Q. A block of mass M is suspended from a light spring of force constant k. another mass m moving upwards with velocity $v$ hits the mass $M$ and gets embedded in it. What will be the amplitude of the combined mass?
A) $\frac{mv}{\sqrt{(M - m)k}}$
B) $\frac{Mv}{(M - m)k}$
C) $\frac{mv}{\sqrt{(M + m)k}}$
D) $\frac{Mv}{\sqrt{(M + m)k}}$
Explanation
(c) If $v$ and $v^{'}$ are the velocities of the block of mass $M$ and $(M + m)$ while passing from the mean position when executing SHM
Using law of conservation of linear momentum, we have
$mv = (M + m)v^{'}\ \ or\ \ \ v^{'} = mv/(M + m)$
Also, maximum PE= maximum KE
$\therefore\ \frac{1}{2}k\ {A^{'}}^{2} = \frac{1}{2}(M + m){v^{'}}^{2}$
or $A^{'} = \left( \frac{M + m}{k} \right)^{1/2} \times \frac{mv}{(M + m)}$
$= \frac{mv}{\sqrt{(M + m)k}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.