Blocks of masses m, 2m, 4m, and 8m are arranged in a line on a frictionless floor. Another block of a mass m, moving with speed v along the same line (see figure) collides with mass m in a perfectly inelastic manner. All the subsequent collisions are also perfectly inelastic. By the time the last block of a mass 8m starts moving the total energy loss is p% of the original energy. Value of 'p; is close to :
Option: 1 77
Option: 2 94
Option: 3 37
Option: 4 87

Answers (1)

All collisions are perfectly inelastic, so after the final collision, all blocks are moving together
So let the final velocity be \mathrm{v}^{\prime}, so on applying momentum conservation:
\mathrm{mv}=16 \mathrm{m}\mathrm{v}^{\prime} \\ \Rightarrow \mathrm{v}^{\prime}=\mathrm{v} / 16
Now

initial energy=\mathrm{E}_{\mathrm{i}}=\frac{1}{2} \mathrm{mv}^{2}
Final energy =\mathrm{E}_{\mathrm{f}}=\frac{1}{2} \times 16 \mathrm{~m} \times\left(\frac{\mathrm{v}}{16}\right)^{2}=\frac{1}{2} \times \mathrm{~m} \times\left(\frac{\mathrm{v}^2}{16}\right)

\begin{array}{l} \text { Energy loss : } \mathrm{E}_{\mathrm{i}}-\mathrm{E}_{\mathrm{f}} = \frac{1}{2} \mathrm{mv}^{2}-\frac{1}{2} \mathrm{~m} \frac{\mathrm{v}^{2}}{16} = \frac{1}{2} \mathrm{mv}^{2}\left[1-\frac{1}{16}\right] =\frac{1}{2} \mathrm{mv}^{2}\left[\frac{15}{16}\right] \end{array}

\begin{aligned} &\% \mathrm{p}=\frac{\text { Energy loss }}{\text { Original energy }} \times 100=\frac{\frac{1}{2} m v^{2}\left[\frac{15}{16}\right]}{\frac{1}{2} m v^{2}} \times 100=93.75 \%\\ &\Rightarrow \text { Value of } P \text { is close to } 94 \end{aligned}

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