2025 Cambridge IGCSE Physics (0625) Practice Paper with Answers

When the object is first dropped, only weight acts downwards, causing it to accelerate. As its velocity increases, the upward air resistance also increases. This reduces the resultant downward force. Eventually, the air resistance increases to a value equal to the weight, so the net force becomes zero, resulting in zero acceleration (terminal velocity).

1 mark: Identify that weight acts downwards and air resistance acts upwards. 1 mark: State that air resistance increases as velocity increases. 1 mark: Explain that when air resistance equals weight, the resultant force becomes zero, hence acceleration is zero.

The weight of the uniform plank acts at its midpoint, which is \( 1.5\text{ m} \) from either end. The distance from the pivot (at \( 0.5\text{ m} \)) to the center of gravity is \( 1.5\text{ m} - 0.5\text{ m} = 1.0\text{ m} \). The clockwise moment produced by the weight is \( 120\text{ N} \times 1.0\text{ m} = 120\text{ N m} \). The force \( F \) is at the right end, which is at a distance of \( 3.0\text{ m} - 0.5\text{ m} = 2.5\text{ m} \) from the pivot. For equilibrium, the anticlockwise moment must equal the clockwise moment: \( F \times 2.5\text{ m} = 120\text{ N m} \implies F = 48\text{ N} \).

1 mark: Calculate the distance from pivot to center of gravity (1.0 m) or the clockwise moment (120 N m). 1 mark: Use the principle of moments (anticlockwise moment = clockwise moment). 1 mark: Obtain the correct final value of 48 N with unit.

The critical angle is the angle of incidence in the optically denser medium (plastic) for which the angle of refraction in the less dense medium (air) is exactly \( 90^\circ \). For total internal reflection to occur: 1. The light must be traveling from an optically denser medium to an optically less dense medium. 2. The angle of incidence must be greater than the critical angle.

1 mark: Define critical angle as the angle of incidence producing a 90-degree refraction angle. 1 mark: State that light must travel from a more dense to a less dense medium. 1 mark: State that the angle of incidence must exceed the critical angle.

Taking the direction to the right as positive: Initial momentum of Ball A: \( p_A = 0.20\text{ kg} \times 4.0\text{ m/s} = 0.80\text{ kg m/s} \). Initial momentum of Ball B: \( p_B = 0.30\text{ kg} \times (-2.5\text{ m/s}) = -0.75\text{ kg m/s} \). Total initial momentum: \( p_{\text{initial}} = 0.80 - 0.75 = 0.05\text{ kg m/s} \). Total mass after collision: \( M = 0.20 + 0.30 = 0.50\text{ kg} \). By conservation of momentum: \( p_{\text{final}} = p_{\text{initial}} \implies 0.50 \times v = 0.05 \implies v = 0.10\text{ m/s} \). Since the velocity is positive, the direction of motion is to the right.

1 mark: Calculate initial momentum of Ball A (0.80) and Ball B (-0.75). 1 mark: Calculate net initial momentum (0.05 kg m/s). 1 mark: Equate to final momentum and calculate magnitude of velocity (0.10 m/s). 1 mark: State correct direction (to the right).

First, find the equivalent resistance of the two \( 30\ \Omega \) parallel resistors: \( R_p = \frac{30 \times 30}{30 + 30} = 15\ \Omega \). Next, calculate the total resistance of the circuit by adding the series resistor: \( R_{\text{total}} = R_p + 15\ \Omega = 15 + 15 = 30\ \Omega \). Finally, apply Ohm's law to find the total current: \( I = \frac{V}{R_{\text{total}}} = \frac{12\text{ V}}{30\ \Omega} = 0.40\text{ A} \).

1 mark: Calculate parallel equivalent resistance of 15 ohms. 1 mark: Calculate total resistance of 30 ohms. 1 mark: Apply Ohm's law (I = V/R). 1 mark: Obtain 0.40 A with correct unit.

The student fills the metal cube with boiling water so both the dull black and polished silver sides are at the same high temperature. An infrared detector is placed at a fixed distance from the dull black side, and the emission reading is noted. The detector is then placed at the exact same distance from the polished silver side, and its reading is noted. The expected result is a significantly higher reading from the dull black surface, showing that dull black surfaces are far better emitters of thermal radiation than polished silver surfaces.

1 mark: Fill cube with hot water to ensure equal temperatures and place detector at equal distances from both sides. 1 mark: State that the dull black side gives a higher detector reading. 1 mark: Explain that dull black surfaces are better/more effective emitters of thermal (infrared) radiation than shiny/polished silver surfaces.

The number of protons is given by the atomic number, which is \( 83 \). The number of neutrons is the mass number minus the atomic number: \( 210 - 83 = 127 \). In a beta-minus decay, a neutron decays into a proton and an electron (beta particle). This increases the proton number by 1 (to 84) while the mass number remains unchanged (210). Polonium is represented as \( ^{210}_{84}\text{Po} \). The beta particle is represented as \( _{-1}^{0}\beta \) (or \( _{-1}^{0}\text{e} \)). The equation is: \( ^{210}_{83}\text{Bi} \rightarrow ^{210}_{84}\text{Po} + \ _{-1}^{0}\beta \).

1 mark: State correct number of protons (83) and neutrons (127). 1 mark: Show the correct reactant (Bi-210) and beta particle symbol. 1 mark: Determine correct nucleon number (210) and proton number (84) for Polonium. 1 mark: Correctly balanced equation.

