2025 IB DP Physics Practice Paper with Answers

The luminosity of a star is given by Stefan-Boltzmann's law as \(L = 4\pi R^2 \sigma T^4\).

Therefore, the ratio of the luminosities is:

\(\frac{L_Y}{L_X} = \left(\frac{R_Y}{R_X}\right)^2 \left(\frac{T_Y}{T_X}\right)^4\)

Substituting the given values:

\(\frac{L_Y}{L_X} = (3)^2 \times (2)^4 = 9 \times 16 = 144\).

Award 1 mark for the correct answer D.

From the ideal gas equation, \(PV = nRT\). Since the volume \(V\) and the pressure \(P\) are constant, the product \(nT\) must also be constant:

\(n_1 T_1 = n_2 T_2\)

We are given \(T_2 = 1.5 T = \frac{3}{2} T\), so:

\(n_1 T = n_2 \left(\frac{3}{2} T\right) \implies n_2 = \frac{2}{3} n_1\)

Therefore, the fraction of the original number of moles remaining is \(\frac{2}{3}\).

Award 1 mark for the correct answer C.

The change in momentum is equal to the impulse, which is represented by the area under the force-time graph.

The graph is a triangle with base \(3.0\text{ s}\) and height \(4.0\text{ N}\):

\(\text{Area} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 3.0\text{ s} \times 4.0\text{ N} = 6.0\text{ N s}\)

Since impulse \(\Delta p = m \Delta v\):

\(6.0\text{ N s} = 0.50\text{ kg} \times (v - 0)\)

\(v = \frac{6.0}{0.50} = 12\text{ m s}^{-1}\).

Award 1 mark for the correct answer D.

The electric potential \(V\) at the midpoint due to the two charges is the sum of the potentials from each charge. The distance from each charge to the midpoint is \(\frac{d}{2}\).

\(V = \frac{1}{4\pi\varepsilon_0} \frac{2Q}{\frac{d}{2}} + \frac{1}{4\pi\varepsilon_0} \frac{-Q}{\frac{d}{2}} = \frac{1}{4\pi\varepsilon_0} \left( \frac{4Q}{d} - \frac{2Q}{d} \right) = \frac{2Q}{4\pi\varepsilon_0 d} = \frac{Q}{2\pi\varepsilon_0 d}\)

The work done \(W\) by an external agent in bringing charge \(+q\) from infinity (where potential is zero) to the midpoint is:

\(W = q \Delta V = q(V - 0) = \frac{qQ}{2\pi\varepsilon_0 d}\).

Award 1 mark for the correct answer B.

The energy required to completely melt the ice of mass \(m\) at \(0\text{ }^\circ\text{C}\) is:

\(Q_{\text{absorbed}} = m L\)

The energy released by the water of mass \(M\) cooling from \(20\text{ }^\circ\text{C}\) to \(0\text{ }^\circ\text{C}\) is:

\(Q_{\text{released}} = M c \Delta T = M c (20 - 0) = 20 M c\)

By conservation of energy, the heat lost by the water equals the heat gained by the ice:

\(m L = 20 M c \implies \frac{m}{M} = \frac{20c}{L}\).

Award 1 mark for the correct answer A.

The observed frequency as the source approaches is:

\(f_{\text{app}} = f_0 \left(\frac{v_s}{v_s - v}\right)\)

The observed frequency as the source recedes is:

\(f_{\text{rec}} = f_0 \left(\frac{v_s}{v_s + v}\right)\)

The difference between the two frequencies is:

\(\Delta f = f_{\text{app}} - f_{\text{rec}} = f_0 v_s \left( \frac{1}{v_s - v} - \frac{1}{v_s + v} \right)\)

Finding a common denominator:

\(\Delta f = f_0 v_s \left( \frac{(v_s + v) - (v_s - v)}{(v_s - v)(v_s + v)} \right) = f_0 v_s \left( \frac{2v}{v_s^2 - v^2} \right) = \frac{2 v v_s f_0}{v_s^2 - v^2}\).

Award 1 mark for the correct answer A.

For a single-slit diffraction pattern, the first minimum occurs at an angle \(\theta\) given by:

\(\sin\theta \approx \theta = \frac{\lambda}{b}\)

With the new wavelength \(\lambda' = \frac{\lambda}{2}\) and the new slit width \(b' = 2b\):

\(\theta' = \frac{\lambda'}{b'} = \frac{\frac{\lambda}{2}}{2b} = \frac{\lambda}{4b} = \frac{\theta}{4}\).

