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 (see figure) collides with mass m in 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 :

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answer is C.

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Detailed Solution

$\begin{array}{l} C o n s e r v a t i o n o f m o m e n t u m, \\ m v = 16 m v' \\ \Rightarrow v' = \frac{v}{16} \\ P e r c e n t a g e l o s s i n k i n e t i c e n e r g y \\ = \frac{K_{i} - K_{f}}{K_{i}} \times 100 \\ = \frac{\frac{1}{2} m v^{2} - \frac{1}{2} (16 m) {(\frac{v}{16})}^{2}}{\frac{1}{2} m v^{2}} \times 100 = \frac{15}{16} \times 100 = 93.75 % \end{array}$