Pulley Challenge

A block of mass 2M and two blocks of mass M are connected to a pulley as shown. While the system is static, the lower block of mass M is flicked up with a velocity v . Assume that the string connecting two blocks of mass M is long enough such that the blocks will not collide. What is the time taken for the rope between the 2 equal blocks be taut again?

Screenshot 2015-01-27 at 20.42.46

From a brilliant.org Problem

When the lower block is flicked, both blocks with mass M starts accelerating. Therefore, the distance between two blocks cannot be calculated directly. Whenever there’s two moving objects, I will follow the relativity strategy. That is, we see one object as static. Relatively, the velocity of the other object should be converted such that the system remains the same.

First, let’s calculate the acceleration of the upper block. Taking that the tension force along the string is the same,

\begin{aligned}  \displaystyle Mg + Ma &= 2Mg - 2Ma \\ 3Ma &= Mg \\ a &= \frac{1}{3}g  \end{aligned}

Taking it as static, then the downward acceleration of the lower block should be increased by \frac{1}{3}g. So, acceleration of the lower block is \frac{4}{3}g. The time taken for the rope between the 2 equal blocks to be taut again is when the lower block get back into its original position, that is, 0 displacement.

\begin{aligned}  \displaystyle vt - \frac{1}{2}at^2 &= 0 \\ vt - \frac{2}{3}gt^2 &= 0 \\ v - \frac{2}{3}gt &= 0 \\ t &= \frac{3v}{2g}  \end{aligned}

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Pulley Challenge

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