2024 CCEA AS-Level Physics 1210 Practice Paper with Answers

(a) Using \( v^2 = u^2 - 2as \), with \(v=0\):
\( 0 = 42^2 - 2a(350) \)
\( a = \frac{42^2}{2 \times 350} = \frac{1764}{700} = 2.52 \text{ m s}^{-2} \)

(b) Using \( v = u - at \), with \(v=0\):
\( t = \frac{u}{a} = \frac{42}{2.52} = 16.7 \text{ s} \)

(c) Using \( s = ut - \frac{1}{2}at^2 \) with \(t=5.0\) s:
\( s = (42)(5.0) - \frac{1}{2}(2.52)(5.0)^2 = 210 - 31.5 = 179 \text{ m} \)

(a) 4 marks: correct equation of motion selected [1], correct rearrangement [1], correct substitution [1], answer 2.52 m s⁻² [1]. (b) 3 marks: correct equation [1], correct substitution (ecf) [1], answer 16.7 s [1]. (c) 5 marks: correct equation selected [1], correct substitution of u, a and t [2], correct arithmetic [1], answer 179 m [1].

(a) At maximum height, the vertical velocity is zero. Using \( v^2=u^2-2gh \):
\( 0 = 15^2 - 2(9.81)h \)
\( h = \frac{15^2}{2\times9.81} = 11.5 \text{ m} \)

(b) The stone returns to its starting height, so using \( v=u-gt=0 \) at the top and doubling (by symmetry), or directly \( s=ut-\frac{1}{2}gt^2=0 \):
\( t = \frac{2u}{g} = \frac{2\times15}{9.81} = 3.06 \text{ s} \)

(c) Using \( v^2=u^2-2gh \) with \(h=8.0\) m:
\( v^2 = 15^2 - 2(9.81)(8.0) = 225-156.96 = 68.04 \)
\( v = \sqrt{68.04} = 8.25 \text{ m s}^{-1} \)

(d) The stone passes through a height of 8.0 m twice: once on the way up (before reaching maximum height of 11.5 m) and once on the way down (after falling back from the maximum height). At both times the stone's speed is the same, 8.25 m s⁻¹, but the velocity (a vector) differs: on the way up, the velocity is directed upwards; on the way down, the velocity is directed downwards (i.e. the two velocities are equal in magnitude but opposite in direction).

(a) 3 marks: correct equation and condition v=0 [1], correct substitution [1], answer 11.5 m [1]. (b) 3 marks: correct method (symmetry or s=0) [1], correct substitution [1], answer 3.06 s [1]. (c) 3 marks: correct equation [1], correct substitution [1], answer 8.25 m s⁻¹ [1]. (d) 3 marks: correctly explains the stone passes through 8.0 m twice (once ascending, once descending) [1], correctly states speed is the same both times [1], correctly states velocity direction is reversed (up vs down) [1].

(a) Newton's second law states that the resultant (net) force acting on an object is directly proportional to its rate of change of momentum; for an object of constant mass, this is expressed as \(F=ma\), where \(F\) is the resultant force, \(m\) is the mass and \(a\) is the acceleration, all in the same direction.

(b) The resultant force is the applied force minus friction:
\( F_{net} = 80-55 = 25 \text{ N} \)
\( a = \frac{F_{net}}{m} = \frac{25}{25} = 1.0 \text{ m s}^{-2} \)

(c) For the crate to move at constant velocity, its acceleration must be zero, which (by Newton's second law) requires the resultant force to be zero. This means the pulling force must exactly balance the frictional force, so the pulling force required is 55 N.

(d) With only friction acting (55 N), the deceleration is:
\( a = \frac{55}{25} = 2.2 \text{ m s}^{-2} \)
Using \( v=u-at \) with \(v=0\), \(u=3.0\) m s⁻¹:
\( t = \frac{u}{a} = \frac{3.0}{2.2} = 1.36 \text{ s} \)

(a) 1 mark: correct statement of Newton's second law (F=ma, resultant force). (b) 3 marks: correct resultant force calculated [1], correct formula a=F/m [1], answer 1.0 m s⁻² [1]. (c) 3 marks: correctly states 55 N [1], correctly links to zero acceleration/constant velocity [1], correctly explains forces must balance [1]. (d) 4 marks: correct deceleration calculated [1], correct equation selected [1], correct substitution [1], answer 1.36 s [1].

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

(b) For balance, the moment of the second child must equal the moment of the first child:
\( 280 \times 1.5 = 350 \times d_2 \)
\( d_2 = \frac{280\times1.5}{350} = \frac{420}{350} = 1.2 \text{ m} \)

(c) Moment due to the first child (280 N at 1.5 m):
\( 280 \times 1.5 = 420 \text{ N m} \)
Moment due to the adult (700 N at 0.60 m):
\( 700 \times 0.60 = 420 \text{ N m} \)
Since the two moments are equal in magnitude (both 420 N m) and act on opposite sides of the pivot (in opposite rotational senses), the resultant moment about the pivot is zero, so the seesaw remains balanced.

(a) 2 marks: correct statement referencing sum of clockwise moments = sum of anticlockwise moments about a point. (b) 4 marks: correct moments equation set up [2], correct rearrangement [1], answer 1.2 m [1]. (c) 5 marks: correct moment for first child (420 N m) [1], correct moment for adult (700×0.60) [2], correct comparison/conclusion that moments are equal [1], correct final statement 'balanced' with reasoning [1].

