2022 OCR A-Level Physics A - H556 Practice Paper with Answers

The density is given by the formula:
\( \rho = \frac{m}{V} = \frac{m}{\pi (d/2)^2 L} = \frac{4m}{\pi d^2 L} \)

To find the overall percentage uncertainty, we add the percentage uncertainties of each component, multiplying by their powers in the formula:

1. Percentage uncertainty in \( m \):
\( \frac{0.6}{24.0} \times 100\% = 2.5\% \)

2. Percentage uncertainty in \( d \):
\( \frac{0.1}{10.0} \times 100\% = 1.0\% \)

3. Percentage uncertainty in \( L \):
\( \frac{1.0}{50.0} \times 100\% = 2.0\% \)

Now, combine these using the uncertainty combination rule:
\( \frac{\Delta \rho}{\rho} \% = \frac{\Delta m}{m} \% + 2\left(\frac{\Delta d}{d} \%\right) + \frac{\Delta L}{L} \% \)
\( \frac{\Delta \rho}{\rho} \% = 2.5\% + 2(1.0\%) + 2.0\% = 6.5\% \)

Award 1 mark for the correct option C.

- Correct calculation of individual percentage uncertainties (Mass: 2.5%, Diameter: 1.0%, Length: 2.0%)
- Correct application of the power rule for combining uncertainties (doubling the uncertainty of diameter)
- Correct total of 6.5%

Since the two gases are in thermal equilibrium in the same container, they are at the same temperature \( T \).

The mean kinetic energy of a gas molecule is:
\( E_k = \frac{3}{2} k_B T = \frac{1}{2} m c^2 \)

Therefore, the root-mean-square speed \( c_{\text{rms}} \) is:
\( c_{\text{rms}} = \sqrt{\frac{3 k_B T}{m}} \)

This shows that the r.m.s. speed is inversely proportional to the square root of the mass of the particles:
\( c_{\text{rms}} \propto \frac{1}{\sqrt{M}} \)

Taking the ratio:
\( \frac{c_{\text{rms, He}}}{c_{\text{rms, O}_2}} = \sqrt{\frac{M_{\text{O}_2}}{M_{\text{He}}}} = \sqrt{\frac{32.0}{4.0}} = \sqrt{8.0} \approx 2.83 \)

This rounds to 2.8.

Award 1 mark for the correct option C.

- Realisation that thermal equilibrium means both gases are at the same temperature, hence same mean kinetic energy
- Correct dependency of r.m.s. speed on the inverse square root of mass
- Correct ratio calculated to be approx 2.8

First, find the effective spring constant, \( k_{\text{eff}} \), of the system.

1. Each parallel pair consists of two identical springs of constant \( k \).
The effective spring constant of each parallel pair is:
\( k_{\text{parallel}} = k + k = 2k \)

2. The system consists of these two parallel pairs connected in series. Therefore, we have two blocks of spring constant \( 2k \) in series:
\( \frac{1}{k_{\text{eff}}} = \frac{1}{2k} + \frac{1}{2k} = \frac{2}{2k} = \frac{1}{k} \)
So, \( k_{\text{eff}} = k \).

3. The total work done in extending this system of effective stiffness \( k \) by a distance \( x \) is given by:
\( W = \frac{1}{2} k_{\text{eff}} x^2 = \frac{1}{2} k x^2 \)

Award 1 mark for the correct option C.

- Correct combination of springs in parallel to give \( 2k \)
- Correct combination of parallel pairs in series to find effective spring constant is \( k \)
- Work formula \( W = \frac{1}{2} k_{\text{eff}} x^2 \) correctly applied

By equating the gravitational force to the centripetal force:
\( \frac{G M m}{r^2} = m \omega^2 r = m \left(\frac{2\pi}{T}\right)^2 r \)

Simplifying for \( T^2 \):
\( T^2 = \frac{4\pi^2 r^3}{G M} \implies T \propto \sqrt{\frac{r^3}{M}} \)

Let the orbital period of the second satellite be \( T_2 \). Substituting the new parameters:
\( T_2 \propto \sqrt{\frac{(2r)^3}{4M}} = \sqrt{\frac{8r^3}{4M}} = \sqrt{2} \sqrt{\frac{r^3}{M}} \)

Therefore, \( T_2 = \sqrt{2} T \).

Award 1 mark for the correct option A.

- Derivation of Kepler's Third Law relationship \( T^2 \propto \frac{r^3}{M} \)
- Substitution of \( 2r \) and \( 4M \) into the ratio
- Simplification to find the multiplier is \( \sqrt{2} \)

In simple harmonic motion, maximum acceleration is given by:
\( a_{\text{max}} = \omega^2 A \)
And maximum velocity is given by:
\( v_{\text{max}} = \omega A \)

From these equations, we can write:
\( \omega = \sqrt{\frac{a_{\text{max}}}{A}} \)

Substituting the values (with amplitude converted to meters: \( A = 0.050 \text{ m} \)):
\( \omega = \sqrt{\frac{20}{0.050}} = \sqrt{400} = 20 \text{ rad s}^{-1} \)

Now, calculate the maximum speed:
\( v_{\text{max}} = \omega A = 20 \times 0.050 = 1.0 \text{ m s}^{-1} \)

Award 1 mark for the correct option A.

- Correct conversions (e.g., converting 5.0 cm to 0.050 m)
- Use of \( a_{\text{max}} = \omega^2 A \) to find \( \omega = 20 \text{ rad s}^{-1} \)
- Correct calculation of \( v_{\text{max}} = 1.0 \text{ m s}^{-1} \)

1. Calculate the total energy \( E \) supplied by the electrical heater in \( 5.0 \text{ minutes} \):
\( t = 5.0 \times 60 = 300 \text{ s} \)
\( E = P \times t = 60 \text{ W} \times 300 \text{ s} = 18000 \text{ J} \)

2. Determine the mass \( m \) of the liquid that actually vaporised:
Only half of the \( 0.20 \text{ kg} \) of liquid vaporised, so:
\( m = 0.10 \text{ kg} \)

3. Use the formula for specific latent heat:
\( E = m L_v \implies L_v = \frac{E}{m} \)
\( L_v = \frac{18000 \text{ J}}{0.10 \text{ kg}} = 1.8 \times 10^5 \text{ J kg}^{-1} \)

Award 1 mark for the correct option B.

- Correct conversion of time to seconds (300 s) and energy calculation (18,000 J)
- Realisation that only half of the mass (0.10 kg) undergoes phase change
- Correct division of energy by mass to find \( 1.8 \times 10^5 \text{ J kg}^{-1} \)

1. From Wien's displacement law (\( \lambda_{\text{max}} T = \text{constant} \)):
\( T \propto \frac{1}{\lambda_{\text{max}}} \)
Therefore, the ratio of temperatures is:
\( \frac{T_{\text{X}}}{T_{\text{Y}}} = \frac{\lambda_{\text{max, Y}}}{\lambda_{\text{max, X}}} = \frac{1}{2} \)

2. From Stefan's Law, the luminosity \( L \) is:
\( L = 4\pi R^2 \sigma T^4 \implies L \propto R^2 T^4 \)

3. Calculate the ratio of the luminosities:
\( \frac{L_{\text{X}}}{L_{\text{Y}}} = \left(\frac{R_{\text{X}}}{R_{\text{Y}}}\right)^2 \left(\frac{T_{\text{X}}}{T_{\text{Y}}}\right)^4 \)
Substituting the ratios:
\( \frac{L_{\text{X}}}{L_{\text{Y}}} = (4)^2 \times \left(\frac{1}{2}\right)^4 = 16 \times \frac{1}{16} = 1.0 \)

Award 1 mark for the correct option B.

- Correct application of Wien's law to find temperature ratio \( T_{\text{X}}/T_{\text{Y}} = 0.5 \)
- Correct application of Stefan's law showing \( L \propto R^2 T^4 \)
- Correct combination of both ratios to yield 1.0

1. Treat both blocks as a single combined mass system to find the acceleration \( a \):
\( F_{\text{total}} = (m_1 + m_2) a \)
\( 12 \text{ N} = (3.0 \text{ kg} + 1.0 \text{ kg}) a \implies a = \frac{12}{4.0} = 3.0 \text{ m s}^{-2} \)

2. Now isolate the second block (mass \( 1.0 \text{ kg} \)). The only horizontal force acting on it is the contact force \( F_{\text{contact}} \) exerted by the first block:
\( F_{\text{contact}} = m_2 a \)
\( F_{\text{contact}} = 1.0 \text{ kg} \times 3.0 \text{ m s}^{-2} = 3.0 \text{ N} \)

According to Newton's third law, the magnitude of the force exerted by the first block on the second is equal to the force exerted by the second on the first, which is \( 3.0 \text{ N} \).

Award 1 mark for the correct option A.

- Correct calculation of the system acceleration (\( 3.0 \text{ m s}^{-2} \))
- Correct application of \( F = ma \) to the \( 1.0 \text{ kg} \) block to determine the contact force of \( 3.0 \text{ N} \)

The density \(\rho\) of the wire is given by:

\(\rho = \frac{m}{V} = \frac{4m}{\pi d^2 L}\)

The percentage uncertainty in the calculated density is found by summing the fractional uncertainties of the independent variables, remembering to multiply the fractional uncertainty of the diameter by 2 (due to the exponent):

\(\frac{\Delta \rho}{\rho} = \frac{\Delta m}{m} + 2\left(\frac{\Delta d}{d}\right) + \frac{\Delta L}{L}\)

Calculating each individual percentage uncertainty:
- Mass: \(\frac{0.02}{2.50} \times 100\% = 0.80\%\)
- Diameter: \(\frac{0.01}{0.80} \times 100\% = 1.25\%\)
- Length: \(\frac{0.5}{40.0} \times 100\% = 1.25\%\)

Combining these together:

\(\%\Delta \rho = 0.80\% + 2(1.25\%) + 1.25\% = 0.80\% + 2.50\% + 1.25\% = 4.55\%\)

To two significant figures, this is \(4.6\%\).

