A mass '$m$' moves with a velocity $v$ and collides inelastically with another identical mass. After collision the $1^{\text {st }}$ mass moves with velocity $\mathrm{v} / \sqrt{ } 3$ in a direction perpendicular to the initial direction of motion. Find the speed of the $2^{\text {nd }}$ mass after collision

JEE Main · 2005 · 08 May 2005 · Offline · Conservation of Momentum

A mass '

m

' moves with a velocity

v

and collides inelastically with another identical mass. After collision the

1^{st}

mass moves with velocity

v / 3

in a direction perpendicular to the initial direction of motion. Find the speed of the

2^{nd}

mass after collision

Answer: (C)

Let the struck mass move off with a component

v_{x}

along the original direction and

v_{y}

across it. Along the original direction momentum conservation gives

m v = m v_{x}

, since the first mass keeps nothing in this direction after leaving perpendicularly, so

v_{x} = v

. Across it,

0 = \frac{m v}{3} - m v_{y}

, so

v_{y} = \frac{v}{3}

. The speed of the second mass is therefore

v^{2} + \frac{v ^{2}}{3} = \frac{4 v ^{2}}{3} = \frac{2 v}{3}

JEE Main · 2005 · 08 May 2005 · Offline · Single correct option