A small block of mass m slides on a frictionless horizontal table. It is constrained to move inside a ring of radius l which is fixed to the table. At t = 0, the block has a tangential velocity v₀. The coefficient of friction between the block and the ring is $μ$ . The velocity of the block at time t is

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$\frac{v_{0}}{1 - (\frac{{μv}_{0} t}{l})}$

$\frac{v_{0}}{1 + \frac{2 {μv}_{0} t}{l}}$

$\frac{v_{0}}{(\frac{{μv}_{0} t}{l})}$

$\frac{v_{0}}{1 + \frac{{μv}_{0} t}{l}}$

answer is A.

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Detailed Solution

Suppose velocity at an instant is y. Then $N = \frac{{mv}^{2}}{r}$ …….(i)

and $- μN = m \frac{dv}{dt}$ ……..(ii)

From Eqs. (i) and (ii), $- \frac{{μv}^{2}}{r} = \frac{dv}{dt}$

$\int_{v_{0}}^{v} \frac{dv}{v^{2}} = - \frac{μ}{r} \int_{0}^{t} dt$

On solving we get v at time t.

${[- \frac{1}{v}]}_{v_{0}}^{v} = - \frac{μt}{r}$

$v = \frac{v_{0}}{1 + \frac{{μv}_{0} t}{l}}$ substituting $r = l$

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