C

First, find the effective spring constant of the parallel combination \(k_p\):
\(k_p = k + k = 200 + 200 = 400\text{ N m}^{-1}\)

Now, calculate the overall effective spring constant \(k_{\text{eff}}\) of the system, where the parallel combination is in series with the third spring:
\(\frac{1}{k_{\text{eff}}} = \frac{1}{k_p} + \frac{1}{k} = \frac{1}{400} + \frac{1}{200} = \frac{3}{400}\text{ m N}^{-1}\)
\(k_{\text{eff}} = \frac{400}{3} \approx 133.3\text{ N m}^{-1}\)

The elastic potential energy \(E_p\) stored under a load of \(F = 30\text{ N}\) is:
\(E_p = \frac{1}{2} \frac{F^2}{k_{\text{eff}}} = \frac{1}{2} \times \frac{30^2}{400/3} = \frac{1}{2} \times \frac{900 \times 3}{400} = 3.375\text{ J} \approx 3.4\text{ J}\)

C

For an ideal gas at a constant volume, the pressure is directly proportional to the absolute temperature (Gay-Lussac's Law):
\(p \propto T\)

Since the pressure increases by a factor of \(1.69\), the absolute temperature must also increase by a factor of \(1.69\).

The r.m.s. speed of gas molecules is given by:
\(c_{\text{rms}} = \sqrt{\frac{3kT}{m}}\)

This shows that \(c_{\text{rms}} \propto \sqrt{T}\).

The new r.m.s. speed \(c_{\text{new}}\) is:
\(c_{\text{new}} = \sqrt{1.69} \times c_{\text{rms}} = 1.30 c_{\text{rms}}\)

B

The total energy of the simple harmonic oscillator is given by \(E = E_k + E_p\).

When the kinetic energy \(E_k\) is equal to the potential energy \(E_p\):
\(E = 2E_p\)

Substituting the definitions of total and potential energy:
\(\frac{1}{2} m \omega^2 A^2 = 2 \left(\frac{1}{2} m \omega^2 x^2\right) \implies A^2 = 2x^2 \implies x = \frac{A}{\sqrt{2}}\)

Since \(x = A \cos(\omega t)\):
\(A \cos(\omega t) = \frac{A}{\sqrt{2}} \implies \cos(\omega t) = \frac{1}{\sqrt{2}}\)

The first non-zero solution occurs when:
\(\omega t = \frac{\pi}{4}\)

Substituting \(\omega = \frac{2\pi}{T}\):
\(\frac{2\pi}{T} t = \frac{\pi}{4} \implies t = \frac{T}{8}\)

A

For a circular orbit, the gravitational force provides the centripetal force:
\(\frac{GMm}{r^2} = \frac{mv^2}{r} \implies v = \sqrt{\frac{GM}{r}}\)

If the orbital radius increases from \(R\) to \(4R\), the new speed \(v'\) is:
\(v' = \sqrt{\frac{GM}{4R}} = \frac{1}{2} \sqrt{\frac{GM}{R}} = 0.5v\)

According to Kepler's Third Law, the period of the orbit is related to the radius by:
\(T^2 \propto r^3 \implies T \propto r^{1.5}\)

If the radius is multiplied by 4, the new period \(T'\) is:
\(T' = 4^{1.5} T = 8T\)

B

First, convert \(H_0\) from \(\text{km s}^{-1}\text{ Mpc}^{-1}\) to SI base units (\(\text{s}^{-1}\)):

\(H_0 = 68\text{ km s}^{-1}\text{ Mpc}^{-1} = \frac{68 \times 10^3\text{ m s}^{-1}}{10^6 \times 3.1 \times 10^{16}\text{ m}} = 2.1935 \times 10^{-18}\text{ s}^{-1}\)

Now calculate the estimated age of the Universe in seconds:

\(t \approx \frac{1}{H_0} = \frac{1}{2.1935 \times 10^{-18}\text{ s}^{-1}} = 4.5588 \times 10^{17}\text{ s}\)

Convert this value into years:

\(t_{\text{years}} = \frac{4.5588 \times 10^{17}\text{ s}}{3.2 \times 10^7\text{ s year}^{-1}} \approx 1.42 \times 10^{10}\text{ years}\)

This matches \(1.4 \times 10^{10}\text{ years}\).

C

Because the block moves at constant velocity, there is zero net force acting on it.

Resolving forces perpendicular to the ramp:
\(N = mg \cos\theta\)
where \(N\) is the normal contact force.

Resolving forces parallel to the ramp:
\(F_f = mg \sin\theta\)
where \(F_f\) is the dynamic frictional force.

Using the dynamic friction relation \(F_f = \mu N\):
\(mg \sin\theta = \mu (mg \cos\theta)\)

Solving for \(\mu\):
\(\mu = \frac{\sin\theta}{\cos\theta} = \tan\theta\)

C

(a) Convert temperature to Kelvin:
\(T = 15.0 + 273.15 = 288.15 \text{ K}\)

Using the ideal gas equation:
\(p V = n R T \implies n = \frac{p V}{R T}\)
\(n = \frac{1.01 \times 10^5 \times 12.5}{8.31 \times 288.15} \approx 527 \text{ mol}\)

(b) Using the ideal gas equation in the form:
\(\frac{p_1 V_1}{T_1} = \frac{p_2 V_2}{T_2}\)

Convert the new temperature to Kelvin:
\(T_2 = -42.0 + 273.15 = 231.15 \text{ K}\)

\(V_2 = \frac{p_1 V_1 T_2}{p_2 T_1} = \frac{1.01 \times 10^5 \times 12.5 \times 231.15}{3.50 \times 10^4 \times 288.15} \approx 28.9 \text{ m}^3\)

(c) As the temperature of the gas decreases, the mean kinetic energy of the helium atoms decreases, meaning their mean speed decreases. This leads to less frequent collisions between the atoms and the walls of the balloon. Additionally, each collision results in a smaller change in momentum. Since pressure is proportional to the rate of change of momentum of the atoms colliding with the walls per unit area, the pressure exerted by the gas decreases.

(a)
- Use of \(T_1 = 288 \text{ K}\) or \(288.15 \text{ K}\) [1 mark]
- Correct rearrangement of ideal gas law: \(n = \frac{pV}{RT}\) [1 mark]
- Calculation to give \(527 \text{ mol}\) (accept range \(527 - 528 \text{ mol}\)) [1 mark]

(b)
- Use of \(T_2 = 231 \text{ K}\) or \(231.15 \text{ K}\) [1 mark]
- Correct rearrangement: \(V_2 = \frac{p_1 V_1 T_2}{p_2 T_1}\) or use of \(V_2 = \frac{nRT_2}{p_2}\) [1 mark]
- Calculation to give \(28.9 \text{ m}^3\) (accept range \(28.8 - 29.0 \text{ m}^3\)) [1 mark]

(c)
- Absolute temperature is proportional to mean kinetic energy / mean square speed of molecules [1 mark]
- Lower speed means fewer collisions per unit time with the container walls [1 mark]
- Lower speed means smaller change in momentum per collision [1 mark]
- Rate of change of momentum decreases, so force per unit area (pressure) decreases [1.625 marks]

(a) Using the equation for the period of a mass-spring system:
\(T = 2\pi \sqrt{\frac{m}{k}} \implies T^2 = 4\pi^2 \frac{m}{k}\)

Rearranging for \(k\):
\(k = \frac{4\pi^2 m}{T^2} = \frac{4\pi^2 \times 0.350}{0.780^2} = 22.71 \text{ N m}^{-1} \approx 23 \text{ N m}^{-1}\).

(b) (i) First, calculate the angular frequency \(\omega\):
\(\omega = \frac{2\pi}{T} = \frac{2\pi}{0.780} \approx 8.055 \text{ rad s}^{-1}\)

The maximum acceleration is:
\(a_{\text{max}} = \omega^2 A = (8.055)^2 \times 0.050 = 64.89 \times 0.050 = 3.24 \text{ m s}^{-2}\)

(ii) The maximum speed is:
\(v_{\text{max}} = \omega A = 8.055 \times 0.050 = 0.403 \text{ m s}^{-1}\)

(c) The kinetic energy is maximum when the mass passes through its equilibrium position (displacement \(x = 0\)).
The elastic potential energy is maximum at the points of maximum displacement (amplitude \(x = \pm 0.050 \text{ m}\)).

(a)
- Recall/use of \(T = 2\pi \sqrt{\frac{m}{k}}\) [1 mark]
- Correct rearrangement to make \(k\) the subject: \(k = \frac{4\pi^2 m}{T^2}\) [1 mark]
- Calculation yielding \(22.7 \text{ N m}^{-1}\) (which rounds to \(23\)) [1 mark]

(b)(i)
- Use of \(\omega = \frac{2\pi}{T}\) to obtain \(8.06 \text{ rad s}^{-1}\) [1 mark]
- Use of \(a_{\text{max}} = \omega^2 A\) (or \(a_{\text{max}} = \frac{k}{m} A\)) [1 mark]
- Correct value of \(3.24 \text{ m s}^{-2}\) (accept range \(3.20 - 3.30\)) [1 mark]

(b)(ii)
- Use of \(v_{\text{max}} = \omega A\) [1 mark]
- Correct value of \(0.403 \text{ m s}^{-1}\) (accept range \(0.40 - 0.41\)) [1 mark]

(c)
- Kinetic energy is maximum at the equilibrium position / where displacement is zero [1 mark]
- Elastic potential energy is maximum at maximum displacement / at amplitude positions [1.625 marks]

(a) The force acting on the wire is its weight:
\(F = m g = 6.00 \times 9.81 = 58.86 \text{ N}\)

Young modulus \(E\) is given by:
\(E = \frac{\text{Stress}}{\text{Strain}} = \frac{F / A}{x / L} = \frac{F L}{A x}\)

Rearranging for extension \(x\):
\(x = \frac{F L}{A E} = \frac{58.86 \times 2.20}{1.50 \times 10^{-7} \times 2.00 \times 10^{11}} = \frac{129.492}{30000} = 4.316 \times 10^{-3} \text{ m} \approx 4.32 \text{ mm}\)

(b) The elastic potential energy stored in the wire is:
\(E_{\text{p}} = \frac{1}{2} F x = \frac{1}{2} \times 58.86 \times 4.316 \times 10^{-3} \approx 0.127 \text{ J}\)

(c) To reduce uncertainties:
1. Use a micrometer screw gauge to measure the diameter of the wire at several different positions and orientations along its length, and calculate the mean to account for non-uniformity of cross-section.
2. Use a long wire (or a reference wire system like Searles apparatus) to ensure that the extension is large enough to be measured with low percentage uncertainty and to eliminate support flexure.

