Questions 1
(a) Topic – 5.c
(b) Topic-5.c
This question is about the electromagnetic spectrum.
(a) The table gives some statements about the electromagnetic spectrum. Place three ticks () in the table to show which statements are correct.
(b) Electromagnetic waves can be useful, but can also be harmful.
(i) Give one use and one harmful effect of microwaves.
(ii) Give one use and one harmful effect of gamma rays.
▶️Answer/Explanation
Ans
(b) (i) microwaves:
one valid use;
• communication /eq
• heating food /eq one valid harmful effect;
• internal heating (of body tissue) / eq
(ii) gamma rays:
one valid use;
• sterilising {food / medical equipment}
• kill microbes or bacteria;
• treating cancer / radiotherapy;
• medical tracing
one valid harmful effect;
• ionisation / mutation of cells /eq
• risk of cancer
Questions 2
(a) Topic – 9.b
(b) Topic-9.b
(c) Topic-9.b
The photograph shows the International Space Station (ISS) in orbit around the Earth.
(a) The ISS orbits the Earth in a circular orbit. Which of these also orbits the Earth?
A a comet
B Mars
C the Moon
D the Sun
▶️Answer/Explanation
C the moon
(b) Which of these forces causes the ISS to orbit the Earth?
A air resistance
B electrostatic
C friction
D gravitational
▶️Answer/Explanation
D gravitational
(c) The ISS completes one orbit of the Earth in a time period of 93 minutes.
(i) The orbital radius of the ISS is \(6.8×10^3 km\). Calculate the orbital speed of the ISS in km/s.
(ii) Show that the ISS completes approximately 15 orbits of the Earth each day.
▶️Answer/Explanation
Ans
(c) (i) substitution into given formula (v= = 2πr/T);
conversion of minutes to seconds; mark independently evaluation;
e.g.
orbital speed = \(2 × π × 6.8 × 10^3 / 93 (×60)\)
93 minutes = 93 × 60 (= 5580 seconds)
(orbital speed =) 7.7 (km/s)
(ii) successful conversion of orbital period and a day into the same unit;
evaluation of ratio to 15.48… to at least 3 sf;
e.g.
1 day = 24 × 60 = 1440 minutes
1440/93 =15.5
Questions 3
(a) Topic – 3.c
(b) Topic-3.c
(c) Topic-3.b
A model electric motor is used to lift a load through a vertical height.
(a) The load has a mass of 400g and gains 3.2J of energy in its gravitational store when lifted.
(i) State the formula linking gravitational potential energy, mass, gravitational field strength (g) and height.
(ii) Calculate the height the load is lifted.
(iii) State the amount of useful work done on the load by the motor when the load is lifted through this height.
(b) The load is lifted at a constant speed. Diagram 1 shows the lifting force acting on the load as it is lifted. Draw a labelled arrow on diagram 1 to show the other force acting on the load. Ignore the effects of air resistance.
(c) A joulemeter measures the amount of energy transferred electrically to the motor as the motor lifts the load. The joulemeter displays a reading of 11.0J when the load has gained 3.2J of energy in its gravitational store.
(i) Calculate the efficiency of the motor.
(ii) Justify why 7.8J of energy must be dissipated into the thermal store of the surroundings as the load is lifted.
(iii) Diagram 2 is an incomplete Sankey diagram. Complete the Sankey diagram to show the energy transferred by the motor.
▶️Answer/Explanation
Ans
(a) (i) GPE = mass × g × height;
(ii) substitution;
rearrangement; in either order
evaluation;
e.g.
3.2 = 0.40 × 10 × h
h = 3.2 / 0.40 × 10
(h =) 0.80 (m)
(iii) 3.2 (J);
(b) downward arrow labelled “weight”/”W”/”mg”;
vertically downward arrow drawn equal in length to lifting force arrow;
(c) (i) recall of efficiency formula; may be implied from
substitution
substitution;
evaluation;
e.g.
