Two blocks A and B of mass m and 2m are intially at rest. Length of block B is L and the block A is placed at the right end corner of block B and the friction coefficient between them is `mu=1//2`. At t=0 a constant force F `=(5 mg)/2` begins to act on block B towards right. just when the block A leaves B, wind begins to below along y-direction which exerts a constant force `(mg)/2` on A. assume that size block A is small compared to B and neglect any rotational effects and toppling of block B. (given h=1/2m, L=1 m and `g=10 m//s^(2)`)

The magnitude of relative acceleration of A with respect to B ( in `m//s^(2)`) just after the block A leaves B is (assume wind does not effects motion of B)

Options

A. `sqrt(10)g`
B. `(sqrt(29)g)/4`
C. `(gsqrt(5))/4`
D. `(3sqrt(5))/4 g`

Explanation

For block `B`:
`2ma_(B)=F-(mg)/(2)`
For block `A`,
`ma_(A)=mg`
`a_(A)=g//2`
`a_(AB)=-g//2`
`L=(1)/(2)(g)/(2).t_(1)^(2)`
`t_(1)=sqrt((2)/(5)s)`
time of flight `t_(2)=sqrt((2h)/(g))=(1)/(sqrt(10))s`
Velocity when `A` leaves `B`:
`V_(A)=g//2t_(1)=g//2xxsqrt((4L)/(g)=sqrt(10)m//s`
`S_(x)=V_(A)t_(2)=1m`
`S_(y)=(1)/(2)(g)/(2)(2h)/(g)=(1)/(4)m`
`(S_(x))/(S_(y))=4`
`veca_(A)=(g)/(2)hatj-ghatk`
`veca_(B)=(5g)/(4)hati,|veca_(AB)|=(sqrt((1)/(4)+1+(25)/(16)))g`