InterviewSolution
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InterviewSolution
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Two unequal masses are connected on two sides of a light and smooth pulley as shown in figure. The system is released from rest. The larger mass is stopped `1.0` second after the system is set into motion and then released immediately. The time elapsed before the string is tight again is: Take `g=10m//s^(2)` |
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Answer» Net pulling force `=2g - 1g =10N` Mass being pulled `=2+1 =3kg` Acceleration of the system is `a =(10)/(30)m//s^(2)` Velocity of both the blocks at `t =1` s will be `v_(0) = at = ((10)/(3)) (1) = (10)/(3) m//s`. Now at this momnet velocity of `2kg` block becomes zero, while that of `1kg` block is `(10)/(3)m//s` upwards Hence, string becomes tight again when displacement of `1kg` block =displacement of `1kg` block =displacement of `2kg` block `v_(0) t- (1)/(2) g t^(2)= (1)/(2)g t^(2) rArrg t^(2)=v_(0)t` `t =(v_(0))/(g) = ((10//3))/(10) = (1)/(3) s` . |
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