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

Fig. 6.2 shows the variation with time t of the current Ip in the primary coil. There is no current in the secondary coil.

Question 2

An ideal iron-cored transformer is illustrated in Fig. 6.1.

Fig. 6.1

(a) Explain why

(i) the supply to the primary coil must be alternating current, not direct current, [2]

(ii) for constant input power, the output current must decrease if the output voltage

increases. [2]

(b) Fig. 6.2 shows the variation with time t of the current Ip in the primary coil. There is no current in the secondary coil.

Fig. 6.2

Fig. 6.3

Fig. 6.4

(i) Complete Fig. 6.3 to show the variation with time t of the magnetic flux Φ in the

core. [1]

(ii) Complete Fig. 6.4 to show the variation with time t of the e.m.f. E induced in the

secondary coil. [2]

(iii) Hence state the phase difference between the current Ip in the primary coil and the

e.m.f. E induced in the secondary coil. [1]

Reference: Past Exam Paper – June 2005 Paper 4 Q6

Solution:

(a)

(i) The flux/field in the core must be changing so that an e.m.f./current is induced in the secondary.

(ii)

Power = VI

Since the output power is constant, if the (secondary) output voltage V_S increases, the (secondary) current I_S must decrease.

(b)

(i)

Graph has same shape and phase as I_P

(ii)

Graph:

same frequency, correct phase w.r.t. Fig. 6.3

{emf induced E = - dϕ / dt

The graph of ϕ is a sine curve. Differentiating sine gives cos. But since we have a negative sign, the cosine curve is inverted.}

(iii)

½π rad or 90°

A hammer with 10 J of kinetic energy hits a nail and pushes it 5.0 mm into a plank.

Question 16

A hammer with 10 J of kinetic energy hits a nail and pushes it 5.0 mm into a plank.

Both the hammer and nail come to rest after the collision.

What is the average force that acts on the nail while it moves the 5.0 mm?

A 0.050 N                    B 2.0 N                        C 50 N                         D 2000 N

Reference: Past Exam Paper – June 2011 Paper 12 Q19 & June 2016 Paper 11 Q19

Solution:

Answer: D.

The kinetic energy is converted into work done by the hammer in pushing the nail into the plank.

Since both the hammer and nail come to rest after the collision, all of the KE of the hammer has become ‘work done’. 

[Remember than both ‘work done’ and ‘kinetic energy’ has unit ‘joule’.]

Work done = Force × distance

10 = F × 5×10^-3

F = 10 / (5×10^-3) = 2000 N