(a) \( h = L\sin\theta = 40 \times \sin15° = 10.4 \text{ m} \)

(b) \( \Delta PE = mgh = 20 \times 9.81 \times 10.4 = 2030 \text{ J} \)

(c) \( W_{friction} = F \times d = 30 \times 40 = 1200 \text{ J} \)

(d) By the work-energy theorem, the kinetic energy gained equals the PE lost minus the work done against friction (since the sledge starts from rest):
\( KE = \Delta PE - W_{friction} = 2030-1200 = 831 \text{ J} \)
\( v = \sqrt{\frac{2\times831}{20}} = 9.12 \text{ m s}^{-1} \)

(e) Fraction converted \( = \frac{KE}{\Delta PE} = \frac{831}{2030} = 0.409 \) (40.9%)

(a) 2 marks: correct formula h=Lsinθ [1], answer 10.4 m [1]. (b) 2 marks: correct formula and substitution (ecf) [1], answer 2030 J [1]. (c) 1 mark: correct value 1200 J. (d) 4 marks: correct application of work-energy theorem [1], correct KE (ecf) [1], correct rearrangement for v [1], answer 9.12 m s⁻¹ [1]. (e) 2 marks: correct method [1], answer 0.409 (or 40.9%) [1] (ecf).

(a) Since the gun was initially at rest, total momentum before firing is zero; by conservation of momentum, the total momentum after firing must also be zero:
\( m_{bullet}v_{bullet} = m_{gun}v_{gun} \)
\( v_{gun} = \frac{m_{bullet}v_{bullet}}{m_{gun}} = \frac{0.020\times300}{4.0} = 1.5 \text{ m s}^{-1} \)

(b) The impulse delivered to the gun equals its change in momentum:
\( J = m_{gun}v_{gun} = 4.0\times1.5 = 6.0 \text{ kg m s}^{-1} \)

(c) The shoulder must deliver an impulse of 6.0 kg m s⁻¹ to bring the gun to rest in 0.050 s:
\( F = \frac{J}{t} = \frac{6.0}{0.050} = 120 \text{ N} \)

(a) 3 marks: correct statement/use of conservation of momentum [1], correct substitution [1], answer 1.5 m s⁻¹ [1]. (b) 3 marks: correct identification J=change in momentum [1], correct substitution (ecf) [1], answer 6.0 kg m s⁻¹ [1]. (c) 5 marks: correct formula F=J/t [1], correct identification that J from (b) applies here [2], correct substitution [1], answer 120 N [1].

(a) \( A_A = \pi\left(\frac{0.60\times10^{-3}}{2}\right)^2 = \pi(0.30\times10^{-3})^2 = 2.83\times10^{-7} \text{ m}^2 \)

(b) \( R_A = \frac{\rho l}{A} = \frac{(4.9\times10^{-7})(2.0)}{2.83\times10^{-7}} = 3.47 \text{ Ω} \)

(c) \( A_B = \pi\left(\frac{1.20\times10^{-3}}{2}\right)^2 = \pi(0.60\times10^{-3})^2 = 1.13\times10^{-6} \text{ m}^2 \)
\( R_B = \frac{(4.9\times10^{-7})(1.0)}{1.13\times10^{-6}} = 0.433 \text{ Ω} \)

(d)(i) In series: \( R_{series} = R_A+R_B = 3.47+0.433 = 3.90 \text{ Ω} \)
(ii) In parallel: \( R_{parallel} = \frac{R_AR_B}{R_A+R_B} = \frac{3.47\times0.433}{3.90} = 0.385 \text{ Ω} \)

(a) 2 marks: correct formula A=πr² with correct radius [1], answer 2.83×10⁻⁷ m² [1]. (b) 2 marks: correct formula R=ρl/A [1], answer 3.47 Ω [1]. (c) 3 marks: correct area for B [1], correct resistance formula and substitution [1], answer 0.433 Ω [1]. (d) 4 marks: correct series formula and answer 3.90 Ω [2, ecf], correct parallel formula and answer 0.385 Ω [2, ecf].

(a) \( R = \frac{\rho l}{A} = \frac{(1.10\times10^{-6})(4.0)}{0.20\times10^{-6}} = 22.0 \text{ Ω} \)

(b) \( I = \frac{V}{R} = \frac{12}{22.0} = 0.545 \text{ A} \)

(c) \( P = VI = 12\times0.545 = 6.55 \text{ W} \) (or equivalently \(P=\frac{V^2}{R} = \frac{144}{22.0} = 6.55\) W)

(d) For the 2.0 m length (half the original length, same cross-sectional area, same material):
\( R_{new} = \frac{\rho l_{new}}{A} = \frac{(1.10\times10^{-6})(2.0)}{0.20\times10^{-6}} = 11.0 \text{ Ω} \)
\( P_{new} = \frac{V^2}{R_{new}} = \frac{144}{11.0} = 13.1 \text{ W} \)
Halving the length halves the resistance; since power (at constant voltage) is inversely proportional to resistance (\(P=V^2/R\)), halving the resistance doubles the power dissipated — this is consistent with the new power (13.1 W) being (almost exactly) double the original power (6.55 W).