(a)
- Correct calculation of tension force \(F = m g = 58.86 \text{ N}\) (accept \(58.9 \text{ N}\) or use of \(g=9.8 \text{ m s}^{-2}\)) [1 mark]
- Correct rearrangement of Young Modulus equation for extension: \(x = \frac{FL}{AE}\) [1 mark]
- Final answer: \(4.32 \times 10^{-3} \text{ m}\) (accept \(4.3 \text{ mm}\)) [1 mark]

(b)
- Recall/use of energy formula: \(E_p = \frac{1}{2} F x\) [1 mark]
- Substitution of tension force and extension values [1 mark]
- Final answer: \(0.127 \text{ J}\) (accept \(0.13 \text{ J}\)) [1 mark]

(c)
- Suggestion to measure diameter at multiple points/orientations [1 mark]
- Use of micrometer screw gauge and calculate average [1 mark]
- Suggestion of using a reference wire (e.g. Searles apparatus) to eliminate errors from support yielding, or using a long test wire to increase extension [2.625 marks]

(a) Wien's displacement law states:
\(\lambda_{\text{max}} T = 2.90 \times 10^{-3} \text{ m K}\)

\(\lambda_{\text{max}} = \frac{2.90 \times 10^{-3}}{5800} = 5.00 \times 10^{-7} \text{ m}\) (or \(500 \text{ nm}\))

(b) Stefan's law for luminosity \(L\):
\(L = 4 \pi r^2 \sigma T^4\)

Substitute the values:
\(L = 4 \pi \times (7.00 \times 10^8)^2 \times (5.67 \times 10^{-8}) \times 5800^4\)
\(L = 4 \pi \times (4.90 \times 10^{17}) \times (5.67 \times 10^{-8}) \times (1.1316 \times 10^{15})\)
\(L \approx 3.95 \times 10^{26} \text{ W}\), which is approximately \(4.0 \times 10^{26} \text{ W}\).

(c) Radiant flux intensity (apparent brightness) \(I\) is given by:
\(I = \frac{L}{4 \pi d^2}\)

Rearranging for distance \(d\):
\(d = \sqrt{\frac{L}{4 \pi I}}\)

\(d = \sqrt{\frac{3.953 \times 10^{26}}{4 \pi \times 1360}} = \sqrt{\frac{3.953 \times 10^{26}}{17090.3}} = \sqrt{2.313 \times 10^{22}} \approx 1.52 \times 10^{11} \text{ m}\)

(a)
- State/use Wien's law formula: \(\lambda_{\text{max}} T = \text{constant}\) [1 mark]
- Correct calculation to give \(5.00 \times 10^{-7} \text{ m}\) (or \(500 \text{ nm}\)) [2 marks]

(b)
- State/use Stefan's law: \(L = 4\pi r^2 \sigma T^4\) [1 mark]
- Substitution of values into formula including \(\sigma = 5.67 \times 10^{-8}\) [1 mark]
- Correct calculation showing \(3.95 \times 10^{26} \text{ W}\) (matching the approximate value of \(4.0 \times 10^{26} \text{ W}\)) [1 mark]

(c)
- State/use equation relating intensity, luminosity, and distance: \(I = \frac{L}{4\pi d^2}\) [1 mark]
- Correct rearrangement: \(d = \sqrt{\frac{L}{4\pi I}}\) [1 mark]
- Substitution of values to find distance [1 mark]
- Final answer: \(1.52 \times 10^{11} \text{ m}\) (accept range \(1.50 \times 10^{11} \text{ m}\) to \(1.53 \times 10^{11} \text{ m}\)) [1.625 marks]

(a) The free-body force diagram should show two vertical vectors acting on the helicopter:
1. An arrow pointing vertically upwards representing the Lift Force \(L\).
2. An arrow pointing vertically downwards representing the Weight \(W\).
Since the helicopter is accelerating upwards, the upward Lift arrow must be drawn longer than the downward Weight arrow.

(b) First, calculate the acceleration \(a\):
\(a = \frac{v - u}{t} = \frac{8.00 - 0}{4.00} = 2.00 \text{ m s}^{-2}\)

Using Newton's second law:
\(F_{\text{net}} = L - W = m a\)
\(L = m g + m a = m (g + a)\)
\(L = 1600 \times (9.81 + 2.00) = 1600 \times 11.81 = 18896 \text{ N} \approx 1.89 \times 10^4 \text{ N}\)

(c) When climbing at a constant speed of \(30.0 \text{ m s}^{-1}\), the acceleration is zero, so the net vertical force is zero.
\(F_{\text{net}} = L - W - F_{\text{drag}} = 0 \implies L = W + F_{\text{drag}}\)

First, calculate the drag force:
\(F_{\text{drag}} = k v^2 = 1.15 \times (30.0)^2 = 1.15 \times 900 = 1035 \text{ N}\)

Weight of the helicopter:
\(W = m g = 1600 \times 9.81 = 15696 \text{ N}\)

Therefore, the required lift force is:
\(L = 15696 + 1035 = 16731 \text{ N} \approx 1.67 \times 10^4 \text{ N}\)

(a)
- Upward arrow labelled 'lift' or 'upthrust' and downward arrow labelled 'weight' or 'gravitational force' [1 mark]
- Upward arrow drawn visibly longer than the downward arrow [1 mark]

(b)
- Calculate acceleration: \(a = 2.00 \text{ m s}^{-2}\) [1 mark]
- Express Newton's second law: \(L - mg = ma\) [1 mark]
- Calculate lift force: \(1.89 \times 10^4 \text{ N}\) (accept \(1.88 \times 10^4 \text{ N}\) using \(g=9.8\)) [1 mark]

(c)
- Recognize that at constant speed, \(a=0\), so \(L = W + F_{\text{drag}}\) [1 mark]
- Correct calculation of drag force: \(F_{\text{drag}} = 1035 \text{ N}\) [1 mark]
- Correct calculation of weight: \(W = 15696 \text{ N}\) [1 mark]
- Final total lift force: \(1.67 \times 10^4 \text{ N}\) (accept range \(1.66 \times 10^4 \text{ N}\) to \(1.68 \times 10^4 \text{ N}\)) [2.625 marks]

(a) Calculate the energy required to raise the temperature:
\(\Delta \theta = 100.0 - 18.0 = 82.0^\circ\text{C}\)
\(Q = m c \Delta \theta = 1.10 \times 4180 \times 82.0 = 377036 \text{ J}\)

Minimum time required:
\(t = \frac{Q}{P} = \frac{377036}{2400} \approx 157.1 \text{ s}\), which is approximately \(150 \text{ s}\) (more precisely \(1.6 \times 10^2 \text{ s}\)).

(b) Energy supplied in \(90.0 \text{ s}\):
\(Q_{\text{vap}} = P \times t = 2400 \times 90.0 = 216000 \text{ J}\)

Using \(Q = m L_{\text{v}}\):
\(m = \frac{Q_{\text{vap}}}{L_{\text{v}}} = \frac{216000}{2.26 \times 10^6} \approx 0.0956 \text{ kg}\)

(c) During boiling, the thermal energy supplied does not increase the kinetic energy of the molecules. Instead, the energy is used to break/overcome the strong intermolecular bonds (forces of attraction) holding the molecules in the liquid state. This increases the potential energy of the molecules as they separate. Since temperature depends only on the mean kinetic energy of the molecules, the temperature remains constant during this phase change.

(a)
- Calculate temperature change \(\Delta \theta = 82 \text{ K}\) [1 mark]
- Apply heat formula: \(Q = mc\Delta\theta\) to find \(Q = 377000 \text{ J}\) [1 mark]
- Use of \(t = \frac{Q}{P}\) to obtain \(157 \text{ s}\) (showing it is close to \(150 \text{ s}\)) [1 mark]

(b)
- Calculate energy supplied during boiling: \(Q = Pt = 2400 \times 90 = 2.16 \times 10^5 \text{ J}\) [1 mark]
- State/use formula: \(Q = mL_v\) [1 mark]
- Correct rearrangement and calculation: \(m = \frac{2.16 \times 10^5}{2.26 \times 10^6} = 0.0956 \text{ kg}\) [2 marks]

(c)
- Supplied thermal energy goes into breaking/overcoming intermolecular bonds [1 mark]
- This results in an increase in the potential energy of the molecules [1 mark]
- Temperature is proportional to the mean kinetic energy of the molecules, which remains constant [1.625 marks]

(a) Charge stored:
\(Q_0 = C V_0 = (470 \times 10^{-6} \text{ F}) \times 12.0 \text{ V} = 5.64 \times 10^{-3} \text{ C}\) (or \(5.64 \text{ mC}\))

(b) Time constant:
\(\tau = R C = (15.0 \times 10^3 \ \Omega) \times (470 \times 10^{-6} \text{ F}) = 7.05 \text{ s}\)
This is approximately \(7.1 \text{ s}\).

(c) Potential difference after time \(t = 10.0 \text{ s}\):
\(V = V_0 e^{-t / R C}\)
\(V = 12.0 \times e^{-10.0 / 7.05} = 12.0 \times e^{-1.4184} = 12.0 \times 0.2421 = 2.91 \text{ V}\)

(d) Energy remaining in the capacitor:
\(E = \frac{1}{2} C V^2\)
\(E = 0.5 \times (470 \times 10^{-6}) \times (2.905)^2 = 2.35 \times 10^{-4} \times 8.44 = 1.98 \times 10^{-3} \text{ J}\) (or \(1.98 \text{ mJ}\) / \(1.99 \text{ mJ}\) depending on rounding).