The phenomenon is called redshift. It indicates that the distant galaxy is moving away from the Earth (observer). Because almost all distant galaxies exhibit redshift, and those further away show larger redshifts (meaning they are receding faster), this implies the entire Universe is expanding. Extrapolating this expansion backwards in time leads to the conclusion that the Universe originated from a single, extremely hot and dense state, which is the core premise of the Big Bang Theory.

1 mark: Identify the phenomenon as redshift. 1 mark: State that it indicates the galaxy is moving away from the observer/Earth. 1 mark: Explain that the expansion of the universe (demonstrated by redshift of distant galaxies) supports the origin from a single point/dense state (Big Bang).

Let the direction of the rebound be positive:
Initial velocity, \(u = -25\text{ m/s}\)
Final velocity, \(v = +35\text{ m/s}\)

(i) Change in momentum, \(\Delta p = m(v - u) = 0.060\text{ kg} \times (35\text{ m/s} - (-25\text{ m/s})) = 0.060 \times 60 = 3.6\text{ kg m/s}\).

(ii) Average force, \(F = \frac{\Delta p}{\Delta t} = \frac{3.6\text{ kg m/s}}{4.0 \times 10^{-3}\text{ s}} = 900\text{ N}\).

1 mark: \(\Delta p = m \Delta v\) formula or showing \(0.060 \times (35 - (-25))\)
1 mark: \(3.6\text{ kg m/s}\) (or \(-3.6\text{ kg m/s}\) depending on chosen sign convention)
1 mark: \(F = \frac{\Delta p}{\Delta t}\) formula or showing \(\frac{3.6}{4.0 \times 10^{-3}}\)
1 mark: \(900\text{ N}\)

(i) Snell's law: \(n = \frac{\sin(i)}{\sin(r)}\)
\(\sin(r) = \frac{\sin(40^\circ)}{1.52} = \frac{0.6428}{1.52} = 0.4229\)
\(r = \arcsin(0.4229) \approx 25.0^\circ\)

(ii) Calculate the critical angle \(c\):
\(\sin(c) = \frac{1}{n} = \frac{1}{1.52} = 0.6579\)
\(c = \arcsin(0.6579) \approx 41.1^\circ\)
Since the angle of incidence (\(50^\circ\)) is greater than the critical angle (\(41.1^\circ\)), the light cannot refract out into the air. Instead, it undergoes total internal reflection inside the plastic.

1 mark: Use of \(n = \frac{\sin(i)}{\sin(r)}\)
1 mark: \(25^\circ\) (accept range \(24.8^\circ - 25.2^\circ\))
1 mark: Use of \(\sin(c) = \frac{1}{n}\) to calculate critical angle \(c \approx 41^\circ\)
1 mark: Statement that the angle of incidence is greater than the critical angle
1 mark: Final conclusion of total internal reflection

(i) Dull black surfaces are the best absorbers of infrared radiation from the Sun, maximizing heat absorption.
(ii) Shiny silver-colored surfaces are excellent reflectors of infrared radiation. Any radiation that misses the pipe initially is reflected back onto it.
(iii) Air trapped between the double-glazed glass panels is a poor thermal conductor, which reduces conduction and prevents convection currents, minimizing heat losses to the colder external environment.

1 mark: Dull black is a very good / the best absorber of infrared radiation
1 mark: Shiny silver is a very good reflector / poor absorber of radiation
1 mark: Explaining silver plate reflects radiation back onto the pipe
1 mark: Trapped air / double-glazing is a poor conductor of thermal energy
1 mark: Reduces heat loss by conduction and convection to the outside

(i) Number of half-lives in \(48\text{ hours}\) is \(N = \frac{48}{12} = 4\).
Number of remaining nuclei = \(4.8 \times 10^{14} \times \left(\frac{1}{2}\right)^4 = 4.8 \times 10^{14} \times \frac{1}{16} = 3.0 \times 10^{13}\).

(ii) Number of half-lives in \(36\text{ hours}\) is \(N = \frac{36}{12} = 3\).
Count rate after 3 half-lives = \(1600 \times \left(\frac{1}{2}\right)^3 = \frac{1600}{8} = 200\text{ counts/minute}\).

1 mark: Correctly identifying 4 half-lives for part (i)
1 mark: \(3.0 \times 10^{13}\) (or \(3 \times 10^{13}\))
1 mark: Correctly identifying 3 half-lives for part (ii)
1 mark: \(200\text{ counts/minute}\)

(i) Current \(I = \frac{Q}{t} = \frac{15\text{ C}}{1.2 \times 10^{-3}\text{ s}} = 12500\text{ A} = 1.25 \times 10^4\text{ A}\).

(ii) Energy \(E = Q \times V = 15\text{ C} \times 1.2 \times 10^{8}\text{ V} = 1.8 \times 10^{9}\text{ J}\).

1 mark: Use of \(I = \frac{Q}{t}\) or showing substitution \(\frac{15}{1.2 \times 10^{-3}}\)
1 mark: \(12500\text{ A}\) (or \(1.25 \times 10^4\text{ A}\))
1 mark: Use of \(E = QV\) or equivalent energy formula
1 mark: \(1.8 \times 10^9\text{ J}\) (or \(1800\text{ MJ}\))

(i) Wavelength increases towards the red end of the spectrum, which is redshift.

(ii) Since the observed wavelength is longer than the reference laboratory wavelength, the galaxy is moving away from Earth.