Award 1 mark for the correct answer A.

We can find the final velocity using energy conservation or by resolving the motion into horizontal and vertical components.

Using components:
- Horizontal velocity remains constant: \(v_x = u\).
- Vertical motion: \(v_y^2 = u_y^2 + 2g\Delta y = 0^2 + 2gh = 2gh\).

The magnitude of the final velocity \(v\) is:

\(v =
\sqrt{v_x^2 + v_y^2} = \sqrt{u^2 + 2gh}\).

Award 1 mark for the correct answer A.

The distance to the star is calculated from its parallax angle: \(d = \frac{1}{p} = \frac{1}{0.025} = 40\text{ pc}\). Converting this distance into meters gives \(d = 40 \times 3.09 \times 10^{16}\text{ m} = 1.236 \times 10^{18}\text{ m}\). The relationship between apparent brightness, luminosity, and distance is \(b = \frac{L}{4 \pi d^2}\). Rearranging to solve for luminosity gives \(L = 4 \pi d^2 b = 4 \pi \times (1.236 \times 10^{18}\text{ m})^2 \times (4.0 \times 10^{-10}\text{ W m}^{-2}) \approx 7.7 \times 10^{27}\text{ W}\).

Award [1] for the correct answer. Correctly calculate distance in pc as 40. Correctly convert distance to meters as \(1.24 \times 10^{18}\text{ m}\). Correctly apply the apparent brightness formula to obtain \(7.7 \times 10^{27}\text{ W}\).

According to the Jeans criterion, gravitational collapse is favored when the gravitational potential energy overcomes thermal kinetic energy. Decreasing the radius \(R\) brings particles closer together, increasing gravitational attraction. Decreasing the temperature \(T\) reduces the random thermal motion of the gas molecules. Both changes make collapse more likely.

Award [1] for the correct answer. Identifies that decreasing temperature reduces thermal pressure. Identifies that decreasing radius increases gravitational force density. Concludes that decreasing both \(T\) and \(R\) is correct.

First, convert the temperatures to Kelvin: \(T_i = 27 + 273 = 300\text{ K}\) and \(T_f = 327 + 273 = 600\text{ K}\). The relationship between pressure, volume, and temperature for a fixed amount of ideal gas is \(\frac{p_i V_i}{T_i} = \frac{p_f V_f}{T_f}\). We are given that \(V_f = 2 V_i\). Substituting this and solving for the pressure ratio gives \(\frac{p_f}{p_i} = \frac{V_i}{V_f} \times \frac{T_f}{T_i} = \frac{1}{2} \times \frac{600}{300} = 1.0\).

Award [1] for the correct answer. Converts temperatures to Kelvin correctly. Correctly applies the ideal gas law to find the ratio as 1.0.

The root-mean-square speed is given by \(v_{\text{rms}} = \sqrt{\frac{3RT}{M}}\). In thermal equilibrium, both gases are at the same temperature \(T\). Therefore, the ratio of the speeds is \(\frac{v_{\text{rms, He}}}{v_{\text{rms, O}_2}} = \sqrt{\frac{M_{\text{O}_2}}{M_{\text{He}}}} = \sqrt{\frac{32.0}{4.0}} = \sqrt{8} \approx 2.8\).

Award [1] for the correct answer. Identifies the inverse square root dependency on molar mass. Correctly evaluates \(\sqrt{8}\) as approximately 2.8.

With the right as the positive direction, the initial velocity is \(u = +20\text{ m s}^{-1}\) and the final velocity is \(v = -30\text{ m s}^{-1}\). The change in momentum is \(\Delta p = m(v - u) = 0.15 \times (-30 - 20) = -7.5\text{ kg m s}^{-1}\). The magnitude of the average force is \(F = \frac{|\Delta p|}{\Delta t} = \frac{7.5}{8.0 \times 10^{-3}\text{ s}} = 9.4 \times 10^2\text{ N}\).

Award [1] for the correct answer. Accounts for direction to find momentum change of 7.5 N s. Divides by correct time in seconds to obtain \(9.4 \times 10^2\text{ N}\).