(a)
- State/use: \(Q = CV\) [1 mark]
- Correct calculation of charge: \(5.64 \times 10^{-3} \text{ C}\) [1 mark]

(b)
- State/use: \(\tau = RC\) [1 mark]
- Substitute values: \(15.0 \times 10^3 \times 470 \times 10^{-6}\) to show \(7.05 \text{ s}\) [1 mark]

(c)
- State/use: \(V = V_0 e^{-t/RC}\) [1 mark]
- Substitute \(t = 10.0 \text{ s}\) and \(\tau = 7.05 \text{ s}\) [1 mark]
- Correct calculation: \(2.91 \text{ V}\) (accept range \(2.90 - 2.93 \text{ V}\)) [1 mark]

(d)
- State/use: \(E = \frac{1}{2} C V^2\) or \(E = E_0 e^{-2t/RC}\) [1 mark]
- Substitute correct potential difference (either from (c) or using exact expression) [1 mark]
- Correct calculation of energy: \(1.98 \times 10^{-3} \text{ J}\) to \(1.99 \times 10^{-3} \text{ J}\) [1.625 marks]

(a) Electric field strength:
\(E = \frac{V}{d} = \frac{1500 \text{ V}}{0.0250 \text{ m}} = 6.00 \times 10^4 \text{ V m}^{-1}\) (or \(\text{N C}^{-1}\))

(b) (i) Electrostatic force on the electron:
\(F = e E = (1.60 \times 10^{-19} \text{ C}) \times (6.00 \times 10^4 \text{ V m}^{-1}) = 9.60 \times 10^{-15} \text{ N}\)

(ii) Work done by the field equals kinetic energy gained:
\(e V = \frac{1}{2} m_{\text{e}} v^2\)
\(v = \sqrt{\frac{2 e V}{m_{\text{e}}}} = \sqrt{\frac{2 \times 1.60 \times 10^{-19} \times 1500}{9.11 \times 10^{-31}}} = \sqrt{\frac{4.80 \times 10^{-16}}{9.11 \times 10^{-31}}} = \sqrt{5.269 \times 10^{14}} \approx 2.30 \times 10^7 \text{ m s}^{-1}\)

(c) The proton experiences a constant vertically downward electrostatic force (towards the negative plate). This causes a constant downward acceleration. Since its horizontal velocity remains constant, the proton will follow a parabolic path curving downwards.

(a)
- State/use: \(E = \frac{V}{d}\) [1 mark]
- Correct calculation: \(6.00 \times 10^4 \text{ V m}^{-1}\) [1 mark]

(b)(i)
- State/use: \(F = q E\) [1 mark]
- Correct substitution of electron charge \(1.60 \times 10^{-19} \text{ C}\) [1 mark]
- Correct force calculation: \(9.60 \times 10^{-15} \text{ N}\) [1 mark]

(b)(ii)
- State/use conservation of energy: \(eV = \frac{1}{2} m v^2\) or equation of motion [1 mark]
- Rearrangement to find \(v\) [1 mark]
- Correct calculation: \(2.30 \times 10^7 \text{ m s}^{-1}\) (accept range \(2.29 \times 10^7 \text{ m s}^{-1}\) to \(2.31 \times 10^7 \text{ m s}^{-1}\)) [1 mark]

(c)
- Constant vertical force/acceleration downwards [1 mark]
- Constant horizontal velocity results in a parabolic trajectory downwards [1.625 marks]

The total energy of the oscillator is proportional to the square of the amplitude, \(E_{\text{total}} = \frac{1}{2} k A^2\). The potential energy at displacement \(x = \frac{1}{3}A\) is given by \(E_p = \frac{1}{2} k x^2 = \frac{1}{2} k \left(\frac{1}{3}A\right)^2 = \frac{1}{9} E_{\text{total}}\). The kinetic energy is therefore \(E_k = E_{\text{total}} - E_p = \frac{8}{9} E_{\text{total}}\). The ratio of kinetic energy to potential energy is \(\frac{E_k}{E_p} = \frac{8/9}{1/9} = 8\).

Correct option D selected. [1 mark]

The root-mean-square speed of an ideal gas molecule is given by \(v_{\text{rms}} = \sqrt{\frac{3kT}{m}}\), where \(k\) is the Boltzmann constant, \(T\) is the absolute temperature, and \(m\) is the mass of a molecule. Thus, \(v_{\text{rms}} \propto \sqrt{T}\). Changes in volume and pressure do not alter this direct dependence as long as the temperature is specified. Since the absolute temperature increases by \(21\%\), the final temperature is \(1.21 T_{\text{initial}}\). Therefore, the ratio of the new r.m.s. speed to the initial r.m.s. speed is \(\sqrt{1.21} = 1.10\).

Correct option A selected. [1 mark]

A \(\pi^+\) meson has a charge of \(+1e\) and a baryon number of \(0\). A neutron (\(udd\)) has a charge of \(0\) and a baryon number of \(+1\). Since weak interactions conserve charge and baryon number, the parent particle \(X\) must have a charge of \(+1e\) and a baryon number of \(+1\) (indicating it is a baryon composed of three quarks). The quark combination \(uud\) (proton) has a charge of \(+2/3 + 2/3 - 1/3 = +1e\) and is a baryon. Thus, \(uud\) is the correct composition.

Correct option A selected. [1 mark]

Let the initial activity of Q be \(A_0\), meaning the initial activity of P is \(2A_0\). At time \(t\), the activity of P is \(A_P(t) = 2A_0 \times 2^{-t/3.0}\) and the activity of Q is \(A_Q(t) = A_0 \times 2^{-t/6.0}\). Setting these equal: \(2A_0 \times 2^{-t/3.0} = A_0 \times 2^{-t/6.0} \implies 2 \times 2^{-t/3.0} = 2^{-t/6.0}\). Dividing both sides by \(2^{-t/3.0}\) gives \(2 = 2^{-t/6.0 + t/3.0} = 2^{t/6.0}\). This yields \(t/6.0 = 1\), which means \(t = 6.0\text{ hours}\).

Correct option B selected. [1 mark]

According to the Stefan-Boltzmann law, the luminosity of a star is given by \(L = 4\pi R^2 \sigma T^4\). Since both stars have the same surface temperature \(T\), their luminosity is proportional to the square of their radius: \(L \propto R^2\). Thus, the ratio of the radii is the square root of the ratio of their luminosities: \(\frac{R_{\text{Alpha}}}{R_{\text{Beta}}} = \sqrt{1.6 \times 10^5} = \sqrt{160000} = 400\).

Correct option A selected. [1 mark]

The fringe spacing is given by the formula \(x = \frac{\lambda D}{d}\). The new fringe spacing is \(x' = \frac{\lambda' D'}{d'} = \frac{(1.2\lambda)(0.5D)}{2d} = \frac{0.60}{2} \frac{\lambda D}{d} = 0.30x\).

Correct option A selected. [1 mark]

In a PET scan, a radioactive tracer that emits positrons is introduced into the patient. The emitted positrons travel a short distance before annihilating with electrons in the body's tissues. This annihilation process converts their mass-energy into two gamma-ray photons travelling in opposite directions, which are then detected by a ring of detectors to construct the image.

Correct option B selected. [1 mark]

The energy of a single photon is given by \(E = \frac{hc}{\lambda} = \frac{(6.63 \times 10^{-34}\text{ J s}) \times (3.00 \times 10^8\text{ m s}^{-1})}{633 \times 10^{-9}\text{ m}} \approx 3.142 \times 10^{-19}\text{ J}\). The number of photons emitted per second \(N\) is the power divided by the photon energy: \(N = \frac{P}{E} = \frac{5.0 \times 10^{-3}\text{ W}}{3.142 \times 10^{-19}\text{ J}} \approx 1.59 \times 10^{16}\text{ s}^{-1}\). Rounding to two significant figures gives \(1.6 \times 10^{16}\text{ s}^{-1}\).

Correct option B selected. [1 mark]

The power \(P\) of a laser is given by the total energy of the photons emitted per second: \(P = N E_{\text{photon}} = N \frac{hc}{\lambda}\) where \(N\) is the number of photons emitted per second. Rearranging for \(N\) gives \(N = \frac{P \lambda}{hc}\). Since \(h\) and \(c\) are constants, \(N \propto P \lambda\). The ratio is \(\frac{N_A}{N_B} = \frac{P_A \lambda_A}{P_B \lambda_B} = \frac{12 \times 10^{-3} \times 400 \times 10^{-9}}{18 \times 10^{-3} \times 600 \times 10^{-9}} = \frac{4800}{10800} = \frac{4}{9}\).

1 mark: Correctly identifies that \(N \propto P \lambda\), substitutes values to obtain \(\frac{4}{9}\), and selects option A.

First, find the equivalent capacitance of the parallel combination: \(C_p = C + C = 2C\). This combination is in series with the third capacitor \(C_3 = C\). The total equivalent capacitance \(C_{\text{eq}}\) is: \(C_{\text{eq}} = \frac{C_p C_3}{C_p + C_3} = \frac{2C \times C}{2C + C} = \frac{2}{3}C\). The total charge drawn from the supply of potential difference \(V\) is: \(Q = C_{\text{eq}} V = \frac{2}{3}CV\). In a series circuit, the charge stored on the third capacitor is equal to the total charge: \(Q_3 = \frac{2}{3}CV\). The energy stored in the third capacitor is: \(E_3 = \frac{Q_3^2}{2C} = \frac{(\frac{2}{3}CV)^2}{2C} = \frac{2}{9}CV^2\).

1 mark: Calculates total equivalent capacitance as \(\frac{2}{3}C\), charge as \(\frac{2}{3}CV\), and energy as \(\frac{2}{9}CV^2\) to select option B.

The intensity \(I\) of the X-ray beam after passing through the composite shield is: \(I = I_0 e^{-\mu_A x_A} e^{-\mu_B x_B} = I_0 e^{-(\mu_A x_A + \mu_B x_B)}\). Calculate the total exponent: \(\mu_A x_A + \mu_B x_B = (0.50 \times 1.8) + (0.20 \times 3.5) = 0.90 + 0.70 = 1.60\). The fraction of intensity transmitted is \(\frac{I}{I_0} = e^{-1.60} \approx 0.2019\). This corresponds to approximately 20\%.

1 mark: Computes the sum of attenuation exponents as 1.60, finds the transmitted percentage as 20\%, and selects option B.

The fringe width is given by \(x = \frac{\lambda D}{d}\). Since \(\lambda = \frac{c}{f}\), we can write \(x = \frac{c D}{f d}\). With the modifications, the new fringe width is \(x' = \frac{c (2D)}{(2f)(0.5d)} = 2 \frac{c D}{f d} = 2x\).

1 mark: Expresses fringe width in terms of frequency, scales the variables appropriately to show that \(x' = 2x\), and selects option C.

Initially in equilibrium: \(q E = mg \implies q \frac{V}{d} = mg\) with the electric force directed upwards. With the changes, the new electric field strength is \(E' = \frac{1.5 V}{0.75 d} = 2 E\). The new upward force is \(F_E' = q E' = 2 q E = 2 mg\). The net vertical force is \(F_{\text{net}} = F_E' - mg = 2 mg - mg = mg\) upwards. Therefore, the acceleration is \(a = \frac{F_{\text{net}}}{m} = g\) upwards.

