A long conveyer belt moves with a constant velocity of `8 ms^(-1)`.Two blockA and B each of mass `2kg` are placed gently on the belt with B on A, initail velocity of block is zero Coefficient of friction between A and belt is `0.1` There is no friction between A and B length of A is `4m`

If the coefficient of friction between the block B and belt is `0.4` Find the separation between the two block when B come to rest w.r.t. belt

Options

A. `8 m`
B. `6 m`
C. `2 m`
D. None of these

Answer

c

Explanation

Just when B falls off A take this instant to be `t = 0`

At ` t = 0`
Velocity of A : `u_(A) = 4ms^(-1)`
Velocity of B : `u_(B) = 0`
acceleration of A :`a_(1) = m_(2)g = 0.1 xx 10 = 1 ms^(-2)`
acceleration of B :`a_(2) = m_(2)g = 0.4 xx 10 = 4 ms^(-2)`
Let as see when A come to rest (w.r.tbelt):
For this `v_(A) = u_(A) + a_(1)t`
`rArr 8 = 4 + 1 xx 1`
`rArr t = 4s`
Let as see when B come to rest (w.r.tbelt):
For this `v_(B) = u_(B) + a_(2)t`
`rArr 8 = 0 + 1 xx 4t`
`rArr t = 2s`
So be comes to rest earlier and till that A coninues to move with acceleration `1 ms^(-2)`
So when have to find separation the block at `t = 2s` At `t = 2s`
`S_(A) = 4 xx 2 + (1)/(2) xx 1xx (2)^(2) = 10m`
`S_(B) = 0 xx 2 + (1)/(2) xx 4 xx (2)^(2) = 8m`
Separation `= S_(A) - S_(B) = 10 - 8= 10 = 2m`