A trolley of mass 300kg carrying a sand bag of 25kg is moving uniformly with speed of 27km/h on a frictionless track. After a while, sand starts leaking out of a hole on the floor of the trolley at the rate of 0.05 kg/s.

What is the speed of the trolley after the entire sand bag is empty?

I was so surprised when I read this question. It doesn't make sense to me. I can't comprehend how the loss of sand creates an external unbalanced force on the trolley such that it affects its velocity.

Maybe I haven't analyzed the question enough but I find this a bit conceptually challenging for me. Maybe I have to consider how the sand particles affect the back wheels of the trolley or maybe consider the sand to be a propellant?

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    $\begingroup$ Have you considered the possibility it is a trick question and that you have already grasped the trick? $\endgroup$ – DWin Dec 22 '13 at 23:49
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    $\begingroup$ I read the question title and thought "Ooh that must be about rockets!". Nope :-/ $\endgroup$ – Brandon Enright Dec 23 '13 at 0:14
  • $\begingroup$ @BrandonEnright: I thought so too. That's why I asked it here. $\endgroup$ – Nick Dec 23 '13 at 13:50
  • $\begingroup$ @Pulsar: Not homework!! $\endgroup$ – Nick Dec 23 '13 at 13:52
  • $\begingroup$ @DWin: I'm not usually that smart. But can you suggest a change in the situation such that the parameters given have significance? $\endgroup$ – Nick Dec 23 '13 at 14:00

The answer is... 27 km/h. It is a trick question, the net force on the trolley is always zero. People might be tricked into blindly applying momentum conservation to find an increase in velocity but this would be incorrect. As the sandbag decreases in weight the momentum carried by the trolley-sandbag system decreases.

  • $\begingroup$ I just performed an experiment with a toy cart filled with water. The cart had a hole on it's rear which was loosely closed by a cork. As the cork fell off halfway through it's motion, the speed of the cart increased. Can you explain that observation and how it's different from the trolley? $\endgroup$ – Nick Dec 23 '13 at 13:58
  • $\begingroup$ If the surface wasn't frictionless and the trolley was either being pushed (or on an inclined surface) in order to keep the speed constant, then I'd expect this to happen. Otherwise, I can't explain what you observed. Basically, If there is a net force on the system, then as the weight on the trolley changes the trolley will behave differently under that force. But in the absence of any net force, there is nothing to change the speed of the trolley. In the case you described there is no net force, the trolleys speed is constant due to the surface being frictionless. $\endgroup$ – UserB Dec 23 '13 at 17:23
  • $\begingroup$ @Nick: It depends on which way the water exits the cart, because it's behaving like a small rocket. $\endgroup$ – Mike Dunlavey Dec 23 '13 at 18:28
  • $\begingroup$ There would need to be a lot of water pressure in the cart for the water flow to produce any noticeable force on a table-top experiment. $\endgroup$ – UserB Dec 23 '13 at 18:37
  • $\begingroup$ @MikeDunlavey: and this could happen with the trolley as well, right? $\endgroup$ – Nick Dec 24 '13 at 8:24

This is no trick question. It is simply an application of the fact that the velocity of the Center of Mass (COM) of a body is unaffected by internal forces. You would have been correct in saying the velocity of COM cannot change in absence of external forces. But this does not apply to the trolley alone.

So what's going on here? Initially the COM of the trolley + sandbag system is moving at $27$ kmph. Now the sand is falling out of the sandbag backwards by some means of an internal force. Since the velocity of the COM must, must be constant in the absence of external forces, the trolley speeds up so the net momentum of the system is conserved.

With some formulas:

$v_c$ = $(m_tv_t + m_sv_s)$ $/$ $(m_t + m_s)$

..where $v_c$ is the velocity of the COM. This will not change even after all the sand falls out.

This is basically just another way of writing linear conservation of momentum, but I want to bring your attention to the significance of the fact that it's the velocity of the COM which dictates the velocity of the sand and the trolley.

To give you another example of where exactly the same happens, think of rocket propulsion. How does the rocket keep gaining speed by simply ejecting fuel at a constant velocity? Despite what you said in the comments, your interpretation is actually correct :)

  • $\begingroup$ My original comment was intended to draw attention to the fact that the sand was dropping with the same velocity as it originally had and that it only changed velocity after it had left the trolley. The manner in which the experiment described by Nick is different than one in which the sand was dropping, is that in the situation where water was flowing out of an orifice at the back of the trolley, it was leaving the orifice with a velocity in the backward direction. If the experiment were arranged with the hole and the cork point directly downward. there would then be no change in velocity. If $\endgroup$ – DWin Dec 23 '13 at 16:52
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    $\begingroup$ Note that it is not just the direction of flow, but also the position of the nozzle which can affect the answer. $\endgroup$ – dmckee --- ex-moderator kitten Dec 23 '13 at 17:54
  • $\begingroup$ Okay, yeah, I think I screwed up here :( The position of the nozzle indeed matters. But won't the trolley tend to move upwards if the orifice is pointing towards the bottom? Like a jetpack! (ofc a non-functioning one) $\endgroup$ – shortstheory Dec 28 '13 at 7:47

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