(i) Using Newton's second law for A and B:
For A: \(T - 1.2g = 1.2a\)
For B: \(2g - T = 2a\)
Solving these equations simultaneously:
\(T - 1.2 \times 9.8 = 1.2a\)
\(19.6 - T = 2a\)
Add the equations: \(19.6 - 1.2 \times 9.8 = 3.2a\)
\(a = 2.5 \text{ m/s}^2\)
Substitute \(a\) back to find \(T\):
\(T = 1.2 \times 9.8 + 1.2 \times 2.5 = 15 \text{ N}\)
(ii) (a) Gain in potential energy:
\(\Delta PE = mgh = 1.2 \times 9.8 \times 1.5 = 18 \text{ J}\)
(b) Work done by tension:
\(W = T \times d = 15 \times 1.5 = 22.5 \text{ J}\)
(c) Gain in kinetic energy:
\(\Delta KE = W - \Delta PE = 22.5 - 18 = 4.5 \text{ J}\)
(iii) Time for string to become taut again:
Velocity of A when B hits the floor:
\(v = u + at = 0 + 2.5 \times 1.6 = 4 \text{ m/s}\)
Time for A to stop and return:
\(v = u - gt\)
\(0 = 4 - 9.8t\)
\(t = \frac{4}{9.8} \approx 0.4 \text{ s}\)
Total time: \(0.4 \times 2 = 0.8 \text{ s}\)