2023 CCEA AS-Level Physics 1210 Practice Paper with Answers

Displacement is defined as the distance moved in a specified direction. Because it has both magnitude and direction, it is a vector quantity, unlike distance, which is a scalar. The SI unit of displacement is the metre (m).

1 mark: correct definition (distance moved in a specified direction / vector quantity referencing direction). 1 mark: correct SI unit, metre (m). Reject 'distance' alone with no reference to direction.

Resistivity \( \rho \) is a property of a material defined by \( \rho = \frac{RA}{l} \), where \(R\) is the resistance of a sample of the material, \(A\) is its cross-sectional area, and \(l\) is its length. Equivalently, resistivity is the resistance between opposite faces of a 1 m cube of the material. Its SI unit is the ohm metre, \( \Omega \text{m} \).

1 mark: definition in terms of \( \rho = \frac{RA}{l} \) or equivalent wording (resistance per unit length per unit cross-sectional area). 1 mark: correct SI unit \( \Omega \text{m} \). ecf if unit stated as \( \Omega \text{m}^{-1} \) is rejected — must be \( \Omega \text{m} \).

The electromotive force (e.m.f.), \(E\), of a source is the total (electrical) energy transferred per unit charge driven around a complete circuit by the source, including energy dissipated inside the source itself. This distinguishes e.m.f. from terminal potential difference, which is the energy transferred per unit charge in the external circuit only. The SI unit of e.m.f. is the volt (V).

1 mark: energy transferred (from chemical/other form to electrical) per unit charge, driven round the whole circuit / including internal resistance. 1 mark: unit volt (V). Accept 'energy given to each unit (coulomb) of charge by the source'.

(a) Resolving the initial velocity:
\( u_x = u\cos\theta = 18 \times \cos 40° = 13.8 \text{ m s}^{-1} \)
\( u_y = u\sin\theta = 18 \times \sin 40° = 11.6 \text{ m s}^{-1} \)

(b) Vertically, at maximum height the vertical velocity is zero. Using \( v^2 = u_y^2 - 2gh \):
\( 0 = (11.57)^2 - 2(9.81)h \)
\( h = \frac{(11.57)^2}{2 \times 9.81} = 6.82 \text{ m} \)

(c) The ball returns to launch height (vertical displacement = 0), so using \( s = u_yt - \frac{1}{2}gt^2 = 0 \):
\( t = \frac{2u_y}{g} = \frac{2 \times 11.57}{9.81} = 2.36 \text{ s} \)

(d) The horizontal velocity is constant (no air resistance), so:
\( R = u_x \times t = 13.79 \times 2.36 = 32.5 \text{ m} \)

(e) Using \( v_y = u_y - gt \):
\( v_y = 11.57 - (9.81 \times 2.36) = -11.6 \text{ m s}^{-1} \) (i.e. 11.6 m s⁻¹ downward)
The horizontal component is unchanged at 13.8 m s⁻¹, so the landing speed is:
\( v = \sqrt{u_x^2 + v_y^2} = \sqrt{13.79^2 + 11.57^2} = 18.0 \text{ m s}^{-1} \)
This equals the initial launch speed, as expected since the ball lands at the same height it was launched from (energy conservation).

(a)(i) 1 mark: correct method (18cos40°); answer 13.8 m s⁻¹. (a)(ii) 1 mark: correct method (18sin40°); answer 11.6 m s⁻¹. (b) 3 marks: correct use of \(v^2=u^2-2gh\) with subs [1], correct rearrangement [1], answer 6.82 m (± 0.1) with unit [1]. (c) 3 marks: recognises symmetry / uses \(s=u_yt-\frac12gt^2=0\) [1], correct substitution [1], answer 2.36 s [1]. (d) 2 marks: correct method \(R=u_xt\) [1], answer 32.5 m (ecf from (a)(i) and (c)) [1]. (e) 3 marks: correct \(v_y\) calculation using \(v=u-gt\) [1], correct combination of components (Pythagoras) [1], answer 18.0 m s⁻¹ [1]. ecf applied throughout for arithmetic errors carried from earlier parts.

Taking the direction of \(F_1\) (east) as the positive x-direction, and the perpendicular ('north') direction as the positive y-direction.

(a) \(F_2\) acts at 120° anticlockwise from \(F_1\), so:
\( F_{2x} = 25\cos(120°) = -12.5 \text{ N} \)
\( F_{2y} = 25\sin(120°) = 21.7 \text{ N} \)

(b) Resultant of \(F_1\) (40 N along +x) and \(F_2\):
\( R_x = 40 + (-12.5) = 27.5 \text{ N}, \quad R_y = 0 + 21.7 = 21.7 \text{ N} \)
\( R = \sqrt{27.5^2 + 21.7^2} = 35.0 \text{ N} \)
\( \theta = \tan^{-1}\left(\frac{21.7}{27.5}\right) = 38.2° \) anticlockwise from \(F_1\)

(c) For equilibrium, \(F_3\) must be equal in magnitude and exactly opposite in direction to the resultant of \(F_1\) and \(F_2\):
\( F_3 = 35.0 \text{ N} \), acting at \(38.2° + 180° = 218.2°\) anticlockwise from \(F_1\).

(d) If the resultant force were not zero, by Newton's second law (\(F=ma\)) the point would experience a net (unbalanced) force and therefore accelerate, so it could not remain at rest (in equilibrium). Only when the vector sum of all forces is exactly zero does the point remain stationary or move with constant velocity.

(a) 3 marks: correct angle identification for F2 [1], correct Fx component with correct sign [1], correct Fy component [1]. (b) 5 marks: correct Rx [1], correct Ry [1], correct method for magnitude (Pythagoras) [1], correct magnitude 35.0 N [1], correct angle 38.2° with reference direction stated [1]. (c) 3 marks: recognises F3 equal magnitude to resultant [1], correct magnitude 35.0 N (ecf) [1], correct direction (180° opposite, stated clearly) [1]. (d) 2 marks: reference to Newton's second law / unbalanced force causing acceleration [1], conclusion that zero resultant is required for the point to stay at rest / constant velocity [1]. Accept scale-drawing methods reaching equivalent numerical answers for full credit.

