Two masses $M$ and $m$ are connected by a light inextensible string which passes over a small pulley as shown in the diagram. If the mass $m$ is moving downward with a velocity $v$ when the string makes an angle of $45^{o}$ with the horizontal, find the total KE of the two masses. Assume that the mass $M$ moves horizontally.

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Solution

Let $y$ be the depth of the mass $m$ below the top of the pulley. If the total length of string is $l$ , the length of string from $M$ to the pulley is $(l - y)$
The distance of $M$ from the edge of the table $= x = (l - y) cos 45^{o}$
It is given that, $(\frac{d y}{d t}) = v$
The velocity of $M = \frac{d x}{d t} = \frac{d}{d t} (l - y) cos 45^{o} = - (\frac{d y}{d t}) cos 45^{o} = - \frac{1}{\sqrt{2}} v$
(The minus sign tell us that $x$ decreases as $y$ increases)
The net $K . E = \frac{1}{2} m v^{2} + \frac{1}{2} m {(\frac{d x}{d t})}^{2} = \frac{1}{2} m v^{2} + \frac{1}{2} m {(- \frac{1}{\sqrt{2}} v)}^{2} = \frac{1}{2} (m + \frac{M}{2}) v^{2}$

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