Two block `A` and `B` of masses `1kg` and`2kg` respectively are connected by a string, passing over a light frictionless pulley `B` as shown. Another string connect the center of pulley. Both the blocks are resting on a horizontal floor and the pulley is help such that string remains just taut. At moment `t = 0`, a force `F = 20 t` starts acting on the pully along vertically upwards direction as shown in figure.Calculate

(a) velocity of `A` when `B` loses contact with the floor.
(b) height raised by the pulley upto that instant.
(Take =`g =10 m//s^(2))`

Options

A. `5ms^(-1)`
B. `5//4ms^(-1)`
C. `4ms^(-1)`
D. `7//3ms^(-1)`

Answer

A

Explanation

`T_(3)=20 t, T_(1)=T_(2)=10 t`
For A to lose contact: `10t = 1g implies 1s`
For B to lose contact: `10t = 2g implies 2s`
For C to lose contact: `20t = 3g implies 1.5s`
`a_(A)=(T_(1)-1g)/(1)` velocity of A when B loses contact.
`V_(1)=int_(1)^(2) a_(A)dt = int_(1)^(2) (10t - g)dt = 5 ms^(-1)`
At `t=2s, a_(B)=0, a_(A)=(10 xx 2 xx -10)/(1) = 10ms^(-2)`
`a_(A//B)=a_(A)-a_(B)=10-0 = 10ms^(-2)`.