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JEE Main 20198 AprilEvening Shift

Exam: JEE Main
Year: 2019
Date: 8 April
Shift: Evening Shift
Subject: Physics
Chapter: Center of Mass, Momentum and Collision

JEE Main 2019 · 8 April · Evening Shift · Physics · Center of Mass, Momentum and Collision

A body of mass m _ 1 moving with an unknown velocity of v _ 1 ^ i ,...

A body of mass

{m}_{1}

moving with an unknown velocity of

{v}_{1}\overset{^}{i},

undergoes a collinear collision with a body of mass

{m}_{2}

moving with a velocity

{v}_{2}\overset{^}{i} .

After the collision,

{m}_{1}

and

{m}_{2}

move with velocities of

{v}_{3}\overset{^}{i}

and

{v}_{4}\overset{^}{i},

respectively. If

{m}_{2} = 0.5{m}_{1}

and

{v}_{3} = 0.5{v}_{1},

then

{v}_{1}

is:

Options

A ${v}_{4} - \dfrac{{v}_{2}}{4}$
B ${v}_{4} - \dfrac{{v}_{2}}{2}$
C ${v}_{4} + {v}_{2}$
D ${v}_{4} - {v}_{2}$

Correct answer

D. ${v}_{4} - {v}_{2}$

Step-by-step solution

From the conservation of linear momentum,

{m}_{1}{v}_{1} + {m}_{2}{v}_{2} = {m}_{1}{v}_{3} + {m}_{2}{v}_{4}

{m}_{1}{v}_{1} + \left(0.5{m}_{1}\right){v}_{2} = {m}_{1}\left(0.5{v}_{1}\right) + \left(0.5{m}_{1}\right){v}_{4}

0.5{v}_{1} + 0 . 5{v}_{2} = 0.5{v}_{4}

{v}_{4} = {v}_{1} + {v}_{2}

{v}_{1} = {v}_{4} - {v}_{2}

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