1 a) Between t = 30 and t = 45 mins
b) 7.5 mins
c) i) distance travelled = area under graph between t = 0 and t = 12½ mins
ii) average speed = total distance travelled
total time for journey
= total area under graph
2 a) ∆v = 32 m/s a =10 m/s²
t = ∆v = 32 = 3.2 s
3 a) OP – constant acceleration
PQ – constant acceleration (greater than OP)
QR – constant speed
RS – constant deceleration
b) O and S
c) 6 m/s
d) 70 s
e) Total distance travelled = area under graph
4 a) u = d = 25 = 12.5 s
b) speed is decreasing
c) time for tree 3 to tree 4 is greater than for time for tree 2 to tree 3
5 a) Stage 1: 20 s
Stage 2: t = 4800 = 400 s
Stage 3: 80 s
c) Total distance travelled = (½ × 20 × 12) + (400 × 12) + (½× 80 × 12)
= 120 + 4800 + 480 = 5400m
d) average speed = total distance = 5400 = 10.8 m/s
time taken 500
1 Weigh the car using the Newton meter.
Find the mass from mass = weight
(for rest of question see description on p. 16)
2 (For first part see description on p.16)
To find the density of cork the mass, M, could still be found using the balance.
But the volume must be calculated by finding its dimensions and using
cross-sectional area × length to calculate the volume.
Area = area of a circle, πr²
d = 2 × radius = 2r
Area = πr²
Volume = πr²l
Density = πr²l
3 a) Volume = Reading Q – Reading P
b) Mass = Reading S – Reading R
c) i) P = m
ii) p = 57.5 = 2.3 g/cm³
1 l = 3.6 cm
2 A = πr² - π(d)² = π(0.6)² = 0.28 cm²
3 V = πr²l = 1.0 cm³
Volume of bundle – 10 × 10 – 10.0 cm³
P = m = 59.1 – 5.91 g/cm³
1 a) length = 17.5 cm
b) load = 2.25 N
c) extension = length – initial length = 15 – 11
= 4 cm
2 a) It has both size and direction
b) f = m × a
= 15 N
3 a) 11 cm
b) 19 cm
c) 19 − 11 = 8 cm
4 a) Q = elastic limit
b) The extension is directly proportional to the force applied
c) In the region QR, the extension is greater for the same increase in applied force
d) k = F/x = gradient of graph = 8/x = 4 N/m
5 c) Extension = 67 – 40 = 27 cm
d) 2.5 N
1 m/g θ/°
2 Graph is a straight line through (0,0)
1 i) 1: force 2: moment
ii) F = F1 + w + F2
2 a) Resultant force = 0 Resultant moment = 0
b) 6 × 40 = F × 30
F = 240 = 8.0 N
c) Force = 0.5 N downwards
1 Use a set square
2 1/d ÷ 1/m
3 c) G = 0.53 Nm
4 w = 0.53 = 1.1 N
1 a) Line of action of weight of box acts outside its base
b) i) less than
ii) The centre of mass is higher and so the line of action of the weight will act outside
the base at a longer angle
2 a) Plastic will rotate clockwise
b) Plastic will not move
10 N ←−−−−−−
1 a) chemical, gravitational PE, kinetic, internal
c) gravitational PE
2 i) kinetic + gravitational potential
ii) kinetic energy is dissipated as heat due to frictional forces
1 a) gravitational
1 a) WD = f × d
b) i) WD = 100 × 8 = 800 J
ii) power = WD = 800 = 160 W
iii) Increasing internal energy
2 a) i) WD = high = 50 × 10 × 4 = 200 J
ii) P = WD = 200 = 100 W
b) gravitational potential → kinetic → heat
3 a) PE = mgh = 1×10×7 = 70J
b) mgh = ½mv²
70 = 0.5v²
V = [symbol] 140
= 1.2 m/s
c) energy dissipated as heat
1 a) Skate, since area is smaller for the same weight so pressure is greater.
b) The area of the vehicle is large and so for a constant wind pressure, the face is high
and could blow the vehicle over
2 a) Pressure = pgh
= 1000 x 10 x 2 = 20000 N/m²
b) Area = force = 50 = 0.0025m²
c) gravitational potential to kinetic
1. distance; time; m/s; gradient; direction; magnitude; rate of change; m/s²
16. 36000 N/m²
17. 0.6 N/cm²
1 a) Speed increases with decreasing acceleration between P and R. Between R and S
acceleration is zero, speed is steady.
c) forces are balanced; resultant force = 0
d) distance between l and T = 40x120 = 4800m
distance between P and Q = ½x6x50 = 150 m
2 a) Time several complete oscillations. Divide time by number of oscillations and then
divide by 2 to find the time between P and Q.
b) i) weight and tension
ii) Towards centre of circle
c) PE = mgh = 0.2×10×0.05 = 0.1J
3 b) a = ∆v = 12 = 1.5 m/s²
c) distance = ½×5×12 = 30m
d) f=mxa = 4000 × 1.2 = 4800N
e) For the same accelerating force, there is a greater mass due to the extra passengers
and so acceleration is less.
