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a) Acceleration \(a = \frac{v - u}{t} = \frac{24\text{ m/s} - 0}{6.0\text{ s}} = 4.0\text{ m/s}^2\).

b) (i) The gradient (slope) of the line on a speed-time graph represents the acceleration.
(ii) The area under the speed-time graph represents the total distance travelled.

c) The journey is divided into three parts:
- Part 1 (acceleration): \(d_1 = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 6.0\text{ s} \times 24\text{ m/s} = 72\text{ m}\).
- Part 2 (constant speed): \(d_2 = \text{base} \times \text{height} = 10\text{ s} \times 24\text{ m/s} = 240\text{ m}\).
- Part 3 (deceleration): \(d_3 = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 4.0\text{ s} \times 24\text{ m/s} = 48\text{ m}\).

Total distance \(d = 72\text{ m} + 240\text{ m} + 48\text{ m} = 360\text{ m}\).
Alternatively, using the area of a trapezium: \(d = \frac{1}{2} \times (a + b) \times h = \frac{1}{2} \times (10\text{ s} + 20\text{ s}) \times 24\text{ m/s} = 360\text{ m}\).

a)
- Uses formula \(a = \frac{\Delta v}{t}\) or correct substitution seen: [1 mark]
- Correct calculation: \(4.0\text{ m/s}^2\) (unit required): [1 mark]

b)
- (i) gradient or slope: [1 mark]
- (ii) area (under line / graph): [1 mark]

c)
- Calculates distance during acceleration phase (\(72\text{ m}\)) OR deceleration phase (\(48\text{ m}\)): [1 mark]
- Calculates distance during constant speed phase (\(240\text{ m}\)): [1 mark]
- Sums all three phases correctly (or uses trapezium formula with correct values): [1 mark]
- Final answer \(360\text{ m}\) (unit required): [1 mark]

a) Momentum is defined as the product of mass and velocity (\(p = mv\)). Its SI unit is kilogram metre per second (\(\text{kg m/s}\) or \(\text{N s}\)).

b) According to the principle of conservation of momentum:
Total initial momentum = Total final momentum
\(m_1 u_1 + m_2 u_2 = (m_1 + m_2) v\)
\((12000 \times 3.0) + (24000 \times 0) = (12000 + 24000) \times v\)
\(36000 = 36000 \times v\)
\(v = 1.0\text{ m/s}\).

c) Initial kinetic energy:
\(E_{ki} = \frac{1}{2} m_1 u_1^2 = \frac{1}{2} \times 12000 \times (3.0)^2 = 54000\text{ J}\).

Final kinetic energy:
\(E_{kf} = \frac{1}{2} (m_1 + m_2) v^2 = \frac{1}{2} \times 36000 \times (1.0)^2 = 18000\text{ J}\).

Loss of kinetic energy:
\(\Delta E_k = 54000\text{ J} - 18000\text{ J} = 36000\text{ J}\) (or \(36\text{ kJ}\)).

a)
- mass \(\times\) velocity: [1 mark]
- \(\text{kg m/s}\) or \(\text{N s}\): [1 mark]

b)
- Use of momentum conservation equation: \(m_1 u_1 = (m_1 + m_2)v\): [1 mark]
- Correct substitution of values: \(12000 \times 3.0 = 36000 v\): [1 mark]
- Final answer: \(1.0\text{ m/s}\): [1 mark]

c)
- Calculates initial kinetic energy (\(54000\text{ J}\)): [1 mark]
- Calculates final kinetic energy (\(18000\text{ J}\)): [1 mark]
- Calculates correct loss of kinetic energy: \(36000\text{ J}\) (or \(36\text{ kJ}\)): [1 mark]

a) Volume of the block \(V = 0.50\text{ m} \times 0.40\text{ m} \times 1.2\text{ m} = 0.24\text{ m}^3\).
Mass \(m = \rho \times V = 2400\text{ kg/m}^3 \times 0.24\text{ m}^3 = 576\text{ kg}\).

b) Weight of the block \(W = m \times g = 576\text{ kg} \times 9.8\text{ N/kg} = 5644.8\text{ N}\).
To exert maximum pressure, the block must rest on its smallest face (minimum contact area).
Smallest area \(A_{\min} = 0.50\text{ m} \times 0.40\text{ m} = 0.20\text{ m}^2\).
Maximum pressure \(P_{\max} = \frac{W}{A_{\min}} = \frac{5644.8\text{ N}}{0.20\text{ m}^2} = 28224\text{ Pa}\) (or \(2.8 \times 10^4\text{ Pa}\) to 2 s.f.).

