• 9702 November 2011 Paper 22 Worked Solutions | A-Level Physics

Question 1

{Detailed explanations for this question is available as Solution 629 at Physics 9702 Doubts | Help Page 124 - http://physics-ref.blogspot.com/2015/04/physics-9702-doubts-help-page-124.html}

Question 2

(a)

(i)

Force is defined as the rate of change of momentum.

(ii)

Work done is defined as the product of the force and the distance moved in the direction of the force.

(b)

Force R acts on mass m along straight line for distance s. Acceleration = a and speed changes from u to v.

(i)

Work W done by F = Fs         or W = mas     

or W = m(v²-u²)/2                    or W = force x distance s

(ii)

Use answer from (i) and an equation of motion to show

Kinetic energy = ½ x mass x (speed)²:

as = (v²-u²) / 2           

W = mas          hence               W = m(v²-u²)/2

Right-hand side represents terms of energy               or with u = 0, KE = ½ mv²

(c)

Resultant force of 3800N causes car of mass 1500kg to accelerate from 15ms^-1 to 30ms^-1

(i)

Distance moved by car during this acceleration:

Work done = ½ (1500)[(30)² – (15)²] = 506 250

Distance = Work Done / F = 506 250 / 3800 = 133m

Or

F = ma                         a = 2.533ms^-2

v² = u² +2as                 s = 133m

 

(ii)

Same force used to change speed of car from 30ms^-1 to 45ms^-1. Why distance moved not same as that calculated:

The change is kinetic energy is greater

OR

The work done by the force has to be greater; hence the distance is greater (for the same force)

Question 3

{Detailed explanations for this question is available as Solution 634 at Physics 9702 Doubts | Help Page 126 - http://physics-ref.blogspot.com/2015/04/physics-9702-doubts-help-page-126.html}

Question 4

(a)

Electric field strength is defined as the force per unit positive charge.

(b)

2 horizontal metal plates 20mm apart in vacuum. P.d of 1.5kV applied across plates. Charged oil drop of mass 5.0x10^-15kg held stationary by the electric field.

(i)

Lines to represent electric field between plates:

Draw at least 3 equally spaced parallel vertical lines

Direction down

(ii)

Electric field strength between plates:

E = 1500 / (20x10^-3) = 75 000Vm^-1

(iii)

Charge on drop:

F = qE

(W = mg          and so,)           qE = mg

q = mg / E = [(5x10^-15)(9.81)] / 75000 = 6.5x10^-19C

{Electric field line is from +ve to –ve. Since the weight of the oil drop acts downwards, the electric force acting on it should be upwards, so that the drop is stationary. Since the upper plate is +ve, the charge of the drop should be –ve so that it is attracted upwards.}

 Negative charge

(iv)

Potential of upper plate is increased. Subsequent motion of drop:

Electric force, F > mg             OR F now greater

The drop will move upwards

Question 5

A potentiometer circuit that is used as a means of comparing potential differences is shown in Fig. 5.1.

Question 6

(a)

Principle of superposition states that when waves overlap, the (resultant) displacement is the sum of the displacements of each of the waves.

(b)

Arrangement used to determine speed of sound in air.

Sound waves of constant frequency emitted from loudspeaker L and reflected from point S on a hard surface. Loudspeaker is moved away from S until a stationary wave is produced.

How sound waves from L give rise to stationary wave between L and S:

The waves travelling in opposite directions overlap / the incident and reflected waves overlap. If the waves have the same speed and frequency, a stationary wave is formed.

(c)

Microphone connected to CRO is positioned between L and S.

Trace obtained on CRO: amplitude = 2cm and period = 4cm

Time-base setting on CRO = 0.10mscm^-1.

(i)

Frequency of sound wave:

Time period, T = 4(0.1) = 0.4ms

F = 1 / T = 1 / (4x10^-4) = 2500Hz

(ii)

Microphone now moved towards S along LS. When moved 6.7cm, trace seen on CRO varies from a maximum amplitude to a minimum and then back to a maximum.

1.

Use properties of stationary waves to explain these changes in amplitude:

The microphone is at an antinode and goes to a nodes and then to an antinode / The maximum amplitude occurs at an antinode and minimum amplitude at node.

2.

Speed of sound:

[Distance between 2 successive antinodes = λ / 2]

λ / 2 = 6.7cm

v = f λ = 2500 x (13.4x10^-2) = 335ms^-1

Question 7

(a)

Experimental observations that show radioactive decay is:

(i)

Spontaneous:

The half-life / count rate / rate of decay / activity is the same no matter what external factors / environmental factors or 2 named factors such as temperature and pressure changes are applied

(ii)

Random:

The observations of the count rate / rate of decay / activity / radioactivity during decay show fluctuations / variations.

(b)

Complete table:

Property          α-particle         β-particle                     γ-particle

Charge             (+)2e                - e                                0

Mass                4u                    9.11x10^-31kg                0

Speed              0.01 to 0.1c     up to 0.99c                  c

(c)

Process by which α-particles lose energy when they pass through air:

Collision with the molecules causes ionization (of the molecules) / electron is removed