A bullet of mass M is fixed with a velocity of `50ms^(-1)` at an angle `theta` with the horizontal. At the highest point trajectory it collides with a bob of mass 3M suspended vertically by a massless string of length `10/3m` and g embedded into it. After the collision the string moves through an angle `120^@`, what is the angle of firing `theta`

Options

A. `cos^(-1)2/5`
B. `cos^(-1)3/5`
C. `cos^(-1)4/5`
D. `cos^(-1)1/5`

Answer

C

Explanation

Velocity at highest point= u `costheta`
Then `Mucostheta=(M+3M)v` or `v=(ucostheta)/4` and `4Mgh=1/2x4M((ucostheta)/4)^2`
Using the given values we get, `cos^2theta=16/25` or `costheta = 4/5, theta=cos^(-1)4/5`