c) The block should be placed on its largest face with dimensions \(0.50\text{ m} \times 1.2\text{ m}\) (area \(0.60\text{ m}^2\)).
Since pressure \(P = \frac{F}{A}\) and the force (weight) remains constant, a larger area of contact distributes the force over a wider surface, which reduces the pressure.

a)
- Volume calculation: \(0.24\text{ m}^3\): [1 mark]
- Mass calculation: \(576\text{ kg}\): [1 mark]

b)
- Weight calculation: \(576 \times 9.8 = 5644.8\text{ N}\): [1 mark]
- Identifies smallest face area: \(0.50 \times 0.40 = 0.20\text{ m}^2\): [1 mark]
- Use of \(P = \frac{F}{A}\) formula: [1 mark]
- Correct pressure calculation: \(28000\text{ Pa}\) (or \(2.8 \times 10^4\text{ Pa}\); accept \(28224\text{ Pa}\)): [1 mark]

c)
- Identifies largest face / dimensions \(0.50\text{ m} \times 1.2\text{ m}\): [1 mark]
- States that pressure is inversely proportional to area / larger area distributes weight, thus reducing pressure: [1 mark]

a) Gas molecules are in continuous, rapid random motion. They collide with the walls of the cylinder, and during these collisions, their direction of motion changes, which means their momentum changes. This rate of change of momentum exerts a force on the walls. Pressure is this force per unit area.

b) (i) Using Boyle's Law for constant temperature:
\(P_1 V_1 = P_2 V_2\)
\(1.0 \times 10^5\text{ Pa} \times 1.5 \times 10^{-3}\text{ m}^3 = P_2 \times 6.0 \times 10^{-4}\text{ m}^3\)
\(150 = P_2 \times 6.0 \times 10^{-4}\)
\(P_2 = \frac{150}{6.0 \times 10^{-4}} = 2.5 \times 10^5\text{ Pa}\).

(ii) Assumption: The mass (or number of molecules) of the gas remains constant (no gas leaks out), or the gas behaves as an ideal gas.

c) When the volume is decreased, the molecules are packed into a smaller space. Therefore, they collide with the walls of the container more frequently (more collisions per second), which increases the average force on the walls, and thus increases the pressure.

a)
- Molecules collide with the walls of the cylinder: [1 mark]
- Change in momentum occurs during these collisions: [1 mark]
- Force is rate of change of momentum (hence force per unit area is pressure): [1 mark]

b)(i)
- Use of Boyle's Law formula: \(P_1 V_1 = P_2 V_2\): [1 mark]
- Correct substitution: \(1.0 \times 10^5 \times 1.5 \times 10^{-3} = P_2 \times 6.0 \times 10^{-4}\): [1 mark]
- Final answer: \(2.5 \times 10^5\text{ Pa}\): [1 mark]

b)(ii)
- No gas escapes / fixed mass of gas / gas behaves as an ideal gas: [1 mark]

c)
- Molecules are closer together / more concentrated, leading to more frequent collisions with the walls: [1 mark]

a) Time \(t = 5.0\text{ minutes} = 5.0 \times 60\text{ s} = 300\text{ s}\).
Thermal energy supplied \(E = P \times t = 800\text{ W} \times 300\text{ s} = 240000\text{ J}\) (or \(2.4 \times 10^5\text{ J}\) or \(240\text{ kJ}\)).

b) (i) Temperature change \(\Delta \theta = 98^\circ\text{C} - 20^\circ\text{C} = 78^\circ\text{C}\).
Using \(E = m c \Delta \theta\):
\(240000 = 2.5 \times c \times 78\)
\(c = \frac{240000}{195} \approx 1230\text{ J/(kg }^\circ\text{C)}\) (or \(1200\text{ J/(kg }^\circ\text{C)}\) to 2 s.f.).

(ii) Explanation: Thermal energy is lost to the surroundings/air. Thus, the actual energy absorbed by the block is less than the energy supplied by the heater. This makes the calculated value of specific heat capacity higher than the true value.
Modification: Wrap insulation (e.g., polystyrene or glass wool) around the block to reduce heat loss.

a)
- Converts time to seconds: \(300\text{ s}\): [1 mark]
- Calculates energy: \(240000\text{ J}\) or \(2.4 \times 10^5\text{ J}\): [1 mark]

b)(i)
- Temperature difference calculation: \(78^\circ\text{C}\): [1 mark]
- Rearranges formula to make \(c\) the subject: \(c = \frac{E}{m \Delta \theta}\): [1 mark]
- Correct calculation: \(1200\text{ J/(kg }^\circ\text{C)}\) (accept \(1230\text{ J/(kg }^\circ\text{C)}\)): [1 mark]

b)(ii)
- Heat is lost to the surroundings / air: [1 mark]
- Meaning actual energy absorbed by block is less than supplied (leading to overestimate of \(c\)): [1 mark]
- Suggests adding insulation / lagging around the block: [1 mark]

a) (i) Using Snell's Law:
\(n = \frac{\sin i}{\sin r}\)
\(1.50 = \frac{\sin 52^\circ}{\sin r}\)
\(\sin r = \frac{\sin 52^\circ}{1.50} = \frac{0.7880}{1.50} \approx 0.5253\)
\(r = \sin^{-1}(0.5253) \approx 31.7^\circ\) (or \(32^\circ\) to 2 s.f.).

