A steel ball falls from a platform on a tower to the ground below, as shown in Fig. 3.1.

Question 30

A steel ball falls from a platform on a tower to the ground below, as shown in Fig. 3.1.

Fig. 3.1

The ball falls from rest through a vertical distance of 192 m. The mass of the ball is 270 g.

(a) Assume air resistance is negligible.

(i) Calculate

1. the time taken for the ball to fall to the ground, [2]

2. the maximum kinetic energy of the ball. [2]

(ii) State and explain the variation of the velocity of the ball with time as the ball falls to the ground. [1]

(iii) Show that the velocity of the ball on reaching the ground is approximately 60 m s^-1. [1]

(b) In practice, air resistance is not negligible. The variation of the air resistance R with the velocity v of the ball is shown in Fig. 3.2.

Fig. 3.2

(i) Use Fig. 3.2 to state and explain qualitatively the variation of the acceleration of the ball with the distance fallen by the ball. [3]

(ii) The speed of the ball reaches 40 m s^-1. Calculate its acceleration at this speed. [2]

(iii) Use information from (a)(iii) and Fig. 3.2 to state and explain whether the ball reaches terminal velocity. [2]

Reference: Past Exam Paper – November 2015 Paper 23 Q3

Solution:

(a)

(i)

1.

{Equation for uniformly accelerated motion:}

s = ut + ½ at²

{Initial speed of the ball: u = 0}

192 = ½ × 9.81 × t²

t = 6.3 (6.26) s

2.

{Conservation of energy:

As the ball falls, its GPE is converted to KE. The KE is maximum when the GPE is least – that is, at the ground.

Max KE (at ground) = GPE at top of tower}

max Ek (= mgh) = 0.27 × 9.81 × 192       

max Ek = 510 (509) J                         

(ii) The velocity of the ball increases at a constant rate as the acceleration is constant.

(iii)

v = u + at

v = 0 + at = 6.26 × 9.8

v = 61.4 m s^-1

(b)

(i)

The air resistance R increases with velocity.

The resultant force is given by mg – R. So, the resultant force decreases.

This results in a reducing acceleration.

(ii)

{From the graph,} at v = 40 m s–1, R = 0.6 (N)

{Resultant force = ma

mg – R = ma}

0.27×9.8 – 0.6 = 0.27 × a

a = 7.6 (7.58) m s^-2                            

(iii)

Air resistance R = weight for terminal velocity

either weight requires velocity to be about 80 m s–1

or at 60 m s–1, R is less than weight

so does not reach terminal velocity

{From (a)(iii), the velocity at the bottom is about 60 m s^-1. The velocity for terminal velocity is when air resistance R = weight = 0.27 × 9.81 = 2.65 N. From the graph, this corresponds to v being about 80 m s^-1.

But the final velocity reached is 60 m s^-1, so terminal velocity is not reached as the air resistance is not equal to the weight at this speed.}