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Two particles $A$ and $B$ seperated 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 $a = \frac{72 V^{2}}{25 π R}$ . Then $A$ collides $B$ after certain time and angle covered by $A$ .

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The time lapse for the two bodies collide is $\frac{5 π R}{6 V}$

The angle covered by $A$ is $\frac{11 π}{6}$

Angular velocity of $A$ is $\frac{11 V}{5 R}$

The radial acceleration of $A$ is $\frac{289 V^{2}}{5 R}$

answer is A, B.

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Detailed Solution

$α = \frac{a_{t}}{R} = \frac{72 V^{2}}{25 π R^{2}}; {(ω_{0})}_{A / B} = 0$

$θ = {(ω_{0})}_{A / B} t + \frac{1}{2} α t^{2}; π = 0 + \frac{1}{2} \cdot \frac{72 V^{2}}{25 π R^{2}} t^{2}$

$t^{2} = \frac{50 π^{2} R^{2}}{72 V^{2}}; t = \frac{5}{6} \frac{π R}{V}; θ_{A} = ω_{A} t + \frac{1}{2} α t^{2}$

$θ_{A} = \frac{V}{R} \cdot \frac{5}{6} \frac{π R}{V} + \frac{1}{2} \times \frac{72 V^{2}}{25 π R^{2}} \cdot \frac{50 π^{2} R^{2}}{72 V^{2}}$

$= \frac{5}{6} π + π = \frac{11 π}{6}$

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