Q 11-05-126JEE MainJEE Main 2022 (26 Jul, Shift 1)Medium
As per the given figure, two blocks each of mass $250\ \text{g}$ are connected to a spring of spring constant $2\ \text{N m}^{-1}$. If both are given velocity $v$ in opposite directions, then the maximum elongation of the spring is
Answer: (B) $\dfrac v2$
By symmetry the centre of mass stays at rest, so at maximum elongation both blocks are momentarily at rest and all the kinetic energy is stored in the spring:
$$2\times\frac12(0.25)v^2 = \frac12(2)x^2\ \Rightarrow\ x^2 = \frac{v^2}{4}\ \Rightarrow\ x = \frac v2$$
Solution by Sreeraj P, M.Sc Physics