(a) By conservation of linear momentum:
\( m_Au_A + m_Bu_B = (m_A+m_B)v \)
\( (0.60)(4.0) + (0.40)(0) = (1.00)v \)
\( v = 2.4 \text{ m s}^{-1} \)

(b) Kinetic energy before:
\( KE_i = \frac{1}{2}m_Au_A^2 + \frac{1}{2}m_Bu_B^2 = \frac{1}{2}(0.60)(4.0)^2 = 4.8 \text{ J} \)
Kinetic energy after:
\( KE_f = \frac{1}{2}(m_A+m_B)v^2 = \frac{1}{2}(1.00)(2.4)^2 = 2.88 \text{ J} \)

(c) Since \(KE_f \neq KE_i\) (kinetic energy decreased from 4.8 J to 2.88 J), the collision is inelastic. Momentum is always conserved in a collision, but kinetic energy is conserved only in an elastic collision; here 1.92 J of kinetic energy was converted to other forms (e.g. heat and sound as the trolleys coupled).

(d) The impulse delivered to trolley B equals its change in momentum:
\( J = m_B(v - u_B) = 0.40 \times (2.4 - 0) = 0.96 \text{ kg m s}^{-1} \)
Average force: \( F = \frac{J}{t} = \frac{0.96}{0.050} = 19.2 \text{ N} \)

(e) Newton's third law states that when object A exerts a force on object B, object B exerts an equal and opposite force on object A. Therefore trolley B exerts a force of 19.2 N on trolley A, in the opposite direction (i.e. decelerating trolley A).

(a) 3 marks: correct statement of conservation of momentum [1], correct substitution [1], answer 2.4 m s⁻¹ [1]. (b) 3 marks: correct KE_i = 4.8 J [1], correct method for KE_f [1], KE_f = 2.88 J [1]. (c) 2 marks: correct identification 'inelastic' [1], valid reason referencing KE not conserved / KE decreased [1]. (d) 3 marks: correct impulse = change in momentum of B [1], correct substitution [1], answer 19.2 N [1]. (e) 2 marks: correct statement of Newton's third law [1], correct application giving 19.2 N in opposite direction on A [1]. ecf applied for (d)/(e) from (a).

(a) The principle of moments states that for a body in rotational equilibrium, the sum of the clockwise moments about any point equals the sum of the anticlockwise moments about that same point.

(b) For complete equilibrium: (i) the resultant (vector sum) of all forces acting on the body is zero, and (ii) the resultant moment about any point is zero.

(c) Taking moments about A (so that the reaction \(F_A\) has no moment about A):
Clockwise moments (from downward forces) = anticlockwise moment (from \(F_B\)):
\( F_B \times 8.0 = (4000 \times 4.0) + (6000 \times 3.0) \)
\( F_B \times 8.0 = 16000 + 18000 = 34000 \)
\( F_B = \frac{34000}{8.0} = 4250 \text{ N} \)

(d) For vertical equilibrium of forces:
\( F_A + F_B = 4000 + 6000 \)
\( F_A = 10000 - 4250 = 5750 \text{ N} \)

(e) Taking moments about B instead: the perpendicular distances from B are (8.0 − 4.0) = 4.0 m for the girder's weight and (8.0 − 3.0) = 5.0 m for the load, and \(F_A\) acts at a distance of 8.0 m from B.
\( F_A \times 8.0 = (4000 \times 4.0) + (6000 \times 5.0) \)
\( F_A \times 8.0 = 16000 + 30000 = 46000 \)
\( F_A = \frac{46000}{8.0} = 5750 \text{ N} \)
This agrees exactly with the value found in (d), confirming the answer is correct.

(a) 1 mark: correct statement of the principle of moments (sum of clockwise moments = sum of anticlockwise moments about any point, for equilibrium). (b) 2 marks: 1 mark each for (i) resultant force zero and (ii) resultant moment zero, about any point. (c) 5 marks: correct moment equation set up about A [2: 1 for each term correctly placed], correct substitution [1], correct rearrangement [1], answer 4250 N [1]. (d) 4 marks: correct equilibrium equation \(F_A+F_B=10000\) [2], correct substitution using (c) (ecf) [1], answer 5750 N [1]. (e) 2 marks: correct alternative moment equation about B [1], answer 5750 N matching (d), confirming consistency [1].

(a) For resistors in parallel: \( \frac{1}{R_p} = \frac{1}{R_1} + \frac{1}{R_2} = \frac{1}{6.0} + \frac{1}{6.0} \), so \( R_p = \frac{R_1R_2}{R_1+R_2} = \frac{6.0 \times 6.0}{12.0} = 3.0 \text{ Ω} \)

(b) Total resistance: \( R_{total} = R_p + R_3 + r = 3.0 + 2.0 + 1.0 = 6.0 \text{ Ω} \)

(c) By \( E = I(R_{total}) \): \( I = \frac{E}{R_{total}} = \frac{12.0}{6.0} = 2.0 \text{ A} \)

(d) Terminal p.d.: \( V = E - Ir = 12.0 - (2.0 \times 1.0) = 10.0 \text{ V} \)

(e) The p.d. across \(R_3\) is \( I \times R_3 = 2.0 \times 2.0 = 4.0 \text{ V} \), so the p.d. across the parallel combination is \( V - V_{R_3} = 10.0 - 4.0 = 6.0 \text{ V} \). Since \(R_1 = R_2\), the current splits equally:
\( I_1 = \frac{V_{parallel}}{R_1} = \frac{6.0}{6.0} = 1.0 \text{ A} \)

(f) Power dissipated in internal resistance: \( P_r = I^2r = (2.0)^2 \times 1.0 = 4.0 \text{ W} \)