\(efficiency = \frac{useful \ energy \ output}{total \ energy \ output}\)
efficiency = 3.2 / 11.0 (×100%)
efficiency = 0.29 or 29%
(ii) idea that energy must be conserved;
demonstration that 7.8 + 3.2 = 11(.0);
(iii) only one additional arrow drawn pointing to the right;
labelled “useful output (energy)”
width of arrow drawn = 8 small squares;
e.g.,
Questions 4
(a) Topic – 7.b
(b) Topic-7.b
(c) Topic-7.b
(d) Topic-7.c
This question is about magnets.
(a) Which of these substances is not attracted to a bar magnet?
A cobalt
B copper
C iron
D nickel
▶️Answer/Explanation
B copper
(b) Diagram 1 shows a bar magnet.
Draw magnetic field lines on diagram 1 to show the shape and direction of the magnetic field around the bar magnet.
(c) Some bar magnets are made of steel. Explain why steel is a good material for making bar magnets
(d) Diagram 2 shows a cross-section through a wire placed between two magnetic poles. The direction of the current in the wire is out of the page.
(i) Draw an arrow on diagram 2 to show the direction of the force on the wire due to the magnetic field. Assume that the magnetic field is uniform.
(ii) State two changes that could be made that would decrease the magnitude of the force on the wire in diagram 2.
▶️Answer/Explanation
Ans
(b) field line connecting one pole to the other;
at least two complete field lines, but none touching / crossing;
all directions shown on field lines correct (N to S);
(c) steel is magnetic / eq;
(therefore) magnet stays magnetised (for a long period of time) /eq ;
(d) (i) arrow drawn is horizontal;
arrow drawn is to the left;
(ii) Any two from:
MP1 reference to weaker field
MP2 moving magnets further apart
MP3 use weaker magnets
MP4 reference to lower current
MP5 decreasing diameter of wire
MP6 decrease voltage (of supply)
Questions 5
Topic – 5.b
A car is travelling in a straight line along a road. The car passes a person standing at the side of the road.
Before passing the person, the driver of the car presses the car’s horn. The horn makes a loud sound of constant frequency. The horn continues to make a sound until after the car has passed the person. Discuss the differences in the frequencies of the sound heard by
• the driver of the car
• the person at the side of the road
▶️Answer/Explanation
Ans
at least one from:
in relation to driver:
MP1. (frequency) does not change;
MP2. no (relative) movement between driver and horn;
PLUS up to five from:
in relation to person at the side of the road:
MP3. recognition that the Doppler effect applies;
MP4. frequency heard by person at side of the road is different to that heard by driver;
MP5. frequency is higher as car approaches;
MP6. because wavefronts become closer together;
MP7. frequency is lower as car moves away;
MP8. because wavefronts become further apart;
MP9. speed of sound remains constant;
MP10. relevant mention of v = f × λ;
Questions 6
(a) Topic – Appendix 7: Electrical circuit symbols
(b) Topic- 2.b
(c) Topic- 2.b
A student investigates how the current in a 60Ω resistor varies with the voltage across the resistor.
(a) The student has access to this equipment
• 12 V battery
• ammeter and voltmeter
• 60Ω resistor
• variable resistor
• switch
• connecting wires
Draw a circuit diagram to show how the student could connect this equipment to carry out the investigation.
(b) Describe a suitable method the student could use for this investigation.
(c) (i) Complete the current–voltage graph by drawing a line that shows the expected results of the investigation.
(ii) The student repeats their investigation with a 120Ω resistor. Explain how a current–voltage graph for a 120Ω resistor compares with the current–voltage graph for the 60Ω resistor.
▶️Answer/Explanation
Ans
(a) resistor, battery, voltmeter, ammeter all present in a complete circuit
variable resistor connected in series with resistor;
ammeter in series with resistor;
voltmeter in parallel with 60 ohm resistor;
(b) any four from:
MP1. measure voltage and current;
MP2. idea of varying voltage (across resistor);
MP3. take repeat readings and average (at each voltage);
MP4. switch off circuit in between readings;
MP5. other reasonable safety measure relating to equipment heating up
(c) (i) line passes through origin;
line is straight throughout;
line passes/would pass through the point (12,0.20);
(ii) any three from:
MP1. line will be same shape / straight line through origin / both components are resistors;
MP2. line (for 120Ω resistor) will have a lower gradient;
MP3. line (for 120Ω resistor) will have half the gradient;
MP4. (because) larger resistance will result in a lower current in the circuit;
Questions 7
(a) Topic – 8.b
(b) Topic- 8.b
Protactinium is an element with several different radioactive isotopes.