(a) 2 marks: correct formula and substitution [1], answer 22.0 Ω [1]. (b) 2 marks: correct formula I=V/R [1], answer 0.545 A [1]. (c) 2 marks: correct formula (either P=VI or P=V²/R) [1], answer 6.55 W [1]. (d) 5 marks: correct new resistance 11.0 Ω [2], correct new power 13.1 W [2], correct comment relating the doubling of power to the halving of resistance at constant voltage [1].

(a) In each case, the e.m.f. equals the current multiplied by the total circuit resistance (external plus internal):
Case 1: \( E = 1.20(5.0+r) \)
Case 2: \( E = 0.75(10.0+r) \)

(b) Setting the two expressions for \(E\) equal:
\( 1.20(5.0+r) = 0.75(10.0+r) \)
\( 6.0+1.20r = 7.5+0.75r \)
\( 1.20r-0.75r = 7.5-6.0 \)
\( 0.45r = 1.5 \)
\( r = \frac{1.5}{0.45} = 3.33 \text{ Ω} \)

(c) Substituting back into either equation, e.g. Case 1:
\( E = 1.20(5.0+3.33) = 1.20\times8.33 = 10.0 \text{ V} \)

(a) 2 marks: 1 mark each for correct equation in each case. (b) 6 marks: correct method of equating the two expressions [1], correct expansion [2], correct collection of terms [2], answer r=3.33 Ω [1]. (c) 3 marks: correct substitution back into one equation (ecf) [2], answer E=10.0 V [1].

(a) A standing wave pattern has antinodes (positions of maximum amplitude/heating) separated by half a wavelength; the hot spots on the food correspond to antinodes of the standing microwave pattern inside the oven, which is why they are half a wavelength apart rather than a full wavelength apart.

(b) \( \lambda = 2 \times 6.0 \text{ cm} = 12.0 \text{ cm} = 0.12 \text{ m} \)

(c) \( v = f\lambda = (2.45\times10^9)(0.12) = 2.94\times10^{8} \text{ m s}^{-1} \)

(d) This calculated speed, \(2.94\times10^8\) m s⁻¹, is very close to the accepted speed of light in a vacuum, \(c=3.00\times10^8\) m s⁻¹ (within about 2%, consistent with experimental measurement uncertainty in locating the hot spots precisely). This confirms that microwaves, like light, are a form of electromagnetic radiation, travelling at (very close to) the speed of light.

(a) 3 marks: correctly identifies hot spots as antinodes [1], correctly states antinodes are half a wavelength apart in a standing wave [2]. (b) 2 marks: correct method (wavelength=2×spacing) [1], answer 0.12 m [1]. (c) 3 marks: correct formula v=fλ [1], correct substitution [1], answer 2.94×10⁸ m s⁻¹ [1]. (d) 3 marks: correctly notes the value is close to c [1], with a sensible comment on the small discrepancy [1], correctly identifies microwaves as electromagnetic waves [1].

(a) For a string fixed at both ends, both ends must be nodes; the fundamental (lowest-frequency) mode has a single antinode in the middle, so the length of the string is half a wavelength: \( L = \frac{\lambda}{2} \)

(b) \( \lambda = 2L = 2\times0.65 = 1.30 \text{ m} \)

(c) \( v = f\lambda = 220\times1.30 = 286 \text{ m s}^{-1} \)

(d) Rearranging \( v=\sqrt{\frac{T}{\mu}} \) to make \(T\) the subject:
\( T = v^2\mu = (286)^2\times(6.5\times10^{-4}) = 81796\times(6.5\times10^{-4}) = 53.2 \text{ N} \)

(a) 1 mark: correct relationship L=λ/2. (b) 2 marks: correct method [1], answer 1.30 m [1]. (c) 3 marks: correct formula v=fλ [1], correct substitution [1], answer 286 m s⁻¹ [1]. (d) 5 marks: correct rearrangement of the given formula [2], correct substitution [2], answer 53.2 N [1].

(a) Constructive interference occurs at a point where the path difference between the waves arriving from the two sources is a whole number of wavelengths: \( \text{path difference} = n\lambda \), where \(n=0,1,2,\ldots\)

(b) \( \text{path difference} = 5.10-3.74 = 1.36 \text{ m} \)

(c) \( \text{number of wavelengths} = \frac{1.36}{0.680} = 2.00 \)
Since the path difference is exactly a whole number (2) of wavelengths, P is a point of constructive interference (the waves arrive in phase, reinforcing each other, so a loud sound is heard).

(d) If the two speakers were no longer coherent (i.e. the phase relationship between them was continuously and randomly changing, rather than being constant), the interference pattern would not be stable. At P, the loudness would no longer be consistently loud or consistently quiet, but would instead fluctuate randomly and rapidly over time, as the phase relationship between the two arriving waves varies unpredictably — a fixed, observable interference pattern requires a constant phase difference between the sources, i.e. coherence.

(a) 2 marks: correct condition stated, path difference = whole number of wavelengths. (b) 2 marks: correct subtraction, answer 1.36 m. (c) 4 marks: correct division [1], answer 2.00 [1], correctly identifies this as a whole number [1], correct conclusion 'constructive' with reasoning [1]. (d) 3 marks: correctly states no stable pattern / loudness fluctuates [2], correct reasoning referencing the loss of a constant phase relationship (coherence) [1].

