In the arrangement shown in the figure, initially the block of mass $m_{1}$ , is in limiting equilibrium. The coefficient of friction between the block of mass $m_{1}$ and floor is $μ$ . Now the particle of mass $m_{2}$ is given a horizontal velocity $v_{0}$ . The initial distance of the block of mass $m_{2}$ from the pulley is $l$ . The masses of pulley and threads are negligible. Choose the correct option(s):

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Acceleration of $m_{1}$ just after the velocity given is $(\frac{μ}{1 + μ}) \frac{v_{0}^{2}}{l}$

Acceleration of $m_{1}$ just after the velocity given is $(\frac{μ}{1 + μ}) \frac{v_{0}^{2}}{2 l}$

Radius of curvature of path of $m_{2}$ just after the velocity given is $l (1 + μ)$

Radius of curvature of path of $m_{2}$ just after the velocity given is $2 l (1 + μ)$

answer is A, C.

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Detailed Solution

Lets observe the motion of $m_{2}$ from an observer fixed at P (point on string)
Question Image
$T + m_{2} a - m_{2} g = \frac{m_{2} v_{0}^{2}}{l}$ ……….. (1)
For $m_{1}$
$T - μ m_{1} g = m_{1} a$ ……….. (2)
From (1) and $m_{1} a + μ m_{1} g - m_{2} a - m_{2} g = \frac{m_{2} v_{0}^{2}}{l}$
$a = \frac{m_{2} \frac{v_{0}^{2}}{l}}{(m_{1} + m_{2})} = (\frac{μ}{1 + μ}) \frac{v_{0}^{2}}{l}$
ROC of $m_{2}$
$T - m_{2} g = μ m_{1 g} + (\frac{μ m_{1}}{1 + μ}) \frac{v_{0}^{2}}{l} - m_{2} g$