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Two particles $A$ and $B$ separated by a distance $2 R$ are moving counter clockwise along the same circular path of radius $R$ each with uniform speed $v$ . At time $t = 0, A$ is given a tangential acceleration of magnitude $a = \frac{32 v^{2}}{25 π R}$ in the same direction of initial velocity

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Angular velocity of $A$ is $\frac{9 V}{4 R}$

The angle covered by $A$ is $\frac{9 π}{4}$

Radial acceleration of $A$ is $\frac{348 v^{2}}{5 R}$

The time lapse for the two bodies to collide is $\frac{5 π R}{3 V}$

answer is C.

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Detailed Solution

As when they collide $\frac{1}{2} (\frac{32 v^{2}}{25 π R}) t^{2} = π R; t = \frac{5 π R}{4 V}$ angle covered $A$

$θ = π + \frac{v t}{R}; θ = \frac{9 π}{4}$

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