# Why isn't momentum conserved in this pulley problem?

I have some conceptual doubt about method of solving this problem.

24. A block of mass $$m$$ and a pan of equal mass are connected by a string going over a smooth light pulley as shown in figure (9-W17). Initially the system is at rest when a particle of mass $$m$$ falls on the pan and sticks to it. If the particle strikes the pan with a speed $$v$$ find the speed with which the system moves just after the collision.

Solution: Let the required speed be $$V$$.

As there is a sudden change in the speed of the block, the tension must change by a large amount during the collision.

Let $$N$$ = magnitude of the contact force between the particle and the pan $$T = \text{tension in the string}$$ Consider the impulse imparted to the particle. The force is $$N$$ in upward direction and the impulse is $$\int N\,dt$$. This should be equal to the change in its momentum. Thus, $$\int N\,dt = mv - mV.\tag{i}$$ Similarly considering the impulse imparted to the pan, $$\int(N - T)dt = mV\tag{ii}$$ and that to the block, $$\int T\, dt = mV.\tag{iii}$$ Adding (ii) and (iii), $$\int N\, dt = 2mV.$$ Comparing with (i), $$mv - mV = 2mV$$ or, $$V = v/3.$$

But, total initial momentum of the system = $$mv$$ downwards.

And final downwards momentum of the system = $$mV + mV - mV = mV = mv/3$$

1. So, is this solution wrong? I think the final downwards velocity should still be $$v$$ (I can get this by making final and initial momentum equal). But I could not find any technical mistake in this solution.
2. If it is correct, why is momentum not conserved in this case. I understand that kinetic energy is already conserved as there has been a plastic collision.

The pulley (and the attachment to the ceiling) are part of the system here. Because of this, you cannot simply use conservation of momentum on the three given masses.

If the final velocity were $v$, then the total energy of the system would have increased since both the pan and counterweight would be moving and the other mass would not have slowed.

You can't use conservation of momentum equations on only a portion of a system. If you imagine a ball bouncing on the floor, you can't say the momentum of the ball is conserved before and after the bounce. You have to consider the change in momentum of the floor as well.

In your problem, the change in momentum of the ceiling will be small, but relevant.

$$\Delta p_{m1} + \Delta p_{m2} + \Delta p_{pan} + \Delta p_{ceiling} = 0$$

As you do not know the change in this final component, you can't use conservation to solve for the remaining momentum of the other three masses.

Let's change the situation to make this more explicit. Instead of a counterweight, consider two pans and two weights.

Let's imagine the pulley and string to be massless, so the two pans and two weights have a total mass of $4m$. If the balls have a velocity $v$, then the total momentum of the pulley inside the room is $1mv + 1mv = 2mv$.

But by symmetry, we can see that the pulley isn't going to turn. If we imagine the pans at rest after the collision, we find the momentum is now $0mv$.

If the pulley's connection to the ceiling/room/earth is not part of the system, then we say that forces from that connection were external and changed the total momentum. We cannot use conservation of momentum due to external forces.

If the ceiling/room/earth are part of the system, then after the collision, they have gained $2mv$ downward momentum, so the total system does not change. If we picture the box as nearly massless and in a spaceship instead of earth, the entire box would be moving downward at $v/2$ after the balls hit the pans. (assuming completely inelastic collision). The more massive the box, the slower it moves to retain the velocity. Consider it attached to a building/earth, and the momentum still changes, but the velocity change is no longer measurable.

Your calculation of final momentum after the collision has a sign error in it. The pulley serves to change the direction of the motion. This means that a mass moving upward on the left side of the pulley is given a mathematical sign of "+" for the associated velocity. As the string goes over the pulley, the direction of the motion changes such that a downward velocity on the right side of the pulley is given a "+" mathematical sign. Thus, $$mV + mV + mV = 3mV$$. Since $$V = \frac {v}{3}$$, $$3mV = mv$$, and momentum is conserved.

