A pendulum consists of a wooden bob of mass `m` and of length `L`. A bullet of mass `m_(1)` is fired towards the pendulum with a speed `v_(1)`. The bullet emerges out of the bob with a speed `v_(1)//3` and the just completes motion along a vertical circle. Then `v_(1)` is:

Options

A. `((m)/(m_(1))) sqrt(5gl)`
B. `(3)/(2)((m)/(m_(1))) sqrt(5gl)`
C. `(2)/(3)((m)/(m_(1))) sqrt(5gl)`
D. `((m_(1))/(m)) sqrt(gl)`

Answer

B

Explanation

Applying the law of conservationof momentum
`m_(1) v_(1) = m_(1)(v_(1))/(3) + mv or v = (3 m_(1) v_(1))/(3m)`
To describe a vertical `v` should be `sqrt(5gl)`
So `(2 m_(1) v_(1))/(3m) = sqrt(5gl) or v_(1) ((m)/(m_(1))) (3)/(2) sqrt(5gl)`