The force per unit length is given by \(F = \frac{\mu_0 I_1 I_2}{2 \pi d}\), which means \(F \propto \frac{I_1 I_2}{d}\). If both currents are doubled and the distance is doubled, the new force per unit length is proportional to \(\frac{(2)(2)}{2} = 2\) times the original force per unit length. Thus, the new force per unit length is \(2F\).

Award [1] for the correct answer. Identifies relationship between force, currents, and distance. Calculates the factor of 2 change correctly.

During vaporization, the energy supplied is \(Q = P \times t = 150\text{ W} \times 300\text{ s} = 45000\text{ J}\). This energy equals \(m L_v\). Therefore, \(L_v = \frac{Q}{m} = \frac{45000\text{ J}}{0.20\text{ kg}} = 2.3 \times 10^5\text{ J kg}^{-1}\).

Award [1] for the correct answer. Calculates total energy during the phase change as 45 kJ. Correctly divides by mass to yield \(2.3 \times 10^5\text{ J kg}^{-1}\).

The Doppler equation for an approaching source is \(f' = f \left(\frac{v}{v - v_s}\right)\). Substituting \(f' = 1.20f\) gives \(1.20 = \frac{v}{v - v_s}\). Rearranging leads to \(1.20(v - v_s) = v\), which simplifies to \(0.20v = 1.20v_s\). Therefore, the ratio is \(\frac{v_s}{v} = \frac{0.20}{1.20} = \frac{1}{6} \approx 0.17\).

Award [1] for the correct answer. Uses correct Doppler formula. Solves the equation to obtain the ratio of approximately 0.17.

The Jeans mass \(M_J\) is the minimum mass a gas cloud must have to overcome its internal thermal pressure and collapse gravitationally. The thermal kinetic energy of the particles is proportional to \(N k_B T \propto \frac{M}{m} T\), and the gravitational potential energy is proportional to \(\frac{G M^2}{R}\). Under critical balance, \(\frac{G M^2}{R} \sim \frac{M T}{m} \implies M \propto T R\). Since density \(\rho \propto \frac{M}{R^3}\), we have \(R \propto \left(\frac{M}{\rho}\right)^{1/3}\). Substituting this into the mass scaling gives \(M \propto T \left(\frac{M}{\rho}\right)^{1/3} \implies M^{2/3} \propto T \rho^{-1/3} \implies M_J \propto T^{3/2} \rho^{-1/2}\).

[1 mark] for identifying the correct dependency on temperature and density: \(M_J \propto T^{3/2} \rho^{-1/2}\).

The main sequence lifetime \(\tau\) of a star is proportional to the total amount of nuclear fuel available (which is proportional to its mass \(M\)) and inversely proportional to the rate at which this fuel is consumed (which is the luminosity \(L\)). Thus, \(\tau \propto \frac{M}{L}\). Substituting \(L \propto M^{3.5}\) gives \(\tau \propto \frac{M}{M^{3.5}} = M^{-2.5}\).

[1 mark] for recognizing that lifetime is proportional to \(M/L\) and correctly performing the exponent subtraction to obtain \(\tau \propto M^{-2.5}\).

The root-mean-square speed of ideal gas molecules is given by \(v_{\text{rms}} = \sqrt{\frac{3 k_B T}{m}}\), meaning the absolute temperature \(T\) is proportional to \(v_{\text{rms}}^2\). If \(v_{\text{rms}}\) is doubled, the absolute temperature increases by a factor of \(2^2 = 4\). Since the volume \(V\) is constant, from the ideal gas law \(P \propto T\). Thus, the pressure also increases by a factor of 4, becoming \(4P\).

[1 mark] for identifying that absolute temperature quadruples when rms speed doubles, and using the constant-volume gas relation to conclude the pressure becomes \(4P\).

The average force is given by Newton's second law in terms of momentum: \(F_{\text{avg}} = \frac{\Delta p}{\Delta t}\). Taking the direction towards the wall as positive, the initial velocity \(u = +20\text{ m s}^{-1}\) and the final velocity \(v = -15\text{ m s}^{-1}\). The change in momentum is \(\Delta p = m(v - u) = 0.15 \times (-15 - 20) = -5.25\text{ kg m s}^{-1}\). The magnitude of this change is \(5.25\text{ N s}\). The magnitude of the average force is therefore \(F_{\text{avg}} = \frac{5.25\text{ N s}}{0.050\text{ s}} = 105\text{ N}\).

