Consider an ideal pulley with two blocks of same mass, say $m$, in equilbrium. ($T$ represents tension force).

Pulley-mass system in equilbrium

One of the blocks is lifted up to a certain height, say $h$, while the other block remains at rest. During this time, the string would get untaut, so the tension in string would be zero. An external force is applied to ensure that the other block remains in equilibrium. one of the blocks is lifted up

The block is then released. When it reaches its original level, the string would get taut again. Just before this instant, its velocity is $u = \sqrt{2gh}$. The external force that was needed to keep the other block in equilbrium is now removed. just before string gets taut

During the instant the string gets taut, an impulsive tension force acts on both blocks. $mg$ also acts on them but it would produce negligible impulse during this time. After this, they would acquire a common velocity, say $v$.

just after string gets taut

$$ \large I_{\ by\ string\ on\ right\ block}= I_{\ by\ string\ on\ left\ block} \\ m(-v-(-u))=mv \\ v=\sqrt{\cfrac{gh}{2}}$$

If the blocks+string is considered as a system, net external impulse should be approximately zero (from the time just before to the time just after the string gets taut). However, before the string gets taut, the linear momentum of this system was $p_i=m\sqrt{2gh}$ downwards. After the string gets taut, the linear momentum is $p_f=mv-mv=0$.

The fact that the linear momentum does change implies there is some external impulse acting on the system. Where does this come from?

  • 1
    $\begingroup$ The flaw is: The moment the string ist not taut. the second mass will accelerate down, when they meet again both have fallen the same amount , so have the same velocity and will stop each other. $\endgroup$
    – trula
    Jan 7 at 20:26
  • $\begingroup$ Indeed, task description hasn't mentions do that external force keeping second block in place is removed when first block is going down or not. If removed,- then yes, blocks will not reach the previous position anymore, but rather keep same relative distances falling down. In not removed,- first block will return at the previous position and depending on the nature of external force which keeps second block in place,- may induce some upwards momentum to the second block, so balance will be changed to the other side. Unless that external force keeps second block pinned to the same position. $\endgroup$ Jan 8 at 14:48

1 Answer 1


If you choose as your system the rope and masses, then the system is not closed because the pulley then acts a source of external forces, not to mention that there is the presence of the gravitational force.


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