Question 95 of 183

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

A$\dfrac{1}{3}$B$\dfrac{2}{3}$C$\dfrac{1}{4}$D$\dfrac{1}{8}$

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Answer: (D) $\dfrac{1}{8}$

$m\times 40=\dfrac{2}{3}mv+\dfrac{m}{3}\times 60$
$120=2v+60$
$v=30{\text{m s}}^{-1}$
Now ${K}_{i}=\dfrac{1}{2}\times m\times {40}^{2}=800m$
${K}_{f}=\dfrac{1}{2}\times \dfrac{2m}{3}(30{)}^{2}+\dfrac{1}{2}\times \dfrac{m}{3}\times (60{)}^{2}$
$=\dfrac{1}{6}{m(1800+3600)}=\dfrac{m(5400)}{6}=900m$
$\dfrac{\Delta KE}{{K}_{i}}=\dfrac{900m-800m}{800m}=\dfrac{1}{8}$