let bigger block moves by 'x' m till smalll block comes to its bottom,

as F ext. on system = 0

CM remains at rest

10m * x  = m * (2.2 - x)

11x  = 2.2

x = 0.2m = 20 cm................................

is it correct..., rate if useful

as block is moving with constant velocity

resultant force = 0

F cos45 = f  = (1/4) N  ( f = frictional force )

 F sin45 + N = mg

solving,we get,

N = 14.4 Newton , and , F = 5.09116 Newton

work done by 'F' = F cos45 * d = 3.6 * 2 = 7.2 J

work done by 'f'  = -(work done by 'F' ) = -7.2 J

please rate if useful.......................

let the cue hits the solid sphere by a force F at a distance 'h' from centre,

for min value of h  ( friction  = 0)

F = ma (a = acceleration of CM )  ......(i)

torque = Fh = (2mr^2)/5 * @  (   @ = ang. acc. )............(ii)

 as for pure rolling   a = @ r  .......(iii)

from  (i) , (ii), & (iii)

h = 2r/5

rate if useful............

rearranging the inequation,

or,

or,

so by graph of sin 4@, or otherwise, we can see that,

rate if usefulllllllll..........

As, x,y,z are real positive no's.

solving RHS,

multiplying all 3 equn, and taking its square root, we get,

solving LHL,

.......(ii)

similarly,

........(iii)

taking AM greater than equal to GM of (ii) and (iii), we get,

as, LHL & RHL ar both greater than root (xyz)

so we cannot comment on sign between LHL and RHL.................

PLEASE CHECK WEATHER I AM CORRECT OR NOT...................

Let the velocity and angular velocity of ball (solid sphere) at the instant it starts pure rolling be 'u' and  respectively.

let after 't' seconds, it starts pure rolling.

 force= ....(i)

torque.......(ii)

from the three equn of motion.......

solving, we get,

so ball will start pure rolling whin its speed drops to (5/7)v.

Hey, my answer is comming  (c) n/2 - 1. if u want its solution, i can post..........