(ii) Using the formula for refractive index and wave speed:
\(n = \frac{c}{v}\)
\(1.50 = \frac{3.0 \times 10^8}{v}\)
\(v = \frac{3.0 \times 10^8}{1.50} = 2.0 \times 10^8\text{ m/s}\).

b) The critical angle is the angle of incidence in an optically denser medium for which the angle of refraction in the less dense medium is \(90^\circ\).
Two conditions for total internal reflection:
1. The light must be travelling from an optically denser medium towards a less dense medium (e.g., from glass to air).
2. The angle of incidence must be greater than the critical angle.

a)(i)
- Snell's Law stated or used: \(n = \frac{\sin i}{\sin r}\): [1 mark]
- Rearrangement to \(\sin r = \frac{\sin 52^\circ}{1.50}\): [1 mark]
- Correct calculation: \(32^\circ\) or \(31.7^\circ\): [1 mark]

a)(ii)
- Formula used: \(n = \frac{c}{v}\) or substitution: [1 mark]
- Correct calculation: \(2.0 \times 10^8\text{ m/s}\): [1 mark]

b)
- Definition of critical angle: angle of incidence that gives an angle of refraction of \(90^\circ\): [1 mark]
- Condition 1: Light travels from denser to less dense medium: [1 mark]
- Condition 2: Angle of incidence is greater than critical angle: [1 mark]

a) (i) Combined resistance \(R_p\) of resistors in parallel:
\(\frac{1}{R_p} = \frac{1}{R_1} + \frac{1}{R_2} = \frac{1}{6.0} + \frac{1}{12} = \frac{2}{12} + \frac{1}{12} = \frac{3}{12}\)
\(R_p = \frac{12}{3} = 4.0\ \Omega\).

(ii) Total current \(I\):
\(I = \frac{V}{R_p} = \frac{12\text{ V}}{4.0\ \Omega} = 3.0\text{ A}\).

(iii) Time \(t = 10\text{ minutes} = 600\text{ s}\).
Electrical energy \(E = V \times I \times t = 12\text{ V} \times 3.0\text{ A} \times 600\text{ s} = 21600\text{ J}\) (or \(22\text{ kJ}\)).

b) When the temperature increases, the resistance of the thermistor decreases. Since the parallel branch resistance decreases, the overall combined resistance of the circuit decreases, causing the total current drawn from the supply to increase.

a)(i)
- Formula used: \(\frac{1}{R_p} = \frac{1}{R_1} + \frac{1}{R_2}\): [1 mark]
- Correct calculation: \(4.0\ \Omega\): [1 mark]

a)(ii)
- Formula used: \(I = \frac{V}{R_p}\) or substitution: [1 mark]
- Correct calculation: \(3.0\text{ A}\): [1 mark]

a)(iii)
- Converts 10 minutes to \(600\text{ s}\): [1 mark]
- Correct calculation: \(21600\text{ J}\) or \(22\text{ kJ}\): [1 mark]

b)
- Resistance of thermistor decreases as temperature increases: [1 mark]
- Since overall resistance decreases, the total current increases: [1 mark]

a) (i) Beta-minus decay converts a neutron into a proton, emitting an electron. The mass number remains \(24\), and the atomic number increases by 1 to became \(12\) (Magnesium):
\(^{24}_{11}\text{Na} \rightarrow ^{24}_{12}\text{Mg} + ^{0}_{-1}\text{e}\) (or \(^{24}_{11}\text{Na} \rightarrow ^{24}_{12}\text{Mg} + \beta\)).

(ii) A \(\beta^-\)-particle is a high-speed electron emitted from the nucleus of an atom.

b) (i) Determine the fraction of activity remaining:
\(\frac{100\text{ Bq}}{800\text{ Bq}} = \frac{1}{8}\).

Since \(\frac{1}{8} = \left(\frac{1}{2}\right)^3\), exactly \(3\) half-lives have elapsed.
Let \(T_{1/2}\) be the half-life:
\(3 \times T_{1/2} = 45\text{ hours}\)
\(T_{1/2} = 15\text{ hours}\).