[1 mark] for correctly calculating the change in momentum (accounting for the sign change due to rebound) and dividing by the contact time to get \(105\text{ N}\).

For the droplet to remain stationary, the upward electric force must equal the downward gravitational force: \(F_E = F_g \implies q E = m g\). The electric field \(E\) between parallel plates is related to the potential difference \(V\) and separation \(d\) by \(E = \frac{V}{d}\). Substituting this in, we get \(q \left(\frac{V}{d}\right) = m g \implies q = \frac{m g d}{V}\).

[1 mark] for equating gravitational and electric forces and substituting \(E = \frac{V}{d}\) to solve for \(q\).

Using the conservation of thermal energy, we have \(\Delta Q_{\text{lost}} = \Delta Q_{\text{gained}}\). Let \(T_f\) be the final equilibrium temperature. The heat lost by copper is \(m_c c_c (100 - T_f)\) and the heat gained by water is \(m_w c_w (T_f - 20)\). Setting them equal: \(2.0 \times 400 \times (100 - T_f) = 1.0 \times 4200 \times (T_f - 20) \implies 800(100 - T_f) = 4200(T_f - 20) \implies 80000 - 800 T_f = 4200 T_f - 84000 \implies 164000 = 5000 T_f \implies T_f = 32.8^\circ\text{C}\).

[1 mark] for writing down the heat exchange equation and correctly solving for \(T_f\).

The diffraction grating equation is \(d \sin\theta = m \lambda\), where \(d\) is the slit spacing and \(m\) is the order of the maximum. Since there are \(n\) lines per metre, the spacing between adjacent slits is \(d = \frac{1}{n}\) metres. For the third-order maximum, \(m = 3\). Substituting these values into the grating equation gives \(\frac{1}{n} \sin\theta = 3 \lambda \implies \lambda = \frac{\sin\theta}{3n}\).

[1 mark] for identifying that \(d = \frac{1}{n}\) and substituting this and \(m = 3\) into the grating equation to obtain the correct expression for \(\lambda\).

According to the Doppler effect for a source moving away from a stationary observer, the observed frequency \(f\) is given by: \(f = f_0 \left(\frac{c_s}{c_s + v_{\text{source}}}\right)\). Here, the source is moving away, so we use the positive sign in the denominator. With \(v_{\text{source}} = 0.10 c_s\), we get: \(f = f_0 \left(\frac{c_s}{c_s + 0.10 c_s}\right) = f_0 \left(\frac{c_s}{1.10 c_s}\right) = \frac{f_0}{1.10} \approx 0.91 f_0\).

[1 mark] for correctly selecting the Doppler formula for a source moving away, substituting \(v = 0.10 c_s\), and calculating the resulting ratio as \(\approx 0.91 f_0\).

From the ideal gas equation, \(PV = N k_B T\), the pressure is given by \(P = \frac{N k_B T}{V}\). When the changes occur, the new values are: \(N' = (1 - 0.40)N = 0.60N\), \(V' = (1 - 0.20)V = 0.80V\), and \(T' = (1 + 0.20)T = 1.20T\). Substituting these into the formula for the new pressure \(P'\): \(P' = \frac{N' k_B T'}{V'} = \frac{0.60N \cdot k_B \cdot 1.20T}{0.80V} = \frac{0.72}{0.80} \frac{N k_B T}{V} = 0.90 P\). Therefore, the correct choice is B.

Award 1 mark for the correct application of proportional changes using \(P \propto \frac{NT}{V}\) to arrive at the correct final option B.

From the ideal gas equation, \(PV = N k_B T\), the pressure is given by \(P = \frac{N k_B T}{V}\). When the changes occur, the new values are: \(N' = (1 - 0.40)N = 0.60N\), \(V' = (1 - 0.20)V = 0.80V\), and \(T' = (1 + 0.20)T = 1.20T\). Substituting these into the formula for the new pressure \(P'\): \(P' = \frac{N' k_B T'}{V'} = \frac{0.60N \cdot k_B \cdot 1.20T}{0.80V} = \frac{0.72}{0.80} \frac{N k_B T}{V} = 0.90 P\). Therefore, the correct choice is B.

