JEE Main · 2019 · 08 Apr 2019 · Shift 2 · Conservation of Momentum

A body of mass

m_{1}

moving with an unknown velocity of

v_{1} \hat{i}

undergoes a collinear collision with a body of mass

m_{2}

moving with a velocity

v_{2} \hat{i}

. After collision

m_{1}

and

m_{2}

move with velocities of

v_{3} \hat{i}

and

v_{4} \hat{i}

respectively. If

m_{2} = 0.5 m_{1}

and

v_{3} = 0.5 v_{1}

then

v_{1}

is:

Answer: (C)

Solution

This is one-dimensional conservation of momentum:

m_{1} v_{1} + m_{2} v_{2} = m_{1} v_{3} + m_{2} v_{4}

. Substitute the given relations

m_{2} = 0.5 m_{1}

and

v_{3} = 0.5 v_{1}

to get

m_{1} v_{1} + 0.5 m_{1} v_{2} = 0.5 m_{1} v_{1} + 0.5 m_{1} v_{4}

. Divide right through by

m_{1}

and collect the

v_{1}

terms,

v_{1} - 0.5 v_{1} = 0.5 v_{4} - 0.5 v_{2}

, that is

0.5 v_{1} = 0.5 (v_{4} - v_{2})

. So

v_{1} = v_{4} - v_{2}

JEE Main · 2019 · 08 Apr 2019 · Shift 2 · Single correct option