Q 11-05-173JEE MainJEE Main 2020 (4 Sep, Shift 1)Medium
Blocks of masses $m$, $2m$, $4m$ and $8m$ are arranged in a line on a frictionless floor. Another block of mass $m$, moving with speed $v$ along the same line, collides with the block of mass $m$ in a perfectly inelastic manner. All the subsequent collisions are also perfectly inelastic. By the time the last block of mass $8m$ starts moving the total energy loss is $p\%$ of the original energy. Value of $p$ is close to:
Answer: (B) $94$
Finally all blocks move together: total mass $16m$, and momentum $mv$ is conserved, so the final speed is $\dfrac{v}{16}$.
$$\frac{K_f}{K_i} = \frac{\tfrac12(16m)(v/16)^{2}}{\tfrac12mv^{2}} = \frac1{16}$$
Energy lost $= \dfrac{15}{16} = 93.75\% \approx 94\%$.
Solution by Sreeraj P, M.Sc Physics