IIT JEE , AIEEE Forum - Physics , Mechanics

waterdemon

Here's the solution:

Let the acceleration of Wedge be "a₁".
Considering the motion of the block in frame of the wedge.

The Forces acting on the block will be:
1)N - Normal Force
2)mg - Downward
3)ma₁- forward (pseudo force)

The Block is sliding down the inclined plane.
we have:

ma₁Cos@ + mgSin@ = ma
a = a₁Cos@ + gSin@ ...........(1)

also,
N + ma₁Sin@ = mgCos@........(2)

Now we will consider the motion of the Wedge frm the
ground fram.No Pseudo force is neede as the frame is
Inertial.
1)Mg =Downward.
2)N =Normal to the incline by the block
3)N' =Normal upward by the horizontal ground.

We have,
NSin@ = Ma₁
N = Ma₁/Sin@...................(3)

Putting the value from equation(3) in Equation(2)

Ma₁/Sin@ + ma₁Sin@ = mgCos@
a₁ = mgSin@Cos@
       M+mSin²@
So acceleration of Wedge = mgSin@Cos@
                                 M+mSin²@

In equation (1)
a = a₁Cos@ + gSin@
a = (mgSin@Cos@*Cos@/M+mSin²@) + gSin@

On solving the above expression we get:
Acceleration of Block:
a = (M+m)gSin@
      M+mSin²@

Hope you find it useful.
Cheers !!!!!!!!!!!!!!!!!!!!!!!!!!!!!

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