Two blocks `m_(1)` and `m_(2)` are allowed to move without friction Block `m_(1)` is on block `m_(2)` and `m_(2)` slides on smooth fixed incline as shown The angle of inclination of inclined plane is `theta`
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The acceleration of `m_(2)` with respect to ground is .

Options

A. `((m_(1)+m_(2))gsin^(2)theta)/(m_(2)+m_(1)sin^(2)theta)`
B. `((m_(1)+m_(2))gsin^(2)theta)/(m_(2)+m_(1)sin^(2)theta)`
C. `((m_(1)+m_(2))gsin^(2)theta)/(m_(2)-m_(1)sin^(2)theta)`
D. `((m_(1)+m_(2))gsin^(2)theta)/(m_(2)-m_(1)sin^(2)theta)`

Answer

B

Explanation

`m_(1)g-N_(1)=m_(1)a_(1)("Vertically down")"___"(1)`
`(N_(1)+m_(2)g)sintheta=m_(2)a_(2)("along incline")"_________"(2)`
From (1) and (2)
`[(m_(1)g-m_(1)a_(1))+m_(2)g]sintheta=m_(2)a_(2)`
`(m_(1)+m_(2)) g sin theta -m_(1)a_(1) sin theta = m_(2) a_(2)"___"(3)`
`sin theta = (x)/(y)`
`x =y sin theta"____"(4)`
From (3) and (4)
`(m_(1)+m_(2))gsintheta=a_(2)[m_(2)+m_(1)sin^(2)theta]`
`a_(2)=(m_(1)+m_(2))/(m_(2)+m_(1)sin^(2)theta)"___"(5)`
`:.a_(1)=((m_(1)+m_(2))gsintheta)/(m_(2)+m_(1)sin^(2)theta)`

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