Inelastic Collision Problem: Speed of Earth and People

[Arcarius]

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.

 

[gneill]

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.

Your work and result of the first part look okay. For the second part you've got the right idea but there's a better (technical) description you should be familiar with. Hint: What do you know about the center of mass of an isolated system (no external forces acting)?

 

[Arcarius]

Your work and result of the first part look okay. For the second part you've got the right idea but there's a better (technical) description you should be familiar with. Hint: What do you know about the center of mass of an isolated system (no external forces acting)?

Thanks for the reply!
Well, I know that the formula is m1x1 + m2x2... / (m1 + m2...)
I guess using this formula in my response will help reinforce it?

 

[gneill]

Thanks for the reply!
Well, I know that the formula is m1x1 + m2x2... / (m1 + m2...)
I guess using this formula in my response will help reinforce it?

Your phase: "the masses are still concentrated around the same areas" is too sloppy; there's no way to interpret it precisely as a physics equation or principle.

Frame your response in terms of appropriate physics concepts such as conservation laws and laws of motion. For example: For an isolated system, due to conservation of momentum the motion of the center of mass conforms to Newtons first law of motion. If it is considered to be at rest initially, then it must remain at rest unless some external force acts on the system. No external forces are acting here (because the system is isolated), so the result follows.

 

[Nickie]

However, the distribution of the mass may have changed slightly due to the people jumping off the ladders, but this would be a negligible change in the overall center of mass. Additionally, the Earth's rotation and orbit would also remain unchanged by this small inelastic collision.