A block moving horizontally on a smooth surface with a speed of 40   m   s - 1 splits into two…

1. $\frac{1}{3}$
2. $\frac{2}{3}$
3. $\frac{1}{4}$
4. $\frac{1}{8}$

Solution

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

Asked in: JEE Main 2021 (31 Aug Shift 2)

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