirm ............ thanks a lot...!!!!