(g) Total power delivered by battery: \( P_{total} = EI = 12.0 \times 2.0 = 24.0 \text{ W} \). Power delivered to external circuit: \( P_{ext} = P_{total} - P_r = 24.0 - 4.0 = 20.0 \text{ W} \) (equivalently \(VI = 10.0 \times 2.0 = 20.0\) W).
Efficiency \( = \frac{20.0}{24.0} \times 100\% = 83.3\% \)

(a) 2 marks: correct parallel formula used [1], answer 3.0 Ω [1]. (b) 2 marks: correctly adds Rp+R3+r [1], answer 6.0 Ω [1]. (c) 3 marks: correct formula E=IR [1], correct substitution [1], answer 2.0 A [1]. (d) 2 marks: correct use of V=E-Ir [1], answer 10.0 V [1]. (e) 3 marks: correct p.d. across R3 or parallel section [1], correct method for I1 [1], answer 1.0 A [1]. (f) 3 marks: correct formula P=I²r [1], correct substitution [1], answer 4.0 W [1]. (g) 2 marks: correct method (Pext/Ptotal) [1], answer 83.3% [1]. ecf applied throughout for values carried from earlier parts.

(a) Using the potential divider equation:
\( V_{out} = V \times \frac{R_2}{R_1+R_2} = 9.0 \times \frac{6.0}{3.0+6.0} = 6.0 \text{ V} \)

(b) Combined (parallel) resistance:
\( R_{combined} = \frac{R_2 R_L}{R_2+R_L} = \frac{6.0 \times 6.0}{6.0+6.0} = 3.0 \text{ kΩ} \)

(c) New total resistance of circuit: \( R_1 + R_{combined} = 3.0 + 3.0 = 6.0 \text{ kΩ} \)
\( I = \frac{V}{R_{total}} = \frac{9.0}{6.0} = 1.5 \text{ mA} \)

(d) New output voltage: \( V_{out} = I \times R_{combined} = 1.5 \text{ mA} \times 3.0 \text{ kΩ} = 4.5 \text{ V} \)

(e) Connecting the load resistor in parallel with \(R_2\) reduces the combined resistance of that section of the circuit (from 6.0 kΩ to 3.0 kΩ). Since \(R_1\) is unchanged, a larger proportion of the total (fixed) supply voltage is now dropped across \(R_1\) rather than across the \(R_2\)/\(R_L\) combination, so the output voltage falls. Equivalently, some of the current now flows through \(R_L\) as well as \(R_2\), increasing the total current drawn and increasing the voltage dropped across \(R_1\), leaving less for the output.

(f) The loading effect can be reduced by making the load resistance \(R_L\) much greater than \(R_2\) (so that \(R_2 \| R_L \approx R_2\)), or equivalently by using much smaller values of \(R_1\) and \(R_2\) relative to \(R_L\), or by using a buffer/voltage follower (op-amp with very high input impedance) between the divider and the load.

(a) 3 marks: correct formula [1], correct substitution [1], answer 6.0 V [1]. (b) 3 marks: correct parallel formula [1], correct substitution [1], answer 3.0 kΩ [1]. (c) 3 marks: correct total resistance found (ecf) [1], correct formula I=V/R [1], answer 1.5 mA [1]. (d) 3 marks: correct method [1], correct substitution (ecf) [1], answer 4.5 V [1]. (e) 3 marks: identifies combined resistance decreases [1], links to more p.d. dropped across R1 / less across output [1], coherent overall explanation [1]. (f) 2 marks: valid suggestion (RL >> R2, or reduce divider resistances, or use a buffer) [1] with brief correct justification [1].

(a) In a transverse wave, the particles of the medium oscillate perpendicular to the direction in which the wave (and its energy) travels.

(b) \( f = \frac{1}{T} = \frac{1}{0.080} = 12.5 \text{ Hz} \)

(c) \( v = f\lambda = 12.5 \times 0.40 = 5.0 \text{ m s}^{-1} \)

(d) The speed of a transverse wave on a string increases as the tension in the string increases. Since \(v = f\lambda\) and the frequency is fixed (set by the source), an increase in \(v\) means the wavelength \(\lambda\) must also increase.

(e) A stationary wave is produced when: (i) two progressive waves of the same frequency (and wavelength) travelling in opposite directions along the same line superpose, and (ii) the two waves have similar (comparable) amplitude, so that there is significant cancellation and reinforcement at fixed points. (This typically occurs when a wave reflects off a boundary and interferes with the incident wave.)

(a) 2 marks: oscillation perpendicular to direction of travel/energy transfer [1], reference to direction of wave travel [1]. (b) 2 marks: correct formula f=1/T [1], answer 12.5 Hz [1]. (c) 3 marks: correct formula v=fλ [1], correct substitution [1], answer 5.0 m s⁻¹ [1]. (d) 3 marks: states v increases with tension [1], correct reasoning using v=fλ with f fixed [1], correct conclusion λ increases [1]. (e) 3 marks: any two of: two waves of the same frequency/wavelength [1]; travelling in opposite directions along the same line [1]; comparable/similar amplitude [1] (max 3, at least two distinct valid points required for full marks).

(a) At the first resonance in a tube closed at one end, the length of the air column \(L\) is a quarter of a wavelength: \( L = \frac{\lambda}{4} \)

(b) The closed end of the tube (at the water surface) is a displacement node, because the air cannot move longitudinally at that boundary. The open end of the tube is (approximately) a displacement antinode, where the air molecules oscillate with maximum amplitude.

(c) \( \lambda = 4L = 4 \times 0.190 = 0.760 \text{ m} \)

(d) \( v = f\lambda = 440 \times 0.760 = 334 \text{ m s}^{-1} \)

(e) At the first resonance, \(L_1 = \frac{\lambda}{4}\). At the next resonance (the next position where an antinode again forms at the open end), the air column length has increased by half a wavelength, so \(L_2 = L_1 + \frac{\lambda}{2}\).
\( \lambda = 2(L_2 - L_1) = 2 \times (0.590 - 0.190) = 2 \times 0.400 = 0.800 \text{ m} \)
\( v = f\lambda = 440 \times 0.800 = 352 \text{ m s}^{-1} \)

(f) The method in (e) uses the difference between two resonance lengths rather than a single length measurement. Because the antinode does not form exactly at the open end of the tube in practice, using a single length (as in (d)) carries a small systematic error. This systematic error is the same at both resonances, so it cancels out when the difference \(L_2-L_1\) is taken, making the value of the speed of sound found in (e) more reliable.

