A coconut of mass 1.5 kg falls from the top of a coconut tree onto the wet sand on a beach. The height of the tree is 10 m. On impact, the coconut comes to rest by making a depression in the sand.

(A) Calculate the velocity of the coconut just before it hits the sand.

(B) Assume that the average resistive force of sand is 3000 N and all of the coconut's energy is used to create the depression in the sand. Calculate the depth of the depression the coconut makes in the sand. Assume g = 10 ms⁻².

Given:\nMass of coconut, $m = 1.5\,\mathrm{kg}$\nHeight, $h = 10\,\mathrm{m}$\nAcceleration due to gravity,\n$g = 10\,\mathrm{m\,s^{-2}}$\n(A) Velocity just before hitting the sand:

Using,

$$mgh = \frac{1}{2}mv^2$$

$$1.5 \times 10 \times 10 = \frac{1}{2} \times 1.5 \times v^2$$

$$150 = 0.75v^2$$

$$v^2 = 200$$

$$v = \sqrt{200}$$

$$v = 14.1 \text{ ms}^{-1}$$

Therefore, the velocity of the coconut just before hitting the sand is $14.1 \text{ ms}^{-1}$

(B) Depth of the depression:

Kinetic energy of the coconut just before impact:

$$K.E. = mgh$$

$$= 1.5 \times 10 \times 10$$

$$= 150 \text{ J}$$

This energy is used to do work against the resistive force of sand.

$$W = F \times s$$

$$150 = 3000 \times s$$

$$s = \frac{150}{3000}$$

$$s = 0.05 \text{ m}$$

Therefore, the depth of the depression made in the sand is 0.05 m or 5 cm