Question 1
(a) Topic – 1.1 Physical quantities and measurement techniques & 1.6 Momentum
(b) Topic – 1.6 Momentum
(a) (i) State the difference between a scalar quantity and a vector quantity.
(ii) Define momentum.
(b) A test car crashes into a barrier to test the safety features. The test car has a total mass of 950 kg. It is moving with constant velocity from time t = 0 for 4.0 s. At t = 4.0 s, the car hits the barrier.
Fig. 1.1 shows the car as it hits the barrier.
(i) During the test crash, the resultant force acting on the car is 27 000 N. The car takes 1.5 s to come to rest. The deceleration is uniform.
Calculate the initial velocity of the car.
initial velocity = …………………………………………………
(ii) On Fig. 1.2, sketch a speed–time graph to show the motion of the car from time t = 0 until the car becomes stationary.
▶️Answer/Explanation
1(a) (i) (scalar) does not have direction or vector has direction
(ii) momentum = mass × velocity
1(b)(i) 43 m/s
FΔt = Δ(mv) OR F = Δ(mv) / Δt OR F = ma OR a = ∆v / ∆t
u= Ft / m OR u = (27 000 × 1.5) / 950
(ii) straight horizontal line from y axis extending to t = 4.0 s
straight line from t = 4.0 s with negative gradient meeting the x axis (less than half-way between 4.0 and the end of the time axis)
Question 2
(a) Topic – 1.5.1 Effects of forces
(b) Topic – 1.5.1 Effects of forces
(a) Describe an experiment to determine the spring constant of a spring.
State:
- the apparatus you need
- details of how to take measurements
- how to calculate the spring constant
You may use the space below to draw a labelled diagram as part of your answer.
(b) Fig. 2.1 shows a baby in a baby bouncer. The baby bouncer consists of a holder suspended from a spring. The baby pushes his feet on the ground and bounces gently up and down.
(i) Two springs Q and R are tested to determine their spring constants.
Each spring is tested up to its limit of proportionality.
Define ‘limit of proportionality’.
(ii) Table 2.1 shows the results of the tests.
The total weight of the baby and the holder is 120 N.
Calculate the extension of each spring for this weight.
extension of spring Q = ………………………………………………………
extension of spring R = ………………………………………………………
(iii) The unstretched length of each spring is 25 cm.
State and explain which spring would be more suitable for the baby bouncer in Fig. 2.1.
▶️Answer/Explanation
2(a) hang mass / weight on the bottom of the spring
use a ruler / metre rule AND measure / calculate extension
use of k = F / x to calculate the spring constant
OR k is the gradient of load-extension graph
any one of:
• repeat measurements using different masses / weights and plot a load against extension graph
• measure final length and initial length of spring and subtract to obtain (magnitude of) extension / change in length
• use a pointer on the bottom of spring to obtain more accurate measurements on ruler
2(b)(i) maximum load that can be applied when the extension is proportional to load
OR the maximum force up to which the extension is proportional to the load
(ii) extension of spring Q = 15 cm or 0.15 m
AND extension of spring R = 110 cm or 1.1 m
(iii) Q AND extension produced by Q is the right length to allow baby to just reach the floor with feet
OR Q AND extension produced by R is too big (for gentle bounces in the space provided) OR R would make the baby collapse on floor (so unsafe) OR bouncer would have to be hung higher if R was used
Question 3
(a) Topic – 2.3.3 Radiation
(b) Topic – 1.7.3 Energy resources
(c) Topic – 2.2.2 Specific heat capacity
Fig. 3.1 shows a portable shower used on a campsite. The bag is filled with water. The water is heated using infrared radiation from the Sun.
(a) (i) Explain why the shower bag is painted black.
(ii) Explain a disadvantage of radiation from the Sun being the only source to heat the water.
(b) Solar energy is a renewable energy resource.
State two other renewable energy resources.
(c) During the day, the Sun shines on the shower bag and some of the energy in the infrared radiation from the Sun transfers to the thermal energy stores of the water.
The water absorbs 60% of the energy incident on the bag. The temperature of the water rises from 10°C to 43°C.
The mass of the water in the bag is 40 kg. The specific heat capacity of water is 4200 J / (kg °C).
