Question

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 :

• A. 77
• B. 37
• C. 87
• D. 94

Answer 1

Answer: (D)

All collisions are perfectly inelastic, so after the final collision, all blocks are moving together
So let the final velocity be $v^{\prime }$, so on applying
momentum conservation:
$mv=16m{v}^{\prime }$
$\Rightarrow v^{\prime }=v/16$
Now initial energy $E_i=\frac{1}{2}m{v}^2$
Final energy : $E_f=\frac{1}{2}\times 16m\times {\left(\frac{v}{16}\right)}^2$
$\Rightarrow E_f=\frac{1}{2}m\frac{v^2}{16}$
Energy loss : $E_i-E_f$
$\Rightarrow \frac{1}{2}m{v}^2-\frac{1}{2}m\frac{v^2}{16}$
$\Rightarrow \frac{1}{2}m{v}^2\left[1-\frac{1}{16}\right]$
$\Rightarrow \frac{1}{2}m{v}^2\left[\frac{15}{16}\right]$
$\%p=\frac{\text{ Energy loss }}{\text{ Original energy }}\times 100$
$=\frac{\frac{1}{2}m{v}^2\left[\frac{15}{16}\right]}{\frac{1}{2}m{v}^2}\times 100=93.75\%$
$\Rightarrow$ Value of $P$ is close to $94.$