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)

Explanation

u=100 m/s , v=0 , ` s =6 xx 10^(-2) m , a=? ,v^(2)-u^(2)=2as`
`0^(2)-(100)^(2) = 2 xx a xx 6 xx 100^(-2)`
` a=(-100 xx 100)/(2 xx 6 xx 10^(-2)), a=(-1)/(12) xx 10^(6) m//s`
`F=ma = 5 xx 10^(-3) xx [(-1)/(12) xx 10^(-6)],F=(-5000)/(12)=-417 N`
Retarding force F= 417 N