A student investigates the motion of a small polystyrene sphere as it falls from rest alongside a vertical scale marked in centimetres.

Question 16

(a) One of the equations of motion may be written as

v² = u² + 2as.

(i) Name the quantity represented by the symbol a.

(ii) The quantity represented by the symbol a may be either positive or negative. State

the significance of a negative value.

[2]

(b) A student investigates the motion of a small polystyrene sphere as it falls from rest

alongside a vertical scale marked in centimetres. To do this, a number of flash

photographs of the sphere are taken at 0.1 s intervals, as shown in Fig. 1.1.

Fig. 1.1

The first photograph is taken at time t = 0.

By reference to Fig. 1.1,

(i) briefly explain how it can be deduced that the sphere reaches a constant speed,

(ii) determine the distance that the sphere has fallen from rest during a time of

1. 0.7 s,

2. 1.1 s.

 [4]

(c) The student repeats the experiment with a lead sphere that falls with constant

acceleration and does not reach a constant speed.

Determine the number of flash photographs that will be observed against the 160 cm

scale.

Include in your answer the photograph obtained at time t = 0. [3]

Reference: Past Exam Paper – November 2003 Paper 2 Q1

Solution:

(a)

(i) acceleration           

(ii) A negative value indicates that the velocity is decreasing or force/acceleration is in negative direction                                                     

(b)

(i) when the separation of the dots becomes constant/does not continue to increase (must make a reference to the diagram)

(ii)

1.

{this corresponds to the 7^th dot (excluding the starting point) as each dot is produced after 0.1 s.}

distance = 132 cm

2.

{distance between 7^th and 8^th (last) dot = 157 – 132 = 25 cm

Distance between 6^th and 7^th dot = 132 – 107 = 25 cm

So, a constant speed is reached as the distance fallen by the sphere after each 0.1 s is 25 cm.}

at constant speed, distance travelled in 0.1 s = 25 cm

{Distance 0.7 s, the distance travelled = 132 cm

Remaining time = 1.1 – 0.7 = 0.4 s

As seen above, a constant speed of 25 cm per 0.1 s has already been reached by the time of 0.7 s.

Distance travelled in 0.1 s = 25 cm

Distance travelled during the remaining 0.4 s = (4 × 25) cm}

distance = 132 + (4 x 25) = 232 cm

(c)

{The sphere may be considered to be in free fall with acceleration g.}

s = ut + ½ at²

1.6 = ½ x 9.8 x t²        (allow g = 10 m s^-2 )                           

t = 0.57 s

{The sphere takes 0.57 s to fall the distance of 160 cm. Since we have a flash photograph after each 0.1 s, there would be 5 flash photographs. Including the photograph obtained at time t = 0, we have a total of 6 photographs.}

hence 6 photographs