A block of mass m is suspended from one end of a light spring as shown. The origin O is considered at distance equal to natural length of the spring from the ceiling and vertical downwards direction as positive y-axis. When the system is in equilibrium a bullet of mass `(m)/(3)` moving in vertical up wards direction with velocity `v_0` strikes the block and embeds into it. As a result, the block (with bullet embedded into it) moves up and start oscillating. Based on the given information, answer the following question:
Q. The time taken by the block bullet system to move from `y=(mg)/(k)` (initial equilibrium position) to `y=0` (natural length of spring) is (A represents the amplitude of motion)

Options

A. `sqrt((4m)/(3k)) [cos^(-1)((mg)/(3kA))cos^(-1)((4mg)/(3kA))]`
B. `sqrt((3m)/(4m)) [cos^(-1)((mg)/(3kA))cos^(-1)((4mg)/(3kA))]`
C. `sqrt((4m)/(6k)) [cos^(-1)((4mg)/(3kA))-sin^(-1)((mg)/(3kA))]`
D. None of the above

Answer

A

Explanation

Initially in equilibrium let the elongation in spring by `y_(0)`, then `rArr y_(0) = (mg)/(k)`

As the bullet strikes the block with velocity `v_(0)` and gets embedded into it, the velocity of the combined mass can be computed bu using the principle of momentum conservation.
`(m)/(3) V_(0) =(4m)/(3)V rArr v = (v_(0))/(4)`
Let new mean position is at distance `y` form origin, then `ky = (4m)/(3)g rArr y = (4mg)/(3k)`.Now, the block executes `S.H.M.` about mean position defined by `y =(4mg)/(3k)`. At `t = 0`, the combined mass is at a displacement of `y -y_(0)` from mean position and is moving with velocity `v`, then by using `v = omega sqrt(A^(2) -x^(2))`, we can fined the amplitude of motion.
`((v_(0))/(4))^(2) = (3k)/(4m) [A^(2) -(y-y_(0))^(2)] = (3k)/(4m) [A^(2) -((mg)/(3k))^(2)]`
`rArr (mv_(0)^(2))/(12k) = A^(2) - ((mg)/(3k))^(2) rArr A = sqrt((mv_(0)^(2))/(12k)+((mg)/(3k))^(2))`
To compute the time taken by the combined mass from `y = (mg)/(k)` to `y = 0`, we can either go for equation method or circular motion projection method:

Required time, `t = (theta)/(omega) =(alpha-beta)/(omega)`
`cos alpha =(y -y_(0))/(A) = (mg)/(3kA), cos beta = (y)/(A) = (4mg)/(3kA)`
so, `t = (cos^(-1)((mg)/(3kA))-cos^(-1)((4mg)/(3kA)))/(omega)`
`= sqrt((4m)/(3k)) [cos^(-1) ((mg)/(3kA)) -cos^(-1) ((4mg)/(3kA))]`