A particle of mass $m$ is dropped from a height $h$ above the ground. At the same time another particle of same mass is thrown vertically upwards from the ground with a speed of $\sqrt{2 \mathrm{gh}}$. If they collide head-on completely inelastically, the time taken for the combined mass to reach the ground, in units of $\sqrt{\frac{h}{g}}$ is:

JEE Main · 2020 · 08 Jan 2020 · Shift 2 · Conservation of Momentum

A particle of mass

m

is dropped from a height

h

above the ground. At the same time another particle of same mass is thrown vertically upwards from the ground with a speed of

2 gh

.
If they collide head-on completely inelastically, the time taken for the combined mass to reach the ground, in units of

\frac{h}{g}

is:

Answer: (D)

Dropping one particle and projecting the other upward at

2 g h

, the gap

h

closes at the relative speed

2 g h

, so they meet after

t_{1} = \frac{h}{2 g h} = \frac{h}{2 g}

. By then the meeting point is at height

h - \frac{1}{2} g t_{1}^{2} = \frac{3 h}{4}

, and the two velocities there are equal and opposite,

- g h /2

for the falling particle and

+ g h /2

for the rising one, so a completely inelastic collision leaves the combined mass momentarily at rest. It then falls freely from

\frac{3 h}{4}

, taking

t = \frac{2 ( 3 h /4 )}{g} = \frac{3 h}{2 g}

, which in units of

h / g

\frac{3}{2}

JEE Main · 2020 · 08 Jan 2020 · Shift 2 · Single correct option