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) $A\sqrt{\dfrac{M}{M + m}}$
Concept. At the mean position the putty adds no horizontal momentum; the new speed there is the new maximum speed $A'\omega'$.
Step 1. Momentum.
$$MA\sqrt{\frac{k}{M}} = (M + m)v' \;\Rightarrow\; v' = \frac{A\sqrt{kM}}{M + m}$$
Step 2. New amplitude.
$$A' = v'\sqrt{\frac{M + m}{k}} = A\sqrt{\frac{M}{M + m}}$$
Answer. $A\sqrt{\dfrac{M}{M + m}}$ (Option D).
Solution by Sreeraj P, M.Sc Physics