- #1

Arcarius

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## Homework Statement

An unlikely weapon was once proposed which had all of the people in China to jump up and down simultaneously. Let us say there are 2 billion people (2.0 x 10^9) with an average mass of 50 kg. They all climb ladders which are 1 meter tall. At a particular instant they all jump off nad land on the ground simultaneously. This is an inelastic collision with the Earth. Assume the Earth does not move proior to the collision.

Calculate what the speed of the Earth and the people would be after the collision.

(Hint: Treat the people as one object.

Caveat: In actuality the Earth would accelearte slightly towards the people as they fell, we will ignore any initial speed the Earth may have at the time of collision.)

Second Part:

Realistically, taking into account the movement of the Earth, has the center of mass of the People-Earth system changed at all during this collision?

## Homework Equations

##m1v1 + m2v2 = (m1+m2)v3##

##vf^2 = vi^2 + 2ad##

## The Attempt at a Solution

So I let m1 = the mass of the people, m2 = mass of Earth, v1 = velocity of the people, v2 = velocity of Earth, and v3 = velocity of people and Earth (after collision).

In the equation ##m1v1 + m2v2 = (m1+m2)v3##, the m2v2 term cancels out. Here I'm making the assumption that the people and the Earth will be moving the same speed (v3) after the collision. I'm not quite sure if that's right!

I then found v1 = sqrt(2ad) = 4.43 m/s.

Substituting this back into the first equation, I got ##v3 = m1v1 / (m1+m2)##, which yields 7.42 * 10^-14 m/s for both the velocity of the people and the velocity of the Earth after collision. I'd appreciate it if anyone could check my work on this problem, because I'm extremely unconfident about it!

Second Part: I don't believe that the center of mass of the People-Earth system has changed at all, because the masses are still concentrated around the same areas.