Calculate the energy incident on the shower bag during the day.
Show your working.
energy = …………………………………………………
▶️Answer/Explanation
3(a)(i) good/better absorber (of radiation) OR bad / poor / worse reflector (of radiation)
(ii) doesn’t work at night / in cloud cover / when there is no sun OR (sun has) variable output
3(b) any two from:
• hydroelectric
• tidal
• wave
• wind
• geothermal
• biofuels
3(c) 9.2 × \(10^{6}\) J OR 9 200 000 J
temperature rise = 33 (°C) OR 43 – 10
c = ∆E / m∆Θ OR E = m × c × ∆Θ OR E = 40 × 4200 × (43 – 10) OR 5.5 × \(10^{6}\) (J)
5.5 × \(10^{6}\) × (100 / 60) OR 5.5 × \(10^{6}\) × 1.667 OR 5.5 × \(10^{6}\) / 0.6
OR efficiency = \( \frac{usefulenergyoutput}{total energy input}\times 100%\)
Question 4
(a) Topic – 3.2.2 Refraction of light
(b) Topic – 3.2.2 Refraction of light
(a) Fig. 4.1 shows a ray of light as it enters the side of a plastic block. The ray of light passes from air into the plastic.
(i) State how the speed, wavelength and frequency of the wave in the plastic block compare with their values in the air.
(ii) Show that the refractive index of the plastic is 1.4.
Show your working.
(iii) Calculate the critical angle for the plastic.
critical angle = …………………………………………………
(b) Fig. 4.2 shows the same plastic as in (a) used to make an optical fibre. A ray of light is passing along the fibre.
(i) Carefully continue the ray of light P until it reaches the other end of the fibre.
(ii) State two uses for optical fibres.
▶️Answer/Explanation
4(a)(i) any two correct = 1 mark
all three correct = 2 marks
• speed decreases / gets less / slower
• wavelength decreases / gets smaller / shorter
• frequency unchanged / stays the same / no effect
(ii) n = sin 45 / sin 30
(iii) 46° OR 45°
n = 1 / sin c OR c = sin–1 (1 / n) OR c = sin–1 (1 / 1.4) OR sin c = 1 / n
4(b) (i) total internal reflection AND i = r for initial reflection ray reaches end of fibre after a total of 2 or 3 reflections only
(ii) any two from:
• internet transmission / (highspeed) broadband
• telephone networks
• cable TV
• endoscope
• lasers in surgery
• microscopy
• military aircraft wiring
• imaging (cameras) in industry
• inspection of pipes or other hard to reach places
Question 5
(a) Topic – 4.2.1 Electric charge
(b) Topic – 4.2.1 Electric charge
Fig. 5.1 shows a metal sphere S. The sphere has been charged with a negative charge.
(a) (i) There is an electric field around sphere S.
On Fig. 5.1, draw four field lines to show the pattern of the field and indicate the direction of the field with arrows on the lines.
(ii) Fig. 5.2 shows a position X next to sphere S.
A small negatively charged particle is placed at position X.
State the direction of the force on the negatively charged particle at X due to the electric field around sphere S.
(iii) The negatively charged particle at X is released from rest.
Describe the motion of the small negatively charged particle due to the electric field around sphere S.
(b) Fig. 5.3 shows sphere S being spray painted. Sphere S is negatively charged. As the paint particles exit the wide nozzle of the paint sprayer, they become charged with a positive charge.
(i) Explain why the paint particles spread out when they leave the nozzle.
(ii) The sphere can be painted by hand using a paintbrush.
Suggest and explain one advantage to using charged paint from a spray gun to paint sphere S.
▶️Answer/Explanation
5(a) (i) four evenly spaced radial lines touching S AND no lines inside S
at least one arrow pointing towards S and none incorrect
(ii) to the left OR away from the sphere / S
(iii) accelerates (due to a resultant force)
(moves) away from the (centre of the) sphere
5(b) (i) like charges repel
(ii) advantage: more even coat or less chance of paint particles clumping together or less chance of creating thicker patches
explain: positive / paint particles are attracted (by the same amount) to all parts of the negative sphere so they spread out or paint creates a fine mist so spreads out (over the surface)
OR advantage: more durable or better adhesion or cheaper or high transfer efficiency or less paint wasted or fewer drips or less paint sag or less overspray on floor / walls
explain: the positive/paint is attracted to the negative sphere / charges and forms a strong(er) bond
OR advantage: quicker to paint
explain: all surface is painted at the same time without having to turn the sphere / go around the other side owtte
Question 6
(a) Topic – 2.2.3 Melting, boiling and evaporation
(b) Topic – 4.3 Electric circuits
(a) A car windscreen is covered in condensation (small droplets of water). Thermal energy is used to remove the droplets of water. The thermal energy is provided by three resistors on the windscreen.
