Question

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

• A. $\frac{1}{3}$
• B. $\frac{2}{3}$
• C. $\frac{1}{8}$
• D. $\frac{1}{4}$

Answer 1

Answer: (C)

$3M{V}_0=2M{V}_2+M{V}_1$
$3V_0=2V_2+V_1$
$120=2V_2+60$
$\Rightarrow V_2=30m/s$
$\frac{\Delta K.E.}{K.E.}=\frac{\frac{1}{2}M{V}_1^2+\frac{1}{2}2M{V}_2^2-\frac{1}{2}3M{V}_0^2}{\frac{1}{2}3M{V}_0^2}$
$=\frac{V_1^2+2V_2^2-3V_0^2}{3V_0^2}$
$=\frac{3600+1800-4800}{4800}$
$=\frac{1}{8}$