(a) 1 mark: correct condition L=λ/4. (b) 2 marks: closed end = node with reason [1], open end = antinode [1]. (c) 2 marks: correct method λ=4L [1], answer 0.760 m [1]. (d) 3 marks: correct formula v=fλ [1], correct substitution [1], answer 334 m s⁻¹ [1]. (e) 3 marks: correct recognition that ΔL=λ/2 [1], correct calculation of λ=0.800 m [1], answer v=352 m s⁻¹ [1]. (f) 2 marks: reference to the antinode not being exactly at the open end / systematic error [1], explains this error cancels when using the difference in lengths [1].

(a) One ray travels from the top of the object parallel to the principal axis and, after refraction by the lens, passes through the focal point on the far side of the lens. A second ray travels from the top of the object straight through the centre of the lens without changing direction. The point where these two rays intersect (converge) after the lens locates the position of the top of the image.

(b) Using the lens equation \( \frac{1}{f} = \frac{1}{v} + \frac{1}{u} \) (real-is-positive convention):
\( \frac{1}{v} = \frac{1}{f} - \frac{1}{u} = \frac{1}{15} - \frac{1}{25} = \frac{5-3}{75} = \frac{2}{75} \)
\( v = 37.5 \text{ cm} \)

(c) Magnification: \( m = \frac{v}{u} = \frac{37.5}{25} = 1.5 \)

(d) Image height: \( h_i = m \times h_o = 1.5 \times 2.0 = 3.0 \text{ cm} \)

(e) Because \(v\) is positive, the image forms on the opposite side of the lens from the object, where light rays actually converge, so the image is real. Since the object distance (25 cm) is greater than the focal length (15 cm) but less than twice the focal length, a converging lens forms a real, inverted, magnified image in this case; the image is therefore inverted.

(a) 2 marks: correct description of the parallel ray refracting through the focal point [1], correct description of the ray through the centre of the lens undeviated [1]. (b) 3 marks: correct rearrangement of lens equation [1], correct substitution [1], answer v=37.5 cm [1]. (c) 2 marks: correct formula m=v/u [1], answer 1.5 [1]. (d) 2 marks: correct method [1], answer 3.0 cm [1]. (e) 2 marks: correctly identifies real (with reason: positive v / rays actually converge) [1], correctly identifies inverted (with reason) [1].

(a) A diverging (concave) lens is required. The myopic eye focuses light from distant objects in front of the retina because the eye's optical system is too powerful (or the eyeball too long); a diverging lens reduces the overall converging power of the eye, moving the focus back onto the retina.

(b) The far point is the maximum distance at which an object can be seen in sharp focus; for a myopic eye this is less than infinity, because the eye cannot relax its focusing power enough to view more distant objects sharply.

(c) The lens must take parallel rays from a very distant object (\(u = \infty\)) and form a virtual image at the far point, 200 cm from the eye, on the same side as the object. Using the real-is-positive convention, a virtual image distance is negative: \(v = -200\) cm.
\( \frac{1}{f} = \frac{1}{v} + \frac{1}{u} = \frac{1}{-200} + 0 \)
\( f = -200 \text{ cm} = -2.00 \text{ m} \)

(d) Power: \( P = \frac{1}{f(\text{in metres})} = \frac{1}{-2.00} = -0.50 \text{ D} \)

(e) With the corrective lens in place, light from an object at infinity is made to appear to come from the person's (uncorrected) far point, which the eye can focus on normally. So the corrected eye can now see clearly all the way out to infinity — the new effective far point is infinity.

(a) 2 marks: correct lens type 'diverging' [1], valid reason (reduces converging power / eye too powerful) [1]. (b) 2 marks: correct definition referencing maximum distance for a sharp image [1], applied correctly to myopia (less than infinity) [1]. (c) 4 marks: correctly identifies v=-200 cm (virtual, negative) [1], correctly sets 1/u=0 [1], correct substitution into lens equation [1], answer f=-2.00 m [1]. (d) 2 marks: correct formula P=1/f [1], answer -0.50 D [1]. (e) 1 mark: correctly states infinity / can now see distant objects clearly.

(a) \( \lambda = \frac{ay}{D} = \frac{(0.50\times10^{-3})(1.44\times10^{-3})}{1.20} = 6.00\times10^{-7} \text{ m} = 600 \text{ nm} \)

(b) Rearranging, \( y = \frac{\lambda D}{a} \). Since \(\lambda\) and \(D\) are unchanged, decreasing \(a\) increases \(y\): the fringe spacing increases (the fringes become more widely spaced) as the slit separation is decreased.

(c) The two sources must be coherent, meaning they have a constant phase difference (and therefore the same frequency/wavelength); it also helps if the two sources have similar/comparable amplitude, so that the dark fringes show good contrast (near-complete cancellation).

(a) 3 marks: correct substitution into given equation [1], correct arithmetic [2, allow 1 for power-of-ten/unit error only] — answer 600 nm (6.00×10⁻⁷ m). (b) 2 marks: correctly states fringe spacing increases [1], correct reasoning using y=λD/a with λ,D constant [1]. (c) 3 marks: coherent / constant phase difference [1], same frequency/wavelength [1], comparable amplitude for good fringe contrast [1] (max 3; any two well-explained points can gain full marks at examiner discretion, but 'coherent' with correct explanation is essential for at least 1 mark).

