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 `veca_(A//g)=veca_(A//B)+veca_(B//g).(1)`
`[a_(A//g)]_(x)=[a_(A//B)]_(x)+[a_(B//g)]_(x)...(2)`
From `FBD` of A it is clear that Block A cannot accelerate horizontally i.e in x-direaction becasue there is no force in x-direaction Block A can accelerate in y-direaction only. `[a_(A//g)]_(x)=0` There
for `[a_(A//g)]=-[a_(B//g)]_(x)` That means for an observation on wedge block moves only `x gt 0`.
For block `A`, mg -N =n (a sin theta) ....(3)`
For block `B`, (N +mg) sin theta = ma...(4)`
On solving eqns (3) and (4) we get `a=[(2gsintheta)/(1+sin^(2)theta)]`
The acceleration of block A, `a_(A) a sin theta`
`a=[(2gsintheta)/(1+sintheta)]sintheta=[(2gsin^(2)theta)/(1+sin^(2)theta)]`
Displacement of block `A` in `1 s` is
`S =0+(1)/(2)a_(A)t^(2)=(1)/(2)xx[(2gsintheta)/(1+sin^(2)theta]]xx(1)^(2)[(gsin^(2)theta)/(1+sin^(2)theta)]` .