CIE AS/A level Physics Paper 2 Prediction - 2025

CIE AS/A level Physics Paper 2 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 2 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

Water leaves the end of a hose pipe at point P with a horizontal velocity of 6.6m/s, as shown in fig.2.1

Point P is at height h above the ground. The water hits the ground at point Q. The horizontal distance from P to Q is 3.5m. Air resistance is negligible. Assume that the water between P and Q consists of non-interacting droplets of water and that the only force acting on each droplet is its weight.

(a) Explain, briefly, why the horizontal component of the velocity of a droplet of water remains constant as it moves from P to Q.

(b) Show that the time taken for a droplet of water to move from P to Q is 0.53s.

(d) For the movement of a droplet of water from P to Q, state and explain whether the
displacement of the droplet is less than, more than or the same as the distance along its path.

(e) Calculate the magnitude of the displacement of a droplet of water that moves from P to Q.

Answer/Explanation

Ans:

(a) force (on droplet of water) in horizontal direction is zero.

(b) (time taken =) 3.5 / 6.6 = 0.53 (s)

(d) displacement is straight-line distance (from P to Q) so less (than distance along path)
or
displacement is the shortest distance (from P to Q).

(e) $(displacement)^2 = 3.5^2 + 1.4^2$
displacement = 3.8 m

Question 2

A rigid uniform beam of weight W is connected to a fixed support by a hinge, as shown in Fig. 2.1.

A compressed spring exerts a total force of $8.2 \mathrm{~N}$ vertically upwards on the horizontal beam. A block of weight $0.30 \mathrm{~N}$ rests on the beam. The right-hand end of the beam is connected to the ground by a string at an angle of $30^{\circ}$ to the horizontal. The tension in the string is $4.8 \mathrm{~N}$. The distances along the beam are shown in Fig. 2.1.
The beam is in equilibrium. Assume that the hinge is frictionless.
(a) (i) Show that the vertical component of the tension in the string is $2.4 \mathrm{~N}$.[1]

(ii) By taking moments about the hinge, determine the weight $W$ of the beam.

(iii) Calculate the horizontal component of the force exerted on the beam by the hinge.

(b) The spring obeys Hooke’s law and has an elastic potential energy of 0.32J. Calculate the compression of the spring.

(c) The string is cut so that the spring extends upwards. This causes the beam to rotate and launch the block into the air. The block reaches its maximum height and then falls back to the ground.
Fig. 2.2 shows part of the path of the block in the air shortly before it hits the horizontal ground.

The block is at a height of $0.090 \mathrm{~m}$ above the ground when it passes through point $\mathrm{A}$. The block has a kinetic energy of $0.044 \mathrm{~J}$ when it hits the ground at point $\mathrm{B}$. Air resistance is negligible.
(i) Calculate the decrease in the gravitational potential energy of the block for its movement from $A$ to $B$.

(ii) Use your answer in (c)(i) and conservation of energy to determine the speed of the block at point $A$.

(iii) By reference to the force on the block, explain why the horizontal component of the velocity of the block remains constant as it moves from $A$ to $B$. [1]

(iv) The block passes through point $\mathrm{A}$ at time $t_{\mathrm{A}}$ and arrives at point $\mathrm{B}$ at time $t_{\mathrm{B}}$. On Fig. 2.3, sketch a graph to show the variation of the magnitude of the vertical component $v_{\mathrm{Y}}$ of the velocity of the block with time $t$ from $t=t_{\mathrm{A}}$ to $t=t_{\mathrm{B}}$. Numerical values of $v_Y$ are not required.

▶️Answer/Explanation

Ans:

(a)(i) (component $=$ ) $4.8 \sin 30^{\circ}=2.4(\mathrm{~N})$

(a)(ii) $(8.2 \times 0.50)$ or $(W \times 0.60)$ or $(0.30 \times 0.80)$ or $(2.4 \times 1.2)$

$(8.2 \times 0.50)=(W \times 0.60)+(0.30 \times 0.80)+(2.4 \times 1.2)$

W = 1.6 N

(a)(iii) $\begin{aligned} \text { force } & =4.8 \cos 30^{\circ} \\ & =4.2 \mathrm{~N}\end{aligned}$

