A pendulum consists of a wooden bob of mass `M` and length `l`. A bullet of mass `m` is fired towards the pendulum with a speed `v`. The bullet emerges immediately out of the bob from the other side with a speed of `v//2` and the bob starts rising. Assume no loss of mass of bob takes place due to penetration.
If the bob stops where the string becomes horizontal then `v` is

Options

A. `(2M)/m sqrt(3gl)`
B. `(2M)/msqrt(5gl)`
C. `(2M)/msqrt(gl)`
D. `(2M)/msqrt(2gl)`

Answer

D

Explanation

Let velocity of bob after collision be `v_(1)`. Then
`Mv_(1)=(mv)/2impliesv_(1)(mv)/(2M)`
Height rasised by bob `=l `
`l=v_(1)^(2)/(2g)impliesl=(m^(2)v^(2))/(8M^(2)g)impliesv=(2M)/msqrt(2gl)`