A block moving horizontally on a smooth surface with a speed of $40 \mathrm{m} {\mathrm{s}}^{-1}$ splits into two parts with masses in the ratio of $1 : 2.$ If the smaller part moves at $60 \mathrm{m} {\mathrm{s}}^{-1}$ in the same direction, then the fractional change in kinetic energy is :

• A

  $\frac{1}{3}$
• B

  $\frac{2}{3}$
• C

  $\frac{1}{4}$
• Correct answer
  D

  $\frac{1}{8}$

Solution

$m\times 40=\frac{2}{3}mv+\frac{m}{3}\times 60$
$120=2v+60$
$v=30 {\text{m s}}^{-1}$
Now $K_i=\frac{1}{2}\times m\times {40}^2=800m$
$K_f=\frac{1}{2}\times \frac{2 m}{3}(30{)}^2+\frac{1}{2}\times \frac{m}{3}\times (60{)}^2$
$=\frac{1}{6}\{ m(1800+3600\left)\right\}=\frac{m\left(5400\right)}{6}=900m$
$\frac{\mathrm{\Delta }KE}{K_i}=\frac{900m-800m}{800m}=\frac{1}{8}$

Hence, the correct answer is option: 'D'.