InterviewSolution
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InterviewSolution
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After falling from rest through a height `h`, a body of mass `m` begins to raise a body of mass `M (M gt m)` connected to it through a pulley. Determnethe time it will take for the body of mass `M` to return to its original positionA. `(2m)/(M_m)sqrt((2h)/g)`B. `(2m)/(M-m)sqrt((2h)/g)`C. `(2m)/(M-m)sqrt(h/g)`D. `m/(M-m)sqrt((2h)/g)` |
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Answer» Correct Answer - B `u=sqrt(2gh)` is the velocity of `m` just before collision, and `v` is the velocilty of system `M+m` just after collision, then `(M+m)v="mu"` `implies v=(msqrt(2gh))/(M+m)` `a=((M-m)/(M+m))g` `t=(2v)/a=((2m)/(M-m))sqrt((2h)/g)` |
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