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)` A. `1//4s`B. `1//2s`C. `2//3s`D. `1//3s`

Two unequal masses are connected on two sides of a light and smooth pu

1.	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)` A. `1//4s`B. `1//2s`C. `2//3s`D. `1//3s`
Answer» Correct Answer - D Acceleration of the system: `a=(m_(2)-m_(1))/(m_(2)+m_(1))g=(10)/(3)m//s^(2)` Velocity of both the block at `t=1s` will be `v_(0)=at=(10)/(3)xx1=(10)/(3)m//s` Now at this moment, 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 `2kg` block or `v_(0)t-(1)/(2)gt^(2)=(1)/(2)gt^(2)` .

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