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Center of Mass and Collision question

2015 · Q141
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Center of Mass and Collision question

2015 · Q141

NEETPhysicsCenter of Mass and CollisionMCQ+4 / −1
Two particles of masses m1, m2 move with initial velocities u1 and u2. On collision, one of the particles get excited to higher level, after absorbing energy ε\varepsilonε. If final velocities of particles be v1 and v2 then we must have :
  1. A
    12{1 \over 2}21​m1u12_1^212​ + 12{1 \over 2}21​ m2u22_2^222​ −-− ε\varepsilonε = 12{1 \over 2}21​ m1v12_1^212​ + 12{1 \over 2}21​m2v22_2^222​
  2. B
    12{1 \over 2}21​m12_1^212​u12_1^212​ + 12{1 \over 2}21​m22_2^222​u22_2^222​ + ε\varepsilonε = 12{1 \over 2}21​m12_1^212​v12_1^212​ + 12{1 \over 2}21​m22_2^222​v22_2^222​
  3. C
    m12_1^212​u1 + m22_2^222​u2 −-− ε\varepsilonε = m12_1^212​v1 + m22_2^222​v2
  4. D
    12{1 \over 2}21​m1u12_1^212​ + 12{1 \over 2}21​m2u22_2^222​ = 12{1 \over 2}21​m1v12_1^212​ + 12{1 \over 2}21​m2v22_2^222​ −-− ε\varepsilonε
View written solutionFree

Correct answer: A

By law of conservation of energy,

K.Ef = K.Ei – excitation energy (e)

∴\therefore∴ 12m1v12+12m2v22=12m1u12+12m2u22−ε{1 \over 2}m_1v_1^2 + {1 \over 2}m_2v_2^2 = {1 \over 2}{m_1}u_1^2 + {1 \over 2}{m_2}u_2^2 - \varepsilon 21​m1​v12​+21​m2​v22​=21​m1​u12​+21​m2​u22​−ε

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