(a) Rearranging \(d\sin\theta=n\lambda\) for \(n=2\), \(\theta=23.0°\):
\( d = \frac{n\lambda}{\sin\theta} = \frac{2\times(650\times10^{-9})}{\sin23.0°} = \frac{1.30\times10^{-6}}{0.3907} = 3.33\times10^{-6} \text{ m} \)

(b) Lines per mm \( = \frac{1}{d \text{ (in mm)}} = \frac{1}{3.33\times10^{-3}} = 301 \text{ lines per mm} \)

(c) Using \(d\sin\theta=n\lambda\) with \(n=1\):
\( \sin\theta_1 = \frac{\lambda}{d} = \frac{650\times10^{-9}}{3.33\times10^{-6}} = 0.1954 \)
\( \theta_1 = 11.3° \)

(d) A grating with more lines per millimetre has a smaller spacing \(d\) between adjacent lines. Since \( \sin\theta = \frac{n\lambda}{d} \), for a given order \(n\) and fixed wavelength \(\lambda\), a smaller value of \(d\) produces a larger value of \(\sin\theta\), and therefore a larger diffraction angle \(\theta\). So using a grating with more lines per mm would cause all the diffraction maxima to be observed at larger angles (spread further apart) than with the original grating.

(a) 4 marks: correct rearrangement [1], correct substitution [2], answer 3.33×10⁻⁶ m [1]. (b) 2 marks: correct method (reciprocal with unit conversion) [1], answer ≈301 lines/mm [1]. (c) 3 marks: correct rearrangement for n=1 [1], correct substitution (ecf) [1], answer 11.3° [1]. (d) 2 marks: correctly identifies more lines/mm gives smaller d [1], correctly concludes angles increase, with reference to the equation [1].

(a) Snell's law states that \( n_1\sin\theta_1 = n_2\sin\theta_2 \), where \(n_1\) and \(n_2\) are the refractive indices of the first and second media, and \(\theta_1\), \(\theta_2\) are the angles of incidence and refraction respectively, measured from the normal.

(b) Taking air as medium 1 (\(n_1=1.00\)) and water as medium 2 (\(n_2=1.33\)):
\( (1.00)\sin50° = 1.33\sin\theta_2 \)
\( \sin\theta_2 = \frac{\sin50°}{1.33} = \frac{0.766}{1.33} = 0.576 \)
\( \theta_2 = 35.2° \)

(c) For light travelling from water (denser, \(n=1.33\)) into air (\(n=1.00\)), using \( \sin C = \frac{1}{n} \):
\( \sin C = \frac{1}{1.33} = 0.752 \)
\( C = 48.8° \)

(d) At angles of incidence (measured from the normal to the water surface, from underneath) greater than the critical angle of 48.8°, light travelling from the water towards the surface undergoes total internal reflection — none of the light escapes into the air above; all of it is reflected back down into the water. So, looking up at the surface at these steep angles, the fish would not see a view of the world above the water at all; instead, it would see a reflection of the scene underwater (for example, a reflected view of the pool or lake bed, or of itself), just as if the water's surface were acting as a mirror.

(a) 1 mark: correct statement of Snell's law with correct terms defined. (b) 3 marks: correct substitution with n1=1.00 for air [1], correct rearrangement [1], answer 35.2° [1]. (c) 3 marks: correct formula sinC=1/n [1], correct substitution [1], answer 48.8° [1]. (d) 4 marks: correctly identifies total internal reflection occurs at these angles [1], correctly concludes no light escapes/is transmitted to the fish's eye from above [1], correctly describes the fish would see a reflection of the underwater scene [2].

(a) The threshold frequency is the minimum frequency of electromagnetic radiation that can just cause photoelectric emission from a given surface; below this frequency, no photoelectrons are emitted, no matter how intense the light, because each individual photon does not carry enough energy to overcome the work function.

(b) At the threshold frequency, a photon's energy exactly equals the work function: \( hf_0 = \phi \)
\( f_0 = \frac{\phi}{h} = \frac{3.60\times10^{-19}}{6.63\times10^{-34}} = 5.43\times10^{14} \text{ Hz} \)

(c) \( \lambda_0 = \frac{c}{f_0} = \frac{3.00\times10^{8}}{5.43\times10^{14}} = 5.53\times10^{-7} \text{ m} = 553 \text{ nm} \)

(d) Energy of a photon of 600 nm light:
\( E = \frac{hc}{\lambda} = \frac{(6.63\times10^{-34})(3.00\times10^{8})}{600\times10^{-9}} = 3.32\times10^{-19} \text{ J} \)
Since \(3.32\times10^{-19}\) J is less than the work function, \(3.60\times10^{-19}\) J, each photon of 600 nm light does not carry enough energy to remove an electron from the metal surface. This is consistent with the threshold wavelength found in (c), 553 nm: since 600 nm is longer (lower energy/frequency) than the threshold wavelength, no photoelectrons would be emitted, regardless of the intensity of the light.

(a) 2 marks: correct definition referring to minimum frequency for emission [1], correctly links to individual photon energy vs work function [1]. (b) 3 marks: correct equation hf0=φ [1], correct substitution [1], answer 5.43×10¹⁴ Hz [1]. (c) 2 marks: correct formula λ0=c/f0 [1], answer 553 nm [1]. (d) 4 marks: correct calculation of photon energy at 600 nm [2], correct comparison with work function [1], correct conclusion 'no emission' with valid reasoning [1].

