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

\(mv =16 m v ^{\prime} \Rightarrow v ^{\prime}= v / 16\)

Now initial energy \(E _{1}=\frac{1}{2} mv ^{2}\)

Final energy \(: E _{ f }=\frac{1}{2} \times 16 m \times\left(\frac{ v }{16}\right)^{2}\)

\(\Rightarrow E _{ f }=\frac{1}{2} m \frac{ v ^{2}}{16}\)

Energy loss : \(E _{ i }- E _{ f }\)

\(\Rightarrow \frac{1}{2} mv ^{2}-\frac{1}{2} m \frac{ v ^{2}}{16}\)

\(\Rightarrow \frac{1}{2} mv ^{2}\left[1-\frac{1}{16}\right] \Rightarrow \frac{1}{2} mv ^{2}\left[\frac{15}{16}\right]\)

\(\% p=\frac{\text { Energy loss }}{\text { Original energy }} \times 100\)

\(=\frac{\frac{1}{2} mv ^{2}\left[\frac{15}{16}\right]}{\frac{1}{2} mv ^{2}} \times 100=93.75 \%\)

\(\Rightarrow\) Value of \(P\) is close to \(94\)