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A small sphere of mass $10 g$ is attached to a point of a smooth vertical wall by a light strength of length $1 m$ . The sphere is pulled out in a vertical plane perpendicular to the wall so that the string makes an angle of $60^{0}$ with the wall then relaxed. it is found that after first rebound the stings makes an maximum angle of $30^{0}$ with the wall. Calculate the restitution and the loss of K.E. due to impact . If all the energy converted into, heat, find the heat produced by impact

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$0.06 J$

$0.07 J$

$. 08 J$

$0.09 J$

answer is A.

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Detailed Solution

Let $v_{1}$ is the velocity of the sphere just before the collision with the wall, then

$m g h = \frac{1}{2} m {v_{1}}^{2}$

$v_{1} = \sqrt{g}$

If $v_{2}$ is the velocity of the sphere after collision with the wall, then

$\frac{1}{2} m {v^{2}}_{2} = m g (1 - \cos 30)$

$v_{2} = \sqrt{g (2 - \sqrt{3})}$

Now according to the Newton's experimental law

$e = 0.518$

loss of K.E

$= \frac{1}{2} m {v^{2}}_{1} - \frac{1}{2} m {v^{2}}_{2} = 0.06 J$

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