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NEET 2015
Exam
NEET
Year
2015
Subject
Physics
Chapter
Center of Mass, Momentum and Collision

NEET 2015 · Physics · Center of Mass, Momentum and Collision

Two particles of masses m _ 1 , m _ 2 move with initial velocities...

Two particles of masses m1,m2{m}_{1}, {m}_{2} move with initial velocities u1{u}_{1} and u2{u}_{2}. On collision, one of the particles get excited to higher level, after absorbing energy ε.ε. If final velocities of particles be v1{v}_{1} and v2{v}_{2} then we must have:

Options

  1. Am12u1+m22u2−ε=m12v1+m22v2{m}_{1}^{2}{u}_{1} + {m}_{2}^{2}{u}_{2} - ε = {m}_{1}^{2}{v}_{1} + {m}_{2}^{2}{v}_{2}
  2. B12m1u12+12m2u22=12m1v12+12m2v22−ε\dfrac{1}{2}{m}_{1}{u}_{1}^{2} + \dfrac{1}{2}{m}_{2}{u}_{2}^{2} = \dfrac{1}{2}{m}_{1}{v}_{1}^{2} + \dfrac{1}{2}{m}_{2}{v}_{2}^{2} - ε
  3. C12m1u12+12m2u22−ε=12m1v12+12m2v22\dfrac{1}{2}{m}_{1}{u}_{1}^{2} + \dfrac{1}{2}{m}_{2}{u}_{2}^{2} - ε = \dfrac{1}{2}{m}_{1}{v}_{1}^{2} + \dfrac{1}{2}{m}_{2}{v}_{2}^{2}
  4. D12m2u12+12m1u12+ ε =12m2v12+12m1v12\dfrac{\text{1}}{\text{2}} {\text{m}}_{\text{2}} {\text{u}}_{\text{1}}^{\text{2}} \text{+}\dfrac{\text{1}}{\text{2}} {\text{m}}_{\text{1}} {\text{u}}_{\text{1}}^{\text{2}} \text{+ ε =}\dfrac{\text{1}}{\text{2}} {\text{m}}_{\text{2}} {\text{v}}_{\text{1}}^{\text{2}} \text{+}\dfrac{\text{1}}{\text{2}} {\text{m}}_{\text{1}} {\text{v}}_{\text{1}}^{\text{2}}

Correct answer

C. 12m1u12+12m2u22−ε=12m1v12+12m2v22\dfrac{1}{2}{m}_{1}{u}_{1}^{2} + \dfrac{1}{2}{m}_{2}{u}_{2}^{2} - ε = \dfrac{1}{2}{m}_{1}{v}_{1}^{2} + \dfrac{1}{2}{m}_{2}{v}_{2}^{2}

Step-by-step solution

From conservation of energy initial total energy
Ei=12m1u12+12m2u22{E}_{i} = \dfrac{1}{2}{m}_{1}{u}_{1}^{2} + \dfrac{1}{2}{m}_{2}{u}_{2}^{2}
After collision total final energy
Ef=12m1v12+12m2v22+ε{E}_{f} = \dfrac{1}{2}{m}_{1}{v}_{1}^{2} + \dfrac{1}{2}{m}_{2}{v}_{2}^{2} + ε


∴12m1u12+12m2u22=12m1v12+12m2v22+ε∴ \dfrac{1}{2}{m}_{1}{u}_{1}^{2} + \dfrac{1}{2}{m}_{2}{u}_{2}^{2} = \dfrac{1}{2}{m}_{1}{v}_{1}^{2} + \dfrac{1}{2}{m}_{2}{v}_{2}^{2} + ε

12m1u12+12m2u22−ε=12m1v12+12m2v22\dfrac{1}{2}{m}_{1}{u}_{1}^{2} + \dfrac{1}{2}{m}_{2}{u}_{2}^{2} - ε = \dfrac{1}{2}{m}_{1}{v}_{1}^{2} + \dfrac{1}{2}{m}_{2}{v}_{2}^{2}

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