Friction

Hey how is a1 w.r.t ground? I am working from the frame of reference attached on the block M, so there is a pseudo force acting on m towards right, friction towards left. Whatever accln. I find from the frame attached to M will be w.r.t to M right?

I have really doubt in this question.

The system is started towards right with an initial velocity v

How is Pseudo force acting in right then?

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.

isnt this the hcv sum?

anyways
no sawal of pseudo force here
it says the system

just remember
friction is self-adjustable
:P:P

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)