Award 1 mark for the correct application of proportional changes using \(P \propto \frac{NT}{V}\) to arrive at the correct final option B.

Part (a)
Resistance \(R\) is determined using the formula \(R = \frac{V}{I}\), where the potential difference \(V\) across the wire is measured using a voltmeter connected in parallel, and the current \(I\) is measured using an ammeter connected in series with the wire.
To prevent temperature changes due to Joule heating, the current should be kept low, or the circuit should be disconnected using a switch between measurements so that current only flows when taking readings.

Part (b)
(i) The cross-sectional area is given by:
\(A = \frac{\pi d^2}{4} = \frac{\pi \times (0.36 \times 10^{-3}\text{ m})^2}{4} = 1.02 \times 10^{-7}\text{ m}^2\).
(ii) The percentage uncertainty in the diameter \(d\) is:
\(\frac{\Delta d}{d} \times 100\% = \frac{0.02}{0.36} \times 100\% = 5.56\%\).
Since \(A \propto d^2\), the percentage uncertainty in the area is twice that of the diameter:
\(\frac{\Delta A}{A} = 2 \times \frac{\Delta d}{d} = 2 \times 5.56\% = 11.1\%\) (or \(11\%\) to two significant figures).

Part (c)
The relationship between resistance and length is:
\(R = \rho \frac{L}{A} = \left(\frac{\rho}{A}\right) L\).
A graph of \(R\) against \(L\) is a straight line through the origin, where the gradient \(m\) is given by \(m = \frac{\rho}{A}\). Therefore, the resistivity can be determined by multiplying the gradient by the cross-sectional area: \(\rho = m \times A\).

Part (d)
(i) \(\rho = m \times A = 4.82\text{ }\Omega\text{ m}^{-1} \times 1.018 \times 10^{-7}\text{ m}^2 = 4.91 \times 10^{-7}\text{ }\Omega\text{ m}\) (accept \(4.9 \times 10^{-7}\text{ }\Omega\text{ m}\)).
(ii) The percentage uncertainty in the gradient \(m\) is:
\(\frac{\Delta m}{m} \times 100\% = \frac{0.15}{4.82} \times 100\% = 3.11\%\).
The total percentage uncertainty in \(\rho\) is the sum of the percentage uncertainties of \(m\) and \(A\):
\(\frac{\Delta \rho}{\rho} = \frac{\Delta m}{m} + \frac{\Delta A}{A} = 3.11\% + 11.11\% = 14.22\%\).
The absolute uncertainty in \(\rho\) is:
\(\Delta \rho = 14.22\% \times 4.91 \times 10^{-7}\text{ }\Omega\text{ m} = 0.70 \times 10^{-7}\text{ }\Omega\text{ m}\) (accept \(0.7 \times 10^{-7}\text{ }\Omega\text{ m}\)).

Part (a)
• \(R = \frac{V}{I}\) stated with correct measurement method (voltmeter in parallel, ammeter in series). [1 mark]
• Precaution clearly explained (e.g., use a switch to turn off current between readings to prevent heating). [1 mark]

Part (b)
• \(A = 1.02 \times 10^{-7}\text{ m}^2\) (accept range \(1.01 \times 10^{-7}\) to \(1.02 \times 10^{-7}\)). [1 mark]
• Calculates percentage uncertainty in \(d\) as \(5.56\%\) OR percentage uncertainty in \(A\) as \(11.1\%\). [1 mark]
• Clearly multiplies percentage uncertainty in diameter by 2 to get \(11\%\) or \(11.1\%\). [1 mark]

Part (c)
• State or show \(R = \frac{\rho L}{A}\). [1 mark]
• Identifies that \(\text{gradient} = \frac{\rho}{A}\) and thus \(\rho = \text{gradient} \times A\). [1 mark]

Part (d)
• Correct value for resistivity: \(4.9 \times 10^{-7}\) (or \(4.91 \times 10^{-7}\)). [1 mark]
• Correct unit: \(\Omega\text{ m}\). [1 mark]
• Calculates percentage uncertainty in gradient (\(3.11\%\)) and adds it to percentage uncertainty in area (\(11.11\%\)) to get \(14.22\%\). [1 mark]
• Calculates absolute uncertainty in \(\rho\) as \(0.7 \times 10^{-7}\text{ }\Omega\text{ m}\) (or \(0.70 \times 10^{-7}\)). [1 mark]