Fig. 6.1 shows two possible circuits for the three resistors.
The three resistors are identical.
(i) Describe two advantages of using Circuit B.
(ii) Describe, in terms of the water particles, the process by which the water droplets are removed from the car windscreen using the heater.
(b) Fig. 6.2 shows a circuit containing two resistors, P and Q. The circuit is powered by a 12 V battery.
(i) Calculate the current in resistor Q.
current = …………………………………………………
(ii) Calculate the energy transferred electrically when the current calculated in (b)(i) is present in resistor Q for 5 minutes.
energy = …………………………………………………
(iii) Energy is transferred from the battery by the electrical current.
State the energy store in the battery.
(iv) Calculate the total resistance of the circuit.
total resistance = …………………………………………………
▶️Answer/Explanation
6(a) (i) any two from:
• if one resistor fails / breaks the others will still work
• each resistor gets the full voltage / 12 V or lower (total) resistance or higher current
• higher power
(ii) any two from:
• evaporation / water evaporates
• energy (from heater) transfers to particles OR particles gain energy (from the heater)
• more energetic particles escape (from water / droplet)
• particles leave from the surface (of the water / droplet)
6(b) (i) 0.17 A
R = V / I OR I = V / R OR I = 12 / 70
ii) 610 J OR 620 J
E = I × V × t OR E = V × I × t OR E = 12 × 0.17 × 300
300 (s) OR 5 × 60
(iii) chemical
(iv) 39 Ω
1 / \(R_{T}\) = 1 / \(R_{1}\) + 1 / \(R_{2}\) OR 1 / \(R_{T}\) = 1 / 90 + 1 / 70 OR \(R_{T}\) = 1 / (1 /\(R_{1}\)+ 1 / \(R_{2}\)) OR \(R_{T}\) = 1 / (1 / 90 + 1 / 70)
Question 7
Topic – (a) 1.5.2 Turning effect of forces
Topic – (b) 4.5.3 Magnetic effect of a current
Fig. 7.1 shows a barrier at the entrance to a car park. The wooden barrier arm has a weight of 60 N which acts through the centre of gravity at the position shown on Fig. 7.1.
(a) Initially the wooden barrier arm is horizontal.
(i) Using Fig. 7.1, calculate the clockwise moment of the weight of the wooden arm about the pivot.
clockwise moment = …………………………………………… Nm
(ii) The wooden barrier arm is in equilibrium. The mass of the soft iron bar A is 23 kg.
Calculate the distance d between the pivot and the joint holding the soft iron bar A.
distance d = …………………………………………………
(b) Fig. 7.2 shows a coil attached to a power supply placed below the soft iron bar A.
(i) State and explain what happens to the wooden barrier arm when the switch in the coil circuit is closed.
(ii) The switch is opened. An operator decreases the potential difference across the coil and the switch is closed.
State and explain how the effect on the wooden barrier arm compares with the effect in (b)(i).
(iii) A student suggests that the soft iron bar A is replaced by a steel bar. Explain why a steel bar is less effective than a soft iron bar in the barrier.
▶️Answer/Explanation
7(a) (i) 100 (Nm)
(ii) 0.44 m OR 0.45 m
total clockwise moment = total anticlockwise moment
d = 100 / 225 OR d = 100 / (23 × 9.8) OR 100 / (23 × g × d)
7(b) (i) (barrier arm) rotates anticlockwise OR (RHS of barrier arm) moves upwards
(when the switch is closed) coil produces a magnetic field OR coil / core becomes an (electro)magnet
(coil/core/electromagnet) attracts (iron) bar (downwards)
(ii) statement: (barrier arm) moves slower / goes up (more) slowly
explanation: decreases strength of (magnetic) field / smaller force / smaller moment
(iii) steel would become permanently magnetised
(so when switch is opened) barrier would stay up or bar A will stay attracted (to soft iron core)
Question 8
Topic – (a) 5.2.3 Radioactive decay
Topic – (b) 5.2.2 The three types of nuclear emission
Topic – (c) 5.2.5 Safety precautions
An isotope of boron is used in the treatment of cancer in the brain.