(a) The electron moves from the ground state (\(-13.6\) eV) to the \(n=2\) level (\(-3.40\) eV), which is a higher (less negative) energy. Since the electron gains energy in this transition, the atom must absorb a photon to provide this energy (rather than emitting one, which would occur if the electron fell to a lower energy level).

(b) \( \Delta E = -3.40-(-13.6) = 10.2 \text{ eV} \)
\( \Delta E = 10.2 \times (1.60\times10^{-19}) = 1.63\times10^{-18} \text{ J} \)

(c) Using \( \Delta E = \frac{hc}{\lambda} \):
\( \lambda = \frac{hc}{\Delta E} = \frac{(6.63\times10^{-34})(3.00\times10^{8})}{1.63\times10^{-18}} = 1.22\times10^{-7} \text{ m} = 122 \text{ nm} \)

(d) A wavelength of 122 nm is shorter than the visible spectrum (400-700 nm), placing it in the ultraviolet region of the electromagnetic spectrum.

(e) The electron in the atom can only occupy certain fixed (discrete/quantised) energy levels, not any arbitrary energy. For the electron to move from one level to another, it must absorb a photon whose energy exactly matches the energy difference between those two specific levels — any photon with a different energy cannot be absorbed, because there is no available energy level for the electron to move to. This is why only photons of certain precise wavelengths (corresponding to the exact energy gaps between levels) can be absorbed, rather than a continuous range of wavelengths.

(a) 2 marks: correctly states 'absorbs' [1], correct reasoning based on energy increasing [1]. (b) 3 marks: correct energy difference in eV [1], correct conversion to J [1], answer 1.63×10⁻¹⁸ J [1]. (c) 3 marks: correct formula [1], correct substitution (ecf) [1], answer 122 nm [1]. (d) 1 mark: correctly states ultraviolet. (e) 2 marks: correctly refers to discrete/fixed energy levels [1], correctly explains photon energy must exactly match the gap between two levels [1].

(a) When an object is placed within the focal length of a converging lens, the rays of light diverging from the object are not converged enough by the lens to actually meet on the far side; instead, when traced backwards, the refracted rays appear to diverge from a point on the same side of the lens as the object. This produces a virtual image that is upright and magnified, which is the useful property that allows the lens to work as a magnifying glass. (If the object were placed beyond the focal length, a real, inverted image would form instead, which is not useful for magnifying an object to view directly.)

(b) Using \( \frac{1}{f}=\frac{1}{v}+\frac{1}{u} \):
\( \frac{1}{v} = \frac{1}{f}-\frac{1}{u} = \frac{1}{10}-\frac{1}{6} = \frac{3-5}{30} = -\frac{2}{30} = -\frac{1}{15} \)
\( v = -15 \text{ cm} \)

(c) Magnification \( = \left|\frac{v}{u}\right| = \frac{15}{6} = 2.5 \)
Image height \( = 2.5\times1.5 = 3.75 \text{ cm} \)

(d) Since \(v\) is negative, the image forms on the same side of the lens as the object (where the refracted rays only appear to come from), so the image is virtual. A virtual image formed in this way by a converging lens (object within the focal length) is upright (the correct way up, not flipped).

(a) 2 marks: correct explanation that rays don't actually converge / appear to diverge from a point on the object side [1], correct link to a virtual, upright, magnified image resulting [1]. (b) 4 marks: correct rearrangement of lens equation [1], correct substitution [2], answer v=−15 cm [1]. (c) 3 marks: correct method for magnification (using magnitude) [1], answer 2.5 [1], correct image height 3.75 cm [1]. (d) 2 marks: correctly states virtual (with reason, negative v) [1], correctly states upright [1].

(a) \( z = \frac{v}{c} = \frac{1.2\times10^7}{3.00\times10^8} = 0.0400 \)

(b) Using \( z = \frac{\Delta\lambda}{\lambda} \):
\( \Delta\lambda = z\lambda = 0.0400\times486.1 = 19.4 \text{ nm} \)
\( \lambda_{observed} = \lambda + \Delta\lambda = 486.1+19.4 = 505.5 \text{ nm} \)

(c) Using \( v = H_0d \):
\( d = \frac{v}{H_0} = \frac{1.2\times10^7}{2.4\times10^{-18}} = 5.0\times10^{24} \text{ m} \)

(d) According to the Big Bang model, space itself is expanding uniformly; as light from a distant galaxy travels through this expanding space towards Earth, its wavelength is stretched (increased) along with the space it is travelling through. Light from a more distant galaxy has to travel through a greater amount of expanding space to reach Earth, and so is stretched by a proportionally greater amount than light from a closer galaxy — this is why more distant galaxies show greater cosmological redshifts. This relationship (greater distance corresponding to a greater recession speed, and hence greater redshift) is exactly what is expressed by Hubble's Law, \(v=H_0d\).

(a) 2 marks: correct formula [1], answer 0.0400 [1]. (b) 3 marks: correct rearrangement to find Δλ [1], correct calculation of Δλ [1], correct final observed wavelength 505.5 nm [1]. (c) 3 marks: correct formula d=v/H0 [1], correct substitution [1], answer 5.0×10²⁴ m [1]. (d) 3 marks: correct reference to expansion of space stretching wavelength [1], correct explanation that greater distance means light travels through more expanding space [1], correct link to Hubble's Law as expressing this relationship [1].

