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Two particles `A` and `B` separated by a distance `2R` 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 = (32v^(2))/(25piR)` in the same direction of initial velocityA. The time lapse for the two bodies to collide is `(6piR)/(5V)`B. The angle coverd by A is `(9pi)/(4)`C. Angular velocity of A is `(11V)/(5R)`D. Radial acceleration of A is `(289upsilon^(2))/(5R)` |
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Answer» Correct Answer - B As when they collide `1/2((32v^(2))/(25piR))t^(2)=piR,t(5piR)/(4V)` angle covered by `A` `theta=pi+(vt)/R,theta=(9pi)/4` |
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