4 a) i)
ii) Force increases
ii) The force of friction was not large enough to provide the centripetal force.
c) i) steady speed
ii) circumference = 2TTr = TTd = 20.4cm
iii) increase in speed per second = 25 = 8.3 m/s²
5 a) When clockwise moment due force of steam > anticlockwise moment due to weight,
the beam turns clockwise and W is lifted.
b) i) 12Nm = F × 0/2
f = 60N
ii) P = F = 60 = 200000 N/m²
1 a) Constant random motion
b)As the temperature increases, the velocity of the particles increases. Particles
collide with walls of container more often and with greater force and so the
2 a) Evaporation
b) Particles with greatest ICE escape from surface of water, reducing the volume of
3 a) The level of water in the tube becomes lower.
b) The air in the flask expands as the temperature increases.
c) Below 0°C, water would freeze.
4 a) The dust particle is constantly bombarded by the air molecules, which are in
constant random motion.
b) i) The ones with the greatest kinetic energy.
ii) These have the energy required to overcome the attractive forces between
5 a) i) Constant, random motion
ii) Collides with container walls putting force on walls.
b) i) Increases
ii) Remains constant
c) i) Movement with greater KE in gas.
ii) Separation of molecules in gas is greater.
1 The temperature of thermometer 2 decreases more rapidly than that of thermometer 1,
due to the evaporation of the water.
2 The temperature of thermometer 3 decreases more rapidly than that of thermometer 2,
because acetone evaporates more rapidly than water.
1 a) Choose a different liquid, one with a larger range of temperature between its
freezing and boiling point.
b) Increase the number of scale divisions and have a thinner thread in the container.
2 a) The piston must move to the right.
b) The heat supplied to the piston is used to do work in pushing the piston to the
right. The temperature remains constant.
3 a) They vibrate with greater kinetic energy and move further apart.
b) i) Thermometers
ii) Railway tracks
1 a) Mass of aluminium on balance.
Change in temperature with thermometer
energy delivered to the aluminium.
c) Some heat is lost to the surroundings.
2 a) Evaporation occurs over a range of temperatures. Boiling occurs at a specific
temperature (the boiling point).
Evaporation decreases the temperature of the liquid. The temperature of the liquid
remains constant during boiling.
b) Energy is required to break the bonds between the liquid particles. All the heat
energy is used to do work against the surroundings, not to increase the internal
energy of the liquid.
c) Є = P × t = 100 × 20 × 60 = 120000J
120000 = ΔML
= 0.05 L
L = 2.4 × 106 J/kg = 2400 J/kg
3 a) i) 20°C, 15°C
ii) As the temperature of the water increases, more heat is lost to the
surroundings and so the temperature rises at a lower rate.
b) C = Є/MΔT = 60 × 210/0.075 × 40 = 4200 J/kg/°C
4 a) Find the mass of the beaker as a balance.
Start the stopclock and simultaneously switch on the heater. After most of the ice
has melted, stop the stopclock. Find the mass of the beaker and the water.
b) Є = ΔML
60 × t = 120 × 340
t = 120 × 340/60 = 680s
c) i) Heat from the surroundings melts the ice.
ii) Insulate the ice from the surroundings.
2 Beaker A: 8°C, Beaker B: 5°C
3 The Bunsen burner loses more heat to the surroundings. The power of the fixed
voltage source is greater than the Bunsen burner.
1 Take the starting temperature on the thermometers. Read the thermometers after a
fixed interval of time. The best absorber of heat radiation will have a higher
2 a) Radiation
1 a) Copper is a good conductor of heat.
b) Plastic is a poor conductor of heat to prevent burns.
c) i) The metal base conducts heat through the increased vibration of the particles.
2 a) i) Copper is a good conductor of heat.
ii) Plastic is an insulator.
b) i) Conduction
ii) Radiation and convection.
c) equal to 40W
3 a) i) Conduction
ii) The particles at A vibrate faster and pass on the increased vibrations to
b) 4 different coloured surfaces are heated at equal distances from a radiant heater
(see diagram page 70). The initial and final temperatures of each of the surfaces
are taken after equal time intervals.