(a) \( a\times m \): \(5.0\times0.40=2.0\); \(4.0\times0.50=2.0\); \(3.0\times0.67=2.01\); \(2.0\times1.00=2.0\) (all in N).

(b) In each of the four trials, the product \(a\times m\) is constant (approximately 2.0 N, matching the constant applied force), within the precision of the data. Since \(a\times m\) is constant, this means \(a=\frac{\text{constant}}{m}\), i.e. \(a\) is inversely proportional to \(m\) — as \(m\) increases, \(a\) decreases proportionally, which is exactly the pattern shown in the data (e.g. doubling \(m\) from 0.50 kg to 1.00 kg halves \(a\) from 4.0 to 2.0 m s⁻²).

(c) Since \(F=ma\), rearranging gives \(a = F\times\frac{1}{m}\); this is of the form \(y=(\text{gradient})x\) when \(a\) is plotted against \(\frac{1}{m}\), so the gradient of this graph represents the constant applied force, \(F\). Based on the data, the gradient (and hence the applied force) is approximately 2.0 N.

(d) Repeating each measurement (for each value of \(m\)) several times and taking a mean value of \(a\) would reduce the effect of random timing/measurement errors and improve the reliability of the results. (Other valid suggestions include: using a wider range of masses, or compensating for friction in the system before taking readings.)

(a) 3 marks: at least 3 of the 4 products correctly calculated [2], all four correct and consistently ≈2.0 N [1]. (b) 3 marks: correctly identifies a×m is constant across trials [1], correctly links this to a∝1/m [2]. (c) 2 marks: correctly identifies the gradient as the applied force F [1], correct value ≈2.0 N stated [1]. (d) 2 marks: valid, relevant suggestion given [1] with correct justification for improved reliability [1].

(a) Rearranging \( v^2=u^2+2gs \):
\( g = \frac{v^2-u^2}{2s} = \frac{(6.405)^2-(1.50)^2}{2\times1.976} = \frac{41.02-2.25}{3.952} = \frac{38.77}{3.952} = 9.81 \text{ m s}^{-2} \)

(b) Rearranging \( v=u+gt \):
\( g = \frac{v-u}{t} = \frac{6.405-1.50}{0.500} = \frac{4.905}{0.500} = 9.81 \text{ m s}^{-2} \)

(c) Both methods give a value of \(g=9.81\) m s⁻², which is in excellent (exact, to 3 s.f.) agreement with the accepted value of \(9.81\) m s⁻². The close agreement between the two independent methods also gives confidence that the measurements (speeds, distance and time) were self-consistent and free from major systematic error.

(d) A possible source of error is air resistance acting on the ball bearing as it falls, which would tend to make the measured value of \(g\) slightly lower than the true value (though for a dense ball bearing falling a short distance, this effect is likely to be very small). This could be reduced by using a denser, more aerodynamic ball bearing, or by minimising the fall distance/using a vacuum tube (though the latter is not practical in a school laboratory). Another possible source of error is the light gates being triggered slightly inaccurately if the ball does not fall centrally through them; this could be reduced by carefully aligning the light gates directly above one another using a plumb line.

(a) 3 marks: correct rearrangement [1], correct substitution [1], answer 9.81 m s⁻² [1]. (b) 3 marks: correct rearrangement [1], correct substitution [1], answer 9.81 m s⁻² [1]. (c) 2 marks: correctly notes both values agree closely with the accepted value [1], valid comment on the significance of the two independent methods agreeing [1]. (d) 2 marks: valid source of error identified [1] with a sensible, correctly linked method of reduction [1].

(a) Using \(R=\frac{V}{I}\):
At \(V=1.0\) V: \(R=\frac{1.0}{0.50}=2.0\) Ω
At \(V=2.0\) V: \(R=\frac{2.0}{0.90}=2.22\) Ω
At \(V=3.0\) V: \(R=\frac{3.0}{1.20}=2.5\) Ω
At \(V=4.0\) V: \(R=\frac{4.0}{1.40}=2.86\) Ω

(b) The resistance of the filament lamp increases as the potential difference (and current) increases.

(c) As the current through the filament increases, more electrical energy is dissipated as heat each second (since power \(=I^2R\)), causing the temperature of the (metal) filament to rise. In a metal, increasing temperature causes the positive ions in the lattice to vibrate with greater amplitude, making it more difficult for the free (conduction) electrons to pass through the lattice without colliding with them; this increased rate of collision increases the resistance of the filament as its temperature rises.

(d) For a metal wire at constant temperature, the graph of \(I\) against \(V\) is a straight line through the origin (since resistance is constant, obeying Ohm's law — this is 'ohmic' behaviour). For the filament lamp, since its resistance increases as \(V\) (and current, and hence temperature) increases, the graph of \(I\) against \(V\) is a curve that becomes progressively less steep at higher values of \(V\) (the gradient, which represents \(\frac{1}{R}\), decreases as \(V\) increases) — this is 'non-ohmic' behaviour.

(a) 3 marks: at least 3 of 4 values correct [2], all four correct [1] (2.0, 2.22, 2.5, 2.86 Ω). (b) 1 mark: correctly states resistance increases. (c) 3 marks: correctly links increasing current to increasing temperature (via power dissipation) [1], correctly explains increased lattice vibration [1], correctly links this to increased electron collision/resistance [1]. (d) 3 marks: correctly describes straight line through origin for a wire at constant temperature [1], correctly describes curve (decreasing gradient) for the filament lamp [1], correct reference to ohmic vs non-ohmic behaviour [1].

