Verified Solution • 2025–26 Session moderate

QJEEM18Ph8520: JEE Main 2018 Physics - Class 11 JEEM18Ph8520 Solution: In a collinear collision, a particle with an initial speed v 0…

In a collinear collision, a particle with an initial speed v0 strikes a stationary particle of the same mass. If the final total kinetic energy is 50% greater than the original kinetic energy, the magnitude of the relative velocity between the two particles, after collision, is :

Multiple Choice Options

  • A v02{{{v_0}} \over {\sqrt 2 }}
  • B v04{{v_0}} \over 4
  • C 2v0\sqrt 2 {v_0}
  • D v02{{v_0}} \over 2

Step-by-Step Verified Solution

From conservation of linear momentum,

mv0 = mv1 + mv2

or v0 = v1 + v2 ........(1)

According to the question,

Kf = 32{3 \over 2} Ki

⇒\Rightarrow 12mv12+12mv22=32×12mv02{1 \over 2}mv_1^2 + {1 \over 2}mv_2^2 = {3 \over 2} \times {1 \over 2}mv_0^2

⇒\Rightarrow v12+v22=32v02v_1^2 + v_2^2 = {3 \over 2}v_0^2

Using eq (1) (v1+v2)2=v02{\left( {{v_1} + {v_2}} \right)^2} = v_0^2

⇒\Rightarrow v12+v22+2v1v2v_1^2 + v_2^2 + 2{v_1}{v_2} = v02v_0^2

⇒\Rightarrow 2v1v22{v_1}{v_2} = v02−32v02v_0^2 - {3 \over 2}v_0^2 = −12v02- {1 \over 2}v_0^2

Now, (v1−v2)2{\left( {{v_1} - {v_2}} \right)^2} = (v1+v2)2−4v1v2{\left( {{v_1} + {v_2}} \right)^2} - 4{v_1}{v_2}

= v02−(−v02)v_0^2 - \left( { - v_0^2} \right) = 2v022v_0^2

∴\therefore v1−v2{{v_1} - {v_2}} = 2v0\sqrt 2 {v_0}

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