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 downward direction as positive \( y \)-axis. When the system is in equilibrium, a bullet of mass \( m / 3 \) moving in vertical upward 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 starts oscillating.
Based on the given information, answer the following questions:
The time taken by the block-bullet system to move from \( y=m g / k \) (initial equilibrium position) to \( y=0 \) (natural length of spring) is ( \( A \) represents the amplitude of motion)
(1) \( \sqrt{\frac{4 m}{3 k}}\left[\cos ^{-1}\left(\frac{m g}{3 k A}\right)-\cos ^{-1}\left(\frac{4 m g}{3 k A}\right)\right] \)
(2) \( \sqrt{\frac{3 k}{4 m}}\left[\cos ^{-1}\left(\frac{m g}{3 k A}\right)-\cos ^{-1}\left(\frac{4 m g}{3 k A}\right)\right] \)
(3) \( \sqrt{\frac{4 m}{6 k}}\left[\sin ^{-1}\left(\frac{4 m g}{3 k A}\right)-\sin ^{-1}\left(\frac{m g}{3 k A}\right)\right] \)
(4) None of the above
š²PW App Link - https://bit.ly/YTAI_PWAP
šPW Website - https://www.pw.live