Verified Solution • 2025–26 Session moderate

QJEEM20Ph393f: JEE Main 2020 Physics - Class 11 JEEM20Ph393f Solution: Particle A of mass m 1 moving with velocity collides with…

Particle A of mass m1 moving with velocity (3i^+j^)ms−1\left( {\sqrt3\widehat i + \widehat j} \right)m{s^{ - 1}} collides with another particle B of mass m2 which is at rest initially. Let V1→\overrightarrow {{V_1}} and V2→\overrightarrow {{V_2}} be the velocities of particles A and B after collision respectively. If m1 = 2m2 and after
collision V1→=\overrightarrow {{V_1}} = (i^+3j^)\left( {\widehat i + \sqrt 3 \widehat j} \right) , the angle between V1→\overrightarrow {{V_1}} and V2→\overrightarrow {{V_2}} is :

Multiple Choice Options

  • A 105o
  • B 15o
  • C -45o
  • D 60o

Step-by-Step Verified Solution

Given m1 = 2m2
So let, m2 = m and m1 = 2m

From momentum conservation

p→i{\overrightarrow p _i} = p→f{\overrightarrow p _f}

⇒\Rightarrow (2m) (3i^+j^)\left( {\sqrt 3 \widehat i + \widehat j} \right) + 0 = 2m (i^+3j^)\left( {\widehat i + \sqrt 3 \widehat j} \right) + m V→2{\overrightarrow V _2}

⇒\Rightarrow V→2{\overrightarrow V _2} = 2 (3i^+j^)\left( {\sqrt 3 \widehat i + \widehat j} \right) - 2 (i^+3j^)\left( {\widehat i + \sqrt 3 \widehat j} \right)

⇒\Rightarrow V→2{\overrightarrow V _2} = (23−2)i^−j^(23−2)\left( {2\sqrt 3 - 2} \right)\widehat i - \widehat j\left( {2\sqrt 3 - 2} \right)

= 2(3−1)(i^−j^)2\left( {\sqrt 3 - 1} \right)\left( {\widehat i - \widehat j} \right)

Also given after collision V1→=(i^+3j^)ms−1\overrightarrow {{V_1}} = \left( {\widehat i + \sqrt3\widehat j} \right)m{s^{ - 1}}

For angle between V→1{\overrightarrow V _1} & V→2{\overrightarrow V _2} ,

cos θ\theta = V→1.V→2∣V→1∣∣V→2∣{{{{\overrightarrow V }_1}.{{\overrightarrow V }_2}} \over {\left| {{{\overrightarrow V }_1}} \right|\left| {{{\overrightarrow V }_2}} \right|}}

= 2(3−1)×1−2(3−1)×32×22(3−1){{2\left( {\sqrt 3 - 1} \right) \times 1 - 2\left( {\sqrt 3 - 1} \right) \times \sqrt 3 } \over {2 \times 2\sqrt 2 \left( {\sqrt 3 - 1} \right)}}

= 2(3−1)(1−3)2×22(3−1){{2\left( {\sqrt 3 - 1} \right)\left( {1 - \sqrt 3 } \right)} \over {2 \times 2\sqrt 2 \left( {\sqrt 3 - 1} \right)}}

= (1−3)22{{\left( {1 - \sqrt 3 } \right)} \over {2\sqrt 2 }}

⇒\Rightarrow θ\theta = 105o

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