Q 11-13-030JEE MainHard
A block of mass $M$ on a smooth floor is attached to a spring and oscillates with amplitude $A$. When it passes through the mean position, a lump of putty of mass $m$ is dropped vertically onto it and sticks. The new amplitude is
Answer: (D) $A\sqrt{\dfrac{M}{M + m}}$
At the mean position the speed is $A\omega = A\sqrt{\dfrac{k}{M}}$. The putty adds no horizontal momentum:
$$MA\sqrt{\frac{k}{M}} = (M + m)v' \;\Rightarrow\; v' = \frac{A\sqrt{kM}}{M + m}$$
New amplitude: $A' = \dfrac{v'}{\omega'} = \dfrac{A\sqrt{kM}}{M + m}\sqrt{\dfrac{M + m}{k}} = A\sqrt{\dfrac{M}{M + m}}$.
Solution by Sreeraj P, M.Sc Physics