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Lets consider two sacks full of sand kept on a stand. One has a hole and the other one doesn't. Put both of them on a very slippery surface (assume $\mu$ to be zero) and give a push.

Now equation of motion for both of will be different, right?

Equation of motion for sack with hole would be: $$F_x = m (\frac{dv}{dt}) + v (\frac{dm}{dt}) = 0 $$ $$F_y = M_{inst} g$$ Similarly for sack without hole would be: $$F_x = m (\frac{dv}{dt}) = 0$$ $$F_y = m g$$ Now when I solve these equations I was doubtful in the case with the hole. I found that solving the first equation above gives me: $$mv = constant $$ That means $v$ is changing w.r.t. time, but what bothers me is physically what is changing its velocity?

Qmechanic
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Vivekanand Mohapatra
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3 Answers3

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I have a solution for this.

We are given a "very slippery surface" so assume that friction is zero. Assume that there is no considerable sand loss during the impulse (during the push) so that both bags are brought to speed $v$ at time $t=0$.

Consider the bag with a hole.
momentum at time $t = (M(t)+\Delta m)v(t)$
momentum at time $t+\Delta t= M(t)v(t+\Delta t) + \Delta m v(t)$

subtract to obtain change in momentum:
$\Delta P= M(t)\Delta v$

divide by $\Delta t$ and take infinitesimal limits to find
$\frac{d}{dt}P=F=M(t)a(t)$.

By assumption after the push $F=0$ so that, having $M \ne 0$ we must have $a(t)=0$. Hence both bags move with the same velocity and zero acceleration after the push.

Note: it is interesting to compare this example with example with example 4.14 ("A leaky freight car") from "An introduction to mechanics", Kleppner,Kolenkow. The question here is: what force is needed to keep a freight car leaking sand moving at constant speed v?

Then for the holed bag $P(t)=m(t)v, v=$ constant and $p \to 0$ as $t$ increaes, but the velocity $v$ is constant.

Best, Matt.

sammy gerbil
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Matt306
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You have to be careful in how you define $m$, the mass of the sack of sand. When you write $$F = \frac{d(mv)}{dt} = m\frac{dv}{dt} + v\frac{dm}{dt} = 0$$ You are implicitly assuming that the sand that falls out of the sack has zero velocity. The second term in the chain-rule expansion, $v\frac{dm}{dt},$ implies that a non-zero force is created by the sand leaving the sack. This is only true if the sand is propelled out the back of the sack so that it has zero velocity with respect to the surface. Since this propulsion would essentially make the sand sack a sand rocket, this is where your acceleration comes from.

The solution is to say that, because there is no external horizontal force acting on the sand leaving the sack, nor on the sack itself, $\frac{dv}{dt}$ must be zero.

Mark H
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Supposing that there is also no friction between the surface and the sand grains, then even as the sack with a hole disintegrates, the entirety of it will keep on travelling at a constant speed. Consider two ice skaters that hold hands. If they let go of each other, it won't affect their velocity. They'll just continue gliding in the same direction next to each other. The same will happen with the sack and sand.

Regarding the motion equations you've written, they are right but their interpretation is slightly flawed. A motion equation in its original sense reads "the change of momentum over time is caused by forces". Then, of course, from zero net force you get that momentum is constant over time, $mv = const$. The problem here is that when you consider only the sack (with some remaining sand in it), it actually loses momentum over time. The sand grains that leave the sack take away their bit of momentum. Since the equation you've written down says directly that momentum is constant in time, it's no wonder that it leads to a nonsense result where, as $m$ decreases, $v$ should increase even though there is nothing pushing the sack.

Ross
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