**(a)**
* Independent variable: Volume \(V\)
* Dependent variable: Pressure \(p\)
*(Both correct for 1 mark)*

**(b) (i)**
Boyle's Law states that for a fixed mass of gas at constant temperature, pressure is inversely proportional to volume: \(pV = \text{constant}\).
We calculate the product \(pV\) for at least three different data points:
* For \(V = 10.0\text{ cm}^3\): \(pV = 10.0 \times 202 = 2020\text{ cm}^3\text{ kPa}\) (or \(2.02\text{ J}\))
* For \(V = 15.0\text{ cm}^3\): \(pV = 15.0 \times 135 = 2025\text{ cm}^3\text{ kPa}\) (or \(2.03\text{ J}\))
* For \(V = 20.0\text{ cm}^3\): \(pV = 20.0 \times 101 = 2020\text{ cm}^3\text{ kPa}\) (or \(2.02\text{ J}\))

The products are constant within experimental uncertainty, which is consistent with Boyle's Law.

**(b) (ii)**
Using the ideal gas equation:
\(pV = nRT \implies n = \frac{pV}{RT}\)

Converting units to SI:
\(pV \approx 2020\text{ cm}^3\text{ kPa} = 2020 \times (10^{-6}\text{ m}^3) \times (10^3\text{ Pa}) = 2.02\text{ Pa m}^3 = 2.02\text{ J}\)

Using \(R = 8.31\text{ J K}^{-1}\text{ mol}^{-1}\) and \(T = 293\text{ K}\):
\(n = \frac{2.02}{8.31 \times 293} = 8.3 \times 10^{-4}\text{ mol}\) (accept range \(8.2 \times 10^{-4}\,\text{mol}\) to \(8.4 \times 10^{-4}\,\text{mol}\)).

To keep the temperature of the gas constant, the volume of the gas must be changed very slowly, and a short pause should be allowed before taking each reading so that the gas can exchange heat with the surrounding environment and remain at room temperature.

**(c) (i)**
According to the ideal gas law:
\(p = \left(\frac{nRT}{V}\right) = (nRT) \cdot \frac{1}{V}\)
Since \(n, R, T\) are constant, this is of the form \(y = mx\) where \(y = p\) and \(x = \frac{1}{V}\). The y-intercept is zero, so the line should be a straight line passing through the origin.

**(c) (ii)**
There is "dead space" inside the tubing connecting the syringe to the pressure sensor.
This means the actual volume of the gas is greater than the volume \(V\) read from the syringe scale:
\(V_{\text{actual}} = V_{\text{measured}} + V_{\text{dead}}\)
Therefore, the recorded volume \(V\) is systematically smaller than the true volume. When plotting \(p\) against \(\frac{1}{V}\), the values of \(\frac{1}{V}\) are systematically larger than \(\frac{1}{V_{\text{actual}}}\), which shifts the data points to the right on the \(\frac{1}{V}\) axis, resulting in a positive y-intercept on the pressure axis when a straight line is fitted.

**(a)**
* Both variables identified correctly: independent is Volume \(V\) and dependent is Pressure \(p\). **[1]**

**(b) (i)**
* Calculations of at least three values of the product \(pV\) shown (or equivalent calculations showing \(p \propto 1/V\)). **[1]**
* Clear concluding statement that the products are constant within the limits of experimental error. **[1]**

**(b) (ii)**
* Correct calculation of the number of moles: \(8.3 \times 10^{-4}\) (accept range \(8.2 \times 10^{-4}\) to \(8.4 \times 10^{-4}\)). **[1]**
* Correct unit: \(\text{mol}\) or \(\text{moles}\). **[1]**
* Outline of method to keep temperature constant: Change volume slowly OR allow time for thermal equilibrium / heat transfer to occur between readings. **[1]**

**(c) (i)**
* Relates \(p = \frac{nRT}{V}\) to \(y = mx\) or states that \(p\) is directly proportional to \(\frac{1}{V}\) (with no additive constant). **[1]**

