A ball of mass `m = 1 kg` is hung vertically by a thread of length `1=1.50 m`. Upper end of the thread is attached to the ceiling of a trolley of mass `M = 4kg`. Initially, the trolley is stationary and it is free to move along horizontal rails with out friction. A shell of mass `m=1 kg`, moving horizontally with velocity `v_(0) = 6m//s` collides with thread starts to deflect towards right

The maximum deflection of the thread with the vertical is

Options

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

Answer

A

Explanation

When shell strikes the ball and gets stuck with it, combined body of mass `2m` starts to move to the right. Let velocity of the combined body (just after collision) be `v_(1)`. acoording to law of conservation of momentum,
`(m+m) v_(1) = m v_(o)`, `v_(1) = (v_(0))/(2) = 3m//s`
As soon as the combined body starts to move rightward, thread becomes inclined to the vertical. Horizontal component of its tension retards the combined body while trolley accelerates rightward due to the same component of tension. Inclination of thread with the vertical continues to increase till velocities of both (combined body and trolley) become identical or combined body comes to rest relative to the trolley.
Let velocity at that instant of maximum inclnation of thread be `v`.
Acoording to law of conservation of momentum,
`(2m+M) v = 2mv , v = 1m//s`
During collision of ball and shell, a part of energy is lost. but after that there is no lose of energy. Hence, after collision, kinetic energy lost is used up in increasing gravitational potential energy of the combined body.
If maximum inclination of thread with the vertical is `theta`, then according to law of conservation of energy.
`(1)/(2)(2m)v_(1)^(2)-(1)/(2)(2m+M)v^(2) = 2mgl(1- cos theta)`
`cos theta = 0.8` (or) `theta = 37^(@)`