A pendulum consists of a wooden bob of mass in 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 bob 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_(1))/m)sqrt(5gl)`
D. `((m_(1))/m)sqrt(gl)`

Answer

B

Explanation

Applying the law of conservation of momentum,
`m_(1)v_(1)=m_(1)(v_(1))/3=mv` or `v=(2m_(1)v_(1))/(3m)`
To describe a vertical circle `v` should be `sqrt(5gl).` so
`(2m_(1)v_(1))/(3m)=sqrt(5gl)` or `v_(1)(m/m_(1))3/2sqrt(5gl)`