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(a) Density is defined as mass per unit volume (\(\rho = \frac{m}{V}\)).

(b)(i) First, pour water into a measuring cylinder and record the initial volume \(V_1\). Lower the rock carefully into the cylinder using a thin thread so it is completely submerged. Record the new volume \(V_2\). The volume of the rock is given by \(V = V_2 - V_1\). Apparatus needed: measuring cylinder, water, and thread.
(ii) Read the scale at the bottom of the meniscus at eye level to avoid parallax error, or ensure no air bubbles are trapped on the surface of the rock.

(c) Volume of rock, \(V = 115\text{ cm}^3 - 55\text{ cm}^3 = 60\text{ cm}^3\).
Density, \(\rho = \frac{m}{V} = \frac{165\text{ g}}{60\text{ cm}^3} = 2.75\text{ g/cm}^3\).

(a) [1 mark] Mass per unit volume (or mass divided by volume).

(b)(i) [3 marks total]:
- [1] Measure and record initial water level in the measuring cylinder.
- [1] Fully submerge the rock in the water and record the final water level.
- [1] Subtract the initial volume from the final volume to get the volume of the rock.

(b)(ii) [1 mark]:
- [1] Read bottom of meniscus / read at eye level to avoid parallax error / ensure no water splashes out when inserting the rock / ensure no air bubbles are trapped on the rock.

(c) [3 marks total]:
- [1] Correct volume of the rock: \(60\text{ cm}^3\).
- [1] Correct formula: \(\rho = \frac{m}{V}\) or correct substitution: \(\frac{165}{60}\).
- [1] Correct final value and unit: \(2.75\text{ g/cm}^3\) (or \(2750\text{ kg/m}^3\) with correct unit).

(a) Momentum is a vector quantity because it has both magnitude and direction.

(b)(i) Total momentum before collision = \(m_A u_A + m_B u_B = (2.0 \times 4.5) + (3.0 \times 0) = 9.0\text{ kg m/s}\).
According to the principle of conservation of momentum, total momentum before = total momentum after:
\(9.0 = (m_A + m_B) v = (2.0 + 3.0) v = 5.0 v\).
\(v = \frac{9.0}{5.0} = 1.8\text{ m/s}\).

(b)(ii) Initial momentum of trolley A, \(p_i = 2.0 \times 4.5 = 9.0\text{ kg m/s}\).
Final momentum of trolley A, \(p_f = 2.0 \times 1.8 = 3.6\text{ kg m/s}\).
Change in momentum of trolley A, \(\Delta p = p_f - p_i = 3.6 - 9.0 = -5.4\text{ kg m/s}\) (or \(5.4\text{ kg m/s}\) in the opposite direction).

(a) [2 marks total]:
- [1] Vector.
- [1] It has both magnitude and direction.

(b)(i) [3 marks total]:
- [1] Correct calculation of total initial momentum: \(9.0\text{ kg m/s}\).
- [1] Correct conservation of momentum equation: \(9.0 = 5.0 v\).
- [1] Correct final velocity: \(1.8\text{ m/s}\).

(b)(ii) [3 marks total]:
- [1] Correct calculation of final momentum of A: \(3.6\text{ kg m/s}\).
- [1] Calculation of change: \(3.6 - 9.0\) or \(9.0 - 3.6\).
- [1] Correct value with unit: \(5.4\text{ kg m/s}\) (or \(5.4\text{ N s}\), accept negative sign).

(a)(i) In a gas, molecules move in a random, high-speed, and continuous motion in all directions.
(ii) In a solid, molecules vibrate about fixed positions.

(b)(i) Gas molecules continuously collide with the inner walls of the container. During each collision, the molecules undergo a change in momentum. According to Newton's second law, this change in momentum produces a force on the wall. The average force per unit area exerted by the molecules is the pressure.
(ii) As temperature increases, the average kinetic energy (and thus speed) of the gas molecules increases. This causes them to collide with the container walls more frequently, and with a greater change of momentum (or greater force) per collision. As a result, the total average force on the walls increases, leading to an increase in pressure.

(a)(i) [1 mark]:
- [1] Random / rapid / haphazard motion in all directions.
(a)(ii) [1 mark]:
- [1] Vibrate about fixed positions.