(b) $E=1 / 2 F x$

$0.32=1 / 2 \times 8.2 \times x$
$x=0.078 \mathrm{~m}$

(c)(i) $((\Delta) E)=m g(\Delta) h$ or $W(\Delta) h$

$\begin{aligned} & =0.30 \times 0.090 \\ & =0.027 \mathrm{~J}\end{aligned}$

(c)(ii) $E=1 / 2 m v^2$

$E=0.044-0.027(=0.017)$

$\begin{aligned} & v^2=(2 \times 0.017) /(0.30 / 9.81) \quad(\text { where } v=\text { speed at } \mathrm{A}) \\ & v=1.1 \mathrm{~m} \mathrm{~s}^{-1}\end{aligned}$

$E=1 / 2 m v^2$

$\begin{aligned} & 0.044=1 / 2 \times(0.30 / 9.81) \times v^2(\text { where } v=\text { speed at } \mathrm{B}) \\ & \left(v^2=2.88\right) \\ & v^2=u^2+2 \text { as }(\text { where } u=\text { speed at } \mathrm{A}) \\ & 2.88=u^2+2 \times 9.81 \times 0.090\end{aligned}$

$u=1.1 \mathrm{~ms}^{-1}$

(c)(iii) (gravitational / resultant) force / weight is vertical

(c)(iv) straight line with positive gradient starting from non-zero value of $v_Y$ at time $t_{\mathrm{A}}$ to a time $t_{\mathrm{B}}$

Question 3

A block is pulled in a straight line along a rough horizontal surface by a varying force X, as shown in Fig. 3.1.

Air resistance is negligible. Assume that the frictional force exerted on the block by the surface is constant and has magnitude 2.0N.
The variation with time t of the momentum p of the block is shown in Fig. 3.2.

(a) State Newton’s second law of motion.

(b) Use Fig. 3.2 to determine, for the block at time t = 2.0s, the magnitude of:

(i) the resultant force on the block
(ii) the force X.

▶️Answer/Explanation

Ans:

(a)(resultant) force (on an object) is proportional to / equal to the rate of change of momentum

(b)(i) $\begin{aligned} \text { resultant force } & =\text { e.g. } 6.0 / 4.0 \\ & =1.5 \mathrm{~N}\end{aligned}$

(b)(ii) force $X=1.5+2.0$ = 3.5 N

from $t=0$ to $t=4.0 \mathrm{~s}$ : horizontal line at $X=3.5 \mathrm{~N}$

from $t=4.0 \mathrm{~s}$ to $t=6.0 \mathrm{~s}$ : horizontal line at $X=2.0 \mathrm{~N}$

Question 4

(a) A uniform metal bar, initially unstretched, has sides of length w, x and y, as shown in Fig. 3.1.

The bar is now stretched by a tensile force F applied to the shaded ends. The changes in
the lengths x and y are negligible. The bar now has sides of length x, y and z, as shown in
Fig. 3.2.

Determine expressions, in terms of some or all of F, w, x, y and z, for:
(i) the stress σ applied to the bar by the tensile force
(ii) the strain ε in the bar due to the tensile force
(iii) the Young modulus E of the metal from which the bar is made.

(b) A copper wire is stretched by a tensile force that gradually increases from 0 to 280N. The
variation with extension of the tensile force is shown in Fig. 3.3.

(i) State the maximum extension of the wire for which it obeys Hooke’s law.
(ii) Use Fig. 3.3 to determine the strain energy in the wire when the tensile force is 120N.
(iii) Explain why the work done in stretching the wire to an extension of 12mm is not equal to
the energy recovered when the tensile force is removed.

Answer/Explanation

(a) (i) σ = F / xy
(a) (ii) ε = (z –w) / w
(a) (iii) E =σ / ε
= Fw / xy(z–w)
(b) (i) extension = 2.2 mm (allow 2.0 –2.4 mm)
(b) (ii) strain energy = area under graph/line or ½Fx or ½kx²
= ½ × 120 × 1.4 × 10–3 or ½ × 8.6 × 10⁴ × (1.4 × 10^–3)²
= 0.084 J
(b) (iii) (some of the) deformation of the wire is plastic/permanent/not elastic
or
wire goes past the elastic limit/enters plastic region
energy (that cannot be recovered) is dissipated as thermal energy/becomes internal energy

Question 5

(a) For a progressive wave, state what is meant by wavelength.

(b) A light wave from a laser has a wavelength of 460nm in a vacuum. Calculate the period of the wave.

(c) The light from the laser is incident normally on a diffraction grating.
Describe the diffraction of the light waves at the grating.

