Annoying Wedge

• $a:$ acceleration of the wedge

• $b :$ acceleration by which $P_{1} P_{2}$ is decreasing.

• $T:$ Tension in the string

(All accelerations shown in the image are in ground frame)

By applying string constraint we can say that $b=\dfrac{a}{2}$ .

Now use the approach of system of masses for getting the equations.

• System 1 - Along the string on the wedge

$\Rightarrow 2g \sin60 -T=2(\dfrac{3}{2}a-a\cos60)$

• System 2 - Along the horizontal direction on combined system of block and wedge

$\Rightarrow \dfrac{3}{2}T=8a+2(a-\dfrac{3}{2}a\cos60)$

Solving these equations we get $T=\dfrac{17}{3}a$ and $a=\dfrac{30\sqrt{3}}{23}$

So the answer is $30+23+3=56$

I Did this way

Consider The wedge to be moving with an acceleration a1 towards the right.

Call the tension in the string throughout to be T and normal reaction with wedge and block to be N.

So we can write T+Nsin(60) = Ma1

Next to analyse motion of block lets go into the frame of wedge (Non Inertial frame)

Call the acceleration of block along wedge to be a2 in frame of wedge

Writing equations for the block along wedge and perpendicular to wedge (Do not forget to apply Pseudo force as we are working in non inertial frame of reference)

N+ma1sin(60) = mgcos(60)

T-ma1cos60 -mgsin(30) = ma2

Notice that acceleration of block along the wedge wrt ground is a2 + a1/2

Applying Virtual work method for constraints we obtain

a2+3a1/2 = 0

Solving above system of equations we get required answer

A Really nice problem on Laws of motion :)