Q 11-05-140JEE MainJEE Main 2021 (31 Aug, Shift 2)Easy
A block moving horizontally on a smooth surface with a speed of $40\ \text{m s}^{-1}$ splits into two parts with masses in the ratio of $1 : 2$. If the smaller part moves at $60\ \text{m s}^{-1}$ in the same direction, then the fractional change in kinetic energy is:
Answer: (D) $\dfrac18$
Masses $m$ and $2m$. Momentum: $3m(40) = m(60) + 2mv \Rightarrow v = 30\ \text{m s}^{-1}$.
$K_i = \dfrac12(3m)(1600) = 2400m$; $K_f = \dfrac12m(3600) + \dfrac12(2m)(900) = 2700m$.
$$\frac{\Delta K}{K_i} = \frac{300m}{2400m} = \frac18$$
Solution by Sreeraj P, M.Sc Physics