Consider the system shown in figure. We know the masses of bodies : `m = 1 kg, M = 3 kg`. The distance `AB = BC = H//2` are covered in equal time intervals by the body of mass m and at B it undergoes an elastic collision with a stationary body of mass `m_(0)` (unknown). Time after this collision when the string in pu,,ey 2 becomes tight again is `t_(0). t_(0)` is second. `t_(0)^(2)` is equal to `1//x`. Value of `(x)/(10)` is. (Given : `H = 1m` )

Answer

2