(iii) Wavelength change \(\Delta \lambda = 658.2\text{ nm} - 656.3\text{ nm} = 1.9\text{ nm}\).
Using \(\frac{\Delta \lambda}{\lambda_0} = \frac{v}{c}\):
\(\frac{1.9}{656.3} = \frac{v}{3.0 \times 10^8}\)
\(v = 3.0 \times 10^8 \times \frac{1.9}{656.3} \approx 8.685 \times 10^5\text{ m/s} \approx 8.7 \times 10^5\text{ m/s}\).

1 mark: Redshift
1 mark: Explaining the galaxy is moving away because observed wavelength is longer
1 mark: Calculating \(\Delta \lambda = 1.9\text{ nm}\)
1 mark: Correct substitution into redshift formula \(\frac{\Delta \lambda}{\lambda} = \frac{v}{c}\)
1 mark: \(8.7 \times 10^5\text{ m/s}\) (allow \(8.68 \times 10^5\text{ m/s}\) to \(8.70 \times 10^5\text{ m/s}\))

(i) Human hearing range is \(20\text{ Hz}\) to \(20\text{000 Hz}\) (\(20\text{ kHz}\)). A frequency of \(25\text{ kHz}\) is above the upper limit of human hearing and is therefore ultrasonic.

(ii) The sound travels twice the depth: \(2d = v \times t\).
\(2d = 1500 \times 0.80 = 1200\text{ m}\).
Depth \(d = 600\text{ m}\).

(iii) Wavelength \(\lambda = \frac{v}{f} = \frac{1500\text{ m/s}}{25 \times 10^3\text{ Hz}} = 0.060\text{ m}\) (or \(6.0\text{ cm}\)).

1 mark: Stating range of human hearing is up to \(20\text{ kHz}\) and noting \(25\text{ kHz}\) is outside/above this
1 mark: Use of \(2d = vt\) (accounting for double path)
1 mark: Depth = \(600\text{ m}\)
1 mark: Use of \(v = f \lambda\) rearranged as \(\lambda = \frac{v}{f}\)
1 mark: \(0.060\text{ m}\) (or \(6.0\text{ cm}\))

(i) Using the transformer equation:
\(\frac{V_p}{V_s} = \frac{N_p}{N_s} \implies \frac{240}{V_s} = \frac{1200}{60}\)
\(V_s = 240 \times \frac{60}{1200} = 12\text{ V}\).

(ii) Power in primary = Power in secondary (since 100% efficient):
\(P_p = V_p \times I_p = 24\text{ W}\)
\(240 \times I_p = 24\)
\(I_p = 0.10\text{ A}\).

1 mark: Correct use of transformer ratio equation
1 mark: \(12\text{ V}\)
1 mark: equating primary input power to secondary output power (\(V_p I_p = P_s\))
1 mark: \(0.10\text{ A}\)

Using the equation of motion for constant acceleration starting from rest: \(s = ut + \frac{1}{2}at^2\). Since it starts from rest, \(u = 0\). Therefore, \(s = \frac{1}{2} \times 1.5 \times (4.0)^2 = 0.5 \times 1.5 \times 16 = 12\text{ m}\).

1 mark for using the correct formula or calculating the final velocity (6.0 m/s); 0.238 marks for the correct final distance of 12 m with unit.

By conservation of momentum: \(m_A u_A + m_B u_B = m_A v_A + m_B v_B\). Taking the initial direction of car A as positive: \(u_A = 2.5\text{ m/s}\), \(u_B = 0\), and \(v_A = -0.30\text{ m/s}\) (rebound). Substituting the values: \(120 \times 2.5 + 80 \times 0 = 120 \times (-0.30) + 80 \times v_B\). This simplifies to \(300 = -36 + 80 v_B\), which gives \(336 = 80 v_B\). Thus, \(v_B = 4.2\text{ m/s}\).

1 mark for stating or using the principle of conservation of momentum with correct sign for rebounds: \(120 \times 2.5 = 120 \times (-0.30) + 80 v_B\) (or equivalent); 0.238 marks for the correct final speed of 4.2 m/s with unit.

Conduction relies on the collision of particles (or free electrons) to pass kinetic energy from one to another. In solid glass, particles are closely packed, allowing relatively fast energy transfer. In air, which is a gas, the particles are much further apart. Collisions between particles occur much less frequently, which significantly reduces the rate of thermal energy transfer by conduction.

1 mark for explaining that air is a gas and its particles are much further apart than in solid glass; 0.238 marks for explaining that this separation leads to much less frequent collisions between particles, thereby reducing conduction.

Using Snell's law: \(n = \frac{\sin(i)}{\sin(r)}\). Substituting the angles: \(n = \frac{\sin(48^\circ)}{\sin(29^\circ)} \approx \frac{0.7431}{0.4848} \approx 1.533\). To two decimal places, the refractive index is 1.53.

1 mark for stating and using Snell's law: \(n = \frac{\sin(48^\circ)}{\sin(29^\circ)}\); 0.238 marks for the correct refractive index of 1.53 (or 1.5) with no unit.

The resistance of a wire is given by \(R = \rho \frac{L}{A}\). For the first wire: \(R_1 = \rho \frac{L_1}{A_1} = 8.0\text{ }\Omega\). For the second wire: \(L_2 = 3L_1\) and \(A_2 = 2A_1\). Thus, the resistance of the second wire is: \(R_2 = \rho \frac{L_2}{A_2} = \rho \frac{3L_1}{2A_1} = 1.5 \left(\rho \frac{L_1}{A_1}\right) = 1.5 \times 8.0 = 12\text{ }\Omega\).

