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This is a question on my test. I got it right by guessing. Can someone explain it, in terms of equations of momentum.

I first thought of the equation of conservation of momentum: $$(m_{sand}+m_{cart})v_i=m_{sand}v_{sand,final}+m_{cart}v_{cart,final}$$

The right side of this equation appliances when the sand is falling out of the cart. Both $v_{sand,final}$ and $v_{cart, final}$ is the same, since the sand is falling out from the cart, so the x-component of the sand's velocity will also be that of the car. But as the sand falls out, $m_{sand}$ is decreasing.

But this reasoning is erroneous. If either mass changes, velocity got to somehow change too.

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Full question:

A cart of sand is rolling on a frictionless surface as a hole in the bottom of the cart allows sand to fall out at a constant rate. As the cart rolls and the sand falls out, the speed of the cart will:

Answer:

remain the same

most venerable sir
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  • What makes you think that "if either mass changes, velocity got to somehow change too"? If I bolt two carts together on adjacent railway lines, send them hurtling down the tracks, and unbolt them as they're travelling, should their speed change? Same thing here. – Emilio Pisanty Jan 23 '16 at 16:49

2 Answers2

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$F = ma$ and the only forces on the cart are gravity and the reaction force of the surface that balance. So the cart does not accelerate: it keeps moving at constant velocity.

Andrea
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Let us consider the cart as one ball and the sand as another ball moving together. The answer could easily be derived using conservation of momentum but since you have doubts, I will begin with this explanation. When the two ball are separated, the sand ball does not change its speed but continues with the same speed. Imagine this, two balls moving closely with the same speed, when slightly moved, will continue moving independently, without any change in speed. I feel so bad not providing a mathematical base for the answer, but I hope this helps.

N.S.JOHN
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