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Gravtiational PE

Two masses M ...

Question

Two masses M and m (with M > m) are connected by means of a pulley as shown in the figure. The system is released from rest. At the instant when mass M has fallen through a distance h, the velocity of mass m will be

$\sqrt{2 g h}$

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$\sqrt{\frac{2 g h M}{m}}$

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$\sqrt{\frac{2 g h (M - m)}{(M + m)}}$

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$\sqrt{\frac{2 g h (M + m)}{(M - m)}}$

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Solution

The correct option is C
$\sqrt{\frac{2 g h (M - m)}{(M + m)}}$

If mass m falls through a distance h, mass m rises up through the same distance h. Let v be the common velocity of the masses when this happens.
Now, Loss in PE = Gain in KE
$\Rightarrow M g h - m g h = \frac{1}{2} (M + m) v^{2}$
$\Rightarrow v = \sqrt{\frac{2 g h (M - m)}{(M + m)}}$
Hence, the correct choice is (c).

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The correct option is C √2gh(M−m)(M+m) If mass m falls through a distance h, mass m rises up through the same distance h. Let v be the common velocity of the masses when this happens. Now, Loss in PE = Gain in KE ⇒Mgh−mgh=12(M+m)v2 ⇒v=√2gh(M−m)(M+m) Hence, the correct choice is (c).