CIE AS/A level Physics Paper 4 Prediction - 2025

CIE AS/A level Physics Paper 4 Prediction- 2025

To excel in A level Physics Exam, consistent practice with CIE AS & A Level Revision resources is key. CIE AS/A level Physics Paper 4 Prediction will guide you for exam pattern.

IITian Academy offers a vast collection of questions that can aid your understanding of specific topics and solidify your concepts. By practicing regularly and focusing on these key areas, you’ll be well-prepared for the AS/A level Physics exam

Question 1

An object is suspended from a vertical spring as shown in Fig. 3.1.

The object is displaced vertically and then released so that it oscillates, undergoing simple harmonic motion.
Fig. 3.2 shows the variation with displacement $x$ of the energy $E$ of the oscillations.

The kinetic energy, the potential energy and the total energy of the oscillations are each represented by one of the lines $P, Q$ and $R$.
(a) State the energy that is represented by each of the lines P, Q and R.
(b) The object has a mass of $130 \mathrm{~g}$. Determine the period of the oscillations.

(c) (i) State the cause of damping.
(ii) A light card is attached to the object. The object is displaced with the same initial amplitude and then released. During each complete oscillation the total energy of the system decreases by 8.0% of the total energy at the start of that oscillation. Determine the decrease in total energy, in mJ, of the system by the end of the first 6 complete oscillations.
(iii) State, with a reason, the type of damping that the card introduces into the system.

▶️Answer/Explanation

Ans:

P: total energy
Q: potential energy
R: kinetic energy

(b) $E=1 / 2 m \omega^2 x_0^2$ or $E=1 / 2 m v_0^2$ and $v_0=\omega x_0$

$\begin{aligned} & 6.4 \times 10^{-3}=1 / 2 \times 0.130 \times \omega^2 \times 0.015^2 \\ & \left(\omega^2=438\right) \\ & (\omega=20.9)\end{aligned}$

$T=2 \pi / \omega$

$\begin{aligned} & =2 \pi / 20.9 \\ & =0.30 \mathrm{~s}\end{aligned}$

(c)(i) resistive forces

(c)(ii) $0.92^6$

$\begin{aligned} \text { decrease in energy } & =6.4-\left(6.4 \times 0.92^6\right) \\ & =2.5 \mathrm{~mJ}\end{aligned}$

(c)(iii) light damping because the amplitude of oscillations gradually reduces
or
light damping because the system still oscillates

Question 2

(a) With reference to velocity and acceleration, describe uniform circular motion.
(b) Two cars are moving around a horizontal circular track. One car follows path X and the other follows path Y, as shown in Fig. 1.1.

The radius of path X is 318m. Path Y is parallel to, and 27m outside, path X. Both cars have mass 790kg. The maximum lateral (sideways) friction force F that the cars can experience without sliding is the same for both cars.

(i) The maximum speed at which the car on path X can move around the track without sliding is 94ms^–1.
Calculate F

(ii) Both cars move around the track. Each car has the maximum speed at which it can move without sliding.
Complete Table 1.1, by placing one tick in each row, to indicate how the quantities indicated for the car on path Y compare with the car on path X

Answer/Explanation

Ans

(a) constant speed or constant magnitude of velocity
acceleration (always) perpendicular to velocity

(b) (i) $F = mv^2 / r$
or
v = rω and F = mrω²
F = 790 × 94² / 318
= 22000 N
(b) (ii) centripetal acceleration: same
maximum speed: greater
time taken for one lap of the track: greater

Question 3

(a) Define the tesla.

(b) A stiff metal wire is used to form a rectangular frame measuring $8.0cm × 6.0cm$. The frame is open at the top, and is suspended from a sensitive newton meter, as shown in Fig. 8.1.

The open ends of the frame are connected to a power supply so that there is a current of 5.0A in the frame in the direction indicated in Fig. 8.1. The frame is slowly lowered into a uniform magnetic field of flux density B so that all of side PQ is in the field. The magnetic field lines are horizontal and at an angle of 50° to PQ, as shown in Fig. 8.2.

When side PQ of the frame first enters the magnetic field, the reading on the newton meter changes by 1.0mN.

(i) Determine the magnetic flux density B, in mT.

(ii) State, with a reason, whether the change in the reading on the newton meter is an increase or a decrease.

(iii) The frame is lowered further so that the vertical sides start to enter the magnetic field. Suggest what effect this will have on the frame.

Answer/Explanation

Ans

(a) newton per ampere per metre
where current/wire is perpendicular to magnetic field

(b) (i) F = BILsin θ
B = 1.0 / (5.0 × 0.060 × sin 50°)
= 4.4 mT

(b) (ii) (from Fleming’s left-hand rule) force on wire is upwards, so reading decreases

(b) (iii) frame will rotate (so that PQ becomes perpendicular to the field)

Question 4

(a) State what is meant by an ideal gas.

(b) A fixed amount of helium gas is sealed in a container. The helium gas has a pressure of $1.10 \times 10^5 \mathrm{~Pa}$, and a volume of $540 \mathrm{~cm}^3$ at a temperature of $27^{\circ} \mathrm{C}$.

