Railroad Car: Man's Weight Affects Velocity Change

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A railroad car moving at speed V1 experiences a change in velocity when a man of weight w moves off its left end with a relative speed Vrel. The formula for this change in velocity is given by change in v = w(v1 + Vrel)/W, highlighting the relationship between the man's weight and his movement. The discussion emphasizes the principle of conservation of momentum, indicating that the man's actions influence the car's velocity. Specifically, a heavier man or a faster movement off the car results in a greater change in the car's velocity. Thus, the man's weight significantly affects the railroad car's velocity change.
skiboka33
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?? railroad car

doesnt make sense... here goes...

A railroad car of weight W can roll without friction along a straight horizontal track. A man od weight w is standing on the acr which is moving at a speed of V1. Prove that change in v = w(v1 + Vrel)/W if Vrel is the speed of the man moving off the left end relative to the car. (car is going right, man is going left) thanks!
 
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Conservation of momentum

Use conservation of momentum.
 


This statement is discussing the relationship between the weight of a man standing on a railroad car and the resulting change in velocity of the car. It is stated that the car is initially moving at a speed of V1, and the man is standing on the car and moving off the left end with a relative speed of Vrel. The formula provided in the statement, change in v = w(v1 + Vrel)/W, suggests that the change in velocity of the car is directly proportional to the weight of the man and the relative speed between the man and the car. This means that the heavier the man is, and the faster he is moving off the left end, the greater the change in velocity of the car will be. This is because the man's weight and movement will create a force on the car, causing it to accelerate or decelerate. Therefore, the weight of the man does have an impact on the velocity change of the railroad car.
 
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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