Calculating Final Speed of a Car with Many Jumpers

In summary: That could be the case. Or maybe they just made a mistake in the answer. Either way, the concept is clear. In summary, the final speed V of a stationary car after k people jump off with a velocity v relative to the car can be calculated using the formula V= mv/[M+m(k)]+mv/[M+m(k-1)]+...+mv/[M+m] where m is the mass of each person and M is the mass of the car. This is based on the principle of conservation of momentum, where the final momentum of the car is equal to the initial momentum plus the sum of all the impulses given by each person jumping off.
  • #1
Zolo
9
0

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...
 
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  • #2
Zolo said:

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
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
Zolo said:
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
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
Zolo said:
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
 

1. How do you calculate the final speed of a car with many jumpers?

The final speed of a car with many jumpers can be calculated using the formula V = √(V₀² + 2ad), where V₀ is the initial velocity of the car, a is the acceleration due to gravity (9.8 m/s²), and d is the distance traveled by the car during the jump.

2. What factors affect the final speed of a car with many jumpers?

The final speed of a car with many jumpers is affected by several factors, including the initial velocity of the car, the acceleration due to gravity, the distance traveled by the car during the jump, and the mass of the jumpers.

3. How does the mass of the jumpers impact the final speed of the car?

The mass of the jumpers has a direct impact on the final speed of the car. The greater the mass of the jumpers, the greater the force exerted on the car during the jump, resulting in a higher final speed.

4. Can you use the same formula to calculate the final speed of a car with different initial velocities?

Yes, the formula V = √(V₀² + 2ad) can be used to calculate the final speed of a car with different initial velocities. However, it is important to use consistent units for all variables in the formula (e.g. meters per second for velocity and meters for distance) to ensure accurate results.

5. Is air resistance a factor when calculating the final speed of a car with many jumpers?

Yes, air resistance is a factor that can affect the final speed of a car with many jumpers. However, its impact may be negligible depending on the distance of the jump and the aerodynamics of the car and jumpers. In most cases, it is safe to assume that air resistance can be ignored when calculating the final speed.

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