(a) Protactinium-234 has a half-life of 6.7 hours. A sample of protactinium-234 has an initial activity of 800 units.
(i) Give a suitable unit for activity.
(ii) On the axes below, sketch a graph for the decay of the sample of protactinium-234 during its first three half-lives.
(iii) When protactinium-234 undergoes beta \((β^–)\) decay it becomes uranium-234. The incomplete nuclear equation shows this process.
Complete the nuclear equation to show the beta decay of protactinium-234. Write your answers in the dashed boxes.
(b) A student suggests an experiment to determine the type of radiation emitted by a different isotope of protactinium, protactinium-231. This is the suggested method.
Step 1 connect a suitable radiation detector to a radiation counter
Step 2 place a source of protactinium-231 at a fixed distance of 3cm from the radiation detector
Step 3 record the count of detected radiation for a time of one minute
Step 4 place a sheet of paper between the source and detector
Step 5 record the count of detected radiation for a time of one minute
Step 6 repeat Steps 4 and 5 using a sheet of aluminium and then a sheet of lead instead of the sheet of paper
The table shows the results of the investigation when it is done by a teacher.
(i) Which of these is the dependent variable in the investigation?
A count measured by the detector
B distance between source and detector
C material between source and detector
D time the count is measured
(i) A (count measured by the detector);
(ii) The student’s method does not allow for background radiation. Describe how the student’s method should be modified to allow for background radiation.
(iii) Describe how the student’s method could be modified to improve the reliability of the results.
(iv) Evaluate the data from the experiment to conclude the type of radiation emitted by protactinium-231.
▶️Answer/Explanation
Ans
(a) (i) becquerel(s);
(ii) evidence that sketch starts at (0,800)
evidence sketch passes through (6.7,400)
smooth curve decreases with decreasing steepness
(iii) both numbers for beta correct;
atomic number of protactinium = 91;
(b) (ii) idea of removing source (from the experiment);
measure count(for a minute);
subtract background count from results;
(iii) idea of repeating measurements (of count);
to determine a mean value;
(iv) count decreases (significantly) using paper;
no (additional) effect on the count when using aluminium AND lead / eq;
radiation must be alpha consistent with candidate’s discussion;
Questions 8
(a) Topic – 1.c
(b) Topic-1.b
Diagram 1 shows a set of masses attached to a spring, which is suspended from a support.
(a) After the masses are added, the length of the spring is 14.6cm. The student measures the extension of the spring as 11.5cm.
(i) Calculate the original length of the spring.
(ii) The student removes the masses and notices that the spring does not show elastic behaviour. Predict a value for the new length of the spring after the masses have been removed.
(b) The student puts the masses back on the spring. The student then pulls the masses down and releases them. The masses vibrate up and down in a vertical direction, as shown in diagram 2.
The distance–time graph shows how the distance between the top of the masses and the support changes with time as the masses vibrate.
(i) Explain how the gradient of the graph shows that the masses accelerate as they vibrate.
(ii) Add crosses (X) to the distance–time graph to show all the times when the masses are not moving.
▶️Answer/Explanation
Ans
(a) (i) 3.1 (cm);
(ii) any value above candidate’s answer for (a)(i) up to and including 14.6cm;
(b) (i) idea that speed is the gradient/slope of the graph;
gradient is not constant;
(therefore) speed is not constant;
(ii) any cross drawn at a peak/trough on the curve;
crosses drawn at all three peaks and all three troughs;
Questions 9
(a) Topic – 1.b
(b) Topic-1.c
(c) Topic-1.c
The driver of a racing car makes a pit stop during a race to change the tyres on the racing car. The area where the tyres are changed is called the pit lane.
(a) Before entering the pit lane, the speed of the car must decrease for safety reasons.
(i) The mass of the racing car is 830kg. The maximum braking force is 41000N. Show that the maximum deceleration of the racing car is approximately 50m/s2.
