Q 11-05-139JEE MainJEE Main 2021 (31 Aug, Shift 1)Easy
A block moving horizontally on a smooth surface with a speed of $40\ \text{m s}^{-1}$ splits into two equal parts. If one of the parts moves at $60\ \text{m s}^{-1}$ in the same direction, then the fractional change in the kinetic energy will be $x : 4$ where $x$ = ______.
Numerical value type. Enter your answer.
Answer: 1
Momentum: $m(40) = \dfrac m2(60) + \dfrac m2v \Rightarrow v = 20\ \text{m s}^{-1}$.
$K_i = \dfrac12m(40)^2 = 800m$; $K_f = \dfrac12\cdot\dfrac m2(3600 + 400) = 1000m$.
$$\frac{\Delta K}{K_i} = \frac{200m}{800m} = \frac14 \Rightarrow x = 1$$
Solution by Sreeraj P, M.Sc Physics