A block of mass m is placed on the inclined sufrace of a wedge as shown in fig. Calculate the acceleration of the wedge and the block when the block is released. Assume all surfaces are frictionless.

Explanation

Method 1: Analysis form ground frame: Let the acceleration of wedge be A and that of block be a (w.r.t wedge).
Then, acceleration of m w.r.t ground is
`vec(a)_(m)=vec(a)_(m,M)+vec(a)_(M)=(a cos theta hat(i)- a sin theta hat(j))-Ahat(i)`
`=(a cos theta-A)hat(i)-a sin theta hat(j)`...(i)

For M: `N sin theta = MA` ...(ii)
For m: `mg-N cos theta=ma sin theta` ...(iii)
`N sin theta = m(a cos theta-A)`...(iv)
Solve (ii), (iii) and (iv) to get
`A=(mg sin theta cos theta)/(M+m sin^(2)theta)` and `((M+m)g sin theta)/(M+m sin^(2)theta)`
Now acceleration of block can be found by putting the values of a in (i)
Method 2: If we consider the motion of block parallel and perpendicular to sloping surface w.r.t. ground, acceleration of m can also be written as :
`vec(a)_(m)=vec(a)_(m,M)_vec(a)_(M)=-a hat(i)+(A cos theta hat(i)+A sin theta hat(j))`

`vec(a)_(m)=(A cos theta-a)hat(i) +A sin theta hat(j)` ...(v)
For M: `N sin theta=MA` ...(vi)
For m: `-mg sin theta=m(A cos theta-a)` ...(vii)
`mg cos theta-N=mA sin theta` ...(viii)
From above equation we can get the valuve of
`A=(mg sin theta cos theta)/(M+m sin^(2)theta)` and `((M+m)g sin theta)/(M+m sin^(2)theta)`
Method 3: Analysis from non-inertial frame: With the help of pseudo force. If we consider the motion of block parallel and perpendicular to ground.

let us make teh FBD of m w.r.t. M.
For M: `N sin theta=MA`...(ix)
For m: `N sin theta+mA=ma cos theta` ...(x)
`mg-N cos theta=ma sin theta` ...(xi)
From above equations, we can get the value of
`A=(mg sin theta cos theta)/(M+m sin^(2)theta)` and `((M+m)g sin theta)/(M+m sin^(2)theta)`
Method 4: Analyses from non-inertial frame: With the help of pseudo force.

When the forces are resolved in directions along the plane and perpendicular to the plene:
For M: `N sin theta=MA`
for m: `mg sin theta+mA cos theta=ma`
`mg cos theta=N+mA sin theta`
From above equations, we can get the value of
`A=(mg sin theta cos theta)/(M+m sin^(2)theta)` and `((M+m)g sin theta)/(M+m sin^(2)theta)`
Method 5: Taking system together:
Considering wedge and block together as a system as shown in fig.

When the system is released, the wedge will start moving towards left with acceleration A and the block will starts sliding on the sloping surface of the wedge with acceleration a with respects to wedge as shown in fig.

Now analyzing the system in the horizontal directon, the system is not having any external force in horizontal direction. So we can apply
`Sigma vec(F)_(x)=Sigmam_(i)vec(a)_(i)=mvec(a)_(m)+Mvec(a)_(M)`
`Sigma vec(F)_(x)=0 implies -MA + m(a cos theta-A)=0`
`implies a=((m+M)A)/(m cos theta)` ..(i)
Now considering the motion of block parallel to the sloping side. Writing the equation of motion with respect to ground.
`Sigma vec(F)_(x')=Sigma mvec(a)_(x')`
`implies mg sin theta=ma'_(x)`
`=m(a-A cos theta)` ...(ii)
Solve (i) and (ii) to get
`A=(mg sin theta cos theta)/(M+m sin^(2)theta)` and `((M+m)g sin theta)/(M+m sin^(2)theta)`