JEE Main · 2020 · 04 Sep 2020 · Shift 1 · Conservation of Momentum

In a perfectly inelastic collision the moving mass sticks to the one it strikes, and momentum is conserved throughout. The incident block of mass $m$ at speed $v$ carries momentum $mv$, and this momentum is handed on unchanged through the whole chain of sticking collisions. By the time the heaviest block has been set moving, the incident block has merged successively with the line of stationary blocks and the whole assembly moves together with combined mass $16m$, so its common speed is $v_f=\frac{mv}{16m}=\frac{v}{16}$. The kinetic energy left in the system is $\frac{1}{2}(16m){\left(\frac{v}{16}\right)}^2=\frac{1}{16}\left(\frac{1}{2}m{v}^2\right)$, only $\frac{1}{16}$ of the original $\frac{1}{2}m{v}^2$. The fraction of energy lost is therefore $1-\frac{1}{16}=\frac{15}{16}$, which is about $94\%$ of the initial energy.