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JEE Main 20208 JanuaryMorning Shift
Exam
JEE Main
Year
2020
Date
8 January
Shift
Morning Shift
Subject
Physics
Chapter
Center of Mass, Momentum and Collision

JEE Main 2020 · 8 January · Morning Shift · Physics · Center of Mass, Momentum and Collision

A body A of mass m = 0.1kg has an initial velocity of 3 ^ i m s ^ -...

A body AA of mass m=0.1kgm = 0.1kg has an initial velocity of 3\overset{^}{i}m{s}^{- 1}. It collides elastically with another body BB of the same mass which has an initial velocity of 5\overset{^}{j}m{s}^{- 1}. After the collision, AA moves with a velocity \overset{\to}{v} = 4\left(\overset{^}{i} + \overset{^}{j}\right)m{s}^{- 1}. The energy of BB after the collision is written as x10J\dfrac{x}{10}J. The value of xx is

Correct answer

0

Step-by-step solution

Using the principle of conservation of linear momentum

0 . 1 \times 3\overset{^}{i} + 0 . 1 \times 5\overset{^}{j} = 0 . 1 \times 4\left(\overset{^}{i} + \overset{^}{j}\right) + 0 . 1 \times \overset{\to}{v}

\overset{\to}{v} = \left(- \overset{^}{i} + \overset{^}{j}\right)m{s}^{- 1}

v=∣v→∣=2ms−1v = \left|\overset{\to}{v}\right| = \sqrt{2}m{s}^{- 1}

KB=12×0.1×v2=110J{K}_{B} = \dfrac{1}{2} \times 0 . 1 \times {v}^{2} = \dfrac{1}{10}J

x=1x = 1

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