(a) The grating has 300 lines per mm, so the spacing between lines is:
\( d = \frac{1}{300 \text{ mm}^{-1}} = 3.33\times10^{-3} \text{ mm} = 3.33\times10^{-6} \text{ m} \)

(b) Using \( d\sin\theta = n\lambda \) with \(n=1\):
\( \sin\theta_1 = \frac{\lambda}{d} = \frac{589\times10^{-9}}{3.33\times10^{-6}} = 0.177 \)
\( \theta_1 = 10.2° \)

(c) The maximum possible order occurs when \(\sin\theta = 1\) (i.e. \(\theta = 90°\)):
\( n_{max} = \frac{d}{\lambda} = \frac{3.33\times10^{-6}}{589\times10^{-9}} = 5.66 \)
Since \(n\) must be a whole number and \(\sin\theta \leq 1\), the highest order that can actually be observed is \(n = 5\) (as \(n=6\) would require \(\sin\theta > 1\), which is impossible).

(a) 2 marks: correct method (reciprocal of lines per unit length, with unit conversion) [1], answer 3.33×10⁻⁶ m [1]. (b) 3 marks: correct rearrangement of dsinθ=nλ [1], correct substitution [1], answer 10.2° [1]. (c) 3 marks: recognises condition sinθ≤1 / θ≤90° [1], correct calculation of d/λ = 5.66 [1], correctly rounds down to give n=5 with justification [1].

(a) \( E_{photon} = \frac{hc}{\lambda} = \frac{(6.63\times10^{-34})(3.00\times10^{8})}{400\times10^{-9}} = 4.97\times10^{-19} \text{ J} \)

(b) \( \phi = 2.30 \times (1.60\times10^{-19}) = 3.68\times10^{-19} \text{ J} \)

(c) Using Einstein's photoelectric equation, \( \frac{1}{2}mv_{max}^2 = hf - hf_0 = E_{photon} - \phi \):
\( KE_{max} = 4.97\times10^{-19} - 3.68\times10^{-19} = 1.29\times10^{-19} \text{ J} \)

(d) \( \frac{1}{2}m_ev_{max}^2 = KE_{max} \)
\( v_{max} = \sqrt{\frac{2 \times KE_{max}}{m_e}} = \sqrt{\frac{2 \times 1.29\times10^{-19}}{9.11\times10^{-31}}} = 5.33\times10^{5} \text{ m s}^{-1} \)

(e) The maximum kinetic energy of the photoelectrons would be unchanged. Increasing the intensity of light of a fixed wavelength increases the number of photons arriving per second, and hence increases the number of photoelectrons emitted per second (the photocurrent), but each individual photon still carries the same energy \(hf\), so the maximum kinetic energy per photoelectron (given by \(hf-\phi\)) is unaffected.

(a) 3 marks: correct formula E=hc/λ [1], correct substitution [1], answer 4.97×10⁻¹⁹ J [1]. (b) 2 marks: correct conversion factor used [1], answer 3.68×10⁻¹⁹ J [1]. (c) 3 marks: correct statement/use of Einstein's equation [1], correct substitution (ecf) [1], answer 1.29×10⁻¹⁹ J [1]. (d) 3 marks: correct rearrangement for vmax [1], correct substitution (ecf) [1], answer 5.33×10⁵ m s⁻¹ [1]. (e) 1 mark: correct statement that KEmax is unchanged, with valid reasoning (intensity affects photon rate/number of photoelectrons, not photon energy).

(a) The electron can only exist with certain specific (fixed, quantised) values of energy, rather than any value in a continuous range. The electron can move between these fixed levels, but cannot have an energy between them.

(b) \( \Delta E = E_3 - E_2 = -1.51 - (-3.40) = 1.89 \text{ eV} \)
\( \Delta E = 1.89 \times (1.60\times10^{-19}) = 3.02\times10^{-19} \text{ J} \)

(c) Using \( \Delta E = hf \):
\( f = \frac{\Delta E}{h} = \frac{3.02\times10^{-19}}{6.63\times10^{-34}} = 4.56\times10^{14} \text{ Hz} \)

(d) \( \lambda = \frac{c}{f} = \frac{3.00\times10^{8}}{4.56\times10^{14}} = 6.58\times10^{-7} \text{ m} = 658 \text{ nm} \)
This wavelength lies in the red part of the visible spectrum (close to the well-known hydrogen Balmer H-alpha line).

(e) In laser action, more electrons are raised (pumped) into a higher (metastable) energy level than remain in a lower level — this is called a population inversion. When a photon of exactly the right energy passes close to an excited atom, it can stimulate that atom to emit a second photon of identical energy, phase, and direction (stimulated emission), rather than the atom emitting spontaneously in a random direction. Because a population inversion exists, this process cascades, with each emitted photon able to stimulate further emissions, producing a large number of photons that are all in phase (coherent) and travelling in the same direction, forming the laser beam.

(a) 1 mark: correct statement referring to fixed/quantised energy values, not continuous. (b) 3 marks: correct ΔE in eV [1], correct conversion to J [1], answer 3.02×10⁻¹⁹ J [1]. (c) 3 marks: correct formula f=ΔE/h [1], correct substitution (ecf) [1], answer 4.56×10¹⁴ Hz [1]. (d) 3 marks: correct formula λ=c/f [1], answer 658 nm (ecf) [1], correctly identifies red/visible region [1]. (e) 2 marks: correct explanation of population inversion [1], correct explanation of stimulated emission producing coherent photons [1].

(a) Total internal reflection occurs when light travelling in an optically denser medium strikes a boundary with a less dense medium at an angle of incidence greater than the critical angle; all of the light is reflected back into the denser medium, with no light transmitted into the less dense medium.

(b) Using \( \sin C = \frac{1}{n} \) for a glass-air boundary:
\( \sin C = \frac{1}{1.50} = 0.667 \)
\( C = 41.8° \)

(c) For light travelling from the core (\(n_1 = 1.50\)) to the cladding (\(n_2 = 1.45\)), the critical angle is found from \( \sin C = \frac{n_2}{n_1} \):
\( \sin C = \frac{1.45}{1.50} = 0.967 \)
\( C = 75.2° \)

(d) If the core were surrounded directly by air, any scratch, contact, or dirt on the fibre's outer surface would disturb the core-air interface, changing the local refractive index and potentially allowing light to escape (frustrating total internal reflection) at that point, causing significant loss of signal. The cladding protects the core surface (so surface imperfections do not affect the core-cladding boundary) and gives a well-defined, precisely controlled critical angle, ensuring reliable total internal reflection and minimising signal loss along the fibre.