**(c) (ii)**
* Identifies the systematic error as "dead space" or "gas trapped in the tubing/sensor nozzle". **[1]**
* Explains that the actual volume of the gas is greater than the recorded volume \(V\) (or recorded volume is systematically too low), which shifts the data points to the right on the \(\frac{1}{V}\) axis, yielding a positive intercept when a straight line of best fit is drawn. **[1]**

(a) Using conservation of momentum: \(m_A u_A + m_B u_B = (m_A + m_B) v\). Substituting the values: \(2.0 \times 6.0 + 4.0 \times 0 = (2.0 + 4.0) v\), which gives \(12 = 6.0 v\), so \(v = 2.0\text{ m s}^{-1}\). (b) Initial kinetic energy: \(E_{k,i} = \frac{1}{2} m_A u_A^2 = \frac{1}{2} (2.0) (6.0)^2 = 36.0\text{ J}\). Final kinetic energy immediately after collision: \(E_{k,f} = \frac{1}{2} (m_A + m_B) v^2 = \frac{1}{2} (6.0) (2.0)^2 = 12.0\text{ J}\). Kinetic energy lost: \(\Delta E_k = 36.0 - 12.0 = 24.0\text{ J}\). Since kinetic energy is lost, the collision is inelastic. (c) Conservation of energy during compression: all kinetic energy after the collision is converted into elastic potential energy at maximum compression: \(\frac{1}{2} (m_A + m_B) v^2 = \frac{1}{2} k x^2\), so \(12.0 = \frac{1}{2} (600) x^2\), giving \(300 x^2 = 12.0\), \(x^2 = 0.04\), thus \(x = 0.20\text{ m}\). (d) Momentum is not conserved during the compression. An external force is exerted by the wall on the spring, which acts on the block-spring system, changing its total momentum to zero.

(a) [2 marks] Award [1] for correct substitution into conservation of momentum equation: \(2.0 \times 6.0 = 6.0 \times v\). Award [1] for correct final answer: \(2.0\text{ m s}^{-1}\). (b) [2 marks] Award [1] for calculating both initial and final kinetic energies (36 J and 12 J). Award [1] for calculating the energy loss of 24 J and stating it is inelastic. (c) [2 marks] Award [1] for equating kinetic energy after collision to elastic potential energy: \(12 = \frac{1}{2} (600) x^2\). Award [1] for correct final compression: \(0.20\text{ m}\) (accept 20 cm). (d) [2 marks] Award [1] for stating 'not conserved'. Award [1] for stating that the wall exerts an external force on the system.

**Part (a)**
(i) The temperature difference is \(\Delta T = 100^\circ\text{C} - 20^\circ\text{C} = 80\text{ K}\).
Using the thermal conduction equation:
\(\frac{\Delta Q}{\Delta t} = \frac{k A \Delta T}{L}\)
\(\frac{\Delta Q}{\Delta t} = \frac{200 \times (1.5 \times 10^{-4}) \times 80}{0.80} = 3.0\text{ W}\) (or \(\text{J s}^{-1}\))
(ii) As the temperature of the gas increases, the temperature difference \(\Delta T\) between the source and the gas decreases. Since the rate of thermal energy transfer is proportional to the temperature difference, the rate decreases.

**Part (b)**
(i) Using the ideal gas equation \(pV = nRT\):
Convert temperature to Kelvin: \(T = 20 + 273 = 293\text{ K}\).
\(V = \frac{nRT}{p} = \frac{0.12 \times 8.31 \times 293}{1.0 \times 10^5} = 2.92 \times 10^{-3}\text{ m}^3\) (or \(2.9 \times 10^{-3}\text{ m}^3\)).
(ii) At constant volume, \(\frac{p_1}{T_1} = \frac{p_2}{T_2}\).
\(T_2 = 100 + 273 = 373\text{ K}\).
\(p_2 = 1.0 \times 10^5 \times \frac{373}{293} = 1.27 \times 10^5\text{ Pa}\) (or \(1.3 \times 10^5\text{ Pa}\)).

**Part (c)**
- Since the temperature remains constant, the average kinetic energy (and thus the average speed) of the gas molecules remains constant.
- As the volume increases, the number density of the molecules decreases, meaning they must travel further to reach the walls, which leads to a decrease in the frequency of collisions per unit area with the cylinder walls. Hence, the average force per unit area (pressure) decreases.

