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A mass m, lying on a horizontal frictionless surface is connected to mass M as shown in Fig. The system is now released. The velocity of mass m when mass M has descended a distance h is:

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

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

$\sqrt{2 g h}$

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

answer is C.

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Detailed Solution

When M has descended a distance h, loss of PE = Mgh. If v is the common velocity of the masses, then:

$gain in K E = \frac{1}{2} (M + m) v^{2} Hence, \frac{1}{2} (M + m) v^{2} = M g h \Rightarrow v = \sqrt{\frac{2 M g h}{(M + m)}}$ . .

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