(a) 2 marks: correct reference to travelling from denser to less dense medium [1], angle of incidence exceeding the critical angle, so all light reflected [1]. (b) 3 marks: correct formula sinC=1/n [1], correct substitution [1], answer 41.8° [1]. (c) 3 marks: correct formula sinC=n2/n1 [1], correct substitution [1], answer 75.2° [1]. (d) 2 marks: reference to protecting the core surface from damage/contact [1], reference to maintaining a reliable, well-defined critical angle / preventing frustrated TIR and signal loss [1].

(a) \( z = \frac{\Delta\lambda}{\lambda} = \frac{675.0-656.3}{656.3} = \frac{18.7}{656.3} = 0.0285 \)

(b) \( z = \frac{v}{c} \), so:
\( v = zc = 0.0285 \times (3.00\times10^{8}) = 8.55\times10^{6} \text{ m s}^{-1} \)

(c) Using \( v = H_0d \):
\( d = \frac{v}{H_0} = \frac{8.55\times10^{6}}{2.4\times10^{-18}} = 3.56\times10^{24} \text{ m} \)

(d) Doppler red shift arises from the relative motion of a source through space towards or away from an observer. Cosmological red shift arises instead from the expansion of space itself as light travels from a distant galaxy to the observer, stretching the wavelength of the light; the galaxies are not necessarily moving 'through' space in the same sense as a Doppler source.

(a) 2 marks: correct formula z=Δλ/λ [1], answer 0.0285 [1]. (b) 2 marks: correct formula v=zc [1], answer 8.55×10⁶ m s⁻¹ [1]. (c) 3 marks: correct formula d=v/H0 [1], correct substitution (ecf) [1], answer 3.56×10²⁴ m [1]. (d) 1 mark: valid distinction between motion-based Doppler shift and expansion-of-space cosmological shift.

(a) Mean diameter \( = \frac{12.02+12.04+12.03+12.05+12.01}{5} = \frac{60.15}{5} = 12.03 \text{ mm} \)

(b) Radius \( = 6.015 \text{ mm} = 0.6015 \text{ cm} \); using \(L = 80.0 \text{ mm} = 8.00 \text{ cm}\):
\( V = \pi r^2 L = \pi \times (0.6015)^2 \times 8.00 = 9.09 \text{ cm}^3 \)

(c) \( \rho = \frac{m}{V} = \frac{68.20}{9.09} = 7.50 \text{ g cm}^{-3} \) (equivalently \(7500 \text{ kg m}^{-3}\))

(d) Percentage uncertainty in a single reading \( = \frac{0.01}{12.03} \times 100 = 0.083\% \). Taking several repeat readings and using their mean reduces the effect of random error/variation between individual readings (e.g. from the rod not being perfectly uniform, or small differences in how the micrometer is applied), giving a mean value that is more representative and reliable than any single reading, though it does not reduce the resolution uncertainty of the instrument itself.

(a) 2 marks: correct sum/method [1], answer 12.03 mm [1]. (b) 3 marks: correct formula for volume of a cylinder [1], correct substitution with consistent units [1], answer 9.09 cm³ (or equivalent in m³) [1]. (c) 3 marks: correct formula density=mass/volume [1], correct substitution (ecf) [1], answer 7.50 g cm⁻³ [1]. (d) 2 marks: correct percentage uncertainty calculation (~0.08%) [1], valid comment that repeats reduce random error / give a more reliable mean (not resolution uncertainty) [1].

(a) Because the metre rule is uniform, its weight acts at its centre of gravity, which is at the 50.0 cm mark — exactly where the pivot is located. The perpendicular distance from the pivot to the line of action of the rule's weight is therefore zero, so the rule's own weight produces no moment about the pivot and can be ignored in the moments equation.

(b) The 90 g mass is at the 30.0 cm mark, a distance of \(50.0 - 30.0 = 20.0\) cm from the pivot. The mass \(m_x\) is at the 75.0 cm mark, a distance of \(75.0 - 50.0 = 25.0\) cm from the pivot.

(c) By the principle of moments, at balance:
\( (90 \text{ g}) \times (20.0 \text{ cm}) = m_x \times (25.0 \text{ cm}) \)
\( m_x = \frac{90 \times 20.0}{25.0} = 72 \text{ g} \)

(d) The rule should be checked to ensure it is balanced horizontally (e.g. using a spirit level, or checking it is level by eye) before and while readings are taken, and the position of each mass should be read from directly above/at eye level to avoid parallax error. (Any one valid, well-explained precaution is credited.)

(a) 2 marks: correctly identifies the rule's weight acts at its centre of gravity/50 cm mark [1], correctly links this to zero perpendicular distance from pivot / zero moment [1]. (b) 2 marks: 1 mark each for correct distance (20.0 cm and 25.0 cm). (c) 4 marks: correct statement of moments equation [1], correct substitution [1], correct rearrangement [1], answer 72 g [1]. (d) 2 marks: valid precaution stated [1] with correct explanation of how it improves accuracy [1] (e.g. avoiding parallax error when reading mass positions; ensuring the rule is exactly horizontal/balanced before reading; using fine adjustment of the mass's position for a precise balance point).

(a) Radius \( = 0.18 \text{ mm} = 1.8\times10^{-4} \text{ m} \)
\( A = \pi r^2 = \pi \times (1.8\times10^{-4})^2 = 1.02\times10^{-7} \text{ m}^2 \)

(b) \( R = \frac{V}{I} = \frac{1.35}{0.50} = 2.70 \text{ Ω} \)

(c) Using \( R = \frac{\rho l}{A} \):
\( \rho = \frac{RA}{l} = \frac{2.70 \times 1.02\times10^{-7}}{0.850} = 3.23\times10^{-7} \text{ Ω m} \)

(d) Since the diameter appears squared in the area (and hence the resistivity) calculation, its uncertainty has a large effect on the final result. Taking several diameter readings at different points along and around the wire (rotating the micrometer slightly between readings) and using the mean reduces the percentage uncertainty in the diameter compared with using a single reading; using a digital micrometer with a finer resolution would also help.