1 mark: Computes the new field strength as \(2E\), determines the net upward force is \(mg\), calculates the acceleration as \(g\) upwards, and selects option D.

In \(\beta^+\) decay, a proton decays into a neutron: \(p \rightarrow n + e^+ + \nu_e\). The proton consists of \(uud\) quarks and the neutron consists of \(udd\) quarks, so the quark change is \(u \rightarrow d\). To conserve charge at the vertex, the exchange boson must be \(W^+\). The emitted leptons are a positron \(e^+\) and an electron neutrino \(\nu_e\). This matches Row B.

1 mark: Identifies the quark change as \(u \rightarrow d\), exchange boson as \(W^+\), and leptons as \(e^+\) and \(\nu_e\), selecting Row B (option B).

The resistance is \(R = \rho \frac{L}{A}\). Since the volume \(V_{\text{vol}} = A L\) is constant, we can write \(A = \frac{V_{\text{vol}}}{L}\), leading to \(R = \rho \frac{L^2}{V_{\text{vol}}}\). Thus, \(R \propto L^2\). When the length is increased by 10\%, the new length is \(L' = 1.10 L\). The new resistance is \(R' \propto (1.10 L)^2 = 1.21 L^2\), which means \(R' = 1.21 R\).

1 mark: Recognizes that for constant volume \(R \propto L^2\), calculates the multiplier as \(1.10^2 = 1.21\), and selects option B.

(a) Convert temperature to Kelvin: \( T_1 = 27 + 273.15 = 300.15 \text{ K} \) (accept \( 300 \text{ K} \)). Using the ideal gas equation: \( P_1 V_1 = n R T_1 \) so \( n = \frac{1.2 \times 10^5 \times 0.045}{8.31 \times 300} = 2.166 \approx 2.17 \text{ moles} \).
(b) The total volume is now \( V_2 = 0.045 + 0.015 = 0.060 \text{ m}^3 \). The final temperature is \( T_2 = 42 + 273.15 = 315.15 \text{ K} \) (accept \( 315 \text{ K} \)). Using the ideal gas equation: \( P_2 = \frac{n R T_2}{V_2} = \frac{2.166 \times 8.31 \times 315}{0.060} = 9.45 \times 10^4 \text{ Pa} \) (or \( 9.5 \times 10^4 \text{ Pa} \)).
(c) The molar mass in kilograms is \( M = 4.0 \times 10^{-3} \text{ kg mol}^{-1} \). The root-mean-square speed is given by \( v_{\text{rms}} = \sqrt{\frac{3 R T}{M}} = \sqrt{\frac{3 \times 8.31 \times 315}{4.0 \times 10^{-3}}} = \sqrt{1.963 \times 10^6} = 1.40 \times 10^3 \text{ m s}^{-1} \).

(a)
- M1: Conversion of 27 \, ^{\circ}\text{C} to 300 K.
- M1: Substitution of values into \( n = \frac{PV}{RT} \).
- A1: Correct calculation of moles as 2.17 (or 2.16) moles.

(b)
- M1: Identification of final volume as \( 0.060 \text{ m}^3 \).
- M1: Use of \( \frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2} \) or \( P_2 = \frac{n R T_2}{V_2} \).
- A1: Correct value of \( 9.45 \times 10^4 \text{ Pa} \) (accept \( 9.5 \times 10^4 \text{ Pa} \)).

(c)
- M1: Use of \( v_{\text{rms}} = \sqrt{\frac{3 R T}{M}} \) or \( \frac{1}{2} m v^2 = \frac{3}{2} k T \).
- M1: Conversion of molar mass to \( 4.0 \times 10^{-3} \text{ kg mol}^{-1} \).
- A1.44: Correct calculation of speed as \( 1.40 \times 10^3 \text{ m s}^{-1} \) (accept \( 1.4 \times 10^3 \text{ m s}^{-1} \)).

(a) For a series connection, the effective stiffness \( k_s \) is given by: \( \frac{1}{k_s} = \frac{1}{k_1} + \frac{1}{k_2} = \frac{1}{150} + \frac{1}{300} = \frac{3}{300} = \frac{1}{100} \). Therefore, \( k_s = 100 \text{ N m}^{-1} \) (shown).
(b) The tension in the springs at equilibrium is equal to the weight: \( F = m g = 2.5 \times 9.81 = 24.525 \text{ N} \). Total extension of the combination: \( x = \frac{F}{k_s} = \frac{24.525}{100} = 0.24525 \text{ m} \). The total elastic potential energy is \( E_p = \frac{1}{2} F x = \frac{1}{2} \times 24.525 \times 0.24525 = 3.01 \text{ J} \) (or using \( E_p = \frac{F^2}{2 k_s} = 3.01 \text{ J} \)).
(c) Since they are connected in series, the tension (force \( F \)) in both springs is the same. The energy stored in each spring is given by \( E = \frac{F^2}{2 k} \). Therefore: \( \frac{E_A}{E_B} = \frac{F^2 / (2 k_1)}{F^2 / (2 k_2)} = \frac{k_2}{k_1} = \frac{300}{150} = 2.0 \).

(a)
- M1: Use of formula \( \frac{1}{k_s} = \frac{1}{k_1} + \frac{1}{k_2} \).
- A1: Correct working leading to exactly 100 \text{ N m}^{-1}.

(b)
- M1: Calculation of weight \( W = 2.5 \times 9.81 = 24.5 \text{ N} \) (or 25 N using g=10).
- M1: Calculation of total extension \( x = 0.245 \text{ m} \).
- M1: Use of \( E_p = \frac{1}{2} k x^2 \) or \( E_p = \frac{1}{2} F x \).
- A1: Final answer of \( 3.01 \text{ J} \) (accept \( 3.0 \text{ J} \) or \( 3.1 \text{ J} \) if g=10 is used).

(c)
- M1: Recognition that the tension/force is the same in both series springs.
- M1: Use of \( E \propto \frac{1}{k} \) or individual energy calculations (\( E_A = 2.0 \text{ J} \), \( E_B = 1.0 \text{ J} \)).
- A1.44: Correct ratio of 2.0.

(a) Wien's displacement law states that the peak emission wavelength \( \lambda_{\text{max}} \) is inversely proportional to the absolute temperature \( T \) of the black body: \( \lambda_{\text{max}} T = \text{constant} \approx 2.90 \times 10^{-3} \text{ m K} \). Therefore: \( T = \frac{2.90 \times 10^{-3}}{410 \times 10^{-9}} = 7073 \approx 7070 \text{ K} \).
(b) From Stefan's Law: \( L = 4 \pi r^2 \sigma T^4 \). Rearranging for radius \( r \): \( r = \sqrt{\frac{L}{4 \pi \sigma T^4}} = \sqrt{\frac{4.5 \times 10^{27}}{4 \pi \times 5.67 \times 10^{-8} \times (7073)^4}} = \sqrt{\frac{4.5 \times 10^{27}}{1.783 \times 10^9}} = 1.59 \times 10^9 \text{ m} \).
(c) The radiant flux intensity \( F \) is: \( F = \frac{L}{4 \pi d^2} = \frac{4.5 \times 10^{27}}{4 \pi \times (1.2 \times 10^{18})^2} = \frac{4.5 \times 10^{27}}{1.81 \times 10^{37}} = 2.49 \times 10^{-10} \text{ W m}^{-2} \).

(a)
- M1: State that peak wavelength is inversely proportional to absolute temperature.
- M1: Set up \( T = \frac{\text{constant}}{\lambda} \).
- A1: Final temperature of \( 7070 \text{ K} \) (or \( 7100 \text{ K} \) or \( 7050 \text{ K} \) depending on the value of constant used).

(b)
- M1: Recall Stefan-Boltzmann law formula: \( L = 4 \pi r^2 \sigma T^4 \).
- M1: Correct rearrangement for \( r \).
- M1: Substitute values correctly (including Stefan constant \( 5.67 \times 10^{-8} \)).
- A1: Correct calculation of radius \( r = 1.59 \times 10^9 \text{ m} \) (accept \( 1.58 \times 10^9 \) to \( 1.61 \times 10^9 \)).

(c)
- M1: Recall and use \( F = \frac{L}{4 \pi d^2} \).
- A1.44: Correct calculation of radiant flux intensity \( F = 2.49 \times 10^{-10} \text{ W m}^{-2} \) (accept \( 2.5 \times 10^{-10} \)).

(a) The normal reaction force \( R \) is given by: \( R = m g \cos(30^{\circ}) = 4.0 \times 9.81 \times \cos(30^{\circ}) = 33.98 \text{ N} \). Frictional force: \( F_f = \mu R = 0.25 \times 33.98 = 8.495 \approx 8.5 \text{ N} \) (shown).
(b) The component of the gravitational force pulling the block down the incline is: \( F_d = m g \sin(30^{\circ}) = 4.0 \times 9.81 \times 0.5 = 19.62 \text{ N} \). The net force acting down the plane is: \( F_{\text{net}} = F_d - F_f = 19.62 - 8.495 = 11.125 \text{ N} \). The acceleration is: \( a = \frac{F_{\text{net}}}{m} = \frac{11.125}{4.0} = 2.78 \text{ m s}^{-2} \).
(c) Using equations of motion with constant acceleration: \( v^2 = u^2 + 2 a s \). Since the block starts from rest, \( u = 0 \): \( v^2 = 0 + 2 \times 2.781 \times 1.5 = 8.343 \implies v = \sqrt{8.343} = 2.89 \text{ m s}^{-1} \).

(a)
- M1: Use of \( R = m g \cos(\theta) \).
- M1: Use of \( F_f = \mu R \).
- A1: Correct calculation leading to \( 8.495 \text{ N} \) which rounds to \( 8.5 \text{ N} \).

(b)
- M1: Calculation of parallel component of gravity \( F_d = m g \sin(30^{\circ}) = 19.6 \text{ N} \).
- M1: Use of \( F_{\text{net}} = m a \) to set up equation \( m g \sin(30^{\circ}) - F_f = m a \).
- A1.44: Correct calculation of acceleration as \( 2.78 \text{ m s}^{-2} \) (or \( 2.8 \text{ m s}^{-2} \)).