1 t/s, θ/°C
2 Place insulation on the bottom of the beaker. Cover the top of the beakers.
Summary Questions on unit 2
2 melt, 1, bal, 1, temperature, 1
3 Structure – see diagrams page 50.
Arrangement of Close together; vibrating Far part; in constant
molecules in fixed positions random motion
Compressible No No
?? No Yes
Shape Fills the bottom of the Fills the whole
4 Arrows point clockwise.
5 Є = MCΔT
C = E / MΔT = 4500 / 0.5 × 10 = 9000 J/kg/°C
6 Є = ΔML
L = 66 000 / 0.2 = 330000 J/kg
7 The pressure increases since the volume decreases at constant temperature. Collisions
between the particles and walls of the container occur more frequently.
1 reading 2 = mass of beaker + stirrer + thermometer + water
a) reading 3 – mass of beaker + stirrer + thermometer + water + melted ice
b) i) heat lost = mass of water × 20 × specific heat capacity of water
ii) heat gained = mass of ice × specific latent heat of fusion
c) L = 12800 / 0.030 = 4.3 × 105 J/kg = 430000 J/kg
d) The ice may not be at 0°C. Heat may be taken in from the surroundings to melt
2 Temperature decreasing. Volume increasing.
3 a) Vibrate faster; move further apart.
b) i) Thermometer
ii) Railway tracks
4 a) i) Constant random motion
ii) Collide with walls and exert a force.
b) i) increases
ii) remains constant
c) i) gas – constant random motion in all directions
solid – vibrating about fixed positions
ii) separation of molecules in a gas is much higher than in a solid.
b) Count the number of waves passing a point in 30s. Divide number of waves by
30s. Frequency = number of waves per second.
b) Air molecules are vibrating, alternately moving closer together and then further
c) λ = v / f = 330 / 500 = 0.66m
3 a) Separation of waves: should be equal between one wavefront and the next.
Shape of wavefronts should be semicircular, not straight.
c) V = f × λ
= 8 × 0.012
= 0.096 m/s
1 a) Angle of incidence – 0° i.e. along normal
b) Speed decreases, frequency is constant, wavelength decreases.
d) sin30° / sin θ = 0.67
θ = sin-1 (sin30° / 0.67) = 48°
b) i) 90°
iii) n = 1 / sin43° = 1.5
c) speed = 3 × 108 / 1.5 = 2.0 × 108 m/s
b) Virtual, inverted, same size
c) Angle of incidence = 0°
d) speed = 3 × 108 / 1.5 = 2 × 108 m/s
e) angle of incidence at b> critical angle
1 a) infra red, UV, X-ray, (symbol)-ray
2 a) i) (symbol) or x-ray
ii) infra-red or radio
b) f = v / λ - 3 × 108 / 1.0 × 10–14 = 3.0 × 1022 Hz
c) 3 × 108 m/s
b) i) dispersion
1 a) t = d / u = 440 / 330 = 1.3s
b) First clap = 220 / 330 = 0.67s after clap
Second clap = 660 / 330 = 2s after clap
2 a) 2000 / 60 = 33 times per second
b) Frequency = 33 Hz
c) Yes, hearing range is typically 20 Hz – 20 kHz
3 a) i) A building/wall.
ii) d = u × t = 320 × 1.5 = 480 m
b) v = d / t = 6 / 5 = 1.2 m/s
1 longitudinal; transverse; transverse; longitudinal; vibrations; parallel; wavelength;
time period; waves; second; V = f × λ; reflection; refraction; diffraction.
4 a) sin r = sin 56 / 1.5
r = 34°
angle of reflection = 56°
b) sin r = sin 35 / 0.67
r = 59°
angle of reflection = 59°
c) angle of reflection = 60°
There is no refracted ray.
7 a) λ = v / f = 3 × 108 / 1 × 1016 = 3 × 10–8m UV
b) f = v / λ = 3 × 108 / 0.1 = 3 × 109 Hz radio
c) λ = v / f = 3 × 108 / 1 × 1020 = 3 × 10–12m (symbol)
9 Different wavelengths of light refract by different amounts in a denser medium,
causing white light to split into a spectrum.
11 f = v / λ = 330 / 0.1 = 3300 Hz
speed = 3300 × 0.45 = 1485 m/s
12 u = d / t – 2 / 0.00042 = 4760 m/s
1 The image is inverted, virtual, same size as object
2 a) t = d / u – 498 / 332 = 1.5s
b) 0.75s, 2.25s
3 a) They are standing in the path of the reflected second wave.
b) 400 Hz
c) λ = v/f – 330/400 = 0.825m
d) Vibrations of particles are parallel to direction of wave.
