Question

Blocks of masses $m$, $m,2m,4mand8m$ 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 ‘$\text{'}p\text{'}$ is close to:

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

Answer 1

The correct option is D

$94$

Step 1: Given data

Masses of five blocks are $m,m,2m,4m,8m$

Step 2: Calculating final velocity

All the collisions are perfectly inelastic, so after the final collision, all blocks are moving together.

Let the final velocity will be $v^{\text{'}}$.

Total mass after the collision$=16m$

From the conservation of momentum

$Momentumbeforecollision=Momentumaftercollision$

$mv=16m{v}^{\text{'}}$

$v^{\text{'}}=\frac{v}{16}$

Step 3: Finding the value of $p$

Initial energy, $E_i=\frac{1}{2}m{v}^2$

Final energy, $E_f=\frac{1}{2}\times 16m\times {\left(\frac{v}{16}\right)}^2$

$Energyloss=E_i-E_f$

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

$=\frac{1}{2}m{v}^{{}^2}\left(\frac{15}{16}\right)$

$Percentageofenergyloss(p\%)=\frac{Energyloss}{Originalenergy}\times 100$

$p\%=\frac{{\displaystyle \frac{1}{2}}m{v}^2\left({\displaystyle \frac{15}{16}}\right)}{{\displaystyle \frac{1}{2}}m{v}^2}\times 100$

$p\%=94\%$

Therefore, 94% of energy is lost by the time the last block of mass $8m$ starts moving.

Hence, the correct option is $\left(D\right)$.