Jack and Jill jump off a crate?

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

Conservation of momentum:

Initial momentum = 0
= Final momentum
= mjack vjack + mjill vjill + mcrate vcrate

They give you all the masses.

For jumping together, they give you that vjill = vjack = vcrate + 4.1 m/s

So substitute the last two equations for jill's and jack's velocity (in terms of crate velocity)into the conservation of momentum equation. Then solve that for the crate speed. I'd expect it to be negative, because it goes the opposite way from J & J. Do it the same way (in pieces) for the cases where one jumps after the other.

Answer 2

Both jump at the same time:
15.7v = (75.7 + 45.5)(4.1)
15.7v = (120.9)(4.1)
v = 31.573 m/s

Jack jumps first:
(45.5 + 15.7)v1 = 75.7*4.1
61.2v1 = 310.37
v1 = 5.0714 m/s
45.5(5.0714 - 4.1) + 15.7v2 = (45.5 + 15.7)5.0714
15.7v2 = (45.5 + 15.7)5.0714 - 45.5(5.0714 - 4.1)
15.7v2 = (45.5*5.0714 + 15.7*5.0714 - 45.5*5.0714 - 45.5*4.1)
15.7v2 = 15.7*5.0714 + 45.5*4.1)
v2 = 16.954 m/s

Jill jumps first:
(75.7+15.7)v1 = 45.5*4.1
v1 = 2.0410 m/s
15.7v2 = 15.7*2.0410 + 75.7*4.1
v2 = 21.810 m/s

Answer 3

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