4 a) i)
ii) same size, upright, virtual, laterally inverted
5 a) Stop clock, starting period
b) Start stopclock when smoke from pistol is seen
Stop stopclock when sound is heard.
Measure distance between student with starting pistol and observer.
Time of travel is reading on stopclock.
c) speed = distance / time
d) Repeat the experiment several times, eliminate anomalous results and average
e) i) 100 m/s
ii) 1000 m/s
b) sin r = sin 40° / 1.52 = 25°
c) i) 3 × 108 m/s
ii) 3 × 108 m/s
1 a) A south pole is a south seeking pole. It is attracted to a north seeking pole.
ii) The induced magnetism is lost.
2 a) i) iron
ii) 1 nothing, 2 nothing
1 a) 6.3 V; 330 (symbol)
c) iii) A best fit line allows values of time or resistance between the plotted values to
be found with greater accuracy.
d) Temperature is greatest for the lowest value of resistance on the graph.
2 a) i) conductor
iii) Put a potential difference across it.
b) i) insulator
b) Vary resistance of variable resistor to obtain 5 different voltage readings, taken on
voltmeter. Take 5 corresponding values of current on the ammeter.
c) i) R = V / I – 6 / 0.5 = 12 Ώ unknown resistor = 9 Ώ
ii) Q = It – 0.5 × 120 = 60C
iii) P = IV – 0.5 × 1.5 = 0.75W
4 a) Attach plate to earth.
Bring positively charged rod towards plate.
Remove earth lead and then remove rod.
b) i) Q = It = 0.02 × 15 = 0.3C
ii) V = IR = 0.02 × 10000 = 200 V
1 AB 50 cm
AC 75 cm
AD 100 cm
3 AB R = V / I = 2.53 Ώ
AC = 4.00 Ώ
AD = 5.20 Ώ
6 R = 5.2 × 1.5 = 7.8 Ώ
1 a) same
d) i) 1 Ώ
ii) greater than
e) i) increases
b) i) I = P / V = 36 / 12 = 3A
ii) R = V / I = 12 / 3 = 4 Ώ
iii) 2 Ώ
iv) Є = Pt – 36 × 30 = 1080 J
c) lamps A and B have a potential difference of 6V not 12V across each of them.
3 The resistance of the thermistor varies with temperature.
a) As the temperature increases, it takes a smaller share of the voltage, switching on
the transistor and lighting the lamp.
b) Vary the size of resistor R.
c) An indicator light to show the oven has reached the required temperature.
d) R = V / I = 12 / 3 = 4 Ώ
e i) Parallel circuit
ii) 4.0 A
5 a) A – resistor
B – light dependent resistor
C – transistor
D – bulb
b) transistor, C
c) Resistance of LDR increases in the dark, increasing in share of voltage from the
supply. The potential difference across the base is enough to switch on the
collector-emitter current and so the bulb lights.
6 a) Region around an electrostatic charge where another charge experiences a force.
c) I = θ / t
= 0.060 / 30 = 0.002 A
d) Є = VIt – 1500 × 0.0080 × 10 = 120 J
ii) No effect
2 There is an ammeter reading the current.
3 Variable resistor.
4 Increase resistance of variable resistor. Reduce voltage of power source.
5 A since current has to pass through more coils of wire.
1 a) Primary turns > Secondary turns
b) Output voltage < Input voltage
d) The bell would receive too high a pd, which could cause it to overhead and be
2 a) The wire has a magnetic field around it when a current flows through it. The field
due to the magnetic poles interacts with this field and puts a force on the wire.
b) Out of the page.
c) Electrical to kinetic.
d) i) commutator and brushes (see page 135 – dc motor)
ii) When the current flows through the coil a force acts on each side of the coil in
opposite directions, due to the interaction with the magnetic field. The couple
causes a turning effect on the coil. The commutator allows the polarity of the
power supply to swap every half turn to keep the coil turning.
b) i) For an emf to be induced across the secondary coil, there must be a changing
magnetic field in the primary coil.
ii) The changing magnetic field due to the ac current induces an emf across the
c) i) P = IV – 1.5 × 12 = 18W
ii) Є = 18 × 3 = 54J
4 a) An alternating current produces an alternating magnetic field.
In the core an alternating magnetic field links the two coils.
In the secondary coil, an emf is induced due to the changing magnetic field.
b) More turns on secondary coil than on primary.
c) The magnetic field linking the two coils would be too weak.
5 a) Attach the millivoltmeter across the wire. Move the wire between the poles of the
b) The millivoltmeter would give a reading
ii) The positive changes on the positive phase attract the negatively charged electron.
b) i) horizontal deflection plates
ii) vertical deflection plates
2 a) i) A and B
ii) in the filament
iv) straight line from left to right
v) light is emitted
c) Otherwise the electrons would collide with air molecules and lose kinetic energy.
