Two particles of equal mass 'm' have respective initial velocities $u \hat{i}$ and $u\left(\frac{\hat{i}+\hat{j}}{2}\right)$. They collide completely inelastically. The energy lost in the process is:

(C) $\sqrt{\frac{2}{3}} m u^{2}$

JEE Main · 2020 · 09 Jan 2020 · Shift 1 · Conservation of Momentum

Two particles of equal mass 'm' have respective initial velocities

u \hat{i}

and

u (\frac{i ^ + j ^}{2})

. They collide completely inelastically. The energy lost in the process is:

Answer: (B)

In a perfectly inelastic collision the equal masses move off together, so momentum conservation reads

m u \hat{i} + m (\frac{u}{2} \hat{i} + \frac{u}{2} \hat{j}) = 2 m v

, from which the common velocity has components

v_{x} = \frac{3 u}{4}

and

v_{y} = \frac{u}{4}

. The second particle moves with speed

\frac{u}{2}

, so the initial kinetic energy is

\frac{1}{2} m u^{2} + \frac{1}{2} m (\frac{u}{2})^{2} = \frac{3}{4} m u^{2}

. The final kinetic energy is

\frac{1}{2} (2 m) (v_{x}^{2} + v_{y}^{2}) = m (\frac{9 u ^{2}}{16} + \frac{u ^{2}}{16}) = \frac{5 m u ^{2}}{8}

. The energy lost is the difference,

\frac{3}{4} m u^{2} - \frac{5}{8} m u^{2} = \frac{1}{8} m u^{2}

JEE Main · 2020 · 09 Jan 2020 · Shift 1 · Single correct option