can anybody solve this?

Question

Jack and Jill are standing on a crate at rest on the frictionless, horizontal surface of a frozen pond. Jack has a mass of 75.7 kg, Jill has a mass of 45.5 kg, and the crate has a mass of 15.7 kg. They remember that they must fetch a pail of water, so each jumps horizontally from the top of the crate. Just after each jumps, that person is moving away from the crate with a speed of 4.10 m/s relative to the crate.

What is the final speed of the crate if both Jack and Jill jump simultaneously and in the same direction? (Hint: Use an inertial coordinate system attached to the ground.)

What is the final speed of the crate if Jack jumps first and then a few seconds later Jill jumps in the same direction?

What is the final speed of the crate if Jill jumps first and then Jack, again in the same direction?

(all solutions are in m/s)

Answer 1

If the surface is frictionless then both can't jump in the same direction because the moment they reared to jump, the crate would slip against the pond surface.

Answer 2

Its a pretty detailed problem. However I think you just have to remember that in all cases, the momentum of a system before an interaction within that system is the same as the momentum afterwards. So since there is no movement by anything to start with, the momentum of the system is 0. Hence, the momentum after they jump is going to be 0. Work out the momentum of the individual parts of the system after they jump (p=m * v). The sum of these individual parts must equal 0. From there you should be able to get the speed (as you know the masses).

Eg - First case:

total momentum = momentum jill + momentum jack + momentum crate = 0

momentum jill = 45.5 * 4.1 = 186.55
momentum jack = 75.7 * 4.1 = 310.37

So,

186.55 + 310.37 + mom crate = 0
Therefore: mom crate = -496.92 (simple rearranging)

mom. crate = mass crate * velocity crate
therefore velocity crate = mom crate / mass crate
= -496.92 / 15.7
= -31.7m/s (to 3s.f)
or 31.7 m/s in the opposite direction as jack & Jill....

Answer 3

Yes, I can solve it. And I'll tell **you** how to solve it. Remember that momentum is conserved and work with an inertial reference frame that *is* the ground.

First, calculate the total momentum of Jack and jill with respect to the ground
(75.7 + 45.5)*v1 and the momentum of the crate
15.7*v1. Now, remember that v1+v2 = 4.1 m/s

Then go ahead and solve the other two the same way. (Hint: The final speed of the crate with respect to the ground will be the same in all three cases ☺)

Doug

Answer 4

42 m/s