• Despite the fact that someone down-voted my explanation, I teach Atwood machines to AP Physics C high school students (calculus based college level physics) the exact way that I detailed above, and all of my Atwood problems solve properly. While my explanation may not be theoretical enough for this forum, it works. – David White Sep 4 '15 at 22:11
• I'm glad the down-vote didn't put you off teaching. – sammy gerbil Jun 19 '18 at 18:32
• @sammygerbil, that's a "good one". The first requirement to teach 16-18 year-olds, is to have a somewhat thick skin! – David White Jun 19 '18 at 19:23

According to me, the momentum of the mass on the other side shouldn't be negative because the system considered is a connected system, i.e, the momentum transfer to the mass hanging on the other side of the pulley is momentum transferred through the pulley's string. I agree with @BowlOFRed about the contribution of the momentum transferred to the ceiling, however the situation asks us to consider an ideal pulley (no rotational friction at all), which implies that a majority of the momentum transferred across the pulley's string affects the mass on the other end and the contributions towards the ceiling are negligible so we shouldn't care (EDIT: Refer to comments). So when we write down the equations, $$\Delta p_{i}=\Delta p_{f}$$ this should imply: $$mv\approx mV+mV+mV$$ which gives us the required result. Of course, since the pulley is ideal, (theoretically) the error of measurement would be really little.

• An ideal,light pulley will have no energy loss and no rotational inertia. But that has nothing to do with whether or not it transfers momentum into the ceiling. – BowlOfRed Sep 3 '15 at 15:45
• So, all momentum transferred by a string can be calculated using same directional sign? – Archisman Panigrahi Sep 3 '15 at 16:49
• @BowlOfRed : In the case you have mentioned in your answer(about the ball bouncing on a surface), isn't it true that the momentum that the ball imparts onto the floor is imparted back onto the ball, with negligible error? Similarly, in this case, the momentum imparted onto the ceiling should revert onto the pulley system. After all $\Delta p_{ceiling}=0$. We aren't concerned with the momentum imparted but the change of momentum for the ceiling in particular for now. ArchismanPanigrahi : Yes. – Prish Chakraborty Sep 3 '15 at 20:27
• @PrishChakraborty, "After all Δpceiling=0". No, not at all. The mass of the ceiling/earth is so large that you cannot measure the $\Delta v$, but the $\Delta p$ is not zero. – BowlOfRed May 17 '17 at 22:36

The ceiling is applying a total upwards force on the pulley of magnitude $2T$, in order to counteract the force of equal magnitude applied by the masses. So the net force on the system (consisting of three masses and the massless pulley and string) is not zero. Since $\int {Tdt} = mV = mv/3$, $\int 2Tdt = {2 \over 3} mv$ which is the amount of momentum lost.

There is a transient tension $\Delta T$ in the string that will give rise to a change in momentum of the counterweight, the pan, and the mass on the pan. At the ceiling, the pulley transfers twice this force to the support:

The change in momentum of the three components is (with + direction up):

$$\Delta p = m v' + m(v-v') - mv' = m(v-v')$$

This change in momentum is provided by the reaction $2\Delta T \delta t$ at the pulley / ceiling:

For the counterweight:

$$\Delta T \delta t= m v'$$

And for the entire system

$$m(v-v') = 2\Delta T \delta t = 2 m v'$$

Which is consistent with the earlier determination that

$$v' = \frac{v}{3}$$

Note that if your two components in the system were in a straight line (straight string, no pulley), the sign of the momentum change for the counterweight would be reversed and there would be no mystery.

• Do we need to calculate the transition in this cases? – Archisman Panigrahi Sep 24 '15 at 17:41
• What do you mean by transition? – Floris Sep 24 '15 at 17:43
• I mean change - $\Delta$ – Archisman Panigrahi Sep 24 '15 at 17:44
• You don't have to calculate it explicitly; it's the integral $\int F\;dt$ that counts as Umut's answer stated. My diagram is really just an elaborate comment on that answer... – Floris Sep 24 '15 at 17:53

In my opinion, the statement regarding the conservation of momentum says that "The momentum of a system remains conserved if no external force acts on it". I think that the mistake you have committed is that you tried to apply momentum conservation principle along the y-direction, along which gravity(an external force for the system) acts. So according to me momentum conservation is not valid in the situation mentioned

• Gravity does not affect the motion of pan and block (they can contitnue to have any velocity). And the answer to the problem is independent of gravity – Archisman Panigrahi Sep 3 '15 at 17:20

Let I be the impulse transferred to the string due to the fall of the ball on the plank. Now as gravity is a non stop impulsive force so for calculation purpose it could be neglected. And the impulse would be transferred on the other side of the string as well.

Let P' be the final momentum and p be the initial momentum. So,I=P'-P I=2mV-mu. -1 And, -I= mV. -2 Adding 1 and 2 we get 3mV=mu So,V=u/3.

• It's very hard to read your formulae, consider using more spaces or using the "LaTeX"  here... – Ilja Apr 4 '16 at 18:46

## protected by Qmechanic♦Apr 4 '16 at 16:27

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