A particle is projected up a $45 °$ inclined plane with an initial velocity ‘u’. If the coefficient of kinetic friction between the particle and the plane is $μ$ , it returns to the bottom of the plane with a velocity of

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$u \sqrt{\frac{(1 - μ)}{(1 + μ)}}$

$u \sqrt{\frac{2 (1 - μ)}{(1 + μ)}}$

$u \sqrt{\frac{(1 + μ)}{(1 - μ)}}$

$u \sqrt{\frac{(1 - μ)}{2 (1 + μ)}}$

answer is C.

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Detailed Solution

For up the plane journey

$0^{2} - u^{2} = - 2 \times (g \sin 45^{0} + μ g \cos 45^{0}) \times s$

Get s. For the return journey

$v^{2} - 0^{2} = 2 \times (g \sin 45^{0} - μ g \cos 45^{0}) \times s$

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