A spring block pendulum is shown in figure . The system is hanging in equilibrium. A bullet of mass `m//2` moving at a speed u bites the block from downwards direction and gets embedded in it. Find the amplitude of oscillation of the block now. Also find the time taken by the block to reach its upper extreme position after hit by bullet.

Explanation

If block is in equilibrium , then spring must be at some steretch if it is h, have `mg = kh`.
If a bullet of mass `m//2` gets embedded in the block, due to this inelastic impact its new mass becoms `3m//2` and new mean positionof the block will be , say at a depth `h_(1)` from old mean position, then we must have
`(3)/(2) mg = k(h + h_(1))`
`(3m)/(2) g = k(h + h_(1))`
`h_(1) = (mg)/(2k) (as mg = kh)`

just after impact due to inelastic collition if the velocity of block becomes v, we have, according to momentum conservation.
`(m)/(2) mu = (3 m)/(2) nu or nu = (u)/(3)`
Now the block executes SHM and at `t = 0` block is at a distance `h = (mg)/(2k)` above its mean position and having a velocity `u//3`. If amplitude of oscillation is A, we have
`(u)/(3) = omega sqrt(A^(2) - ((mg)/(2 k))^(2))`
or `(u^(2))/(9) = (2k)/(3m) [A^(2) - ((mg)/(2k))^(2)]`
[as for this spring block system omega `= sqrt((k)/((3m//2))) ]`
`A = sqrt((mu^(2))/(6k) + ((mg)/(2k))^(2))`
Now time taken by partical to reach the topmost point can be obtained by circular motion representation as shown in figure. This figure shows the position of block P and its corresponding circular motion partical `p_(0)` at the`t = 0`, Block P will reach its upper extereme position when partical `P_(0)` will traveres the angle `theta` and reach the topmost point as `P_(0)` moves at constant angular velocity `omega`, it will take a time given as
`t = (theta)/(omega) = (cos^(-1) (h_(1)//A))/sqrt((k)/(3m//2)) = sqrt((3m)/(2k)) cos^(-1) ((3 mg)/(2 kA))`