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 (as shown) collides with mass $m$ in a 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. The value of $p$ is close to-

• A. $77$
• B. $94$
• C. $37$
• D. $87$

Answer 1

The correct option is B $94$

According to the question, all collisions are perfectly ineleastic, so after the final collision, all blocks are moving together.


Let the final velocity of the system be $v$, using momentum conservation,

$mv=16m{v}^{{}^{\prime }}$
$\Rightarrow v^{{}^{\prime }}=\frac{v}{16}$

Now initial and final energy of the system is,

$E_i=\frac{1}{2}m{v}^2$
$E_f=\frac{1}{2}m{\left(\frac{v}{16}\right)}^2=\frac{1}{2}\frac{m{v}^2}{16}$

Energy loss: $E_i-E_f=\frac{1}{2}m{v}^2-\frac{1}{2}m\frac{v^2}{16}$
$\Rightarrow \frac{1}{2}m{v}^2[1-\frac{1}{16}]\Rightarrow \frac{1}{2}m{v}^2\left[\frac{15}{16}\right]$

The total energy loss is $p\%$ of the original energy.

$\therefore p\%=\frac{\text{Energy loss}}{\text{Original energy}}\times 100$
$=\frac{\frac{1}{2}m{v}^2\left[\frac{15}{16}\right]}{\frac{1}{2}m{v}^2}\times 100=93.75\%$

Hence, value of $p$ is close to $94$.

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, $\left(B\right)$ is the correct answer.