IIT JEE , AIEEE Forum - Physics , Mechanics

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As the block goes down to the wedge the only external force here is the wt of the block or mg. As there is no external force in the X-axis thus the wedge will proceed towards left (from statics) so that the Centre of mass of the system in X-axis remains the same.

Now let

acc. of m w.r.t M be = a = req answer
acc. of M w.r.t ground = a_o .

Now the horizontal component of a is
a_x = acos@ .............(1)

The vertical component of a is
a_y = asin@ ...............(2)

The wedge will provide a normal rxn on the block N and equal force in opposite direction in also exerted on the wedge by the block. (see the FBD in the diagram)

Now frm FBD of the block (m)

The eq of motion along Y-axis

mg - Ncos@ = ma_y ............ (3)

The absolute acc of m along X-axis is (a_x - a_o)

Hence the eq of motion along X-axis

Nsin@ = m(a_x - a_o) ..................(4)

Frmo FBD of the wedge

Nsin@ = Ma_{o ........................(5)}From 4 and 5 we have

a_o = ma_x/(M+m) ........(6)

Again frm 4

Nsin@ = mMa_x/(M+m) .........(7)

Frm 3 we have

Ncos@ = mg - ma_y ...........(8)

by (8)/(7)

cot@ = m(M+m)(g-a_y)/mMa_x

or cot@ = m(M+m)(g/a_x - a_y/a_x)/mM
or cot@ = (g/acos@ - tan@)(M+m)/M

Solve this to get

a = gsin@(M+m)/(M+msin²@) [proved]

As u can see in the question we had to prove that
a = gsin@ + (mgsin@cos²@)/(M+msin²@)

or a = gsin@[M+mcos²@ + msin²@]/(M+msin²@)

hence a = gsin@(M+m)/(M+msin²@)

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