1 mark for showing that resistance is proportional to length and inversely proportional to cross-sectional area, leading to the multiplier 1.5; 0.238 marks for the correct final resistance of 12 \(\Omega\) (or ohms).

Determine the number of half-lives that have elapsed: \(800 \xrightarrow{1} 400 \xrightarrow{2} 200 \xrightarrow{3} 100\). So, 3 half-lives have elapsed in 18 hours. Let \(T_{1/2}\) be the half-life: \(3 \times T_{1/2} = 18\text{ hours}\), which gives \(T_{1/2} = \frac{18}{3} = 6.0\text{ hours}\).

1 mark for identifying that the activity has halved 3 times (3 half-lives elapsed); 0.238 marks for the correct half-life of 6.0 hours with appropriate unit.

Redshift occurs when the wavelengths of light from distant galaxies are stretched, shifting towards the red end of the spectrum. This tells us that these galaxies are moving away from Earth. Observations show that more distant galaxies have a greater redshift, meaning they are receding faster. This proportional relationship between distance and speed indicates that space itself is expanding in all directions, supporting the Big Bang theory.

1 mark for explaining that redshift indicates galaxies are moving away from us (longer wavelengths observed); 0.238 marks for explaining that the greater redshift of more distant galaxies shows space itself is expanding.

As the magnet falls, the changing magnetic field cuts through the copper tube, inducing an electromotive force (e.m.f.) and thus an electric current in the tube. According to Lenz's law, the direction of this induced current creates its own magnetic field that opposes the motion of the falling magnet. This creates an upward magnetic force (braking force) on the magnet. When this upward magnetic force plus air resistance equals the downward weight of the magnet, the net force becomes zero, and the magnet falls at a constant terminal velocity.

1 mark for explaining that the changing magnetic field of the falling magnet induces an e.m.f. and current in the copper tube; 0.238 marks for applying Lenz's law to explain that the induced current creates a magnetic field that opposes the magnet's fall, resulting in an upward resistive force.

(a) Since air resistance is negligible, the only force acting on the stone is its weight. Therefore, the acceleration is constant throughout the flight, including at the highest point, and is equal to the acceleration of free fall, which is 9.8 m/s^2 downwards. (b) As the stone rises, its speed decreases and its height increases. This means energy is transferred from its kinetic energy store to its gravitational potential energy store.

Part (a): - 9.8 m/s^2 (or 10 m/s^2) [1 mark] - downwards / towards the center of Earth [1 mark] Part (b): - Energy is transferred from the kinetic store (of the stone) [1 mark] - to the gravitational potential store [1 mark]

(a) First, calculate the extension x of the spring: x = 11.0 cm - 8.0 cm = 3.0 cm = 0.030 m. Using Hooke's Law, F = k * x, we rearrange to find k: k = F / x = 6.0 N / 0.030 m = 200 N/m. (b) The limit of proportionality is the point beyond which the extension of the spring is no longer directly proportional to the applied load.

Part (a): - Recall of F = k * x or k = F / x [1 mark] - Calculation of extension x = 3.0 cm or 0.030 m [1 mark] - Correct calculation of spring constant with correct unit, e.g. 200 N/m or 2.0 N/cm [1 mark] Part (b): - Mention of extension and load/force [1 mark] - No longer directly proportional / linear relationship ceases [1 mark]

(a) Let the direction to the right be positive. Before collision, the total momentum p_initial is: p_initial = m1 * u1 + m2 * u2 = (0.50 * 4.0) + (0.30 * -2.0) = 2.0 - 0.60 = 1.40 kg m/s. After collision, the total momentum p_final is: p_final = m1 * v1 + m2 * v2 = (0.50 * 1.0) + (0.30 * v2) = 0.50 + 0.30 * v2. According to the conservation of momentum, p_initial = p_final: 1.40 = 0.50 + 0.30 * v2 => 0.30 * v2 = 0.90 => v2 = 3.0 m/s. Since the result is positive, the sphere moves to the right. (b) Impulse is defined as the change in momentum of an object.

Part (a): - Recall of conservation of momentum: m1 * u1 + m2 * u2 = m1 * v1 + m2 * v2 [1 mark] - Correct substitution with correct signs (e.g. -2.0 m/s for leftward motion) [1 mark] - Evaluation of v2 = 3.0 m/s [1 mark] - Correct direction: to the right [1 mark] Part (b): - Correct definition: change in momentum [1 mark]

(a) The dull black side is a better emitter of infrared radiation than the shiny silver side. Therefore, thermal energy is radiated away from the dull black side at a higher rate, which increases the rate of cooling of the water near that side. (b) Convection involves the upward movement of hot, less dense air and steam. Placing a lid on the container traps this warm air/steam inside, preventing it from escaping and carrying thermal energy away, thereby stopping convection currents from transferring heat to the room.

Part (a): - Dull black is the better emitter of infrared radiation [1 mark] - This increases the rate of energy emission / cooling of water [1 mark] Part (b): - Lid traps warm air / steam / fluid inside the container [1 mark] - This prevents convection currents from carrying energy to the surroundings [1 mark]

(a) The relation between critical angle c and refractive index n is given by: sin(c) = 1 / n = 1 / 1.5 = 0.667. Rearranging for c: c = arcsin(0.667) = 41.8 degrees. (b) Since the angle of incidence (45 degrees) is greater than the critical angle (41.8 degrees), total internal reflection occurs, and all the light is reflected back into the glass at an angle of 45 degrees.