The volume of the container is rapidly decreased to $30.0 \mathrm{~cm}^3$. The pressure of the helium gas increases to $6.70 \times 10^6 \mathrm{~Pa}$ and its temperature increases to $742^{\circ} \mathrm{C}$, as illustrated in Fig. 2.1.

No thermal energy enters or leaves the helium gas during this process.
(i) Show that the helium gas behaves as an ideal gas.
(ii) The first law of thermodynamics may be expressed as

$
\Delta U=q+W .
$

Use the first law of thermodynamics to explain why the temperature of the helium gas increases.[2]
(iii) The average translational kinetic energy $E_{\mathrm{K}}$ of a molecule of an ideal gas is given by
$
E_{\mathrm{K}}=\frac{3}{2} k T
$
where $k$ is the Boltzmann constant and $T$ is the thermodynamic temperature.Calculate the change in the total kinetic energy of the molecules of the helium gas.

(c) The mass of nitrogen gas in another container is $24.0 \mathrm{~g}$ at a temperature of $27^{\circ} \mathrm{C}$. The gas is cooled to its boiling point of $-196^{\circ} \mathrm{C}$. Assume all the gas condenses to a liquid. For this change the specific heat capacity of nitrogen gas is $1.04 \mathrm{~kJ} \mathrm{~kg}^{-1} \mathrm{~K}^{-1}$. The specific latent heat of vaporisation of nitrogen is $199 \mathrm{~kJ} \mathrm{~kg}^{-1}$. Determine the thermal energy, in $\mathrm{kJ}$, removed from the nitrogen gas.

▶️Answer/Explanation

Ans:

(a) gas for which $p V \propto T$

where T is thermodynamic temperature A1

(b)(i) evidence of two temperature conversions between ${ }^{\circ} \mathrm{C}$ and $\mathrm{K}$

two calculations shown, one for each state e.g.

$\frac{1.10 \times 10^5 \times 540 \times 10^{-6}}{(273+27)}=0.198$ and $\frac{6.70 \times 10^6 \times 30 \times 10^{-6}}{(273+742)}=0.198$

(b)(ii) work is done on the gas
internal energy increases (so temperature increases)

(b)(iii) $\begin{aligned} & p V=N k T \text { e.g. } \\ & \begin{aligned} N & =\frac{1.10 \times 10^5 \times 540 \times 10^{-6}}{1.38 \times 10^{-23} \times 300} \\ & =1.435 \times 10^{22}\end{aligned} \\ & \Delta E_k=(3 / 2) k \Delta T N\end{aligned}$

$=(3 / 2) \times 1.38 \times 10^{23} \times(742-27) \times \frac{1.10 \times 10^5 \times 540 \times 10^{-6}}{1.38 \times 10^{-23} \times 300}$

= 212 J

$\Delta \theta=(27+196)$ or 223

$\begin{aligned} E & =0.0240 \times 1.04 \times(27+196)+0.0240 \times 199 \\ & =10.3 \mathrm{~kJ}\end{aligned}$

Question 5

(a) Two long straight wires P and Q are parallel to each other, as shown in Fig. 8.1. There is a current in each wire in the direction shown.
The pattern of the magnetic field lines in a plane normal to wire P due to the current in the wire is also shown.

(i) Draw arrows on the magnetic field lines in Fig. 8.1 around wire P to show the direction of the field.
(ii) Determine the direction of the force on wire Q due to the magnetic field from wire P.
(iii) The current in wire Q is less than the current in wire P. State and explain whether the magnitude of the force on wire P is less than, equal to, or greater than the magnitude of the force on wire Q.

(b) Nuclear magnetic resonance imaging (NMRI) is used to obtain diagnostic information about internal structures in the human body.
Radio waves are produced and directed towards the body. The radio waves affect the protons within the body.
(i) Explain why radio waves are used.
(ii) Explain why the radio waves are applied in pulses.

Answer/Explanation

Ans

(a) (i) at least one anticlockwise arrow and no clockwise arrows
(a) (ii) (force is to the) left
(a) (iii) force is the same
Newton’s third law (of motion)
or force depends on the product of the two currents

(b) (i) frequency of radio waves is equal to natural frequency of protons
resonance of protons occurs / protons absorb energy
(b) (ii) in between pulses / when pulse stops
Any 1 from:
• protons de-excite
• protons emit r.f. pulses
• emitted (r.f.) pulse (from proton) detected

Question 6

(a) State what is represented by an electric field line.

(b) Two point charges P and Q are placed $0.120$ m apart as shown in Fig. 4.1.

(i) The charge of P is $+4.0$ nC and the charge of Q is $–7.2$ nC. Determine the distance from P of the point on the line joining the two charges where the electric potential is zero.

(ii) State and explain, without calculation, whether the electric field strength is zero at the same point at which the electric potential is zero.

(iii) An electron is positioned at point X, equidistant from both P and Q, as shown in Fig. 4.2.

On Fig. 4.2, draw an arrow to represent the direction of the resultant force acting on the electron.

