Question 1
Topic – (a) 1.3 Mass and weight
Topic – (b) 1.2 Motion
Topic – (c) 1.5 Forces (specifically, 1.5.1 Effects of forces)
(a) A rocket has an initial mass of 7.4 × \(10^{6}\) kg.
(i) Calculate the initial weight of the rocket.
weight = …………………………………………………
(ii) Define, in words, the term weight.
(b) Fig. 1.1 shows part of the speed-time graph for the rocket as it leaves the ground and travels into space.
(i) Describe the motion of the rocket:
From O to A ………………………………………………………………………………………………………..
From A to B …………………………………………………………………………………………………………
(ii) Draw a tangent to the graph at time = 400 s and use this to calculate the acceleration of the rocket at this time. Show your working.
acceleration = …………………………………………………
(c) Rockets are used to launch satellites into space. When the satellite is released, the rocket returns to the Earth.
Explain in terms of forces why the rocket reaches terminal velocity as it travels through the atmosphere back to the Earth.
▶️Answer/Explanation
1(a) (i) 7.3 × \( 10^{7}\) N
(ii) (weight is) the gravitational force on a mass / an object (with mass) OR (weight is) the effect of a gravitational field on a mass
1(b) (i) (from O to A) increasing acceleration
(from A to B) constant / uniform acceleration
(ii) tangent drawn at time = 400 s
∆y / ∆x from candidate’s tangent seen AND 17 m / \(s^{2}\) ⩽ acceleration ⩽ 23 m / \(s^{2}\)
1(c) resistive force / air resistance / drag increases as velocity increases until gravitational force is balanced by air resistance (at terminal velocity) OR until resultant / net force is zero (at terminal velocity)
Question 2
Topic – (a) 1.6 Momentum
Topic – (b) 1.6 Momentum
Fig. 2.1 shows a golfer about to hit a golf ball with a golf club. The initial momentum of the golf ball is zero.
(a) Define momentum.
(b) The golf club is in contact with the ball for 5.0 × \(10^{–4}\) s.
The velocity of the golf ball as it leaves the golf club is 41 m / s. The golf ball has a mass of 0.046 kg.
(i) Calculate the impulse on the golf ball.
impulse = …………………………………………………
(ii) Calculate the force applied to the ball by the golf club.
force = …………………………………………………
▶️Answer/Explanation
2(a) (momentum is) mass × velocity
2(b) (i) 1.9 N s OR 1.9 kg m/s
impulse = F∆t = ∆{mv} OR impulse =F∆t = ∆{mv} OR (impulse =) 0.046 × 41
(ii) 3800 N
F=∆p / ∆t OR F= ∆p / (∆)t OR (F=) ∆(mv) / (∆)t OR (F =) impulse / (∆)t OR 1.9 / 0.0005 OR 3.8 × \(10^{N}\) N
Question 3
Topic – (a) 1.7.3 Energy resources
Topic – (b) 1.7.4 Power
Topic – (c) 1.7.3 Energy resources
(a) State two energy resources for which radiation from the Sun is the main source of energy.
(b) A wind turbine is used to generate electricity.
The useful output from the turbine in 1.0 s is 6000 J. The kinetic energy of the wind hitting the turbine in 1.0 s is 11 000 J. The velocity of the wind hitting the turbine is 6.3 m / s.
(i) Show that the mass of air hitting the turbine each second is approximately 550 kg.
(ii) Calculate the efficiency of the turbines. You may assume that all the kinetic energy stored in the wind is transferred to the turbine.
efficiency = ………………………………………………%
(c) Tidal energy and wind energy are both renewable energy resources.
Suggest one reason why tidal energy is a more useful energy resource than wind energy.
Ignore the costs of construction and maintenance.