(b)(i) [3 marks total]:
- [1] Molecules collide with the walls of the container.
- [1] Collisions result in a change in momentum of the molecules.
- [1] Change in momentum per unit time exerts a force (and pressure is force per unit area).

(b)(ii) [3 marks total]:
- [1] Higher temperature increases kinetic energy / speed of molecules.
- [1] More frequent collisions with the walls.
- [1] Harder collisions (greater change of momentum per collision), resulting in a larger average force on the walls.

(a)(i) Wavelength is the distance between two consecutive identical points on a wave, such as from crest to crest.
(ii) Frequency is the number of complete wave cycles passing a fixed point per unit time (or per second).

(b)(i) Wave speed \(v = \frac{\text{distance}}{\text{time}} = \frac{45\text{ cm}}{3.0\text{ s}} = 15\text{ cm/s}\) (or \(0.15\text{ m/s}\)).
(ii) Wavelength \(\lambda = 1.5\text{ cm}\).
Using the wave equation: \(v = f \lambda \implies f = \frac{v}{\lambda} = \frac{15\text{ cm/s}}{1.5\text{ cm}} = 10\text{ Hz}\).

(c) The distance between wavefronts is the wavelength. Since wave speed \(v = f \lambda\) remains constant, wavelength is inversely proportional to frequency. When the frequency is doubled, the wavelength (distance between wavefronts) is halved, becoming \(0.75\text{ cm}\).

(a)(i) [1 mark]:
- [1] Distance between two consecutive crests or troughs (or equivalent).
(a)(ii) [1 mark]:
- [1] Number of waves passing a point per second / per unit time.

(b)(i) [2 marks total]:
- [1] Correct formula or substitution: \(45 / 3.0\).
- [1] Correct value with unit: \(15\text{ cm/s}\) or \(0.15\text{ m/s}\).

(b)(ii) [2 marks total]:
- [1] Correct formula or substitution: \(15 / 1.5\) or \(0.15 / 0.015\).
- [1] Correct answer with unit: \(10\text{ Hz}\).

(c) [2 marks total]:
- [1] The distance between wavefronts (wavelength) is halved / becomes \(0.75\text{ cm}\).
- [1] Explanation: Since wave speed is constant, wavelength is inversely proportional to frequency (or \(v = f \lambda\)).

(a) The circuit diagram must show a single series loop containing: a d.c. power supply (represented by cell/battery symbols), a fixed resistor of \(6.0\ \Omega\), and an LDR (a rectangle with two arrows pointing inwards). A voltmeter must be connected in parallel specifically across the \(6.0\ \Omega\) resistor.

(b)(i) Total resistance in series: \(R_{\text{total}} = R_1 + R_2 = 6.0\ \Omega + 2.0\ \Omega = 8.0\ \Omega\).
(ii) Total current \(I = \frac{V}{R_{\text{total}}} = \frac{12\text{ V}}{8.0\ \Omega} = 1.5\text{ A}\).
Voltmeter reading (potential difference across the \(6.0\ \Omega\) resistor): \(V = I \times R = 1.5\text{ A} \times 6.0\ \Omega = 9.0\text{ V}\).
(Alternatively: \(V = 12 \times \frac{6.0}{6.0 + 2.0} = 9.0\text{ V}\)).

(c) As the light level decreases, the resistance of the LDR increases. This increases the total resistance of the circuit, which reduces the current. Since the resistance of the \(6.0\ \Omega\) resistor remains unchanged, the reduced current results in a lower potential difference across it (since \(V = IR\)), so the voltmeter reading decreases.

(a) [3 marks total]:
- [1] Correct symbol for d.c. power supply / battery.
- [1] Correct symbols for fixed resistor and LDR connected in series.
- [1] Voltmeter connected in parallel across the \(6.0\ \Omega\) resistor only.

(b)(i) [1 mark]:
- [1] \(8.0\ \Omega\).

(b)(ii) [2 marks total]:
- [1] Correct current calculation: \(1.5\text{ A}\) (or correct potential divider formula: \(12 \times \frac{6.0}{8.0}\)).
- [1] Correct voltmeter reading: \(9.0\text{ V}\).

(c) [2 marks total]:
- [1] Voltmeter reading decreases.
- [1] Explanation: Total resistance increases, so current decreases, causing a smaller potential difference across the fixed resistor (or the LDR takes a larger fraction of the supply voltage).

(a) Electromotive force (e.m.f.) is the electrical work done per unit charge (or energy transferred per unit charge) in driving charge around a complete circuit.