Boron sticks to cancer cells in the brain.
(a) The isotope of boron is bombarded with neutrons then undergoes fission to form lithium and alpha‑particles.
(i) Describe one difference between fission and fusion.
(ii) A nucleus of boron (B) contains 5 protons and 5 neutrons. Complete the nuclide equation for this fission reaction.
(b) The alpha‑particles destroy the cancer cells. Suggest and explain one reason why alpha particles are more suitable than gamma radiation for use in this treatment of brain cancer.
(c) Other cancers are treated with gamma radiation. Describe one safety precaution a nurse or radiologist takes during this treatment.
▶️Answer/Explanation
8(a) (i) fission is splitting of nuclei OR fusion is the joining of nuclei
(ii) \(_{5}^{10}\textrm{B} + _{0}^{1}\textrm{n}\to _{3}^{7}\textrm{Li}+_{2}^{4}\textrm{α}\)
B: proton number 5 and nucleon number 10
Li: proton number 3 and nucleon number 7
α: proton number 2 and nucleon number 4
8(b) alpha is less penetrating / has shorter range (than gamma)
alpha more easily absorbed / stopped by cancer / tumour cells or won’t travel beyond cancer cells or so won’t damage other / healthy cells
OR
alpha highly ionising (than gamma) will destroy / damage cancer cells more easily
8(c) any one from:
• reduce exposure time
• increase distance between source and living tissue or stand in another room whilst radiation is emitted
• use shielding or wear a lead apron or stand behind a glass partition / lead barrier
Question 9
Topic – (a) 3.1 General properties of waves
Topic – (b) 3.3 Electromagnetic spectrum
Topic – (c) 3.3 Electromagnetic spectrum
(a) Fig. 9.1 shows a diagram of a transverse wave.
From Fig. 9.1, identify all the lengths which represent one wavelength.
(b) Hydrogen in a very distant galaxy emits electromagnetic radiation which is observed on the Earth.
Scientists on the Earth measure the wavelength of the radiation from the very distant galaxy.
The wavelength is 918 nm.
On the Earth, hydrogen in the laboratory emits electromagnetic radiation of wavelength 656 nm.
Name the effect that the scientists observe and state what this shows about the very distant galaxy.
(c) Table 9.1 shows a wavelength of electromagnetic radiation from hydrogen observed in the laboratory and from three galaxies.
The galaxies are at different distances from the Earth.
Describe what Table 9.1 shows about the motions of the galaxies and state what this suggests is happening to the Universe.
▶️Answer/Explanation
9(a) Q and V
9(b) red shift
(galaxy) moving away / receding (from Earth/us)
9(c) galaxies further away (are receding) with higher speed OR galaxies further away have greater redshift
Universe is expanding
Question 10
Topic – (a) 6.2.2 Stars
Topic – (b) 6.2.3 The Universe
(a) Stars more massive than the Sun can eventually form black holes.
Describe how a black hole can be formed from a more massive star.
(b) The star system V404 Cygni contains a black hole. The system is approximately 7800 light‑years from the Earth.
(i) Describe what is meant by a light‑year.
(ii) Calculate the approximate distance from V404 Cygni to the Earth in km.
distance = …………………………………………… km
▶️Answer/Explanation
10(a) any two from:
• (most of the) hydrogen has been converted to helium or hydrogen begins to run out
• (star expands and) forms a red supergiant
• (red supergiant) explodes as a supernova (forming a nebula)
core (of red supergiant collapses and) forms black hole OR a black hole forms at the centre (of the supernova/nebula)
10(b) (i) the distance travelled (in the vacuum of space) by light in one year
(ii) 7.4 × \(10^{16}\) km
9.5 × \(10^{15}\) (m) or 9.5 × \(10^{12}\) (km)