▶️Answer/Explanation
3(a) any two from:
• fossil fuels / named fossil fuel
• biofuels / named biofuel / biomass
• wave(s)
• wind (turbines)
• Hydroelectric
• solar panels / solar cells
3(b)(i) \(E_{k}\) = ½ \(mv^{2}\)
m = 11000 × 2 ÷ \([6.3]^{2}\) OR m = 11000 × 2 ÷ [39.69] OR 554 kg SEEN
(ii) 55 %
efficiency = useful energy out(put) / (total) energy in(put) × 100 %
OR (6000 ÷ 11000) × 100
3(c) (tides) follow a regular / predictable pattern OR (tides) are reliable ORA for wind
Question 4
Topic – (a) 2.2.3 Melting, boiling and evaporation
Topic – (b) 2.1.3 Gases and the absolute scale of temperature
Fig. 4.1 shows a pressure cooker on an electric heating element. The cooker has a tight-fitting lid.
(a) The pressure cooker is half-full of water. As the water is heated some water evaporates before the water boils.
Describe two differences between evaporation and boiling of the water in the cooker.
(b) As the water is heated, the pressure of the gas inside the cooker increases.
Explain this increase in pressure in terms of particles.
▶️Answer/Explanation
4(a) boiling happens at a specific temperature OR evaporation happens at a range of temperatures OR evaporation happens at any temperature (below the boiling point)
evaporation happens at the surface of the water OR boiling happens throughout the water
4(b) any four from:
1. (as the water is heated) the number of gas particles increases
2. (particles) gain internal / kinetic energy
3. (there are) more frequent collisions between particles and surface / wall / lid of the cooker
4. (each) collision (of particles) is harder / exerts more force (on cooker surface) OR greater change of momentum when particles collide (on cooker surface)
5. pressure ∝ (total) force (of collisions) OR pressure = force / area
Question 5
Topic – (a) 2.3.3 Radiation
Topic – (b) 2.3.2 Convection
Topic – (c) 2.3.2 Convection
Topic – (d) 2.3.3 Radiation
On a sunny day, the temperatures of a black tarmac road and the air above the road increase.
(a) Explain why the surface temperature of the tarmac increases.
(b) State the method of thermal energy transfer from the tarmac to the air immediately above the road.
(c) State the main method of thermal energy transfer from the air immediately above the road to the rest of the air.
(d) Explain why the surface temperature of the tarmac is higher than the surrounding air temperature.
▶️Answer/Explanation
5(a) (tarmac / it) absorbs infrared radiation (emitted from the Sun)
(tarmac / it) absorbs radiation / infrared (emitted from the Sun)
5(b) conduction
5(c) convection
5(d) any two from:
1. black / tarmac is a better absorber (of radiation) than air
2. tarmac is a poor emitter (at low / this temperature)
3. thin layer of tarmac / very large volume / column of air above road
Question 6
Topic – (a) 3.4 Sound
Topic – (b) 3.2.1 Reflection of light
A student plays the violin near the doorway to a large room.
Fig. 6.1 shows a young teacher standing where he can hear the sound but cannot see the student.
(a) (i) State the wave effect that allows the young teacher to hear sounds from the violin at the
position he is standing in Fig. 6.1.
(ii) Calculate the frequency of sound with a wavelength of 0.75 m.
The speed of sound in air is 340 m/s.
frequency = …………………………………………………
(iii) A violin produces sounds in the frequency range 200 Hz–3800 Hz. The width of the open doorway is 0.75 m.
Explain why the young teacher hears the frequency calculated in (a)(ii) clearly but finds a frequency of 3500 Hz much harder to hear.
(b) A plane mirror is placed at point X so that the teacher can see the student.
On Fig. 6.1:
• draw a light ray from the violin to point X and from point X to the teacher
• draw and label the mirror
• add an arrow to the ray to show how the teacher sees the student.
Use a ruler and sharp pencil for this drawing.
▶️Answer/Explanation
6(a) (i) diffraction
(ii) 450 Hz A2
v = f λ OR f = v / λ OR 340 / 0.75
(iii) large diffraction when gap size / doorway is similar to wavelength
high frequency / 3500 Hz has (much) shorter wavelength AND there is less diffraction (with shorter wavelengths)
6(b) ray drawn from the violin to point X AND from point X to the teacher
mirror drawn at point X and labelled AND angle of incidence = angle of reflection
correct arrow on incident ray OR correct arrow on reflected ray
Question 7
Topic – (a) 4.2.3 Electromotive force and potential difference
Topic – (b) 4.2.3 Electromotive force and potential difference
Topic – (c) 4.2.4 Resistance
Topic – (d) 4.3.2 Series and parallel circuits
A washing machine has an electric motor and an electric heater.