(a) 3 marks: correct radius from diameter [1], correct formula A=πr² [1], answer 1.02×10⁻⁷ m² [1]. (b) 2 marks: correct formula R=V/I [1], answer 2.70 Ω [1]. (c) 3 marks: correct rearrangement of R=ρl/A [1], correct substitution (ecf) [1], answer 3.23×10⁻⁷ Ω m [1]. (d) 2 marks: valid suggestion (repeat diameter readings at different points/orientations and average; use a more precise instrument) [1], correct justification linking to reduced percentage uncertainty [1].

(a) \( \Delta PE = mgh = 0.150 \times 9.81 \times 0.800 = 1.18 \text{ J} \)

(b) \( KE = \frac{1}{2}mv^2 = \frac{1}{2} \times 0.150 \times (3.85)^2 = 1.11 \text{ J} \)

(c) Percentage converted \( = \frac{KE}{\Delta PE} \times 100 = \frac{1.11}{1.18} \times 100 = 94.4\% \)

(d) A small amount of the gravitational potential energy is converted into other forms rather than kinetic energy, principally due to air resistance acting on the ball as it falls (which does negative work on the ball, transferring some energy to the surrounding air/heat) and to a lesser extent energy loss to sound as the ball is released.

(a) 2 marks: correct formula ΔPE=mgh [1], answer 1.18 J [1]. (b) 3 marks: correct formula KE=½mv² [1], correct substitution [1], answer 1.11 J [1]. (c) 3 marks: correct method (KE/ΔPE×100) [1], correct substitution (ecf) [1], answer 94.4% [1]. (d) 2 marks: valid physical reason (air resistance / drag doing work against the ball) [1] with correct link to energy not appearing as KE [1]. Accept sound/friction at release point as a minor secondary factor.

(a) Gradient \( = \frac{\Delta x}{\Delta F} = \frac{48.0 \text{ mm} - 0}{6.0 \text{ N} - 0} = 8.0 \text{ mm N}^{-1} = 0.0080 \text{ m N}^{-1} \)

(b) Since \(F = kx\), rearranging gives \(x = \frac{1}{k}F\). The graph plots \(x\) against \(F\), so its gradient is \(\frac{1}{k}\); therefore \(k = \frac{1}{\text{gradient}}\).
\( k = \frac{1}{0.0080} = 125 \text{ N m}^{-1} \)

(c) A stiffer spring has a larger spring constant \(k\). Since the gradient of the extension-force graph equals \(\frac{1}{k}\), a larger \(k\) corresponds to a smaller gradient — the stiffer spring produces less extension for the same applied force, so the new graph would be less steep than the original.

(a) 3 marks: correct method (rise/run using the two given points) [1], correct value 8.0 mm N⁻¹ or 0.0080 m N⁻¹ [1], correct unit stated [1]. (b) 3 marks: correct rearrangement of F=kx to x=(1/k)F showing gradient=1/k [1], correct calculation of k (ecf) [1], answer 125 N m⁻¹ [1]. (c) 2 marks: correctly states gradient decreases for a stiffer spring [1], correct reasoning linking larger k to smaller gradient (=1/k) [1].

(a) \( T = \frac{25.4}{20} = 1.27 \text{ s} \)

(b) The absolute uncertainty due to reaction time (±0.3 s) is essentially the same whether timing 1 oscillation or 20 oscillations, since it depends on the student's reaction time at the start and end of the timing, not on the number of oscillations timed. Because this fixed absolute uncertainty is divided by a much larger total time (20 periods instead of 1), the percentage uncertainty in the period is reduced by a factor of about 20.

(c) Percentage uncertainty in the total time \( = \frac{0.3}{25.4} \times 100 = 1.18\% \). Since the period \(T\) is found by dividing the total time by the exact number 20, the percentage uncertainty in \(T\) is the same as the percentage uncertainty in the total time: 1.18%.

(d) Absolute uncertainty in \(T\) \( = \frac{0.3}{20} = 0.015 \text{ s} \) (equivalently, \(1.18\%\) of 1.27 s ≈ 0.015 s). Rounding to an appropriate number of significant figures: \( T = (1.27 \pm 0.02) \text{ s} \).

(a) 2 marks: correct method (divide by 20) [1], answer 1.27 s [1]. (b) 2 marks: recognises the absolute uncertainty (reaction time) is roughly fixed/independent of number of oscillations [1], correctly links this to a reduced percentage uncertainty when divided by a larger total time [1]. (c) 2 marks: correct percentage uncertainty calculation on total time [1], correctly states this equals the percentage uncertainty in T [1] — answer ≈1.18%. (d) 2 marks: correct absolute uncertainty ≈0.015 s (ecf) [1], correctly rounded final answer quoted with uncertainty, e.g. (1.27±0.02) s [1].

(a) Percentage uncertainty in \(r\) \( = \frac{0.05}{2.50} \times 100 = 2.0\% \)

(b) When a quantity is raised to a power \(n\), the percentage uncertainty in the result is \(n\) times the percentage uncertainty in the original quantity. Since \(V \propto r^3\), the percentage uncertainty in \(V\) is 3 times the percentage uncertainty in \(r\):
\( \text{% uncertainty in } V = 3 \times 2.0\% = 6.0\% \)

(c) \( V = \frac{4}{3}\pi r^3 = \frac{4}{3}\pi (2.50)^3 = 65.4 \text{ cm}^3 \)
Absolute uncertainty in \(V\) \( = 6.0\% \text{ of } 65.4 = 0.060 \times 65.4 = 3.9 \text{ cm}^3 \)
So \( V = (65.4 \pm 3.9) \text{ cm}^3 \).

