What Is the Change in the Velocity of a Railroad Car When a Man Jumps Off?

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The discussion centers on the physics problem involving a railroad flatcar and a man jumping off it, specifically analyzing the change in velocity of the car when the man runs to the left. The conservation of momentum principle is applied, leading to the equation v(W + w) = uW + (u - Vrel)w, where v is the initial velocity, W is the weight of the car, w is the weight of the man, u is the final velocity of the car, and Vrel is the man's relative velocity to the car. The solution reveals that the car's velocity increases as the man runs, demonstrating the effect of momentum transfer. The final formula for the car's velocity is u = v + (Vrel * w) / (W + w).

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As i was doing a chapter review, I came across this question. I have tried to solve it, to now avail, I do not know where to begin. :confused: No solution was given in the textbook, and I would like to know how to solve it. Thanks for taking your time to help me.


Question:
A railroad flatcar of weight W can roll without friction along a straight horizontal track.. A man of weight w is standing on the car, which is moving to the right with a speed v. What is the change in the velocity of the car if the man runs to the left so that his speed relative to the car is Vrel just before he jumps off the left end?
 

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Sounds like a momentum question. The momentum of the system is conserved, so:

v(W + w) = uW + (u - Vrel)w

Where:
v - velocity of the car before the man started running
W - the weight of the car
w - the weight of the man
u - velocity of the car once the man is running
Vrel - relative velocity of the man and car

All you have to do is solve for u.
 
When I solve for u, i get

u = vo + (Vrelw) / (W + W)
 
I assume you meant to divide by (W + w). If so, then yes. You can see that the speed of the car increased. That is because the man "pushed" the car when he began to run, thereby accelerating it further.
 
Thanks for your help...It is much appreciated. I understand now.
 

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