The system shown in the figure consider of a light , inextenslible cord, light, frictionless pulley, and blocks of equal mass. Notice that block `B` is attached to one of the pulleys. The system is initial held at rest so that the ground . The blocks are then released. Find the speed of block `A` at the moment the vertical separation of the blocks is `h`.

Options

A. `sqrt((6 gh)/(15))`
B. `sqrt((8 gh)/(13))`
C. `sqrt((8 gh)/(15))`
D. `sqrt((6 gh)/(13))`

Answer

B

Explanation

When block `B` moves up by `1 cm` block `A` moves down by the `2 cm` and the sepration becomes `3 cm` . We then choose the final point to be when `B` has moved by `(h)/(3)` and has speed `(v_(A))/(2)`. Then `A` has moved down `(2h)/(3)` and has speed `v_(A)` :
`Delta K + Delta U = 0`
`(K_(A) + K_(B) +(U_(g))_(f) - (K_(A) +(K_(B) +U_(g))_(i) = 0`
`(K_(A) +K_(B) + U_(g))_(i) +(K_(A) +(K_(B) + U_(g))_(f)`
`0 + 0 + 0 = (1)/(2) mv_(A)^(2) + (1)/(2)m ((v_(A))/(2))^(2) + (mgh)/(3) - (mg 2h)/(3)`
`(mgh)/(3) = (5)/(8) mv_(A)^(2) implies v_(A) = sqrt((8gh)/(15))`