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JEE Main 20188 April
Exam
JEE Main
Year
2018
Date
8 April
Subject
Physics
Chapter
Center of Mass, Momentum and Collision

JEE Main 2018 · 8 April · Physics · Center of Mass, Momentum and Collision

In a collinear collision, a particle with an initial speed v _ 0...

In a collinear collision, a particle with an initial speed v0{v}_{0} strikes a stationary particle of the same mass. If the final total kinetic energy is 5050 % greater than the original kinetic energy, the magnitude of the relative velocity between the two particles, after the collision, isIn a collinear collision, a particle with an initial speed v _ 0 strikes a stationary particle of the same mass. If theIn a collinear collision, a particle with an initial speed v _ 0 strikes a stationary particle of the same mass. If the

Options

  1. Av02\dfrac{{v}_{0}}{\sqrt{2}}
  2. Bv04\dfrac{{v}_{0}}{4}
  3. C2v0\sqrt{2}{v}_{0}
  4. Dv02\dfrac{{v}_{0}}{2}

Correct answer

C. 2v0\sqrt{2}{v}_{0}

Step-by-step solution

In a collinear collision, a particle with an initial speed v _ 0 strikes a stationary particle of the same mass. If the

KE2=32mv02K{E}_{2} = \dfrac{3}{2}m{v}_{0}^{2}. Conservation of momentum,~v1+v2=v0{v}_{1} + {v}_{2} = {v}_{0}. Conservation of energy,~12mv12+12mv22=12×32mv02\dfrac{1}{2}m{v}_{1}^{2} + \dfrac{1}{2}m{v}_{2}^{2} = \dfrac{1}{2} \times \dfrac{3}{2}m{v}_{0}^{2}
⇒v12+v22=32v02⇒v12+v22+2v1v2=v02⇒32v02+2v1v2=v02⇒ {v}_{1}^{2} + {v}_{2}^{2} = \dfrac{3}{2}{v}_{0}^{2}⇒ {v}_{1}^{2} + {v}_{2}^{2} + 2{v}_{1}{v}_{2} = {v}_{0}^{2}⇒ \dfrac{3}{2}{v}_{0}^{2} + 2{v}_{1}{v}_{2} = {v}_{0}^{2}
⇒2v1v2=−12v02⇒(v1−v2)2=32v02+12v02=2v02⇒ 2{v}_{1}{v}_{2} = - \dfrac{1}{2}{v}_{0}^{2}{⇒ ( {v}_{1} - {v}_{2} )}^{2} = \dfrac{3}{2}{v}_{0}^{2} + \dfrac{1}{2}{v}_{0}^{2}= 2{v}_{0}^{2}
⇒v1−v2=2v0⇒ {v}_{1} - {v}_{2} = \sqrt{2}{v}_{0}.

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