(a) 2 marks: correct formula (absolute uncertainty/value ×100) [1], answer 2.0% [1]. (b) 3 marks: correct statement of the power rule (multiply percentage uncertainty by the power) [1], correct identification that V∝r³ so n=3 [1], answer 6.0% [1]. (c) 3 marks: correct calculation of V=65.4 cm³ [1], correct calculation of absolute uncertainty (ecf) [1], correctly quoted final answer with uncertainty, e.g. (65.4±3.9) cm³ [1].

(a) At the stopping voltage, the maximum kinetic energy of the photoelectrons is exactly removed by the retarding electric field, so \(eV_s = KE_{max}\). Combining with Einstein's photoelectric equation \(KE_{max} = hf - hf_0\):
\( eV_s = hf - hf_0 \)
\( V_s = \left(\frac{h}{e}\right)f - \frac{hf_0}{e} \)
This has the form \(y = mx + c\) with \(V_s\) plotted against \(f\), so the graph is a straight line with gradient \(\frac{h}{e}\).

(b) Gradient \( = \frac{2.31 - 0.65}{(10.0-6.0)\times10^{14}} = \frac{1.66}{4.0\times10^{14}} = 4.15\times10^{-15} \text{ V s} \)

(c) Since gradient \( = \frac{h}{e} \):
\( h = \text{gradient} \times e = (4.15\times10^{-15}) \times (1.60\times10^{-19}) = 6.64\times10^{-34} \text{ J s} \)
(This is very close to the accepted value of the Planck constant, \(6.63\times10^{-34}\) J s.)

(a) 2 marks: correctly equates eVs with KEmax and substitutes into Einstein's equation [1], correctly identifies the gradient as h/e (comparing with y=mx+c) [1]. (b) 3 marks: correct method (rise/run using given points) [1], correct substitution [1], answer 4.15×10⁻¹⁵ V s [1]. (c) 3 marks: correct rearrangement h=gradient×e [1], correct substitution (ecf) [1], answer 6.64×10⁻³⁴ J s [1].

(a) The trolley's velocity increases throughout the 5.0 s, so it is always accelerating (never decelerating) — but the velocity increases more and more slowly as time goes on (the increase in velocity per second gets smaller: from 1.8 m s⁻¹ in the first second down to only 0.2 m s⁻¹ in the last second). This means the trolley's acceleration is not uniform: it is largest at the start and steadily decreases with time, with the trolley's velocity appearing to level off towards a roughly constant value.

(b) The gradient of the tangent at \(t=2.0\) s gives the instantaneous acceleration:
\( a = \frac{4.4-1.6}{3.0-1.0} = \frac{2.8}{2.0} = 1.4 \text{ m s}^{-2} \)

(c) Using the trapezium rule, with each interval of width \(\Delta t = 1.0\) s, the area of each trapezium is \(\frac{1}{2}(v_1+v_2)\Delta t\):
0 to 1.0 s: \(\frac{1}{2}(0+1.8)(1.0) = 0.9\) m
1.0 to 2.0 s: \(\frac{1}{2}(1.8+3.2)(1.0) = 2.5\) m
2.0 to 3.0 s: \(\frac{1}{2}(3.2+4.2)(1.0) = 3.7\) m
3.0 to 4.0 s: \(\frac{1}{2}(4.2+4.8)(1.0) = 4.5\) m
4.0 to 5.0 s: \(\frac{1}{2}(4.8+5.0)(1.0) = 4.9\) m
Total distance \( = 0.9+2.5+3.7+4.5+4.9 = 16.5 \text{ m} \)

(d) As the trolley speeds up, the resistive force acting on it (e.g. air resistance and/or friction) increases with speed, so the net (resultant) forward force on the trolley — the driving force minus the resistive force — decreases. By Newton's second law, \(a = \frac{F_{net}}{m}\), a smaller resultant force produces a smaller acceleration, so the acceleration decreases as the velocity (and hence the resistive force) increases. The trolley's velocity would eventually level off at a terminal velocity, where the resistive force exactly balances the driving force and the acceleration falls to zero.

(a) 2 marks: correctly identifies velocity always increasing / never decelerating [1], correctly identifies acceleration is non-uniform and decreasing with time [1]. (b) 3 marks: correct method (gradient of tangent using given points) [1], correct substitution [1], answer 1.4 m s⁻² [1]. (c) 4 marks: correct method (trapezium rule) applied [1], at least 4 of the 5 individual trapezium areas correct [1], correct summation [1], answer 16.5 m [1]. (d) 1 mark: correctly links increasing resistive force with speed to decreasing net force and hence decreasing acceleration (via F=ma).

(a) Electrical energy \( = VIt = 6.0 \times 1.5 \times 4.0 = 36.0 \text{ J} \)

(b) \( \Delta PE = mgh = 0.50 \times 9.81 \times 1.20 = 5.89 \text{ J} \)

(c) Efficiency \( = \frac{\text{useful energy output}}{\text{energy input}} \times 100 = \frac{5.89}{36.0} \times 100 = 16.4\% \)

(d) The low efficiency indicates most of the electrical energy is lost, for example to friction in the motor's bearings/gears, to resistive heating in the motor's coils, and to raising the kinetic energy of moving parts (string, pulley) rather than only the mass. One useful modification would be to also measure and record the final speed of the mass as it is raised, so that the kinetic energy gained by the mass (and any moving apparatus) can be calculated and included in the 'useful energy output', giving a more accurate (less underestimated) value for the true mechanical efficiency of the motor itself, rather than attributing that lost energy entirely to heat/friction.

(a) 2 marks: correct formula E=VIt [1], answer 36.0 J [1]. (b) 2 marks: correct formula ΔPE=mgh [1], answer 5.89 J [1]. (c) 2 marks: correct method (ratio ×100) [1], answer 16.4% [1] (ecf from (a)/(b)). (d) 3 marks: a valid, specific modification is suggested (e.g. also measuring the final speed to include KE gained; using a lower-friction pulley system; repeating and averaging readings) [1], with a clear, correct explanation of how it improves the accuracy of the efficiency measurement [2, allow 1 for a partial/vague link].