Block A of mass m is placed over a wedge of same mass m. Both the block and wedge are placed on a fixed inclined plane. Assuming all surfaces to be smooth, the displacement of the block A in ground frame in 1s is `(gsin^(2)theta)/(x+sin^(2)theta)` then the value of x is:

Answer

1

Explanation

Let acceleration of wedge in ground frame is a down the plane. The acceleration of block A will be `a sin theta` vertically downward
`vec(a)_(A//g)=vec(a)_(A//B)+vec(a)_(B//g)`....(i)
`[a_(A//g)]_(x)=[a_(A//B)]_(x)+[a_(B//g)]_(x)`......(ii)
From F.B.D. of A it is clear that Block A cannot accelerate horizontally i.e., in x-direction because there is no force in x-direction. Block A can accelerate in y-direction only. `[a_(A//g)]_(x)=0` Therefore `[a_(A//g)]_(x)=-[a_(B//g)]_(x)` That means for an observation on wedge block moves only `x gt 0`
For block A: `mg-N=m(a sin theta)` ....(iii)
For block B: `(N+mg)sin theta=ma` ....(iv)
On solving Eqns (iii) and (iv) we get
`a=[(2g sin theta)/(1+sin^(2)theta)]`
The acceleration of block A,
`a_(A)=asin theta=[(2gsin theta)/(1+sin theta)]sin theta=[(2g sin ^(2)theta)/(1+sin^(2)theta)]`
Displacement of block A in 1 s is
`s=0+1/2a_(A)t^(2)`
`=1/2xx[(2gsin ^(2)theta)/(1+sin^(2)theta)]xx(1)^(2)[(g sin^(2)theta)/(1+sin^(2)theta)]`