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

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The discussion focuses on a physics problem involving a railroad car and a man jumping off. The key point is the conservation of momentum, which is applied to determine the change in the car's velocity when the man runs to the left before jumping off. The equation derived shows that the car's velocity increases as the man runs, due to the momentum transfer. The final formula for the car's velocity after the man starts running is clarified, confirming that the initial misunderstanding about the denominator was corrected. The explanation concludes with an acknowledgment of understanding the concept better after the discussion.
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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.
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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