Two block of A and B of mass 1kg and 2kg are hung from light pulley. Initially the block B is held stationary. At `t=0` block B is given velocity 10m//s in upward direction. String and pulley are light and there is no friction anywhere.

Total mechanical energy lost when the string becomes taut is

Options

A. `100/3`J
B. `100` J
C. `300` J
D. `200/3`J

Answer

A

Explanation

String will be taught when distance travelled by the two blocks become equal i.e.
`l_(A)=l_(B)Rightarrow1/2g t^(2)=10t-1/2g t^(2)Rightarrowt=1sec.`
Before t=1 sec acceleration of both block will be equation t 'g' in downward direction
After =1 sec, there will jerk in string and velocity of A and B will be given by
`mv_(A)+2mv_(B)=m u_(B)`(u_(A)=velocity of A at t=1 sec)
`Rightarrow v_(A)=u_(A)/3 and v_(A)=u_(A)/3(:.v_(A)=v_(B)`after string becomes thought)
`v_(A)=v_(B)=10//3 m//s`
:. velocity of 'B' when 'A' has ascended by 5m is given by
`v=sqrt((10/3)^(2)+2xx10/3xx10)=10/3sqrt(7)m//s` (:. After t=1sec, a=`10/3 ms^(2))`