Fig. 7.1 shows a simplified circuit diagram for the washing machine.
The heater has a resistance of 25 Ω and the power supply has an electromotive force (e.m.f.) of 230 V.
(a) State the meaning of electromotive force.
(b) State the potential difference (p.d.) across the heater.
p.d. = …………………………………………………
(c) Calculate the current in the heater.
Current = …………………………………………………
(d) The current in the motor is 1.6 A.
Determine the reading on the ammeter in Fig. 7.1. Explain your answer.
▶️Answer/Explanation
7(a) (e.m.f. is the electrical) work done (by a source in) moving a unit charge around a (complete) circuit
(electrical) work done moving a charge
7(b) 230 V
7(c) 9.2 A
R = V / I OR I = V / R OR 230 / 25
7(d) ammeter reading = {1.6 + candidate’s answer to 7(c)} A
Sum of currents into a junction = sum of currents out of junction OR total current is sum of the current in the branches owtte
Question 8
Topic – (a) 3.4 Sound
Topic – (b) 3.4 Sound
Topic – (c) 3.4 Sound
A fisherman uses high frequency sound waves to locate fish in the sea.
Fig. 8.1 shows the sound waves emitted from the boat.
(a) State the name of sound waves which have a frequency greater than 20 kHz.
(b) High frequency sound waves travel from the boat through the sea water.
The speed of sound in water is 1500 m/s. The seabed is 22 m below the boat.
Calculate the time taken for the boat to receive the reflected wave from the seabed after the sound is emitted.
time = …………………………………………………
(c) Fig. 8.2 shows a fish below the boat.
Describe and explain how the reflected sound wave received by the boat from the fish differs from the reflected sound wave received from the seabed.
▶️Answer/Explanation
8(a) ultrasound
8(b) 0.029 s
(distance travelled =) 22 × 2 OR 44 SEEN
v = s / t OR t = s / v OR 44 / 1500
8(c) reflected wave is weaker / has smaller amplitude
fish is small(er) OR only small part of wave reflects off fish OR most of sound goes to seabed
Question 9
Topic – (a) 4.5.3 Magnetic effect of a current
Topic – (b) 4.5.6 The transformer
Topic – (c) 4.2.5 Electrical energy and electrical power
Fig. 9.1 shows a wireless charging plate used to charge the battery in a mobile phone (cell phone).
The coil of wire is part of an electric circuit.
The charging plate is connected to an a.c. power supply. The power supply is turned on.
(a) Describe the magnetic field around the charging plate in terms of its magnitude and direction.
(b) A mobile phone is placed on the charging plate as shown in Fig. 9.2. The coil in the mobile phone is part of a separate circuit that charges the battery.
The coil in the charging plate and the coil in the mobile phone act like a transformer.
(i) Explain why there is a current in the secondary coil shown in Fig. 9.2.
(ii) Suggest why the transformer made from the charging plate and mobile phone is not 100% efficient.
(c) The mobile phone battery can be recharged using this charging plate and stores 4.5 × 104 J of energy when fully recharged. The current in the secondary coil is 0.63 A when the output voltage is 12 V.
(i) Calculate the time taken to fully recharge a completely uncharged battery.
time = …………………………………………………
(ii) Calculate the charge passing through the battery in 60 s.
charge = …………………………………………………
▶️Answer/Explanation
9(a) any two from:
1. (magnitude is) constantly changing owtte
2. (magnetic field is) stronger closer to the coil of wire ORA
3. (the magnetic field is) perpendicular to the a.c. current (producing it)
4. (the magnetic field) changes direction owtte
9(b) (i) any one from:
• secondary coil is in changing / varying magnetic field
• secondary coil is in the magnetic field of primary coil
voltage is induced (in the secondary coil)
(ii) any one from:
• There is no iron core (to strengthen field)
• transformer usually has an iron core to connect the two coils
• some energy transferred to thermal energy (in phone / surroundings)
9(c) (i) 6000 s
E = VIt OR t = E / IV OR 4.5 × \(10^{4}\) / [12 × 0.63]
(ii) Q = 38 C
I = Q / t OR Q= It OR 0.63 × 60
Question 10
Topic – (a) 5.1.2 The nucleus
Topic – (b) 5.2.4 Half-life
Topic – (c) 5.2.5 Safety precautions
Carbon-14 (\(_{6}^{14}\textrm{C}\)) is a radioactive isotope of carbon. Carbon-12 (\(_{6}^{12}\textrm{C}\)) is not radioactive.
