I have a situation of a simple mass pulley system with an inelastic string, in which a mass of 15 kg held 5 m above ground is attached to one end of the string and a mass of 7 kg is attached to the other side of the string but sitting on the ground.

When the heavier mass is released then it will move down with a certain acceleration with the lighter mass moving up. Heavier mass will finally hit the ground and have its energy dissipated as sound/heat. The lighter mass will then continue moving up under gravity i.e. downwards acceleration of $9.8 ms^{-2}$. The lighter mass will reach a max height and then start moving down under gravity.

My question is how the system of masses will behave once the lighter mass reaches 5 m above ground? Will it come to rest instantly? We know the heavier mass on ground cannot be lifted up by the lighter mass.

One guess I've made is that the lighter mass will instantly come to rest when it reaches 5 m level during its descent under free fall. Or may be the lighter will oscillate about 5 m level till it finally comes to rest with the heavier mass going up and down till it also comes to a final rest.

UPDATE regarding impulsive force when string becomes suddenly taut

We could apply law of conservation of momentum to mass pulley system when the string suddenly tightens producing a high impulsive tension force in the string on both the lighter and heavier mass. This large impulsive force acts for a very small interval of time on both masses and after this small time interval it stabilizes with tension becoming smaller in accordance with Newton's second law of motion i.e $\vec F_{net} = ma$. This large impulsive force is the reason for the heavier mass moving up of the ground when string becomes suddenly taut.

Given below is an explanation on why the law of conservation of momentum becomes applicable in this scenario.

From impulse momentum equation over a period from $t_1=0$ to $t_2=t_2$ we can say that Total Impulse = change in momentum for each of the masses or mathematically as below. We are taking upwards direction as +ve and downwards as -ve since impulse is a vector and we need to consider its direction. Just after the impulsive force we assume that masses acquire a velocity of $V_2$. Both masses will have same magnitude of velocity.

$$\text{For heavier mass M} \; \int_0^{t2} {\vec F} dt = MV_2 - M (0) $$

$$\text{For lighter mass m} \; \int_0^{t2} {\vec F} dt = (-mV_2) - (-mV_1) $$

$$\text {Since the variable impulsive force F is nothing but the tension} \\ \text{in string which is going to be same for both masses} $$

$$\therefore MV_2 - M (0) = (-mV_2) - (-mV_1) $$

$$\therefore MV_2 + mV_2 = mV_1 $$

$$\text {i.e. Total momentum just after impulsive force =} \\ \text{Total momentum just before impulsive force }$$

$$\text {Above equation is nothing but the law of conservation of momentum} \\ \text {for the system of masses}$$

UPDATE on how the system behaves before coming to rest

Smaller mass would go up and down about the 5m above ground position till it finally slows down and comes to rest at this position.

  • When string suddenly becomes taut the heavier mass would be lifted by high impulsive force of tension.

  • Then heavier mass would move up and lighter mass down till heavier mass comes to rest.

  • After this heavier mass will start moving down while lighter mass will start moving up.

  • Then when heavier mass reaches the ground, the lighter mass would move up under gravity from 5m position and then down under gravity.

  • Once smaller mass reaches 5 m position in its downward free fall, an impulsive force on larger mass would lift it off the ground and the same cycle of motion would repeat till velocity of smaller mass is reduced to 0 at 5 m above ground position.

For each successive cycle of motion described above, the height to which lighter mass rises under gravity will reduce till it finally becomes zero.


Since you assume your string to be perfectly inelastic, the mass will immediately come to a rest.

Of course, in a realistic setting, the string would stretch a little, in essence acting as a spring (probably far from the linear Hooke regime) and you'll get oscillations around the equilibrium position.

For the fun of it, one can also ignore your statement that the large mass can not be lifted from the ground anymore. In that case, the kinetic energy of the small mass going down would be converted into potential energy of the large mass, thus lifting the large mass up a tiny bit. Once that large mass reaches its highest point, that essentially re-sets your question to the original configuration except that now there is much less potential energy available (it was lost to the original "thump" of the large mass hitting the ground). In a realistic setting this probably won't even repeat once, but for sure not more than once.

