A block is thrown with a velocity of `2m//s^(-1)` (relative to ground) on a belt, which is moving with velcoity `4ms^(-1)` in opposite direction of the initial velcoity of block. If the block stops slipping on the belt after `4sec` of the throwing tehn choose teh correct statements `(s)`

Options

A. Displacement with respect to ground is zero after `2.66` and magnitude of displacement with respect to ground is `12m` after `4sec`.
B. Magnitude of displacement with respect to ground in `4sec` is `4m`.
C. Magnitude of displacement with respect to belt in `4sec` is `12m`.
D. Displacement with respect to ground is zero in `8//3sec`.

Answer

B, C, D

Explanation

Blocks stops slipping when its velocity becomes`-4m//s` or equal to velcoity of belt.
`v = u +at rArr - 4 = 2 +a xx 4`
`a=- 1.5 m//s^(2) =- (3)/(2) m//s^(2)`

For displacement w.r.t. ground, `x = ut +(1)/(2)at^(2)`
`0 = 2t +(1)/(2) (-(3)/(2)) t^(2) rArr t = (8)/(3) sec`.
`rArr x_((4sec)) = 2 xx 4 +(1)/(2) (-(3)/(2)) xx (4)^(2)`
`= 8 - 12 =- 4m`
`rArr x_(B//"belt") = (u_(B//"Belt")) t +(1)/(2) a_((B//"Belt")) t^(2)`
`= 6 xx 4+(1)/(2) xx (-(3)/(2)) xx (4)^(2) = 24 - 12 = 12 m`