JEE Main · 2019 · 09 Apr 2019 · Shift 2 · Conservation of Momentum

A particle of mass

m

is moving with speed

2 v

collides with a mass

2 m

moving with speed v in the same direction. After collision, the first mass is stopped completely while the second one splits into two particles each of mass m, which move at angle

4 5^{0}

with respect to the original direction. The speed of each of the moving particle will be:

(A) $v / (22)$
(B) $22 v$ ✓ Correct
(C) $2 v$
(D) $v / 2$

Answer: (B)

Solution

Momentum along the original direction is conserved. Before the collision the total is

m (2 v) + 2 m (v) = 4 m v

. Afterwards the first mass is at rest, and the

2 m

has broken into two pieces of mass

m

flying off symmetrically at

4 5^{\circ}

on either side of the original line, each with speed

v^{'}

. Their sideways components cancel, and along the original direction they give

2 m v^{'} cos 4 5^{\circ} = 2 m v^{'}

. Setting

2 m v^{'} = 4 m v

gives

v^{'} = \frac{4 v}{2} = 22 v

JEE Main · 2019 · 09 Apr 2019 · Shift 2 · Single correct option

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JEE Main · 2019 · 09 Apr 2019 · Shift 2 · Conservation of Momentum

Solution

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