(a) Using \( \frac{1}{f}=\frac{1}{u}+\frac{1}{v} \):
For \(u=15.0, v=30.0\): \( \frac{1}{f}=\frac{1}{15.0}+\frac{1}{30.0}=\frac{2+1}{30.0}=\frac{3}{30.0} \Rightarrow f=10.0\) cm
For \(u=20.0, v=20.0\): \( \frac{1}{f}=\frac{1}{20.0}+\frac{1}{20.0}=\frac{2}{20.0} \Rightarrow f=10.0\) cm
For \(u=30.0, v=15.0\): \( \frac{1}{f}=\frac{1}{30.0}+\frac{1}{15.0}=\frac{3}{30.0} \Rightarrow f=10.0\) cm

(b) Mean \( f = \frac{10.0+10.0+10.0}{3} = 10.0 \) cm

(c) Using a graph of \(\frac{1}{v}\) against \(\frac{1}{u}\) (which, rearranging the lens equation, gives \(\frac{1}{v}=-\frac{1}{u}+\frac{1}{f}\), a straight line of gradient \(-1\) and intercept \(\frac{1}{f}\)) allows a best-fit line to be drawn through all the data points; this averages out the effect of random errors across the whole data set more effectively than simply averaging a small number of individually calculated \(f\) values, and also makes it easy to visually identify any anomalous (outlier) readings that do not fit the trend, so they can be checked or excluded.

(d) A diverging lens always produces a virtual image (for a real object), regardless of the object distance. A virtual image cannot be projected onto and captured on a screen, because the light rays do not actually converge at the image location (they only appear to diverge from it) — a screen can only show an image where light rays actually meet. Since this method relies on capturing a sharp real image on a screen, it cannot be used directly for a diverging lens.

(a) 4 marks: correct method shown for all three pairs [2], all three values of f=10.0 cm correct [2]. (b) 1 mark: correct mean, 10.0 cm (ecf). (c) 2 marks: valid advantage stated (e.g. averages random error across more points / allows identification of anomalies) with correct justification. (d) 3 marks: correctly states a diverging lens always forms a virtual image (for a real object) [1], correctly explains a virtual image cannot be captured on a screen [1], correct overall conclusion linking to why the method fails [1].

(a) \( \text{gradient} = \frac{3.6-0}{0.80-0} = 4.5 \text{ Ω m}^{-1} \)

(b) Since \(R=\frac{\rho L}{A} = \left(\frac{\rho}{A}\right)L\), and for a given wire \(\rho\) (a property of the material) and \(A\) (constant cross-sectional area) are both constant, this equation has the form \(y=mx\) (with \(y=R\), \(x=L\)), representing a straight line through the origin. The gradient of the line, \(m\), is equal to \(\frac{\rho}{A}\).

(c) Since gradient \( = \frac{\rho}{A} \):
\( \rho = \text{gradient} \times A = 4.5 \times (2.0\times10^{-7}) = 9.0\times10^{-7} \text{ Ω m} \)

(a) 4 marks: correct method (rise/run using given point) [2], answer 4.5 [1], correct unit Ω m⁻¹ [1]. (b) 4 marks: correctly rearranges R=ρL/A to y=mx form [1], correctly identifies ρ and A as constant for a given wire [1], correctly concludes straight line through origin [1], correctly identifies gradient=ρ/A [1]. (c) 2 marks: correct rearrangement [1], answer 9.0×10⁻⁷ Ω m (ecf) [1].

(a) \( \%\text{ uncertainty in mass} = \frac{0.5}{250.0}\times100 = 0.2\% \)

(b) \( \%\text{ uncertainty in volume} = \frac{0.5}{45.0}\times100 = 1.11\% \)

(c) Since density \(=\frac{\text{mass}}{\text{volume}}\) is calculated by division, the percentage uncertainties in the mass and volume are added:
\( \%\text{ uncertainty in density} = 0.2\%+1.11\% = 1.31\% \)

(d) \( \text{density} = \frac{250.0}{45.0} = 5.56 \text{ g cm}^{-3} \)
Absolute uncertainty \( = 1.31\%\text{ of }5.56 = 0.0131\times5.56 = 0.073 \text{ g cm}^{-3} \)
So the final answer is \( \text{density} = (5.56\pm0.07) \text{ g cm}^{-3} \)

(a) 2 marks: correct formula [1], answer 0.2% [1]. (b) 2 marks: correct formula [1], answer 1.11% [1]. (c) 2 marks: correct rule identified (addition for division) [1], answer 1.31% (ecf) [1]. (d) 4 marks: correct density calculation [1], correct absolute uncertainty calculation (ecf) [2], correctly and consistently rounded final answer with uncertainty [1].

(a) Squaring both sides of \( T=2\pi\sqrt{\frac{L}{g}} \):
\( T^2 = 4\pi^2\frac{L}{g} = \left(\frac{4\pi^2}{g}\right)L \)
This is of the form \(y=mx\) (with \(y=T^2\), \(x=L\)), so a graph of \(T^2\) against \(L\) is a straight line through the origin, with gradient \( m = \frac{4\pi^2}{g} \).

