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Conservation of momentum

  • Thread starter Zolo
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  • #1
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Homework Statement


A number of k people standing on a stationary car. These people jump off with a velocity of v relative to the car one by one. Assume these people jump to the direction, what is the final speed V of the car.


Homework Equations


initial momentum=final momentum

The Attempt at a Solution


I tried to apply the formula a few times..and I get MV= kmv + m(V1+V2+...+V(k-1))
V1 means the velocity of the car after 1 people jump off. V(k-1) means the velocity of the car after k-1 people jump off.
But i can't simplify it...
 

Answers and Replies

  • #2
Andrew Mason
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Homework Statement


A number of k people standing on a stationary car. These people jump off with a velocity of v relative to the car one by one. Assume these people jump to the direction, what is the final speed V of the car.


Homework Equations


initial momentum=final momentum

The Attempt at a Solution


I tried to apply the formula a few times..and I get MV= kmv + m(V1+V2+...+V(k-1))
V1 means the velocity of the car after 1 people jump off. V(k-1) means the velocity of the car after k-1 people jump off.
But i can't simplify it...
Let's assume the people all have the same mass. Examine each jump in the reference frame of the car before the person jumps. What impulse does each person give to the car when each person jumps off? (impulse = FΔt = Δp). How does the final motion of the car relative to its initial motion relate to the sum of all those impulses?

AM
 
  • #3
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In the reference frame of the car,each people jump off will give an impulse of mv to the car? final momentum of car= initial momentum of car plus all these impulse?
 
  • #4
Andrew Mason
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In the reference frame of the car,each people jump off will give an impulse of mv to the car? final momentum of car= initial momentum of car plus all these impulse?
Ok. But keep in mind that when the car that receives the first impulse it has k-1 people on it and when it receives the second impulse it has k-2 people on it, etc. So now comes the tricky part. Express the change in velocity of the car when the nth person jumps off (k > n > 0). Then all you have to do is add up all those changes in velocity.

AM
 
  • #5
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So the increase in velocity when nth people jump off= mv/[M+m(k-n)]?
Then, the final V=mv/[M+m(k-1)]+mv/[M+m(k-2)]+...+mv/M?

But the answer is V=mv/[M+m(k)]+mv/[M+m(k-1)]+...+mv/[M+m]...
 
  • #6
Andrew Mason
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So the increase in velocity when nth people jump off= mv/[M+m(k-n)]?
Right.
Then, the final V=mv/[M+m(k-1)]+mv/[M+m(k-2)]+...+mv/M?
Ok. When the last person jumps off, n=k so:

[tex]V = \sum_{n=1}^{n=k} Δv_n = \sum_{n=1}^{n=k} mv/[M+m(k-n)][/tex]

But the answer is V=mv/[M+m(k)]+mv/[M+m(k-1)]+...+mv/[M+m]...
It seems to me that the first term in the sum (n=1) is: mv/[M+m(k-1)] and the last term (n=k) is: mv/[M+m(k-k)] = mv/M. So I agree with you.

Maybe they assume that it has k+1 people on it intially and k jump off (the last one, being the driver, stays on the car).

AM
 

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