A block moving horizontally


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

  1. $\frac{1}{3}$

  2. $\frac{2}{3}$

  3. $\frac{1}{8}$

  4. $\frac{1}{4}$

Correct Option: , 3


$3 \mathrm{MV}_{0}=2 \mathrm{MV}_{2}+\mathrm{MV}_{1}$

$3 \mathrm{~V}_{0}=2 \mathrm{~V}_{2}+\mathrm{V}_{1}$

$120=2 \mathrm{~V}_{2}+60 \Rightarrow \mathrm{V}_{2}=30 \mathrm{~m} / \mathrm{s}$

$\frac{\Delta \mathrm{K} . \mathrm{E} .}{\mathrm{K} . \mathrm{E} .}=\frac{\frac{1}{2} \mathrm{MV}_{1}^{2}+\frac{1}{2} 2 \mathrm{MV}_{2}^{2}-\frac{1}{2} 3 \mathrm{MV}_{0}^{2}}{\frac{1}{2} 3 \mathrm{MV}_{0}^{2}}$

$=\frac{\mathrm{V}_{1}^{2}+2 \mathrm{~V}_{2}^{2}-3 \mathrm{~V}_{0}^{2}}{3 \mathrm{~V}_{0}^{2}}$



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