Mass mA rests on a smooth horizontal surface, mB hangs vertically

(a) mA = 11 kg

mB = 7 kg

The formula for the Tension in the cord, T = mAmBg / (mA+mB) = 77*9.8 / 18 = 41.92 N

In case of mA, we have T = mA*a

acceleration of mA, a = T / mA = 41.92 / 11 = 3.81 m/s^2

In case of mB, we have mB*a = mB*g - T

                                   a = (mB*g - T) / mB

                                    a = (7*9.8 - 41.92) / 7 = 3.81 m/s^2

(b) Initial velocity of mA, u = 0

The distance travelled by mA, s = 1.3 m

We have, s = u t + (1/2) a t^2

               s = at^2 / 2

               t = (2s / a)^(1/2) = (2*1.3 / 3.81) (1/2) = 0.82 sec

(c) mB = 1 kg

The acceleration of the system, a = g/100

The acceleration of the system is given by the formula, a = mB*g / (mA+mB)

mA = [mB*g / a] - mB = [ 1*9.8*100/9.8 ] - 1 = 99 kg