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A truck is moving at constant velocity on a friction-less surface. Sand is dropped into the truck at X, and the sand is removed at Y.

What is the change in velocity at X and Y?

My answer was a decrease in velocity at X and an increase in Y.

The right answer is a decrease in velocity at X and at Y it stays the same.

I get that at X, as momentum is constant and the mass increases, the velocity has to decrease.

But why does velocity stay the same at Y? Isn't momentum still constant at Y? Since the mass decreases at Y, wouldn't the velocity increase?

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If the sand is moving at the same velocity as the truck when it enters the truck then no change in v (say, dropped from a plane moving at the same speed overhead).

If the sand is stationary (relative to the ground) the truck is slowed (conservation of momentum).

If sand is moving in opposite direction when it impacts the truck the truck is really slowed (depending on the mass of sand added).

And yes, no change in truck's (post impact) velocity when sand is released from the truck. The sand (because of its velocity) will also slide along with the truck on the frictionless surface after release.

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  • $\begingroup$ Oh... It's much clearer now. Thank you! $\endgroup$
    – imanoob
    Commented Aug 22, 2021 at 16:00

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