Part (a): - Recall of sin(c) = 1 / n [1 mark] - Correct substitution sin(c) = 1 / 1.5 [1 mark] - Correct calculation of c = 41.8 degrees (or 42 degrees) [1 mark] Part (b): - State that total internal reflection occurs [1 mark] - Explain that the angle of incidence is greater than the critical angle [1 mark]

(a) Resistance R is given by R = rho * L / A, where L is length and A is cross-sectional area. Since wire Q has twice the length (2L), its resistance doubles due to length. Since it has half the cross-sectional area (A/2), its resistance doubles again due to area. Thus, the resistance of wire Q is 2 * 2 * R = 4R. (b) In a series connection, the current I through both wires is identical. The rate of development of thermal energy (power) is given by P = I^2 * R. Since wire Q has a higher resistance (4R) than wire P (R), wire Q will develop more thermal energy per second.

Part (a): - State that doubling length doubles resistance [1 mark] - State that halving cross-sectional area doubles resistance [1 mark] - State that the final resistance of Q is 4R [1 mark] Part (b): - Identify that wire Q develops more thermal energy per second [1 mark] - Explain that current is the same in series and power depends on resistance (P = I^2 * R) [1 mark]

(a) Using the transformer ratio equation: Vp / Vs = Np / Ns. Substituting the values: 240 / 12 = 6000 / Ns => 20 = 6000 / Ns => Ns = 6000 / 20 = 300 turns. (b) Transformers operate on the principle of electromagnetic induction. A constant direct current produces a static/constant magnetic field. In order to induce an electromotive force (e.m.f.) in the secondary coil, there must be a continuously changing magnetic field, which is only produced by an alternating current (a.c.).

Part (a): - Recall of Vp / Vs = Np / Ns [1 mark] - Correct substitution of values [1 mark] - Correct calculation of Ns = 300 turns [1 mark] Part (b): - Explain that d.c. produces a constant/static magnetic field [1 mark] - Explain that a changing magnetic field is needed to induce an e.m.f. in the secondary coil [1 mark]

(a) We find the number of half-lives elapsed by halving the count rate step-by-step: 240 -> 120 (1 half-life) -> 60 (2 half-lives) -> 30 (3 half-lives). Thus, 3 half-lives correspond to a total time of 30 minutes. Therefore, one half-life is equal to 30 minutes / 3 = 10 minutes. (b) Two sources of naturally occurring background radiation are cosmic rays from outer space and radioactive radon gas emitted from rocks and soil. (c) Background radiation is always present in the environment from sources other than the sample itself. Subtracting the background count rate ensures that the student measures only the radiation emitted by the sample.

Part (a): - Identify that 3 half-lives have elapsed [1 mark] - Equate 3 half-lives to 30 minutes [1 mark] - Calculate half-life = 10 minutes [1 mark] Part (b): - Identify two natural sources: e.g. cosmic rays, radon gas, rocks, food/potassium-40 [2 marks, 1 mark each] Part (c): - Explain that background radiation comes from other sources / to measure the sample's activity only [1 mark]

1. Calculate the impulse (change in momentum) using \(I = F \Delta t\): \(I = 4.0 \text{ N} \times 1.5 \text{ s} = 6.0 \text{ N s}\). 2. Use the impulse-momentum equation: \(F \Delta t = m \Delta v = m(v_f - v_i)\). 3. Substitute the values: \(6.0 = 0.80 \times (v_f - 12)\). 4. Solve for \(v_f\): \(7.5 = v_f - 12 \Rightarrow v_f = 19.5 \text{ m/s}\).

C1: recall or use of \(F \Delta t = m(v - u)\) OR \(I = F \Delta t\)
C1: calculation of impulse as 6.0 N s OR change in velocity as 7.5 m/s
A1: correct final velocity of 19.5 m/s with unit

1. Use Snell's Law: \(n = \frac{\sin i}{\sin r}\). 2. Substitute the given values: \(1.45 = \frac{\sin 45.0^\circ}{\sin r}\). 3. Rearrange to find \(\sin r\): \
\sin r = \frac{\sin 45.0^\circ}{1.45} = \frac{0.7071}{1.45} \approx 0.4877\). 4. Calculate \(r\): \(r = \sin^{-1}(0.4877) \approx 29.2^\circ\).

C1: recall or use of \(n = \frac{\sin i}{\sin r}\)
C1: rearrangement of formula to make \(\sin r\) the subject \(\sin r = \frac{\sin 45^\circ}{1.45}\)
A1: correct calculation of angle of refraction as \(29.2^\circ\) (allow range 29.1 to 29.3)

1. First, find the resistance \(R\) of the wire using Ohm's Law: \(R = \frac{V}{I} = \frac{9.0 \text{ V}}{1.8 \text{ A}} = 5.0\ \Omega\). 2. Use the resistivity formula: \(R = \frac{\rho L}{A}\), where \(\rho\) is the resistivity, \(L\) is the length, and \(A\) is the cross-sectional area. 3. Rearrange the formula to solve for \(\rho\): \(\rho = \frac{R A}{L}\). 4. Substitute the known values: \(\rho = \frac{5.0 \times 1.5 \times 10^{-7}}{2.0} = 3.75 \times 10^{-7}\ \Omega \text{ m}\).