(d) A diffraction grating is used with different wavelengths of visible light. The angle θ of the
fourth-order maximum from the zero-order (central) maximum is measured for each
wavelength. The variation with wavelength λ of sinθ is shown in Fig. 4.1.

(i) The gradient of the graph is G.
Determine an expression, in terms of G, for the distance d between the centres of two adjacent slits in the diffraction grating.

(ii) On Fig. 4.1, sketch a graph to show the results that would be obtained for the
second-order maxima.

Answer/Explanation

Ans:

(a) distance moved by wavefront/energy during one cycle/oscillation/period (of source)
or
minimum distance between two wavefronts
or
distance between two adjacent wavefronts

(b) v =λ / T
or
v = fλ and f = 1 / T

$T = 460 \times 10^{–9} / 3.00 \times 10^8 $

$= 1.5\times 10^{–15} s$

(c) waves pass through/enter the slit(s)
waves spread (into geometric shadow) B1
(d)(i) n λ = d sin θ
G = sin θ / λ
d = 4 / G

(d)(ii) straight line from 400 nm to 700 nm that is always below printed line
straight line has smaller gradient than printed line and is 5 small squares high at wavelength of 700 nm

Question 6

A battery of electromotive force (e.m.f.) 9.6V and negligible internal resistance is connected in series with two fixed resistors and a thermistor, as shown in Fig. 7.1.

The fixed resistors have resistances of $3400 \Omega$ and $5800 \Omega$. The reading on the voltmeter in the circuit is $6.0 \mathrm{~V}$.
(a) Calculate the current in the resistor of resistance $5800 \Omega$.

(b) Calculate the resistance of the thermistor.

(c) The initial energy stored in the battery is $2.6 \times 10^4 \mathrm{~J}$. Assume that the e.m.f. of the battery is constant.
Determine the final energy stored in the battery after a charge of $330 \mathrm{C}$ has moved through it.

(d) The environmental conditions change causing an increase in the resistance of the thermistor. State whether there is a decrease, increase or no change to:

(i) the temperature of the thermistor

(ii) the current in the thermistor

(iii) the potential difference across the thermistor.

▶️Answer/Explanation

Ans:

$9.6=6.0+(I \times 5800)$ or $3.6=I \times 5800$

$
9.6=6.2 \times 10^{-4} \times(3400+5800+R)
$
or
$
6.0=6.2 \times 10^{-4} \times(3400+R)
$

$I=6.2 \times 10^{-4} \mathrm{~A}$

$R=6.3 \times 10^3 \Omega$

$\begin{aligned} \text { final stored energy } & =2.6 \times 10^4-3170 \\ & =2.3 \times 10^4 \mathrm{~J}\end{aligned}$

(d)(i) decrease

(d)(ii) decrease

(d)(iii) increase

Question 7

(a) Nuclei $X$ and $Y$ are different isotopes of the same element.
Nucleus $X$ is unstable and emits a $\beta^{+}$particle to form nucleus $Z$.
By comparing the number of protons in each nucleus, state and explain whether the charge of nucleus $\mathrm{X}$ is less than, the same as or greater than the charge of:

(i) nucleus $Y$
(ii) nucleus $Z$

(b) Hadrons can be divided into two groups (classes), P and Q. Group P is baryons.

(i) State the name of group $Q$.[1]

(ii) Describe, in general terms, the quark structure of hadrons that belong to group $Q$.

▶️Answer/Explanation

Ans:

Use identity $\sin 2 t=2 \sin t \cos t$

Attempt solution of $y=0$ for $\cos t$

Obtain $\cos t=\frac{2}{3}$

b.Differentiate to obtain at least one of $\frac{\mathrm{d} x}{\mathrm{~d} t}$ and $\frac{\mathrm{d} y}{\mathrm{~d} t}$ correct

Obtain $\frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{6 \cos 2 t-4 \cos t}{k \sec ^2 t}$

Attempt to express $\frac{\mathrm{d} y}{\mathrm{~d} x}$ in terms of $\cos t$

Obtain $\frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{6\left(2 \cos ^2 t-1\right) \cos ^2 t-4 \cos ^3 t}{k}$

c.Substitute value from part (a) in expression for $\frac{\mathrm{d} y}{\mathrm{~d} x}$ involving $k$ and $\cos t$

Equate to $-\frac{10}{9}$ and solve for $k$

Obtain $k=\frac{4}{3}$

CIE AS/A level Physics Paper 2 Prediction

CIE AS/A level Physics Paper 2 Prediction - 2025