A small ball of mass m start at a point A with speed $V_{0}$ and moves along a frictionless track AB as shown. The track BC has coefficient of friction $" μ "$ the ball comes to stop at $" c "$ after travelling a distance $" L "$ which is

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$\frac{2 h}{μ} + \frac{v_{0}^{2}}{2 μ g}$

$\frac{h}{2 μ} + \frac{v_{0}^{2}}{2 μ g}$

$\frac{h}{μ} + \frac{v_{0}^{2}}{2 μ g}$

$\frac{h}{2 μ} + \frac{v_{0}^{2}}{μ g}$

answer is B.

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Detailed Solution

Initial speed at A is $V_{0}$
Speed at B is v=?
$v^{2} - u^{2} = 2 g h$
$v^{2} - v_{0}^{2} = 2 g h$
$v^{2} = v_{0}^{2} + 2 g h$
Let ball travels distance “L” before coming to rest
$L = \frac{v^{2}}{2 a} a = μ g L = \frac{v^{2}}{2 μ g} L = \frac{v_{0}^{2} + 2 g h}{2 μ g} = \frac{2 g h}{2 μ g} + \frac{v_{0}^{2}}{2 μ s} = \frac{h}{μ} + \frac{v_{0}^{2}}{2 μ g}$