C1: calculation of resistance \(R = 5.0\ \Omega\) using Ohm's Law
C1: recall or use of resistance and resistivity relationship \(R = \frac{\rho L}{A}\)
A1: correct final value of \(3.75 \times 10^{-7}\ \Omega \text{ m}\) (accept \(3.8 \times 10^{-7}\ \Omega \text{ m}\)) with correct unit

1. Use the specific heat capacity formula: \(Q = m c \Delta T\). 2. Substitute the given values to find the change in temperature \(\Delta T\): \(18000 = 0.50 \times 900 \times \Delta T \Rightarrow 18000 = 450 \times \Delta T\). 3. Solve for \(\Delta T\): \(\Delta T = \frac{18000}{450} = 40^\circ\text{C}\). 4. Find the final temperature: \(T_f = T_i + \Delta T = 20^\circ\text{C} + 40^\circ\text{C} = 60^\circ\text{C}\).

C1: recall or use of Specific Heat Capacity formula \(Q = m c \Delta T\)
C1: calculation of change in temperature \(\Delta T = 40^\circ\text{C}\)
A1: correct final temperature of \(60^\circ\text{C}\)

1. State Hubble's Law equation: \(v = H_0 d\). 2. Rearrange the equation to solve for distance \(d\): \(d = \frac{v}{H_0}\). 3. Substitute the given values: \(d = \frac{4.2 \times 10^4 \text{ km/s}}{70 \text{ km/s/Mpc}}\). 4. Calculate the distance: \(d = \frac{42000}{70} = 600 \text{ Mpc}\).

C1: recall of Hubble's Law equation \(v = H_0 d\)
C1: substitution or rearrangement of equation to \(d = \frac{v}{H_0}\)
A1: correct distance of 600 Mpc with correct unit

1. Use the transformer turns ratio equation: \(\frac{V_p}{V_s} = \frac{N_p}{N_s}\). 2. Substitute the given values: \(\frac{240}{12} = \frac{1200}{N_s}\). 3. Solve for \(N_s\): \(20 = \frac{1200}{N_s} \Rightarrow N_s = \frac{1200}{20} = 60\).

C1: recall of transformer equation \(\frac{V_p}{V_s} = \frac{N_p}{N_s}\)
C1: rearrangement of equation to make \(N_s\) the subject \(N_s = N_p \times \frac{V_s}{V_p}\)
A1: correct calculation yielding 60 turns

1. Find the distance traveled during acceleration phase: Since acceleration is uniform, the average speed is \(\frac{0 + 24}{2} = 12 \text{ m/s}\). Distance 1: \(d_1 = v_{\text{avg}} \times t_1 = 12 \times 6.0 = 72 \text{ m}\). 2. Find the distance traveled during constant speed phase: Distance 2: \(d_2 = v \times t_2 = 24 \times 10 = 240 \text{ m}\). 3. Sum the distances: \(d_{\text{total}} = d_1 + d_2 = 72 + 240 = 312 \text{ m}\).

C1: calculation of distance during acceleration as 72 m (or area of triangle on a speed-time graph)
C1: calculation of distance during constant speed as 240 m (or area of rectangle on a speed-time graph)
A1: correct total distance of 312 m with unit

1. Find the number of half-lives that have elapsed: \(800 \xrightarrow{\text{1st half-life}} 400 \xrightarrow{\text{2nd half-life}} 200 \xrightarrow{\text{3rd half-life}} 100\). 2. Therefore, 3 half-lives have elapsed. 3. Relate the elapsed time to the half-life: \(3 \times t_{1/2} = 15 \text{ hours}\). 4. Calculate \(t_{1/2}\): \(t_{1/2} = \frac{15}{3} = 5.0 \text{ hours}\).

C1: recognition that the activity has halved 3 times (from 800 to 100)
C1: setting up the equation \(3 \times t_{1/2} = 15\)
A1: correct calculation of half-life as 5.0 hours with correct unit

(a) Take the initial direction of motion of Glider A (to the right) as the positive direction.
Initial total momentum:
\(p_i = m_A u_A + m_B u_B\)
\(p_i = (0.35\text{ kg} \times 2.4\text{ m/s}) + (0.25\text{ kg} \times 0\text{ m/s}) = 0.84\text{ kg m/s}\)

Final total momentum:
\(p_f = m_A v_A + m_B v_B\)
Since Glider A rebounds to the left, its velocity is \(-0.20\text{ m/s}\):
\(p_f = (0.35\text{ kg} \times -0.20\text{ m/s}) + (0.25\text{ kg} \times v_B) = -0.070\text{ kg m/s} + 0.25 v_B\)

Using the conservation of momentum:
\(p_i = p_f\)
\(0.84 = -0.070 + 0.25 v_B\)
\(0.91 = 0.25 v_B\)
\(v_B = \frac{0.91}{0.25} = 3.64\text{ m/s}\)
Since \(v_B\) is positive, the direction of Glider B is to the right.

(b) The average force exerted on Glider B is given by the rate of change of momentum:
\(F = \frac{\Delta p}{\Delta t} = \frac{m_B v_B - m_B u_B}{t}\)
\(F = \frac{0.25\text{ kg} \times 3.64\text{ m/s} - 0}{0.080\text{ s}}\)
\(F = \frac{0.91\text{ kg m/s}}{0.080\text{ s}} = 11.375\text{ N} \approx 11.4\text{ N}\)

(c) The forces are equal in magnitude and opposite in direction. This is because of Newton's Third Law of motion (action and reaction are equal and opposite).

