Sand leaking out of bag and skater "A person skating on a frictionless icy surface is holding a sandbag. The sandbag has a small hole at the bottom, from which the sand starts to leak. As the sand leaks from the sandbag, the speed of the skater..."
The answer was that the speed of the skater stays the same. I thought that the speed of the skater must increase because now the mass of the sandbag is less than the mass before it leaked. What is wrong with my intuition here? The system I'm considering is the skater and sandbag together.
$p_{system}(t_i) = (m_{skater}+m_{sandbag})v_i \\
p_{system}(t_f) = (m_{skater}+m_{sandbag})v_f$
Since the mass of the sandbag is smaller, the $v_f$ must increase, no?
 A: The problem is that you have to consider the full $p_{system}$ when you look at $p_{system} (t_f)$.
The full momentum of the system has to consider every part of the system for it to remain conserved, and that includes the falling sand.  So really you have:
$$p_{system} (t_f) = (m_{skater} + m_{sandbag} + m_{sand}) v_f$$
We can assume the falling sand has the same velocity $v_f$ because it should retain the velocity from when it fell (neglecting slight air drag), and since the surface is frictionless it should keep sliding along.  If the sand were slowed by friction or air drag, that momentum from the sand would go into the ground or the air respectively; but wouldn't take momentum away from the skater or sandbag, or add any momentum to them.
If the sand were being pushed backwards relative to the skater when it fell out (not like here), then the velocity of the skater would increase; and there would be a force due to the sand pushing away from the skater.
A: So what you have done is proceeded with "momentum is always conserved. So $m_i v_i = m_f v_f$ so that $v_f = \frac{m_i}{m_f} v_i$. If you know what the ratio $\frac{m_i}{m_f}$ is, this should answer your question.
But
The net force on skater-bag system is zero. You originally proceeded to  apply momentum conservation and therefore there'd be an increase/decrease in velocity to the skater, but this is not the case. The sand still has its initial momentum when it is released. The mass that is being lost (leaking sand) has the same horizontal velocity as the skater-bag. So the sand carries away horizontal momentum from the system. At the end there is less moving mass but also less horizontal momentum remains in the system. That is, as the bag leaks sand, the sandbag decreases in mass and the total momentum carried by the skater-bag system therefore decreases.
Overall the velocity for the skater must stay the same (since no external forces are acting in the direction of the skater-bag system).
A: A given force will propel a lighter body to a greater velocity.
Once moving, if we could just vanish part of the body's mass, it might speed up.   But we can't.  The sand, although jettisoned, still exists and takes its kinetic energy along with it.
So no speed up.
This ignores air resistance of course.  If the leak changes the shape of the bag, this might affect the slowing effect of air resistance.
This assumes still air of course.  If the body is being propelled by the wind...
A: A quite intuitive visualization just popped into my head. Imagine that the skater is not losing the sand steadily, but simply drops the whole bag next to him. He's not catapulted forward because of that, is he? We'll assume he doesn't throw it backwards but simply drops it. Instead since everything is frictionless, the bag just keeps moving together with the skater next to him, where it was dropped. Actually, this is kinda like floating in zero gravity vacuum. You can release the bag and... it will continue to float next to you where you left it. You'll both keep moving forward with the same speed.
