(a)(i) Free-Body Diagram:

The free-body diagram shows two forces: tension (T) acting upwards along the string towards the point of support, and weight (mg) acting downwards. The tension force is longer than the weight, indicating a net centripetal force towards the center of the circular motion.

(a)(ii) Speed Calculation:

Using conservation of energy, the potential energy at the horizontal position is converted to kinetic energy at the vertical position:

$ m g h = \frac{1}{2} m v^2 $

$ v = \sqrt{2 g h} = \sqrt{2 \times 9.81 \times 0.95} = \sqrt{18.639} \approx 4.32 \, \mathrm{m \, s^{-1}} $

Thus, the speed is approximately 4.3 m/s.

(a)(iii) Tension Calculation:

The net force provides the centripetal force, and the tension must counteract both the weight and provide the centripetal acceleration:

$ T – m g = \frac{m v^2}{r} $

$ T = m g + \frac{m v^2}{r} = 0.800 \times 9.81 + \frac{0.800 \times 4.32^2}{0.95} $

$ T = 7.848 + \frac{0.800 \times 18.6624}{0.95} \approx 7.848 + 15.728 \approx 23.58 \, \mathrm{N} \approx 23.5 \, \mathrm{N} $

Thus, the tension is approximately 23.5 N.

(b)(i) Elastic Collision:

Using conservation of momentum:

$ m_1 u_1 = m_1 v_1 + m_2 v_2 $

$ 0.800 \times 4.32 = 0.800 \times (-v_1) + 2.40 \times 2.16 $

$ 3.456 = -0.800 v_1 + 5.184 $

$ v_1 = \frac{5.184 – 3.456}{0.800} = 2.16 \, \mathrm{m \, s^{-1}} $ (rebound speed)

Calculate initial kinetic energy:

$ KE_i = \frac{1}{2} \times 0.800 \times 4.32^2 = 0.400 \times 18.6624 = 7.465 \, \mathrm{J} $

Calculate final kinetic energy:

$ KE_f = \frac{1}{2} \times 0.800 \times 2.16^2 + \frac{1}{2} \times 2.40 \times 2.16^2 $

$ KE_f = 0.400 \times 4.6656 + 1.20 \times 4.6656 = 1.86624 + 5.59872 = 7.465 \, \mathrm{J} $

Since $ KE_i = KE_f $, the collision is elastic.

(b)(ii) Maximum Height:

Using conservation of energy, the kinetic energy after collision is converted to potential energy at the maximum height:

$ \frac{1}{2} m v^2 = m g h $

$ h = \frac{v^2}{2 g} = \frac{2.16^2}{2 \times 9.81} = \frac{4.6656}{19.62} \approx 0.238 \, \mathrm{m} = 23.8 \, \mathrm{cm} $

Thus, the maximum height is approximately 23.8 cm.

(c) Distance Travelled:

Frictional force: $ f = \mu m g = 0.400 \times 2.40 \times 9.81 = 9.4176 \, \mathrm{N} $

Work done by friction equals initial kinetic energy:

$ f d = \frac{1}{2} m v^2 $

$ 9.4176 \times d = \frac{1}{2} \times 2.40 \times 2.16^2 = 1.20 \times 4.6656 = 5.59872 $

$ d = \frac{5.59872}{9.4176} \approx 0.594 \, \mathrm{m} $

Thus, the distance travelled is approximately 0.594 m.