3 a) Q = It = 3 × 6 = 18C
b) I = Q/t = 0.006 / 100 = 60 μC
c) t = Q / I – IJ / 0.1 = 150s
4 V = IR
a) V = 100 × 0.5 = 50V
b) I = V / R = 10 / 330 = 0.030 A
c) R = V / I = 1.5 / 0.8 – 1.9 Ώ
5 a) P = IV – 3 × 12 – 36W
b) I = P / V = 40 / 230 = 0.17A
c) V = P / I = 100 / 0.5 = 200V
6 If the live wire touches the metal casing of an appliance, the current flows through the
earth wire to ground. The current is large enough to melt the fuse and break the
circuit, protecting the person from an electric shock.
8 See page 122
9 A small current in the coil produces a magnetic field which attracts the iron armature.
The iron armature pivots to close the contacts in the high voltage circuit, switching on
a large pd.
11 field; force; force; left hand; field; current; force; current; strength; magnetic field
12. See page 135-136
13 See page 132
14 See page 132
15 Spin the coil faster
More turns on coil
The transformer requires a changing current to produce a changing magnetic field.
2 a) When the switch is closed a current flows in the circuit, through the metal spokes.
The magnetic field due to this current and due to the fixed magnets interact and
put a force on the spokes, which makes the wheel turn.
c) See page 135
d) The commutator keeps electrical contact with the brushes but the brushes swap
sides of the commutator every half turn. This keeps the coil moving in the same
3 a) I = P / V = 9 / 6 = 1.5A
b) i) 8 Ώ
c) i) brightness decreases
ii) Resistance in the circuit increases and so the current decreases, which reduces
d) i) 4 Ώ
ii) 4 Ώ
4 a) i) Prevents the current flowing in one direction.
ii) Inverts the input
5 a) i) Group 1
ii) Group 2
c) Region around a changed particle where another changed particle experiences a
6 a) electrons
c) i) D
ii) Produces a bright spot where the electrons hit the stream.
d) Deflects the cathode rays vertically.
e) Otherwise the electrons would collide with air molecules and not reach the screen.
7 a) i) Potential divider.
ii) 12 Ώ
iii) I = V / R = 6 / 12 = 0.5 A
b) i) 1:6V, 2:0V
1 a) proton and neutron
2 α - into page
β - out of page
(symbol) – net deflected
3 a) An electron is removed from the atom by collision with the ionising radiation.
b) α radiation consisted of helium nuclei, which are much more massive than β
radiation, which consists of electrons.
1 a) 0
2 a) 4
3 a) 0
1 ½ life – 1.6 min
b) Take count over one minute without the α source present i.e. the background
count. Take count over one minute with the α source very close to the detector.
Take a count over one minute with the α source 3 cm from the detector. Repeat
each reading two more times.
3 See page 154-156
1 Protons 92; newtons: 146
[note printing error – 192 should be 92]
b) The vast majority of the α particles go straight through undetected, which is
evidence that the atom is mainly space. Very few are deflected straight back,
because they approach a gold nucleus head on. The gold nucleus must contain all
of the positive charge of the atom and therefore almost all of the mass.
3 a) 2 protons and 2 neutrons
b) 1 electron
1 alpha; βeta; gamma; nucleus; gamma; alpha; paper; millimetres; deflected; positive;
4 6 8
b) Measure count with detector opposite beam of β particles with pd switched off.
Switch on pd and take new count. Move detector until a high reading is obtained
closer to the positive plate, indicating that the particles have been deflected.
c) The reduction in count rate when the pd was switched on and the increase in
count rate when the detector was moved closer to the positive plate.
d) The negative β particles are attracted to the positive plate.
2 a) 16 min
ii) Take count over 1 minute without the β source.
Take count over 1 minute with β source.
Take count over 1 minute with β source and aluminium sheet in place.
Repeat each of the three readings twice more.
3 a) 92
b) in orbitals around the nucleus
d) in the nucleus
e) decreases by 3
4 a) i) Time taken for half of the radioactive nucleis to decay.
ii) The radiation in the surroundings due to man-made and natural sources.
b) i) 80–25 = 55 counts/minute
ii) 1:15 counts/minute, 2:40 counts/minute
iii) 28 minutes
iv) 25 counts/minute = background count
5 a) i) 2 protons, 2 neutrons
b) 11 protons, 11 electrons, 13 neutrons
c) i) 0
6 a) strontium –90 since it has the greatest half life.