(c)
- M1: Select appropriate kinematic equation \( v^2 = u^2 + 2as \).
- M1: Correct substitution of values with \( u = 0 \).
- A1: Correct final velocity as \( 2.89 \text{ m s}^{-1} \) (or \( 2.9 \text{ m s}^{-1} \)).

(a) The angular frequency is: \( \omega = 2 \pi f = 2 \pi \times 25 = 50\pi \approx 157.1 \text{ rad s}^{-1} \). The maximum acceleration is related to amplitude by: \( a_{\text{max}} = \omega^2 A \implies A = \frac{180}{(157.08)^2} = 7.30 \times 10^{-3} \text{ m} \) (or 7.3 mm).
(b) The maximum speed is given by: \( v_{\text{max}} = \omega A = 157.08 \times 7.295 \times 10^{-3} = 1.15 \text{ m s}^{-1} \).
(c) The restoring force magnitude is: \( F = m a = m \omega^2 x \). Converting displacement to meters: \( x = 1.5 \times 10^{-3} \text{ m} \). Hence: \( F = 0.12 \times (157.08)^2 \times 1.5 \times 10^{-3} = 4.44 \text{ N} \).

(a)
- M1: Correct calculation of \( \omega = 2\pi f \approx 157 \text{ rad s}^{-1} \).
- M1: State and rearrange \( a_{\text{max}} = \omega^2 A \).
- A1: Correct amplitude as \( 7.30 \times 10^{-3} \text{ m} \) (or \( 7.3 \text{ mm} \)).

(b)
- M1: Use of \( v_{\text{max}} = \omega A \) or \( v_{\text{max}} = \sqrt{a_{\text{max}} A} \).
- M1: Correct substitution of values.
- A1: Correct calculation of maximum speed as \( 1.15 \text{ m s}^{-1} \).

(c)
- M1: Recognition that \( F = m \omega^2 x \) or \( a = \omega^2 x \).
- M1: Correct substitution of \( x = 1.5 \times 10^{-3} \text{ m} \).
- A1.44: Correct force magnitude of \( 4.44 \text{ N} \) (accept \( 4.4 \text{ N} \)).

(a) The process consists of two stages:
1. Heating the ice to \( 0 \, ^{\circ}\text{C} \): \( Q_1 = m_i c_i \Delta T = 0.45 \times 2100 \times 15 = 14175 \text{ J} \).
2. Melting the ice to water: \( Q_2 = m_i L_f = 0.45 \times (3.34 \times 10^5) = 150300 \text{ J} \).
Total thermal energy required: \( Q_{\text{total}} = 14175 + 150300 = 164475 \text{ J} \approx 1.64 \times 10^5 \text{ J} \).
(b) Specific heat capacity is the energy required to change temperature, increasing the average kinetic energy of the molecules (vibrations) without changing state or potential energy. Specific latent heat is the energy required to change the state at constant temperature, breaking bonds (increasing potential energy and separation) without changing kinetic energy.
(c) Thermal energy released by the condensing steam and subsequent cooling of the condensed water to \( 0 \, ^{\circ}\text{C} \):
\( Q_{\text{released}} = m_s L_v + m_s c_w \Delta T_w = m_s \times (2.26 \times 10^6) + m_s \times 4180 \times 100 \)
\( Q_{\text{released}} = m_s (2.26 \times 10^6 + 4.18 \times 10^5) = m_s \times 2.678 \times 10^6 \text{ J} \).
Using conservation of energy: \( m_s \times 2.678 \times 10^6 = 164475 \implies m_s = 0.0614 \text{ kg} \) (or 61.4 g).

(a)
- M1: Use of \( Q = mc\Delta T \) for heating ice.
- M1: Use of \( Q = mL_f \) for melting ice.
- M1: Correct addition of both energy components.
- A1: Correct final energy as \( 1.64 \times 10^5 \text{ J} \) (accept \( 1.6 \times 10^5 \text{ J} \)).

(b)
- B1: Specific heat capacity associated with changes in kinetic energy of molecules (temperature change).
- B1: Specific latent heat associated with breaking molecular bonds / changes in potential energy of molecules (state change).

(c)
- M1: Expression for steam condensation: \( m L_v \).
- M1: Expression for hot water cooling: \( m c \Delta T \).
- M1: Setting heat gained equal to heat lost.
- A1.44: Correct mass calculation to give \( 0.0614 \text{ kg} \) (or \( 6.14 \times 10^{-2} \text{ kg} \)).

(a) The nuclear equation is: \( {}_0^1\text{n} \rightarrow {}_1^1\text{p} + {}_{-1}^{\phantom{-}0}\text{e} + \bar{\nu}_e \) (accept \( \text{n} \rightarrow \text{p} + \text{e}^- + \bar{\nu}_e \)).
(b) In terms of quarks, a down quark decays into an up quark: \( \text{d} \rightarrow \text{u} + \text{e}^- + \bar{\nu}_e \). On the left-hand side, the charge of the down quark is \( -1/3 \, e \). On the right-hand side, the total charge is: \( +2/3 \, e \text{ (up quark)} + (-1 \, e \text{ (electron)}) + 0 \text{ (antineutrino)} = -1/3 \, e \). Thus, charge is conserved.
(c) The momentum \( p \) of the electron is related to its kinetic energy \( E_k \) by: \( E_k = \frac{p^2}{2m} \implies p = \sqrt{2 m E_k} = \sqrt{2 \times 9.11 \times 10^{-31} \times 5.2 \times 10^{-14}} = 3.078 \times 10^{-22} \text{ kg m s}^{-1} \). The de Broglie wavelength is: \( \lambda = \frac{h}{p} = \frac{6.63 \times 10^{-34}}{3.078 \times 10^{-22}} = 2.15 \times 10^{-12} \text{ m} \).

(a)
- M1: Neutron on LHS, proton and electron on RHS.
- M1: Electron antineutrino on RHS (with bar over nu).
- A1: Balance of nucleon and proton numbers shown correctly.

(b)
- M1: State quark change as \( \text{d} \rightarrow \text{u} \).
- M1: Identify charges of down quark (\( -1/3 \, e \)) and up quark (\( +2/3 \, e \)).
- A1: Show sum of charges on RHS is \( +2/3 - 1 = -1/3 \, e \), balancing the LHS.

(c)
- M1: Use of \( p = \sqrt{2 m E_k} \) or calculate velocity first using \( E_k = \frac{1}{2} m v^2 \).
- M1: Use of de Broglie equation \( \lambda = \frac{h}{p} \).
- A1.44: Correct calculation of wavelength as \( 2.15 \times 10^{-12} \text{ m} \) (accept \( 2.2 \times 10^{-12} \text{ m} \)).

(a) The distance from the central bright fringe to the 5th bright fringe is equal to 5 fringe widths (\( 5x \)). Hence: \( x = \frac{12.5 \text{ mm}}{5} = 2.5 \text{ mm} = 2.5 \times 10^{-3} \text{ m} \).
(b) Using the double-slit equation \( \lambda = \frac{a x}{D} \) (where slit separation \( a = d \)): \( d = \frac{\lambda D}{x} = \frac{650 \times 10^{-9} \times 1.8}{2.5 \times 10^{-3}} = 4.68 \times 10^{-4} \text{ m} \) (or \( 0.47 \text{ mm} \)).
(c) Since wavelength decreases (from 650 nm to 450 nm), the fringe spacing decreases, so the fringes become narrower and closer together. The new fringe spacing is: \( x_{\text{new}} = \frac{\lambda_{\text{new}} D}{d} = \frac{450 \times 10^{-9} \times 1.8}{4.68 \times 10^{-4}} = 1.73 \times 10^{-3} \text{ m} \). The distance to the 5th bright fringe is: \( 5 \times x_{\text{new}} = 5 \times 1.73 \times 10^{-3} = 8.65 \times 10^{-3} \text{ m} \) (or 8.65 mm).

(a)
- M1: State that the distance is 5 fringe spacings.
- A1: Correctly calculate \( x = 2.5 \text{ mm} \).

(b)
- M1: Recall \( \lambda = \frac{a x}{D} \).
- M1: Correct rearrangement for slit separation \( d = a \).
- M1: Correct substitution of values with consistent units.
- A1: Final answer of \( 4.68 \times 10^{-4} \text{ m} \) (or \( 4.7 \times 10^{-4} \text{ m} \)).

(c)
- B1: State that fringes become closer together/narrower because wavelength decreases.
- M1: Set up calculation for new spacing or use ratio method (e.g., \( 12.5 \times \frac{450}{650} \)).
- A1.44: Correct calculation of the 5th fringe distance as \( 8.65 \text{ mm} \) (accept \( 8.6 \text{ mm} \) or \( 8.7 \text{ mm} \)).

Part (a): Wien's displacement law states that \(\lambda_{\text{max}} T = 2.90 \times 10^{-3}\text{ m K}\). Rearranging for \(T\): \(T = \frac{2.90 \times 10^{-3}}{3.80 \times 10^{-7}} = 7631.6\text{ K}\), which rounds to \(7.63 \times 10^3\text{ K}\) or \(7630\text{ K}\). Part (b): Stefan's law for stellar luminosity is \(L = 4 \pi R^2 \sigma T^4\). Substituting the radius \(R = 1.40 \times 10^9\text{ m}\), Stefan's constant \(\sigma = 5.67 \times 10^{-8}\text{ W m}^{-2}\text{ K}^{-4}\), and the temperature \(T = 7632\text{ K}\): \(L = 4 \pi (1.40 \times 10^9)^2 \times (5.67 \times 10^{-8}) \times (7632)^4 = 4.74 \times 10^{27}\text{ W}\). Part (c): The intensity of the radiation at a distance \(d\) is given by the inverse square law: \(I = \frac{L}{4 \pi d^2}\). Rearranging for \(d\): \(d = \sqrt{\frac{L}{4 \pi I}} = \sqrt{\frac{4.74 \times 10^{27}}{4 \pi \times 1.65 \times 10^{-11}}} = \sqrt{2.286 \times 10^{37}} = 4.78 \times 10^{18}\text{ m}\). To convert this distance to light-years: \(d = \frac{4.78 \times 10^{18}\text{ m}}{9.46 \times 10^{15}\text{ m ly}^{-1}} = 505\text{ ly}\).

