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A solid rubber ball of mass 2 kg and radius 3 cm collides with the horizontal ground with velocity 4.83 m/s and angular velocity 50 rad/s. The coefficient of friction and restitution are 2 and 0.5 respectively. Find the horizontal distance covered by the ball between the first and second impact of the ground in m.

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answer is 7.

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Detailed Solution

$\int N d t = m e v_{0} - (- m v_{0}) .... (i)$

$\int μ n d t = m v .... (i i)$

From (i) and (ii)
$v = v_{0} (1 + e) μ$
Now,
$t = \frac{2 e v_{0}}{g}$
dis tance, d vt 7

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