InterviewSolution
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InterviewSolution
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Two particles of masses `m_(1),m_(2)` move with initial velocities `u_(1)` and `u_(2)`. On collision, one of the particles get excited to higher level, after absording enegry. If final velocities of particles be `v_(1)` and `v_(2)` then we must haveA. `m_(1)^(2)u_(1) + m_(2)^(2)u_(2)-epsilon =- m_(1)^(2)v_(1) + m_(2)^(2)v_(2)`B. `1/2 m_(1)u_(1)^(2) + 1/2m_(2)u_(2)^(2)=1/2m_(1)v_(1)^(2) + 1/2m_(2)v_(2)^(2) - epsilon`C. `1/2 m_(1)u_(1)^(2)+1/2m_(2)u_(2)^(2)-epsilon=1/2m_(1)v_(1)^(2)+1/2m_(2)v_(2)^(2)`D. `1/2m_(1)^(2)u_(1)^(2)+1/2m_(2)^(2)u_(2)^(2) + epsilon=1/2m_(1)^(2)v_(1)^(2)+1/2m_(2)^(2)v_(2)^(2)` |
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Answer» Correct Answer - C Total initial energy =`1/2(m_(1)u_(1)^(2) + 1/2(m_(2)u_(2)^(2)` ltbr Sinc,e after collision one particle absorb energy `epsilon`. `therefore` Total final energy = `1/2(m_(1)v_(1)^(2) + 1/2m_(2)(v_(1)^(2) + epsilon` From conservation of energy `1/2m_(1)u_(1)^(2) + 1/2m_(2)u_(2)^(2) = 1/2(m_(1)v_(1)^(2) + 1/2(m_(2)v_(2)^(2) + epsilon` `rArr 1/2(m_(1)u_(1)^(2) + 1/2(m_(2)u_(2)^(2) -epsilon = 1/2m_(1)v_(1)^(2) + 1/2m_(2)v_(2)^(2)` |
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