Part (a) [2 marks]:
- C1: Wien's law formula used correctly: \(T = \frac{2.90 \times 10^{-3}}{\lambda_{\text{max}}}\)
- A1: Correct calculation of temperature, \(7.6 \times 10^3\text{ K}\) or \(7630\text{ K}\) (allow 7600 to 7640 K).

Part (b) [3 marks]:
- C1: Recall and use of Stefan's law formula: \(L = 4 \pi R^2 \sigma T^4\)
- C1: Correct substitution of values including \(R^2\) and \(T^4\)
- A1: Final answer for luminosity of \(4.7 \times 10^{27}\text{ W}\) to \(4.8 \times 10^{27}\text{ W}\) (value dependent on rounding in part a).

Part (c) [4 marks]:
- C1: Recall and use of inverse square law: \(I = \frac{L}{4 \pi d^2}\)
- C1: Correct rearrangement for distance: \(d = \sqrt{\frac{L}{4 \pi I}}\)
- C1: Correct conversion from metres to light-years by dividing by \(9.46 \times 10^{15}\)
- A1: Correct final distance in light-years: 500 to 510 ly (allow ecf from part b).

(a) By definition, the time constant \(\tau = RC\) is the time taken for the charge, potential difference, or current in a discharging capacitor circuit to fall to \(1/e \approx 36.8\%\) of its initial value.

(b)(i) Using the capacitor discharge equation:
\(V = V_0 e^{-t/\tau}\)
Given \(V / V_0 = 0.25\) at \(t = 8.2\ \text{s}\):
\(0.25 = e^{-8.2/\tau}\)
Taking the natural logarithm of both sides:
\(\ln(0.25) = -\frac{8.2}{\tau}\)
\(-1.3863 = -\frac{8.2}{\tau}\)
\(\tau = \frac{8.2}{1.3863} = 5.915\ \text{s} \approx 5.9\ \text{s}\).

(b)(ii) Using \(\tau = RC\):
\(R = \frac{\tau}{C} = \frac{5.915}{470 \times 10^{-6}} = 12585\ \Omega \approx 1.3 \times 10^4\ \Omega\).

(c)(i) The exponent \(\gamma t\) must be a dimensionless quantity (having no units).
Therefore, \([\gamma] \times [t] = 1 \implies [\gamma] = \text{s}^{-1}\).
Alternatively, since \(\gamma = 1/\tau\) and the unit of \(\tau\) is seconds (\(\text{s}\)), the unit of \(\gamma\) must be \(\text{s}^{-1}\).

(c)(ii) We are given \(\gamma = \frac{1}{\tau} = \frac{1}{5.915\ \text{s}} = 0.1691\ \text{s}^{-1}\).
To find the time \(t\) for the amplitude \(A\) to decay to \(10\%\) of its initial value \(A_0\):
\(A = A_0 e^{-\gamma t} \implies 0.10 = e^{-0.1691 t}\)
\(\ln(0.10) = -0.1691 t\)
\(-2.3026 = -0.1691 t\)
\(t = \frac{2.3026}{0.1691} = 13.62\ \text{s} \approx 14\ \text{s}\) (or \(13.6\ \text{s}\)).

(a) B1: Time taken for the potential difference/charge/current of a discharging capacitor to fall to 1/e (approx. 37%) of its initial value.
(b)(i)
C1: Recalls and uses V = V_0 * e^(-t / tau) or equivalent log form.
C1: Correctly substitutes 0.25 and 8.2 s into the equation.
A1: tau = 5.9 s (accept 5.92 s).
(b)(ii)
C1: Recalls and uses R = tau / C.
A1: R = 1.3 x 10^4 ohms (allow ecf from b(i)).
(c)(i)
B1: States that the exponential argument gamma * t must be dimensionless (or states gamma = 1 / tau).
B1: Shows unit of gamma = s^(-1).
(c)(ii)
C1: Calculates gamma = 0.169 s^(-1) (allow ecf from b(i)).
C1: Sets up the equation 0.10 = e^(-gamma * t).
C1: Rearranges to t = -ln(0.10) / gamma.
A1: t = 14 s (or 13.6 s).

(a) Kepler's third law states that the square of the orbital period \(T\) of a body in orbit is directly proportional to the cube of its orbital radius \(r\) (i.e., \(T^2 \propto r^3\)).

(b)(i) The total orbital radius \(r\) of the spacecraft is the sum of the radius of Mars \(R_{\text{Mars}}\) and its height \(h\) above the surface:
\(r = R_{\text{Mars}} + h = 3.39 \times 10^6\ \text{m} + 2.10 \times 10^6\ \text{m} = 5.49 \times 10^6\ \text{m}\).
The gravitational force provides the centripetal force:
\(\frac{G M m}{r^2} = \frac{m v^2}{r} \implies v = \sqrt{\frac{G M}{r}}\)
\(v = \sqrt{\frac{6.67 \times 10^{-11} \times 6.42 \times 10^{23}}{5.49 \times 10^6}}\)
\(v = \sqrt{7.7998 \times 10^6} = 2.7928 \times 10^3\ \text{m s}^{-1} \approx 2.8 \times 10^3\ \text{m s}^{-1}\).

(b)(ii) The orbital speed is given by \(v = \frac{2\pi r}{T}\). Rearranging for the period \(T\):
\(T = \frac{2\pi r}{v} = \frac{2\pi \times 5.49 \times 10^6}{2792.8} = 12351\ \text{s}\).
To express this in hours:
\(T = \frac{12351}{3600} = 3.43\ \text{hours}\).

(b)(iii) The gravitational potential energy is:
\(E_p = -\frac{G M m}{r} = -\frac{6.67 \times 10^{-11} \times 6.42 \times 10^{23} \times 320}{5.49 \times 10^6} = -2.494 \times 10^9\ \text{J} \approx -2.49 \times 10^9\ \text{J}\).

(c) To escape the gravitational field of Mars completely, the spacecraft's total mechanical energy must be at least zero (\(E_{\text{total}} \ge 0\)).
The minimum escape velocity \(v_{\text{esc}}\) is given by:
\(\frac{1}{2} m v_{\text{esc}}^2 - \frac{G M m}{r} = 0 \implies v_{\text{esc}} = \sqrt{\frac{2 G M}{r}} = \sqrt{2} v\)
\(v_{\text{esc}} = \sqrt{2} \times 2792.8\ \text{m s}^{-1} = 3949.6\ \text{m s}^{-1}\).
The minimum change in speed \(\Delta v\) required is:
\(\Delta v = v_{\text{esc}} - v = 3949.6 - 2792.8 = 1156.8\ \text{m s}^{-1} \approx 1.16 \times 10^3\ \text{m s}^{-1}\) (or \(1.16\ \text{km s}^{-1}\)).

(a) B1: Period squared is proportional to orbital radius cubed (T^2 proportional to r^3).
(b)(i)
C1: Calculates total orbital radius r = 5.49 x 10^6 m.
C1: Equates gravitational force to centripetal force and obtains v = sqrt(GM/r).
A1: Shows value evaluates to 2.79 x 10^3 m/s.
(b)(ii)
C1: Uses T = 2 * pi * r / v.
C1: Divides seconds by 3600 to convert to hours.
A1: T = 3.43 hours (accept 3.4 hours).
(b)(iii)
C1: Uses E_p = -G M m / r.
A1: E_p = -2.49 x 10^9 J (ignore sign if only magnitude is given, but accept fully correct signed value).
(c)
C1: Explains that total mechanical energy must equal zero to escape (or uses escape speed formula v_esc = sqrt(2GM/r)).
C1: Calculates escape speed v_esc = 3.95 x 10^3 m/s (or states v_esc = sqrt(2) * v).
A1: Delta v = 1.16 x 10^3 m/s (or 1.16 km/s).

(a) Standard assumptions include: (1) The gas contains a very large number of molecules in rapid, random motion. (2) The volume of the gas molecules is negligible compared to the volume of the container. (3) Collisions between molecules and the container walls are perfectly elastic. (4) The duration of collisions is negligible compared to the time between collisions. (5) Intermolecular forces are negligible except during collisions.

(b)(i) Converting initial temperature to Kelvin:
\(T_1 = 20 + 273.15 = 293.15\ \text{K}\).
Using the ideal gas equation \(p V = n R T\):
\(p_1 = \frac{n R T_1}{V_1} = \frac{0.082 \times 8.31 \times 293.15}{2.0 \times 10^{-3}} = 99879\ \text{Pa} \approx 1.0 \times 10^5\ \text{Pa}\).

(b)(ii) Using the ideal gas equation for state 2:
\(p_2 V_2 = n R T_2 \implies T_2 = \frac{p_2 V_2}{n R}\)
\(T_2 = \frac{6.1 \times 10^5 \times 5.0 \times 10^{-4}}{0.082 \times 8.31} = \frac{305}{0.68142} = 447.59\ \text{K}\).
Converting to Celsius:
\(\theta_2 = 447.59 - 273.15 = 174.44\ ^∘\text{C} \approx 174\ ^∘\text{C}\).

(c)(i) The thermal energy lost by the gas is equal to its change in internal energy as it cools:
\(\Delta U = \frac{3}{2} n R \Delta T\)
where \(\Delta T = T_2 - T_1 = 447.59 - 293.15 = 154.44\ \text{K}\).
\(\Delta U = 1.5 \times 0.082 \times 8.31 \times 154.44 = 157.86\ \text{J} \approx 158\ \text{J}\).

(c)(ii) Using \(Q = m c \Delta T_{\text{cyl}}\) for the aluminum container:
\(\Delta T_{\text{cyl}} = \frac{Q}{m c} = \frac{157.86}{0.45 \times 900} = \frac{157.86}{405} = 0.390\ \text{K}\) (or \(^∘\text{C}\)).

(a) B1, B1: Any two standard ideal gas assumptions stated clearly.
(b)(i)
C1: Converts temperature to Kelvin (293 K).
A1: Rearranges ideal gas law and shows calculation yielding approx. 1.0 x 10^5 Pa.
(b)(ii)
C1: Recalls and uses p_1*V_1/T_1 = p_2*V_2/T_2 or p_2*V_2 = n*R*T_2.
C1: Calculates T_2 = 448 K.
A1: theta_2 = 174 °C.
(c)(i)
C1: Recalls and uses Delta U = 3/2 * n * R * Delta T.
C1: Identifies temperature change Delta T = 154 K.
A1: Delta U = 158 J (accept 157.9 J).
(c)(ii)
C1: Uses Q = m * c * Delta T.
A1: Delta T = 0.39 K (or °C) (allow ecf from c(i)).