Part (a):
- C1: Statement or formula for conservation of momentum: \(m_A u_A + m_B u_B = m_A v_A + m_B v_B\)
- C1: Correct substitution of values with correct signs: \(0.35 \times 2.4 + 0 = 0.35 \times (-0.20) + 0.25 \times v_B\)
- A1: Correct final value of velocity \(3.64\text{ m/s}\) (or \(3.6\text{ m/s}\)) and direction specified as 'to the right'

Part (b):
- C1: Formula for force as rate of change of momentum: \(F = \frac{\Delta p}{\Delta t}\) or \(F = \frac{m(v-u)}{t}\)
- C1: Correct substitution of momentum change and time: \(\frac{0.91}{0.080}\) (allow ecf from part a)
- A1: Correct final value \(11.4\text{ N}\) (or \(11\text{ N}\)) with appropriate unit

Part (c):
- B1: Forces are equal in magnitude
- B1: Forces are opposite in direction / Reference to Newton's Third Law

(a) As the magnet approaches the copper ring, the magnetic field lines from the magnet cut through the conductor (the copper ring). This causes a change in the magnetic flux linkage of the ring, which induces an electromotive force (e.m.f.) across the ring in accordance with Faraday's Law of electromagnetic induction.

(b) According to Lenz's law, the induced current always flows in a direction to oppose the change producing it. As the North pole of the magnet approaches the ring from above, the induced current must create a magnetic field that opposes this approach. Therefore, the upper face of the ring behaves as a North pole to repel the falling magnet. To produce a North pole on the upper face, the current must flow in an anticlockwise direction when viewed from above (according to the right-hand grip rule).

(c) When the magnet falls through and leaves the bottom of the ring, the South pole of the magnet is moving away from the ring. To oppose this departure, the bottom of the ring must act as a North pole to attract the departing South pole. The direction of the change in magnetic flux is now opposite to when the magnet was entering. Consequently, the direction of the induced e.m.f. and the resulting current are reversed, causing the galvanometer needle to deflect in the opposite direction.

Part (a):
- B1: Mention of magnetic field lines cutting the ring / change in magnetic flux linkage
- B1: Statement that change in magnetic flux induces an e.m.f. (or electromagnetic induction)

Part (b):
- B1: Direction is anticlockwise (when viewed from above)
- B1: Reference to Lenz's law (induced current/field opposes the change producing it)
- B1: Explanation that the top of the ring must become a North pole to repel the approaching North pole

Part (c):
- B1: As the magnet leaves, the South pole is moving away
- B1: The magnetic flux decreases/changes in the opposite direction (or bottom of the ring becomes a North pole to attract the South pole)
- B1: This reverses the direction of the induced e.m.f. and thus reverses the current deflection

(i) Speed \(v = a \times t = 2.5\text{ m/s}^2 \times 4.0\text{ s} = 10\text{ m/s}\).

(ii) Distance during the acceleration phase:
\(s_1 = \frac{1}{2} \times v \times t_1 = \frac{1}{2} \times 10\text{ m/s} \times 4.0\text{ s} = 20\text{ m}\).

Distance during the constant speed phase:
\(s_2 = v \times t_2 = 10\text{ m/s} \times 6.0\text{ s} = 60\text{ m}\).

Total distance climbed:
\(s = s_1 + s_2 = 20\text{ m} + 60\text{ m} = 80\text{ m}\).

(i) [1 mark]
\(v = a \times t\) or \(2.5 \times 4.0 = 10\text{ m/s}\) (B1)

(ii) [3 marks]
\(s_1 = \frac{1}{2} \times 4.0 \times 10 = 20\text{ m}\) (C1)
\(s_2 = 10 \times 6.0 = 60\text{ m}\) (C1)
Total distance = \(20 + 60 = 80\text{ m}\) (A1)

Using the principle of conservation of momentum:

Total initial momentum = Total final momentum

\(m_A u_A + m_B u_B = m_A v_A + m_B v_B\)

\((150\text{ kg} \times 2.4\text{ m/s}) + (120\text{ kg} \times 0) = (150\text{ kg} \times 0.60\text{ m/s}) + (120\text{ kg} \times v_B)\)

\(360\text{ kg m/s} = 90\text{ kg m/s} + 120\text{ kg} \times v_B\)

\(120 \times v_B = 360 - 90 = 270\)

\(v_B = 2.25\text{ m/s}\).

Since the calculated value is positive, bumper car B moves in the direction of the initial motion (to the right).

(i) [3 marks]
Formula for conservation of momentum used: \(m_A u_A = m_A v_A + m_B v_B\) (C1)
Correct substitution: \(150 \times 2.4 = 150 \times 0.60 + 120 \times v_B\) (C1)
Correct calculation of velocity: \(2.25\text{ m/s}\) (A1)

(ii) [1 mark]
To the right / same direction as initial motion of car A (B1)

(i) Convert time to seconds: \(t = 5.0\text{ minutes} = 5.0 \times 60\text{ s} = 300\text{ s}\).
Thermal energy supplied: \(E = \text{Power} \times \text{time} = 48\text{ W} \times 300\text{ s} = 14400\text{ J}\).

(ii) Temperature rise: \(\Delta \theta = 44^\circ\text{C} - 20^\circ\text{C} = 24^\circ\text{C}\).
Using the equation \(E = m c \Delta \theta\):
\(14400\text{ J} = 1.5\text{ kg} \times c \times 24^\circ\text{C}\)
\(14400 = 36 \times c\)
\(c = \frac{14400}{36} = 400\text{ J/(kg }^\circ\text{C)}\).

