In the figure shown, a particle of mass m is released from the position A on a smooth track. When the particle reaches at B, then normal reaction on it by the track is

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$\frac{m^{2} g}{h}$

$\frac{2}{3} mg$

$mg$

$2 mg$

answer is A.

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Detailed Solution

$\begin{array}{l} By conservation of energy \\ mg (3 h) = mg (2 h) + \frac{1}{2} {mv}^{2} (h e r e v = velocity at B) \\ \Rightarrow mgh = \frac{1}{2} {mv}^{2} \\ \Rightarrow v = \sqrt{2 gh} \\ From free body diagram of block at B \end{array}$

$\begin{array}{l} N + mg = \frac{{mv}^{2}}{h} = 2 mg \\ \Rightarrow N = mg \end{array}$

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