(a) The diagram must depict the oil drop with two opposite vertical force vectors:
1. An upward electric force vector, \(F_E\), because the negative charge is attracted towards the positive upper plate.
2. A downward gravitational force vector (weight), \(W = mg\).

(b)(i) The magnitude of the electric field \(E\) between the parallel plates is:
\(E = \frac{V}{d} = \frac{1800\ \text{V}}{12 \times 10^{-3}\ \text{m}} = 1.50 \times 10^5\ \text{V m}^{-1}\).
The electric force \(F_E\) is given by:
\(F_E = q E = (2.4 \times 10^{-18}\ \text{C}) \times (1.50 \times 10^5\ \text{V m}^{-1}) = 3.60 \times 10^{-13}\ \text{N}\).

(b)(ii) The gravitational force (weight) acting on the oil drop is:
\(W = m g = (4.2 \times 10^{-14}\ \text{kg}) \times 9.81\ \text{m s}^{-2} = 4.12 \times 10^{-13}\ \text{N}\).
Since \(W > F_E\), the net vertical force acts downwards:
\(F_{\text{net}} = W - F_E = 4.12 \times 10^{-13}\ \text{N} - 3.60 \times 10^{-13}\ \text{N} = 5.20 \times 10^{-14}\ \text{N}\).
The net vertical acceleration is:
\(a = \frac{F_{\text{net}}}{m} = \frac{5.20 \times 10^{-14}\ \text{N}}{4.2 \times 10^{-14}\ \text{kg}} = 1.238\ \text{m s}^{-2} \approx 1.24\ \text{m s}^{-2}\) downwards.

(c)(i) The horizontal motion of the oil drop has a constant velocity \(u = 0.85\ \text{m s}^{-1}\):
\(t = \frac{L}{u} = \frac{0.050\ \text{m}}{0.85\ \text{m s}^{-1}} = 0.0588\ \text{s} \approx 0.059\ \text{s}\).

(c)(ii) The vertical displacement \(s\) can be found using the equations of motion for constant acceleration:
\(s = \frac{1}{2} a t^2 = 0.5 \times 1.238\ \text{m s}^{-2} \times (0.0588\ \text{s})^2 = 2.14 \times 10^{-3}\ \text{m}\) (or \(2.1\ \text{mm}\)).

(a)
B1: Draws a diagram with a vertical upward force labeled electric force (or F_E).
B1: Draws a downward force labeled weight (or W or gravitational force / mg).
(b)(i)
C1: Recalls and uses E = V / d to find E = 1.5 x 10^5 V/m.
A1: Shows F_E = q * E evaluates to 3.6 x 10^-13 N.
(b)(ii)
C1: Calculates weight W = mg = 4.12 x 10^-13 N.
C1: Subtracts forces to find F_net = 5.2 x 10^-14 N.
A1: Acceleration a = 1.2 m/s^2 (or 1.24 m/s^2) and specifies direction as downwards.
(c)(i)
C1: Recalls and uses t = distance / horizontal speed.
A1: t = 0.059 s (or 0.0588 s).
(c)(ii)
C1: Uses s = 1/2 * a * t^2.
C1: Substitutes values of a and t (allow ecf from previous parts).
A1: s = 2.1 x 10^-3 m (or 2.1 mm).

(a) The extension \(x_0\) of the spring when the mass is in equilibrium is:
\(x_0 = 0.38\ \text{m} - 0.25\ \text{m} = 0.13\ \text{m}\).
At equilibrium, the upward tension equals the downward gravitational force:
\(k x_0 = m g \implies k = \frac{m g}{x_0}\)
\(k = \frac{0.60\ \text{kg} \times 9.81\ \text{m s}^{-2}}{0.13\ \text{m}} = 45.28\ \text{N m}^{-1} \approx 45\ \text{N m}^{-1}\).

(b)(i) According to Hooke's law, the tension is proportional to the extension. When displaced from equilibrium by a distance \(y\), the net restoring force is \(F = m g - k(x_0 + y) = -k y\). Because the restoring force is directly proportional to the displacement and is directed towards the equilibrium position, the motion is simple harmonic.

(b)(ii) The period \(T\) of oscillation for a mass-spring system is:
\(T = 2\pi \sqrt{\frac{m}{k}} = 2\pi \sqrt{\frac{0.60\ \text{kg}}{45.28\ \text{N m}^{-1}}} = 0.723\ \text{s} \approx 0.72\ \text{s}\).

(b)(iii) The maximum velocity is given by \(v_{\max} = \omega A\).
The angular frequency is:
\(\omega = \sqrt{\frac{k}{m}} = \sqrt{\frac{45.28}{0.60}} = 8.687\ \text{rad s}^{-1}\).
The amplitude of the oscillation is \(A = 4.0\ \text{cm} = 0.040\ \text{m}\).
\(v_{\max} = 8.687\ \text{rad s}^{-1} \times 0.040\ \text{m} = 0.347\ \text{m s}^{-1} \approx 0.35\ \text{m s}^{-1}\).

(c) Relative to the equilibrium position:
When the mass is at its lowest point, it has been displaced downward by \(y = -0.040\ \text{m}\).
The change in gravitational potential energy is:
\(\Delta E_g = m g y = 0.60 \times 9.81 \times (-0.040) = -0.2354\ \text{J}\).
The change in elastic potential energy is:
\(\Delta E_e = \frac{1}{2} k (x_0 + A)^2 - \frac{1}{2} k x_0^2\)
\(\Delta E_e = 0.5 \times 45.28 \times (0.13 + 0.04)^2 - 0.5 \times 45.28 \times (0.13)^2\)
\(\Delta E_e = 22.64 \times [0.0289 - 0.0169] = 22.64 \times 0.0120 = 0.2717\ \text{J}\).
The total potential energy change of the system is:
\(\Delta E_{\text{pot}} = \Delta E_g + \Delta E_e = -0.2354\ \text{J} + 0.2717\ \text{J} = 0.0363\ \text{J} \approx 0.036\ \text{J}\).
(Alternatively, using the SHM potential energy formula, \(\Delta E_{\text{pot}} = \frac{1}{2} k A^2 = 0.5 \times 45.28 \times (0.040)^2 = 0.0362\ \text{J}\).)

(a)
C1: Finds extension x_0 = 0.13 m.
A1: Shows calculation of k = mg/x_0 = 45.3 N/m (approximating to 45 N/m).
(b)(i)
B1: States that the net restoring force is proportional to displacement from equilibrium.
B1: States that the force/acceleration is always directed towards the equilibrium position.
(b)(ii)
C1: Recalls T = 2 * pi * sqrt(m / k).
C1: Substitutes mass and spring constant correctly.
A1: T = 0.72 s.
(b)(iii)
C1: Uses v_max = omega * A.
A1: v_max = 0.35 m/s (accept 0.347 m/s).
(c)
C1: Calculates change in GPE = -0.235 J OR states total potential energy change is 1/2 * k * A^2.
C1: Calculates change in elastic energy = 0.272 J OR substitutes correctly into 1/2 * k * A^2.
A1: Total potential energy change = 0.036 J (accept 0.0362 J).

(a) Wien's displacement law relates the surface temperature to peak emission wavelength:
\(\lambda_{\max} T = 2.90 \times 10^{-3}\ \text{m K}\)
\(\lambda_{\max} = \frac{2.90 \times 10^{-3}}{9600} = 3.02 \times 10^{-7}\ \text{m} \approx 3.0 \times 10^{-7}\ \text{m}\) (or \(302\ \text{nm}\)).

(b)(i) According to the Stefan-Boltzmann law:
\(L = 4 \pi R^2 \sigma T^4\)
Substituting the values:
\(L = 4 \pi \times (1.6 \times 10^9\ \text{m})^2 \times (5.67 \times 10^{-8}\ \text{W m}^{-2}\ \text{K}^{-4}) \times (9600\ \text{K})^4\)
\(L = 4 \pi \times (2.56 \times 10^{18}) \times (5.67 \times 10^{-8}) \times (8.4935 \times 10^{15})\)
\(L = 1.549 \times 10^{28}\ \text{W} \approx 1.5 \times 10^{28}\ \text{W}\).

(b)(ii) The radiant flux intensity (intensity at Earth) is given by:
\(I = \frac{L}{4 \pi d^2}\)
\(I = \frac{1.549 \times 10^{28}\ \text{W}}{4 \pi \times (2.37 \times 10^{17}\ \text{m})^2} = \frac{1.549 \times 10^{28}}{7.0583 \times 10^{35}} = 2.195 \times 10^{-8}\ \text{W m}^{-2} \approx 2.2 \times 10^{-8}\ \text{W m}^{-2}\).

(c) First, find the grating spacing \(d_{\text{slit}}\):
\(d_{\text{slit}} = \frac{1.0 \times 10^{-3}\ \text{m}}{600} = 1.667 \times 10^{-6}\ \text{m}\).
Using the diffraction grating equation:
\(d_{\text{slit}} \sin\theta = n \lambda\)
For the second-order maximum (\(n = 2\)) and \(\lambda_{\max} = 3.02 \times 10^{-7}\ \text{m}\):
\(\sin\theta = \frac{2 \times 3.02 \times 10^{-7}\ \text{m}}{1.667 \times 10^{-6}\ \text{m}} = 0.3623\)
\ heta = \sin^{-1}(0.3623) = 21.2^\circ \approx 21^\circ\ (or \(0.37\ \text{rad}\)).

(a)
C1: Recalls Wien's displacement law expression.
A1: Wavelength = 3.0 x 10^-7 m (or 302 nm).
(b)(i)
C1: Recalls Stefan's law formula.
C1: Substitutes correct value of Stefan-Boltzmann constant (5.67 x 10^-8).
A1: Shows clear working leading to L = 1.5 x 10^28 W.
(b)(ii)
C1: Recalls the inverse-square law for intensity, I = L / (4 * pi * d^2).
C1: Correctly substitutes distance d into the formula.
A1: Intensity = 2.2 x 10^-8 W/m^2.
(c)
C1: Calculates grating spacing d_slit = 1.67 x 10^-6 m.
C1: Recalls d_slit * sin(theta) = n * lambda.
C1: Substitutes n = 2 and wavelength from (a).
A1: Angle theta = 21° (allow ecf from (a)).