(i) [2 marks]
Time converted to seconds: \(300\text{ s}\) (C1)
Energy calculated correctly: \(E = 14400\text{ J}\) (A1)

(ii) [2 marks]
\(\Delta \theta = 24^\circ\text{C}\) and formula rearranged: \(c = \frac{E}{m \Delta \theta}\) (C1)
Specific heat capacity calculated correctly: \(400\text{ J/(kg }^\circ\text{C)}\) (A1)

(i) Using Snell's law: \(n = \frac{\sin i}{\sin r}\)
\(1.45 = \frac{\sin(50^\circ)}{\sin r}\)
\
\sin r = \frac{\sin(50^\circ)}{1.45} = \frac{0.766}{1.45} \approx 0.528\)
\(r = \arcsin(0.528) \approx 31.9^\circ\).

(ii) Using the critical angle formula: \(\sin c = \frac{1}{n}\)
\(\sin c = \frac{1}{1.45} \approx 0.690\)
\(c = \arcsin(0.690) \approx 43.6^\circ\).

(i) [2 marks]
Use of \(n = \frac{\sin i}{\sin r}\) (C1)
Angle of refraction calculated correctly: \(31.9^\circ\) or \(32^\circ\) (A1)

(ii) [2 marks]
Use of \(\sin c = \frac{1}{n}\) (C1)
Critical angle calculated correctly: \(43.6^\circ\) or \(44^\circ\) (A1)

(i) Combined resistance of the parallel combination \(R_p\):
\(\frac{1}{R_p} = \frac{1}{6.0\ \Omega} + \frac{1}{12\ \Omega} = \frac{2}{12\ \Omega} + \frac{1}{12\ \Omega} = \frac{3}{12\ \Omega}\)
\(R_p = \frac{12}{3} = 4.0\ \Omega\).

(ii) Total resistance of the circuit \(R_{total}\):
\(R_{total} = R_p + 5.0\ \Omega = 4.0 + 5.0 = 9.0\ \Omega\).

Total current in the circuit \(I\):
\(I = \frac{V}{R_{total}} = \frac{12\text{ V}}{9.0\ \Omega} \approx 1.33\text{ A}\).

(i) [2 marks]
Use of parallel resistor formula: \(\frac{1}{R_p} = \frac{1}{R_1} + \frac{1}{R_2}\) (C1)
Parallel resistance calculated correctly: \(4.0\ \Omega\) (A1)

(ii) [2 marks]
Total resistance calculated correctly: \(9.0\ \Omega\) (C1)
Total current calculated correctly: \(1.3\text{ A}\) or \(1.33\text{ A}\) (A1)

(i) Since there are \(2.0\text{ claps per second}\), the time interval between successive claps is:
\(t = \frac{1.0\text{ s}}{2.0} = 0.50\text{ s}\).
This is the time taken for the sound to travel to the wall and back.

(ii) The sound travels a total distance of \(2d\) in \(0.50\text{ s}\).
\(2d = v \times t\)
\(2d = 340\text{ m/s} \times 0.50\text{ s} = 170\text{ m}\)
\(d = \frac{170}{2} = 85\text{ m}\).

(i) [1 mark]
Time interval calculated correctly: \(0.50\text{ s}\) (B1)

(ii) [2 marks]
Relationship \(2d = v \times t\) used (C1)
Distance calculated correctly: \(85\text{ m}\) (A1)

(i) Recession velocity in \(\text{m/s}\):
\(v = 4.2 \times 10^4\text{ km/s} = 4.2 \times 10^4 \times 1000\text{ m/s} = 4.2 \times 10^7\text{ m/s}\).

(ii) Using Hubble's Law: \(v = H_0 d\):
\(d = \frac{v}{H_0} = \frac{4.2 \times 10^7\text{ m/s}}{2.2 \times 10^{-18}\text{ s}^{-1}} \approx 1.91 \times 10^{25}\text{ m}\).

(i) [1 mark]
Velocity converted to m/s correctly: \(4.2 \times 10^7\text{ m/s}\) (B1)

(ii) [2 marks]
Hubble's Law formula rearranged: \(d = \frac{v}{H_0}\) (C1)
Distance calculated correctly: \(1.9 \times 10^{25}\text{ m}\) or \(1.91 \times 10^{25}\text{ m}\) (A1)

(i) Using the transformer turns-ratio equation:
\(\frac{V_p}{V_s} = \frac{N_p}{N_s}\)
\(\frac{240\text{ V}}{V_s} = \frac{1200}{100} = 12\)
\(V_s = \frac{240}{12} = 20\text{ V}\).

(ii) Current in the secondary circuit \(I_s\):
\(I_s = \frac{V_s}{R} = \frac{20\text{ V}}{4.0\ \Omega} = 5.0\text{ A}\).

Using the conservation of power for a 100% efficient transformer:
\(V_p I_p = V_s I_s\)
\(240\text{ V} \times I_p = 20\text{ V} \times 5.0\text{ A} = 100\text{ W}\)
\(I_p = \frac{100}{240} \approx 0.417\text{ A}\) (or \(0.42\text{ A}\)).

(i) [2 marks]
Use of transformer equation: \(\frac{V_p}{V_s} = \frac{N_p}{N_s}\) (C1)
Secondary voltage calculated correctly: \(20\text{ V}\) (A1)

(ii) [2 marks]
Secondary current calculated correctly (\(5.0\text{ A}\)) or power calculated correctly (\(100\text{ W}\)) (C1)
Primary current calculated correctly: \(0.42\text{ A}\) or \(0.417\text{ A}\) (A1)