(ii) The recorded count rate is higher because of background radiation (from cosmic rays, rocks, etc.), which is detected by the GM counter and added to the count rate from the sample.

a)(i)
- Reactant side correct: \(^{24}_{11}\text{Na}\): [1 mark]
- Product magnesium correct: \(^{24}_{12}\text{Mg}\): [1 mark]
- Beta particle correct: \(^{0}_{-1}\text{e}\) or \(^{0}_{-1}\beta\): [1 mark]

a)(ii)
- High-speed electron (emitted from nucleus): [1 mark]

b)(i)
- Identifies activity halving 3 times (e.g., \(800 \rightarrow 400 \rightarrow 200 \rightarrow 100\)): [1 mark]
- Relates 3 half-lives to 45 hours: [1 mark]
- Correct calculation: \(15\text{ hours}\): [1 mark]

b)(ii)
- Background radiation is present and added to the GM counter reading: [1 mark]

(a) Using the law of conservation of momentum:
\(p_{\text{initial}} = p_{\text{final}}\)
\(m_A u_A + m_B u_B = (m_A + m_B) v\)
\((0.50 \times 4.0) + 0 = (0.50 + 0.30) v\)
\(2.0 = 0.80 v\)
\(v = \frac{2.0}{0.80} = 2.5\text{ m/s}\)

(b) Calculate initial and final kinetic energies:
\(E_{k\text{, initial}} = \frac{1}{2} m_A u_A^2 = \frac{1}{2} \times 0.50 \times 4.0^2 = 4.0\text{ J}\)
\(E_{k\text{, final}} = \frac{1}{2} (m_A + m_B) v^2 = \frac{1}{2} \times 0.80 \times 2.5^2 = 2.5\text{ J}\)
\(\text{Kinetic energy lost} = 4.0\text{ J} - 2.5\text{ J} = 1.5\text{ J}\)

(c) An inelastic collision is a collision in which the total kinetic energy is not conserved. The 'lost' kinetic energy is converted into thermal energy (heating up the coupling mechanism and bodies of the trucks) and sound waves.

(a)
- State or use: \(m_A u_A = (m_A + m_B) v\) [1 mark]
- Correct substitution of values: \(0.50 \times 4.0 = 0.80 v\) [1 mark]
- Correct evaluation of \(v = 2.5\text{ m/s}\) with correct unit [1 mark]

(b)
- Calculation of initial KE (\(4.0\text{ J}\)) OR final KE (\(2.5\text{ J}\)) [1 mark]
- Subtraction of final KE from initial KE [1 mark]
- Correct final answer: \(1.5\text{ J}\) with correct unit [1 mark]

(c)
- Definition of inelastic collision: Kinetic energy is not conserved [1 mark]
- Explanation of energy transformation: Converted into thermal energy/heat and/or sound [1 mark]

(a) Using Boyle's Law for constant temperature:
\(p_1 V_1 = p_2 V_2\)
\(1.5 \times 10^5 \times 0.024 = p_2 \times 0.0080\)
\(p_2 = \frac{1.5 \times 10^5 \times 0.024}{0.0080}\)
\(p_2 = 1.5 \times 10^5 \times 3 = 4.5 \times 10^5\text{ Pa}\)

(b) When the volume of the container decreases, the density of the molecules increases (they are packed closer together). Since the temperature is constant, the average speed of the molecules remains unchanged. Because the molecules are closer together and the wall area is smaller, they collide with the walls of the container more frequently. This greater rate of collisions results in a larger total force per unit area, which is an increase in pressure.

(c) When the piston is pushed, work is done on the gas. This work done increases the internal kinetic energy of the gas molecules, which would cause the temperature to rise. Pushing the piston slowly allows enough time for this excess thermal energy to conduct through the walls of the cylinder to the surroundings, keeping the temperature of the gas constant.

(a)
- State or use formula: \(p_1 V_1 = p_2 V_2\) [1 mark]
- Correct substitution: \(1.5 \times 10^5 \times 0.024 = p_2 \times 0.0080\) [1 mark]
- Correct calculation with units: \(4.5 \times 10^5\text{ Pa}\) [1 mark]

(b)
- Molecules are closer together / higher concentration of molecules [1 mark]
- More frequent collisions with the walls / more collisions per unit area per second [1 mark]
- Molecules keep same average kinetic energy/speed (so individual impact force is constant), leading to a higher overall average force per unit area [1 mark]

(c)
- Compressional work is done on the gas which increases its internal kinetic energy (temperature) [1 mark]
- Compressing slowly allows time for thermal energy (heat) to escape to the surroundings [1 mark]