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Q.

A uniform horizontal rod of length l falls vertically from height h on two identical blocks placed symmetrically below the rod as shown in figure. The coefficients of restitution are $e_{1} a n d e_{2}$ . The maximum height through which the centre of mass of the rod will rise after bouncing off the blocks is

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a

$\frac{h}{e_{1} + e_{2}}$

b

$\frac{{(e_{1} + e_{2})}^{2}}{4} h$

c

$\frac{e_{1} + e_{2}}{2} h$

d

$\frac{{(e_{1} + e_{2})}^{2} h}{2}$

answer is D.

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Detailed Solution

velocity of the two end after collision is $e_{1} \sqrt{2 g h} a n d e_{2} \sqrt{2 g h}$

$V_{c m} = \frac{e_{1} + e_{2}}{2} \sqrt{2 g h}$

maximum height of center of mass= $\frac{{(V_{c m})}^{2}}{2 g}$

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A uniform horizontal rod of length l falls vertically from height h on two identical blocks placed symmetrically below the rod as shown in figure. The coefficients of restitution are e1 and e2. The maximum height through which the centre of mass of the rod will rise after bouncing off the blocks is