In the arrangement shown in fig., at a particular instant, the roller is coming down with a speed of `12ms^(-1)` and C is moving up with `4ms^(-1)`. At the same instant, it is also known that w.r.t. pulley P, block A is moving down with speed `3ms^(-1)` . Determine the motion of block B (velocity ) w.r.t. ground.

Options

A. `4ms^(-1)` in downward direction
B. `3ms^(-1)` in upward direction
C. `7ms^(-1)` in downward direction
D. `7ms^(-1)` in upward direction

Answer

D

Explanation

Using constraint theory
`l_(1)+l_(2)+l_(3)` = constant
`implies v_(1)+2v_(2)+v_(3)=0`
Take downward as positive and upward as -ve.
So `12+2 (-4)+v_(3)=0`
`v_(3)=`velocity of pulley `P=-4 ms^(-1)`
`=4ms^(-1)` in upward direction
`vec(v)_(AP)= - vec(v)_(BP) implies v_(AP)=-(v_(B)-v_(P))`
`v_(B)=v_(P)-v_(AP)=-4-(3)=-7ms^(-1)`
i.e., block B will moves up with speed `7ms^(-1)`.