(b) \( \text{gradient} = \frac{3.62-0}{0.90-0} = 4.02 \text{ s}^2\text{m}^{-1} \)

(c) Rearranging \( \text{gradient}=\frac{4\pi^2}{g} \):
\( g = \frac{4\pi^2}{\text{gradient}} = \frac{4\pi^2}{4.02} = 9.82 \text{ m s}^{-2} \)

(d) The value obtained, \(9.82\) m s⁻², is in excellent agreement with the accepted value of \(9.81\) m s⁻² (a difference of only about 0.1%), which is well within the precision expected from this method, giving confidence in both the experimental technique and the theoretical relationship used.

(a) 3 marks: correct squaring of the equation [1], correctly rearranged into y=mx form [1], correctly identifies gradient=4π²/g [1]. (b) 3 marks: correct method (rise/run) [1], correct substitution [1], answer 4.02 s² m⁻¹ [1]. (c) 3 marks: correct rearrangement [1], correct substitution (ecf) [1], answer 9.82 m s⁻² [1]. (d) 1 mark: valid comment on the close agreement with the accepted value.

(a) The parachutist's velocity increases throughout the whole 10 s period (never decreasing), but the rate of increase becomes smaller and smaller as time goes on: the velocity increases by 15 m s⁻¹ in the first 2 s, but by only 0.5 m s⁻¹ in the final 2 s shown. This means the acceleration is large at the start and steadily decreases towards zero, as the velocity approaches what appears to be a constant (terminal) value of around 30 m s⁻¹.

(b) The gradient of the tangent gives the instantaneous acceleration at \(t=2.0\) s:
\( a = \frac{22.0-10.0}{3.0-1.0} = \frac{12.0}{2.0} = 6.0 \text{ m s}^{-2} \)

(c) Using the trapezium rule with \(\Delta t = 2.0\) s for each interval:
\(0\)–\(2\) s: \(\frac{1}{2}(0+15)(2.0) = 15.0\) m
\(2\)–\(4\) s: \(\frac{1}{2}(15+24)(2.0) = 39.0\) m
\(4\)–\(6\) s: \(\frac{1}{2}(24+28)(2.0) = 52.0\) m
\(6\)–\(8\) s: \(\frac{1}{2}(28+29.5)(2.0) = 57.5\) m
\(8\)–\(10\) s: \(\frac{1}{2}(29.5+30)(2.0) = 59.5\) m
Total distance \( = 15.0+39.0+52.0+57.5+59.5 = 223 \text{ m} \)

(d) As the parachutist's downward speed increases, the upward air resistance (drag) force acting on them and their parachute also increases. Terminal velocity is reached when the upward air resistance force has increased until it exactly equals the parachutist's weight (the downward force due to gravity); at this point, the resultant force on the parachutist is zero, so (by Newton's second law) their acceleration is zero and they continue to fall at this constant maximum speed.

(a) 2 marks: correctly identifies velocity always increasing but at a decreasing rate [1], correctly identifies acceleration decreasing towards zero / approaching terminal velocity [1]. (b) 3 marks: correct method (gradient of tangent) [1], correct substitution [1], answer 6.0 m s⁻² [1]. (c) 4 marks: correct method (trapezium rule) [1], at least 4 of 5 individual areas correct [2], correct total 223 m [1]. (d) 1 mark: correctly explains terminal velocity occurs when air resistance equals weight, giving zero resultant force/acceleration.

(a) Since the micrometer reads \(+0.02\) mm too high even with nothing between the jaws, this amount must be subtracted from every reading it gives:
\( \text{true diameter} = 0.48-0.02 = 0.46 \text{ mm} \)

(b) This is a systematic error, not a random error. A zero error affects every single reading taken with the instrument by the same fixed amount, in the same direction (here, every reading is 0.02 mm too high) — this is the defining feature of a systematic error, whereas a random error would cause readings to scatter unpredictably both above and below the true value.

(c) Area is proportional to diameter squared (\(A=\pi(d/2)^2\)):
Uncorrected area: \( A_{wrong} \propto (0.48)^2 = 0.2304 \)
Corrected (true) area: \( A_{true} \propto (0.46)^2 = 0.2116 \)
Percentage too large \( = \frac{0.2304-0.2116}{0.2116}\times100 = \frac{0.0188}{0.2116}\times100 = 8.9\% \)
(This shows that a relatively small 0.02 mm zero error leads to a much larger, roughly 9%, error in the calculated area — and hence in any resistivity calculated using it — because area depends on the diameter squared.)

(d) Before taking any measurements of the wire, the student should have closed the micrometer's jaws completely (with nothing between them) and checked the reading on the scale. If this reading is not exactly zero, as was the case here (+0.02 mm), this reveals the presence of a zero error, and the value of this error can be recorded and then either subtracted from (or added to, depending on its sign) all subsequent readings, or the micrometer can be physically re-zeroed if it has an adjustable zero.

(a) 2 marks: correct method (subtracting the zero error) [1], answer 0.46 mm [1]. (b) 2 marks: correctly identifies 'systematic' [1], correct reasoning (affects all readings consistently/same direction) [1]. (c) 3 marks: correctly recognises area depends on diameter squared [1], correct calculation of both areas or their ratio [1], answer ≈8.9% [1]. (d) 3 marks: correctly describes checking the reading with the jaws fully closed (before use) [2], correctly explains this reveals/quantifies the zero error [1].