(b)(i) Power, \(P = 1.8\text{ kW} = 1800\text{ W}\).
Using \(P = I V \implies I = \frac{P}{V} = \frac{1800\text{ W}}{240\text{ V}} = 7.5\text{ A}\).
(ii) Using Ohm's law, \(R = \frac{V}{I} = \frac{240\text{ V}}{7.5\text{ A}} = 32\ \Omega\).
(Alternatively, \(R = \frac{V^2}{P} = \frac{240^2}{1800} = 32\ \Omega\)).
(iii) Time \(t = 30 \times 60 = 1800\text{ s}\).
Energy transferred \(E = P \times t = 1800\text{ W} \times 1800\text{ s} = 3.24 \times 10^6\text{ J}\) (or \(3\,240\,000\text{ J}\)).

(a) [2 marks total]:
- [1] Work done / energy transferred per unit charge.
- [1] In driving charge around a complete circuit.

(b)(i) [2 marks total]:
- [1] Correct conversion of power to watts (\(1800\text{ W}\)) and formula/substitution: \(1800 / 240\).
- [1] Correct value and unit: \(7.5\text{ A}\).

(b)(ii) [2 marks total]:
- [1] Correct formula or substitution: e.g., \(240 / 7.5\) or \(240^2 / 1800\).
- [1] Correct value and unit: \(32\ \Omega\).

(b)(iii) [2 marks total]:
- [1] Convert time to seconds: \(1800\text{ s}\).
- [1] Correct energy calculation: \(3.24 \times 10^6\text{ J}\) (or \(3.24\text{ MJ}\) or \(3\,240\,000\text{ J}\)).

(a) Lenz's law states that the direction of an induced current is always such that it opposes the change in magnetic flux that produces it.

(b)(i) As the magnet moves into the solenoid, the magnetic field lines cut across the coils of the solenoid (a change in magnetic flux linkage occurs). This change in magnetic flux induces an electromotive force (e.m.f.) across the solenoid, causing an electric current to flow because it is a closed circuit.
(ii) A North pole is produced at the near end of the solenoid. According to Lenz's law, the induced current must create a magnetic field that opposes the movement of the magnet. Since a North pole is approaching, the solenoid creates a North pole at its entrance to repel the incoming magnet and oppose its motion.
(iii) The magnitude of the induced current can be increased by:
1. Moving the magnet into the solenoid at a higher speed (faster change of flux linkage).
2. Using a stronger bar magnet.
3. Increasing the number of turns on the solenoid.
4. Inserting a soft iron core into the solenoid.

(a) [2 marks total]:
- [1] The direction of an induced current (or e.m.f.).
- [1] Opposes the change that produces it.

(b)(i) [2 marks total]:
- [1] Magnetic field lines cut the coil / change in magnetic flux linkage.
- [1] This induces an e.m.f. (and hence a current because the circuit is complete).

(b)(ii) [2 marks total]:
- [1] North pole.
- [1] To repel the approaching North pole of the magnet / oppose its motion (Lenz's law).

(b)(iii) [2 marks total]:
- [2] Any two correct methods: increase speed of magnet, use stronger magnet, use coil with more turns, insert soft iron core.

(a)(i) In the core of a star, hydrogen nuclei (protons) collide and fuse together at extremely high temperatures and pressures to form helium nuclei. This process of fusion releases a massive amount of energy.
(ii) The two opposing forces are the inward pull of gravity and the outward thermal/radiation pressure from nuclear fusion.

(b) When the hydrogen fuel in the core of a star similar in mass to our Sun is depleted, the core contracts, heating up, while the outer layers expand and cool. The star becomes a Red Giant. In the core of the Red Giant, helium fuses into heavier elements like carbon. Eventually, fusion stops, and the outer layers of the star are ejected into space as a planetary nebula. The hot, dense core remains behind, cooling down over time to become a White Dwarf.

(a)(i) [2 marks total]:
- [1] Hydrogen nuclei (or protons) fuse / join together to form helium nuclei.
- [1] Occurs at high temperature/pressure, releasing a large amount of energy.
(a)(ii) [2 marks total]:
- [1] Inward force of gravity.
- [1] Outward radiation pressure / thermal pressure (due to fusion).