  • $\begingroup$ Would the heavier mass be disturbed from its position on ground if the string was elastic? $\endgroup$
    – Sunil
    Apr 6 at 8:47
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    $\begingroup$ i added some thoughts to this end. whether the string is elastic or not again is a question whether you want to discuss a realistic setting (in which case I'm certain string elasticity is utterly negligible) or a thought experiment (in which case the string elasticity acts as a small energy reservoir that you can use to temporarily park energy, in essence smearing any delta-response out to something smoother) $\endgroup$
    – rfl
    Apr 6 at 8:54
  • $\begingroup$ Regarding heavier mass getting lifted, if we think in terms of forces on the heavier mass when the slack string becomes taut as the lighter mass reaches 5 m position, then we know tension T upwards and normal reaction from ground N pointing upwards would be acting on the heavier mass. Only if T + N > 15g, would the heavier mass move upwards. Would T + N > 15g be the explanation about why heavier mass would get lifted? We would then have 2 equations but 3 unknowns. T + N - 15g = 15a and 7g - T = 7a and the three unknowns would be T,N and a. $\endgroup$
    – Sunil
    Apr 6 at 9:12
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    $\begingroup$ Your question states an inelastic string, so you can simply convert the kinetic energy of the down-moving 7kg mass ($mv^2/2$) into potential energy ($Mgh$) of the resting 15kg mass. Equate these two to get the height $h$ that the heavy mass moves up. $\endgroup$
    – rfl
    Apr 6 at 9:22
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    $\begingroup$ Note that the 15kg mass also needs to hit the ground inelastically. If the ground - 15kg collision were partially elastic, the latter would bounce upon contact and the 7kg mass would briefly drop proportionally to bounce $\endgroup$ Apr 6 at 17:57

When the $7kg$ mass returns to $5m$ and the string becomes tight, use conservation of momentum for an instantaneous point in time, to find how they both move.

As the string is inelastic, it's similar to a collision.

If $v$ is the downward speed of your $7kg$ mass, (same as its previous upwards speed at $5m$)

momentum before = momentum after:

$$7v+(15\times0) = 7v_2+15v_2$$

where $v_2$ is the speed just after the string gets tight. You could imagine this happening on a smooth table, where the two masses are separating and then the string goes tight.

You might have to consider whether it's possible that the 7kg mass rebounds, but as the equation above is giving a sensible answer, maybe you won't need that part...

  • $\begingroup$ If we take the the whole pulley masses as system. Then at the point when 7kg mass is descending at 5 m, we have external forces of 7g, 15g and N I(normal reaction from ground on heavier mass) acting. Momentum would be conserved if net external force was zero. Is my reasoning correct or I am missing something about law of conservation of momentum? $\endgroup$
    – Sunil
    Apr 6 at 9:26
  • $\begingroup$ Also, the right side of your momentum equation should have one of the momentums as negative since velocities will be in opposite directions for the two masses? $\endgroup$
    – Sunil
    Apr 6 at 9:29
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    $\begingroup$ For a collision type of situation you can find a sudden change in velocity as described. $\endgroup$ Apr 6 at 9:33
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    $\begingroup$ Best to use the masses on a table analogy, if the directions are confusing. If you are worried about momentum conservation as viewed, we would have to take into account the momentum of the pully/earth, that has a slight increase downward. $\endgroup$ Apr 6 at 9:37
  • $\begingroup$ I think the reason we can apply conservation of momentum here because the forces involved between masses is far greater than other external forces acting on the system. That is tension when string becomes taut is going to cause a large force of tension. So external forces may be considered negligible when compared to weight of each mass or normal reaction on heavier mass. $\endgroup$
    – Sunil
    Apr 6 at 10:41

I find your question rather confusing, but other answerers seem to find it clear.

Time1: First, you release the 15kg weight. The effective gravitational mass of the system is 15-8=7kg, but its inertial mass is 15+8=23kg so it will accelerate at $\frac {7g}{23}$. Since $v^2=2ad$, the velocity when the large weight reaches the ground will be given by $v^2=\frac {35mg}{23}$, so $v=\sqrt{\frac {35mg}{23}}$. Of course, when we take the square root, we get $\pm$, and we can take the positive solution to represent the smaller weight moving upwards and the negative solution to represent the larger weight moving down.

Time2: At this point, the smaller mass will have no force acting on it other than gravity, so it will enter parabolic flight. It will continue to travel upwards, then fall back down. It will reach the same speed, but opposite direction, on its downward path. That is, it will return to being 5m above the ground, but now have a velocity of $-\sqrt{\frac {35m}g{23}}$.

Time3: Now the string will become taut. The weight of the larger mass is larger than the smaller mass, so the smaller mass will not be able to lift the larger mass. In the real world, there are complications such as elasticity of the string that can cause the larger weight to lift slightly, but if we're assuming a perfect rigidity, it won't move.

The question really seems to me to come across as asking whether at Time1 the system will come to a halt. The answer to that, as I said, is “no”. But other people seem to have taken it to mean whether it will come to a halt at Time3. So apparently your meaning is clearer to those people than it is to me.

  • $\begingroup$ I think you may have misunderstood my question. All I'm asking is what will happen when the string becomes taut suddenly and lighter mass returns to 5 m position. That's quite clear in my question paragraph. $\endgroup$
    – Sunil
    Apr 7 at 19:08

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