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A small ball moving with a velocity 10 m/s, horizontally (as shown in figure) strikes a rough horizontal surface having $μ = 0 .5$ . If the coefficient of restitution is e = 0.4. Horizontal component of velocity of ball after 1^st impact will be $(g = 10 m / s^{2})$

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Detailed Solution

Vertical component of velocity after impact, $e = \frac{v_{y}}{u_{y}} \Rightarrow v_{y} = 0 .4 \times 10 = 4 m / s$

In vertical direction, impulse-momentum relation is

$\int Ndt = {mv}_{y} - {mu}_{y}$

$\Rightarrow \int Ndt = m \times 4 - (- 10 m) = 14 m$

In horizontal direction, impulse-momentum relation is

$- \int μNdt = {mv}_{x} - {mu}_{x}$

$\Rightarrow - 0 .5 \times 14 m = {mv}_{x} - m \times 10$

$\Rightarrow v_{x} = 3 m / s$

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