(a) Explain how an atom of carbon-14 (\(_{6}^{14}\textrm{C}\)) differs from an atom of carbon-12 (\(_{6}^{12}\textrm{C}\)).
(b) All living organisms contain both carbon-12 atoms and carbon-14 atoms. The ratio of carbon-14 to carbon-12 is 1 : 1 × \(10^{12}\).
Carbon-14 has a half-life of 5700 years.
(i) When an organism dies no new carbon is absorbed. The amount of carbon-12 in the dead organism remains fixed.
Describe how the amount of carbon-14 in the dead organism decreases with time.
(ii) A sample of wood contains carbon-14 to carbon-12 atoms in the ratio 1 : 4 × \(10^{12}\).
Calculate how many years ago the tree died.
(c) Other radioactive isotopes have different half-lives.
Suggest a use of a radioactive isotope with a half-life of one hour.
Explain why a short half-life is suitable for this use.
▶️Answer/Explanation
10(a) (nucleus of carbon-14 contains) more neutrons OR (nucleus of) carbon-12 has fewer neutrons
any one from:
• (atom / nucleus of carbon-14) is heavier
• (atom / nucleus of carbon-14 is) not stable
• (nucleus of carbon-14 contains) two more neutrons
10(b) (i) Every 5700 years the (remaining) carbon-14 decreases by half OR amount of C-14 halved every half life
amount of carbon-14 halves in 5700 years OR amount of carbon-14 decreases at a decreasing rate (with time)
(ii) 11,000 (years ago)
2 × half-life have elapsed
25% of C-14 at time of death is still present in tree OR 75% of C-14 has decayed.
10(c) medical tracers OR medical imaging OR medical diagnosis
any one from:
• keep dose low
• doesn’t stay in body too long
• less damage (to body) OR less harmful (to humans)
Question 11
Topic – (a) 6.1.2 The Solar System
Topic – (b) 6.1.2 The Solar System
Topic – (c) 6.1.2 The Solar System
Topic – (d) 6.1.2 The Solar System
(a) (i) State the name of one planet that has an orbit further away from the Sun than Venus.
(ii) State the name of one planet that has an orbit closer to the Sun than Venus.
(b) Venus has an average radius of orbit of 1.1 × \(10^{11}\) m and an orbital period of 220 Earth days.
Calculate the average orbital speed of Venus. Give your answer in m/s.
average orbital speed = ………………………………………….. m/s
(c) State the relationship between the orbital speeds of the planets and their distances from the Sun.
(d) Comets are balls of ice and dust. Some comets orbit the Sun.
State how the speed of a comet changes as it orbits the Sun.
Explain your answer using ideas about the conservation of energy.
You may include a labelled diagram in your answer.
▶️Answer/Explanation
11(a) (i) Earth / Mars / Jupiter / Saturn / Uranus / Neptune
(ii) Mercury
11(b) 3.6 × \(10^{4}\) m/s
T = 220 × 24 × 60 × 60 OR T = 1.9 × 107 s
v = \(\frac{2πr}{T}\) OR v = \(\frac{2π \times 1.1\times 10^{11}}{220 \times 24\times 60\times 60}\)
11(c) The further away from the Sun the slower the orbital speed / ORA
11(d) 1. any one from:
• comet has an elliptical orbit
• speed of comet is faster when it closer to the Sun
• speed of comet is slower when it is further away from the Sun
2. any one from:
• (conservation of energy requires that) transfers between kinetic and gravitational stores (as comet changes speed)
• total energy remains constant
• energy cannot be created or destroyed
3. as radius of orbit decreases, gravitational energy decreases and kinetic energy increases ORA