InterviewSolution
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InterviewSolution
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A body `A` of mass moving with velocity `v` while passing through its mean position collides in perfect inelastically with a body `B` of same mass which is connected to a vertical wall through a spring whose spring constant is `k`. After collision it sticks to `B` and executes `S.H.M.` Find the amplitude of reulting motion:A. `sqrt((m)/(v))v`B. `sqrt((m)/(2k))v`C. `(m)/(k)sqrt(v)`D. `(m)/(2k)sqrt(v)` |
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Answer» Correct Answer - B Assuming the collision lests for a small interval. We can apply conservation of momentum and get the common velocity = `(v)/(2) K.E. =(1)/(2) (2m) ((v)/(2))^(2) = (1)/(4)mv^(2)` This is also the total energy of system as the spring is unstretched at this moment. If the amplitude is `A`, total energy `=(1)/(2) kA^(2) :. (1)/(2) kA^(2) = (1)/(4)mv^(2)` `:.A = sqrt((m)/(2k)).v` |
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