Verified Solution • 2025–26 Session moderate

QJEEM20Ph6da2: JEE Main 2020 Physics - Class 11 JEEM20Ph6da2 Solution: A body A, of mass m = 0.1 kg has an initial velocity of 3 ms -1 …

A body A, of mass m = 0.1 kg has an initial velocity of 3 i^\widehat i ms-1 . It collides elastically with another body, B of the same mass which has an initial velocity of 5 j^\widehat j ms-1. After collision, A moves with a velocity v→=4(i^+j^)\overrightarrow v = 4\left( {\widehat i + \widehat j} \right) . The energy of B after collision is written as x10{x \over {10}} . The value of x is ___________.

Step-by-Step Verified Solution

By conservation of linear momentum :

(0.1)(3 i^\widehat i ) + (0.1)(5 j^\widehat j ) = (0.1)(4)( i^\widehat i + j^\widehat j ) + (0.1) v→\overrightarrow v

⇒\Rightarrow v→=−i^+j^\overrightarrow v = - \widehat i + \widehat j

∴\therefore ∣v→∣\left| {\overrightarrow v } \right| = 2\sqrt 2

KEB = 12×0.1×(2)2{1 \over 2} \times 0.1 \times {\left( {\sqrt 2 } \right)^2} = 110{1 \over {10}} J

∴\therefore x = 1

Related Learning Resources & Exam Preparation

JEE Main PYQ Question Bank
Browse year-wise authentic previous years' questions with solutions
Authentic CBT Mock Tests
Simulate official exam environment with live timer and percentile scoring