The distance s moved by an object in time t may be given by the expression s = ½ at2 where a is the acceleration of the object.

Question 11

(a) The distance s moved by an object in time t may be given by the expression

s = ½ at²

where a is the acceleration of the object.

State two conditions for this expression to apply to the motion of the object. [2]

(b) A student takes a photograph of a steel ball of radius 5.0 cm as it falls from rest. The

image of the ball is blurred, as illustrated in Fig. 2.1.

The image is blurred because the ball is moving while the photograph is being taken.

Fig. 2.1

The scale shows the distance fallen from rest by the ball. At time t = 0, the top of the ball

is level with the zero mark on the scale. Air resistance is negligible.

Calculate, to an appropriate number of significant figures,

(i) the time the ball falls before the photograph is taken, [3]

(ii) the time interval during which the photograph is taken. [3]

(c) The student in (b) takes a second photograph starting at the same position on the scale.

The ball has the same radius but is less dense, so that air resistance is not negligible.

State and explain the changes that will occur in the photograph. [2]

Reference: Past Exam Paper – June 2010 Paper 22 Q2

Solution:

(a)

Choose any 2:

Initial speed is zero

Constant acceleration

Straight line motion

(b)

(i)

s = ½ at²

0.79 = ½ × 9.8 × t²                 

t = 0.40 s

(ii)

{At time = zero, the top of the ball is at the zero mark. Since we are considering the top of the ball initially, we need to continue to consider the distance travelled by the top.

We cannot take the top of the ball as a reference initially, and the bottom of the ball in the final case.}

Total distance travelled by the end of the time interval = 90 cm

0.90 = ½ × 9.8 × t²

So, total time taken for the motion of the ball:

t = 0.43 s (allow 2sf or greater)

Time interval = 0.43 – 0.40 = 0.03 s  

(c)

Air resistance being not negligible means the ball’s speed / acceleration is less. So, the length of the image is shorter.