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# Physics A2 question Watch

1. in october 1999 the united nations announced that the world's population had reached 6 billion (6x10^9) and that the population of the people's republic of china was rapidly approaching 1.3 billion.

imagine a situation in which the entire population of China were gathered together and, all at the same time, they each jumped as high as they were able. estimated the speed that would be (temporarily) imparted to the earth as a resuult of this jump. show all you calculations and reasoning and state clearly any assumptions you made. take the mass of the earth to be 6x 10^24k kg
2. Hey, I have absolutely no idea if this is right but it sounded like a fun problem.

Assuming the average chinaman weighs 70kg and the avrage jump is 20cm, the using E = mgh

E = 1.3 x 10^9 x 70 x 0.2 = 1.8 x 10^11

Then using the conservation of energy with E = 1/2mv^2, assuming all the gravitational potential energy of the jump is converted to kinetic energy of the movement of the earth:

1.8 x 10^11 = 1/2 x 6 x 10^24 x v^2

so v = sqrt((2x1.8x10^11)/(6x10^24)) = 2.5 x 10^-7 m/s

Again no idea of that is correct but that's my attempt.
3. I'm fairly certain that the intention here is to use conservation of momentum.
Find what initial speed the Chinese will need to jump with to reach a height of 20cm, and then their total momentum, mv.
The Earth will receive an amount of momentum in the opposite direction equal to this.
4. (Original post by sam1am)
Hey, I have absolutely no idea if this is right but it sounded like a fun problem.

Assuming the average chinaman weighs 70kg and the avrage jump is 20cm, the using E = mgh

E = 1.3 x 10^9 x 70 x 0.2 = 1.8 x 10^11

Then using the conservation of energy with E = 1/2mv^2, assuming all the gravitational potential energy of the jump is converted to kinetic energy of the movement of the earth:

1.8 x 10^11 = 1/2 x 6 x 10^24 x v^2

so v = sqrt((2x1.8x10^11)/(6x10^24)) = 2.5 x 10^-7 m/s

Again no idea of that is correct but that's my attempt.

(Original post by Stonebridge)
I'm fairly certain that the intention here is to use conservation of momentum.
Find what initial speed the Chinese will need to jump with to reach a height of 20cm, and then their total momentum, mv.
The Earth will receive an amount of momentum in the opposite direction equal to this.
I think both of these would provide you with roughly the same answer. Depending in what module this is, I would want to use conservation of momentum. Makes a little more sense.
5. (Original post by Helsy)
I think both of these would provide you with roughly the same answer. Depending in what module this is, I would want to use conservation of momentum. Makes a little more sense.
Yes in hindsight you are probably right as A2 physics has stuff on momentum. Also using energy doesnt take into account that some would be lost for friction, sound etc.

But mainly I am probably wrong because the I just worked the other method through and the two results differ by a factor 10^-7!
6. just found solution if any1s interested

Estimate average mass of person m = 50 kg (NB Population will contain many children)
Jump height _h = 0.2 m Take-off speed so = 2 m s-1 (accept any reasonable method or blind guess of about 1 to 3 m s-1)
Total upward momentum of all jumpers:
p =1.3 x 10^9 x 50 kg x 2 m s-1 = 1.3 x 10^11 kg m s-1
To conserve momentum, Earth must acquire equal momentum in the opposite direction.
Earth, mass M, acquires speed v = p/M = 1 x 3 x 10^11 / 6 x 10^24 = 2x10^-14 m s-1 i.e. velocity of the Earth is negligible
7. (Original post by Freakonomics123)
just found solution if any1s interested

Estimate average mass of person m = 50 kg (NB Population will contain many children)
Jump height _h = 0.2 m Take-off speed so = 2 m s-1 (accept any reasonable method or blind guess of about 1 to 3 m s-1)
Total upward momentum of all jumpers:
p =1.3 x 10^9 x 50 kg x 2 m s-1 = 1.3 x 10^11 kg m s-1
To conserve momentum, Earth must acquire equal momentum in the opposite direction.
Earth, mass M, acquires speed v = p/M = 1 x 3 x 10^11 / 6 x 10^24 = 2x10^-14 m s-1 i.e. velocity of the Earth is negligible
yes, I agree with that. Roughly anyway, some different numbers. The earth shouldn't move very much, so this sounds more right.

And the reason why there were different answers to each method is because with the kinetic energy, you sqrt the answer gained in the second one, causing the difference.

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