Answer/Explanation

Ans:

(a) direction of force

force on a positive charge

(b) (i) $V$ = $\frac{Q}{4\pi \varepsilon _{0}r}$

$\frac{4.0 \times 10^{-9}}{4\pi \varepsilon _{0}x} + \frac{-7.2\times 10^{-9}}{4\pi \varepsilon _{0}(0.120 – x)}$

$4(0.120 – x)$ = $7.2$ x

x = $0.043$ m

(ii) fields are in the same direction so no

(iii) straight arrow drawn leftwards from X in direction between extended line joining Q and X and the horizontal

Question 7

(a) A Hall probe is placed in a magnetic field. The Hall voltage is zero. The Hall probe is rotated to a new position in the magnetic field. The Hall voltage is now maximum. Explain these observations.
(b) The formula for calculating the Hall voltage $V_H$ as measured by a Hall probe is

$
V_{\mathrm{H}}=\frac{B I}{n t q}

Table 6.1 shows the value of $n$ for two materials.

(i) State the meaning of n
(ii) Explain why a Hall probe is made from silicon rather than copper

(c) A Hall probe gives a maximum reading of $24 \mathrm{mV}$ when placed in a uniform magnetic field of flux density $32 \mathrm{mT}$.

The same Hall probe is then placed in a magnetic field of fixed direction and varying flux density. The Hall probe is in a fixed position so that the angle between the Hall probe and the magnetic field is the same as when the Hall voltage was $24 \mathrm{mV}$.

The variation of the reading $V_{\mathrm{H}}$ on the Hall probe with time $t$ from time $t=0$ to time $t=8.6 \mathrm{~s}$ is shown in Fig. 6.1.

A coil with 780 turns and a diameter of 3.6cm is placed in this varying magnetic field. The plane of the coil is perpendicular to the field lines.
Calculate the magnitude of the maximum electromotive force (e.m.f.) induced in the coil in the time between t = 0 and t = 8.6s.

▶️Answer/Explanation

Ans:

(a) it is zero when (plane of) probe is parallel to the (magnetic) field (lines) it is maximum when (plane of) probe is perpendicular to (magnetic) field (lines) B1

(b)(i) number density of charge carriers

(b)(ii) smaller value of $n$ so greater Hall voltage / $V_H$

use of 1.4 s or (8.6 – 7.2) s

$E=\Delta B A N / \Delta t$

$\begin{aligned} & =\frac{48 \times 10^{-3} \times 0.018^2 \times \pi \times 780}{1.4} \\ & =0.027 \mathrm{~V}\end{aligned}$

Question 8

(a) An isolated metal sphere of radius r is charged so that the electric field strength at its surface is E₀. On Fig. 6.1, sketch the variation of the electric field strength E with distance x from the centre of the sphere. Your sketch should extend from x = 0 to x = 3r.

(c) A radioactive isotope decays with a half-life of 15s to form a stable product. A fresh sample of the radioactive isotope at time t = 0 contains N₀ nuclei and no nuclei of the stable product. On Fig. 6.3, sketch the variation with t of the number n of nuclei of the stable product for time t = 0 to time t = 45s.

Answer/Explanation

Ans:

(a) from x = 0 to x = r: E = 0

from x = r to x = 3r: curve with negative gradient of decreasing magnitude passing through (r, E₀)

line passing through (2r, E₀ / 4) and (3r, E₀ / 9)

(b) from p =p₀ / 2 to p =p₀: curve with negative gradient of decreasing magnitude passing through (p₀, λ₀)

line passing through (1⁄2p₀, 2λ₀)

line passing through (15, 1⁄2N₀)

line passing through (30, 0.75N₀) and (45, 0.88N₀)

Question 9

(a) A student observes different stars from the Earth.Give two reasons why some stars appear brighter than others.

(b) State what is meant by a standard candle

(c) A spectral line from a star within a galaxy is observed to have a wavelength of 660.9nm. The same spectral line measured in the laboratory is observed to have a wavelength of 656.3nm.

(i) Show that the speed of the star relative to the Earth is $2.1 \times 10^6 \mathrm{~m} \mathrm{~s}^{-1}$. [1]
(ii) Calculate the distance to the star. The Hubble constant is $2.3 \times 10^{-18} \mathrm{~s}^{-1}$.
(iii) State and explain what can be concluded about the Universe based on this change in observed wavelength.

▶️Answer/Explanation

Ans:

(a) brighter star could be closer (to Earth)
brighter star could have a greater luminosity (in the visible wavelengths)

(b) object with known luminosity

(c)(i) $\frac{660.9-656.3}{656.3} \approx \frac{v}{3.0 \times 10^8}$ leading to $2.1 \times 10^6 \mathrm{~m} \mathrm{~s}^{-1}$

(c)(ii) $v=H_0 d$

(c)(iii) wavelength has increased / light is redshifted B1
star within galaxy is moving away / receding (from Earth) B1
Universe is expanding

CIE AS/A level Physics Paper 4 Prediction

CIE AS/A level Physics Paper 4 Prediction - 2025