(b) [4 marks total]:
- [1] Hydrogen runs out and star expands to become a Red Giant.
- [1] Helium fusion occurs in the core (forming heavier elements like carbon/oxygen).
- [1] Outer layers of the star are ejected/shed (as a planetary nebula).
- [1] The remaining hot, dense core is left behind as a White Dwarf.

(a) Momentum is given by the formula \(p = mv\). Substituting the given values: \(p = 0.40\text{ kg} \times 2.5\text{ m/s} = 1.0\text{ kg m/s}\).

(b) By the principle of conservation of momentum, the total momentum before the collision equals the total momentum after the collision. The total initial momentum is \(1.0\text{ kg m/s}\) (since car B is stationary). After the collision, the combined mass is \(m_A + m_B = 0.40\text{ kg} + 0.60\text{ kg} = 1.00\text{ kg}\). Therefore, \(1.0\text{ kg m/s} = (1.00\text{ kg}) \times v\), which gives \(v = 1.0\text{ m/s}\).

(c) The impulse acting on car A is equal to its change in momentum: \(\Delta p = m_A v - m_A v_A = (0.40\text{ kg} \times 1.0\text{ m/s}) - 1.0\text{ kg m/s} = -0.60\text{ kg m/s}\). The magnitude of this change in momentum (impulse) is \(0.60\text{ N s}\). Using the relation \(\text{Impulse} = F \Delta t\): \(0.60\text{ N s} = 6.0\text{ N} \times \Delta t\). This yields \(\Delta t = 0.10\text{ s}\).

(a) [1] For using \(p = mv\) or showing substitution \(0.40 \times 2.5\). [1] For correct answer with units: \(1.0\text{ kg m/s}\) (or \(\text{N s}\)).

(b) [1] For stating conservation of momentum or equating initial and final momentum. [1] For calculating the combined mass \(1.00\text{ kg}\) and setting up the equation: \(1.0 = 1.00 \times v\). [1] For correct answer: \(1.0\text{ m/s}\).

(c) [1] For calculating the change in momentum (impulse) of car A: \(0.60\text{ kg m/s}\) (ignore sign). [1] For using \(F \Delta t = \Delta p\) rearranged as \(\Delta t = \Delta p / F\). [1] For correct final calculation: \(0.10\text{ s}\).

(a) Redshift is the shift in the observed wavelength of light towards the longer wavelength (red) end of the spectrum because the light source (galaxy) is moving away from the observer.

(b) Hubble's law states that the recession speed \(v\) of a distant galaxy is directly proportional to its distance \(d\) from Earth. It can be expressed as \(v = H_0 d\), where \(H_0\) is the Hubble constant.

(c)(i) Rearranging the formula for the Hubble constant: \(H_0 = v / d\). Substituting the values: \(H_0 = (6.8 \times 10^{4}\text{ m/s}) / (3.1 \times 10^{22}\text{ m}) = 2.1935 \times 10^{-18}\text{ s}^{-1} \approx 2.2 \times 10^{-18}\text{ s}^{-1}\).

(c)(ii) The age of the Universe \(T\) can be estimated by taking the reciprocal of the Hubble constant: \(T \approx 1 / H_0\). Using the calculated value: \(T = 1 / (2.1935 \times 10^{-18}\text{ s}^{-1}) \approx 4.56 \times 10^{17}\text{ s}\) (or \(4.6 \times 10^{17}\text{ s}\)).

(a) [1] For describing an increase in observed wavelength / shift towards red end of spectrum. [1] For specifying that the source of light is moving away from the observer.

(b) [1] For stating that the recession speed is directly proportional to distance (or providing formula \(v = H_0 d\)). [1] For defining \(v\) as recession speed, \(d\) as distance, and \(H_0\) as the Hubble constant.

(c)(i) [1] For correct substitution of values: \(H_0 = (6.8 \times 10^{4}) / (3.1 \times 10^{22})\). [1] For correct final value with unit: \(2.2 \times 10^{-18}\text{ s}^{-1}\) (allow \(2.19 \times 10^{-18}\text{ s}^{-1}\)).

(c)(ii) [1] For showing the method \(T \approx 1 / H_0\). [1] For correct calculation of age: \(4.6 \times 10^{17}\text{ s}\) (allow range \(4.5 \times 10^{17}\text{ s}\) to \(4.6 \times 10^{17}\text{ s}\) based on rounding of \(H_0\)).