(ii) The racing car is travelling at an initial speed of 72m/s. Calculate the minimum distance needed to decrease the speed of the racing car from 72m/s to 26m/s.
(b) The racing car slows down using its brakes. The brakes work using friction. The brakes become very hot when the racing car slows down. Using ideas about energy, explain why the brakes become hot.
(c) The tyres of the racing car also get very hot during a race. A mechanic has to handle the hot tyres during the pit stop. They wear protective gloves which have several layers of insulating materials. Explain how the layers of insulating materials in the gloves reduce the risk of the mechanic burning their hands on the hot tyres.
▶️Answer/Explanation
Ans
(a) (i) recall of (unbalanced) force = mass × acceleration;
substitution and rearrangement;
evaluation to 2 s.f. or more;
e.g.
F = m × a
a = 41000 / 830
a = 49 (m/s²)
(ii) substitution into \(v^2 = u^2 + 2as\);
rearrangement;
evaluation;
e.g.
\(26^2 = 72^2 + 2×(-50)×s\)
(distance =) 5184-676 / 100
(distance =) 45 (m)
(b) kinetic energy (store) of car decreases;
thermal energy (store) of brake(s) increases;
energy transferred mechanically;
(c) any two from:
MP1. idea that insulating materials are poor conductors;
MP2. layers trap air;
MP3. air itself is a poor conductor/(good) insulator
MP4. (energy transfer due to / rate of) conduction reduces;
MP5. idea increased thickness reduces (rate of) conduction
Questions 10
(a) Topic – 4.b
(b) Topic-4.b
(c) Topic-4.b
A dam is a structure designed to hold water in a reservoir.
(a) The water in the reservoir has a depth of 35m.
(i) State the formula linking pressure difference, height, density and g.
(ii) Atmospheric pressure at the surface of the reservoir is 100kPa. Calculate the total pressure at the bottom of the reservoir. [for water, density = 1000kg/m³]
(b) An underwater camera is used in the water reservoir. The camera lens experiences a force of 430N at a pressure of 260kPa.
(i) State the formula linking pressure, force and area.
(ii) Calculate the area of the camera lens. Give a suitable unit.
(c) Sea water has a density of 1030 kg/m³. Explain how the design of the dam would need to be changed to hold the same depth of sea water safely.
▶️Answer/Explanation
Ans
(a) (i) pressure difference = height × density × g;
(ii) substitution;
evaluation of pressure difference in kPa;
evaluation of total pressure by adding 100 (kPa);
e.g.
(pressure difference =) 35 × 1000 × 10
(pressure difference =) 350 (kPa)
(pressure = 350 + 100 =) 450 (kPa)
(b) (i) pressure = force ÷ area;
(ii) substitution;
rearrangement;
evaluation;
corresponding unit of area;
e.g.
260,000 = 430 / area
(area =) 430 / 260 000
(area =) 0.0017 m²
(c) pressure (at bottom) is greater than before / eq;
wider base /eq;
Questions 11
(a) Topic – 9.b
(b) Topic-9.b
The gravitational field strength of a planet decreases with increasing distance from the planet. The table shows the value of the gravitational field strength of Mars at different distances from the centre of Mars.
(a) A student finds this formula in a textbook, which links distance from the centre of a planet to its gravitational field strength
gravitational field strength × distance² = constant
Use data from the table to justify this formula.
(b) Olympus Mons is the tallest mountain on Mars. The distance between the centre of Mars and the peak of Olympus Mons is 3410km. Calculate the gravitational field strength at the peak of Olympus Mons.
▶️Answer/Explanation
Ans
(a) substitution into given formula;
evaluation of constant;
evaluation of constant for a second set of data;
conclusion consistent with candidate’s evidence;
e.g. calculated value of constant doesn’t change (much) so formula is justified constant decreases so formula isn’t justified
(b) rearrangement of given formula;
substitution of constant and distance;
evaluation; condone 3.7
e.g.
gravitational field strength = constant / distance²
gravitational field strength = 42 700 000 / 3410²
gravitational field strength = 3.67 (N/kg)