Friction

swordfish u hav written

"ma1 = μmg/2"
this is eqn for m in grnd frame or in M's frame ?

Errrm...may be in ground frame......
What will be its equation in the frame of M then?
One doubt...how do you find whether the given equation is from ground frame or frame of M.

@ Vivek...
Think of it this way.......The system is initially started with a velocity v ( assume no friction present....for time being)...........and then the whole system is kept on the surface where friction is present.....so there is a retardation acting on the block M.

No sawal of pseudo force?
Then what causes the smaller block to move forward and finally fall? Some kind of Devilish Force? haha

there is nothing like "pseudo force" in reality

btw swordfish it is better if u accelerations of each m in ground frame and then find relative accn of m wrt M, and then use that kinematics eqn ...

Look for the block M, its acceleration w.r.t ground will be along -i (it is decelerating)
\mu (m+M)g(-\hat{i})+\frac{\mu}{2}mg(\hat{i})=M\vec{a}

\vec{a}=\mu g\left(1+\frac{m}{2M}\right)(-\hat{i})

Now when we observe the block m from ground, it accelerates forward with

\vec{a_1}=\left(\frac{\mu}{2}\right)g(\hat{i})

Now relative acceleration of m w.r.t ground gives wrong answer.
Please help qwerty.

so m will accelerate along - ve x , ( but its accn will be less than accn of M bcz of given value of μ, hence it is able to cover that distance L )

Ma_M =μmg + μMg - μmg/2

so a_M = μg( 1 + m2M) (-i)

a_m = μ2g ( -i )

a_m/M = μg( 1 + m2M) - μ2g ( i ) = μg2(1+mM) (i)

so

s_m/M = u_m/M+12a_m/Mt²

L = 0 + μg4(1+mM)t²

t² = 4LMμg(m+M)