**Part (a) (i)**
- Award **[1]** for calculating \(\Delta T = 80\text{ K}\) or substituting correct values into the conduction equation.
- Award **[1]** for the final answer of \(3.0\text{ W}\) (or \(3\text{ W}\) or \(3.0\text{ J s}^{-1}\)).

**Part (a) (ii)**
- Award **[1]** for stating that \(\Delta T\) decreases, which reduces the rate of conduction.

**Part (b) (i)**
- Award **[1]** for converting temperature to \(293\text{ K}\) and rearranging \(pV = nRT\).
- Award **[1]** for \(2.92 \times 10^{-3}\text{ m}^3\) or \(2.9 \times 10^{-3}\text{ m}^3\).

**Part (b) (ii)**
- Award **[1]** for \(1.27 \times 10^5\text{ Pa}\) (or \(1.3 \times 10^5\text{ Pa}\)).

**Part (c)**
- Award **[1]** for stating that constant temperature means constant average kinetic energy / molecular speed.
- Award **[1]** for stating that larger volume leads to lower collision frequency/rate with the walls, resulting in a lower pressure.

### Part (a)
Point \(P\) is located halfway between the charges, so the distance from each charge to \(P\) is:
\(r = \frac{0.12\text{ m}}{2} = 0.06\text{ m}\)

Let the rightward direction be positive.
* The electric field \(E_1\) due to the positive charge \(q_1\) points away from \(q_1\) (to the right):
\(E_1 = \frac{k |q_1|}{r^2} = \frac{8.99 \times 10^9 \times 4.0 \times 10^{-6}}{(0.06)^2} \approx 9.99 \times 10^6\text{ N C}^{-1}\)

* The electric field \(E_2\) due to the negative charge \(q_2\) points towards \(q_2\) (to the right):
\(E_2 = \frac{k |q_2|}{r^2} = \frac{8.99 \times 10^9 \times 9.0 \times 10^{-6}}{(0.06)^2} \approx 2.25 \times 10^7\text{ N C}^{-1}\)

Since both electric fields point to the right, we add their magnitudes:
\(E_{\text{net}} = E_1 + E_2 = 9.99 \times 10^6 + 2.25 \times 10^7 = 3.25 \times 10^7\text{ N C}^{-1}\) (or \(3.3 \times 10^7\text{ N C}^{-1}\) to two significant figures).

**Direction:** To the right (towards \(q_2\)).

### Part (b)
For the net electric field to be zero, the individual electric fields must be equal in magnitude and opposite in direction.
* Between the charges, both fields point to the right, so they cannot cancel.
* To the right of \(q_2\), the distance to the larger charge \(q_2\) is always smaller than the distance to \(q_1\), so \(E_2 > E_1\) everywhere in this region.
* Therefore, the zero-field point must lie to the left of \(q_1\). Let this point be at a distance \(d\) to the left of \(q_1\).

Setting the magnitudes equal:
\(E_1 = E_2\)
\(\frac{k |q_1|}{d^2} = \frac{k |q_2|}{(d + 0.12)^2}\)

Substitute the charge values:
\(\frac{4.0 \times 10^{-6}}{d^2} = \frac{9.0 \times 10^{-6}}{(d + 0.12)^2}\)

Taking the square root of both sides:
\(\frac{2.0}{d} = \frac{3.0}{d + 0.12}\)

Cross-multiplying:
\(2.0(d + 0.12) = 3.0d\)
\(2.0d + 0.24 = 3.0d\)
\(d = 0.24\text{ m}\) (or \(24\text{ cm}\))

### Part (c)
* A proton has a positive charge, so it experiences a force in the direction of the electric field (to the right, towards \(q_2\)).
* Since there is a net force, the proton accelerates to the right, meaning its velocity increases.
* As the proton moves to the right from point \(P\), it gets closer to the larger-magnitude charge \(q_2\). The increase in the field from \(q_2\) outweighs the decrease in the field from \(q_1\), meaning the net electric field strength increases.
* Since \(F = qE\) and \(a = \frac{F}{m}\), the electric force and therefore the proton's acceleration both increase over time.

### Part (a) [3 marks]
* **[1 mark]** for calculating either \(E_1\) (\(1.0 \times 10^7\text{ N C}^{-1}\)) or \(E_2\) (\(2.3 \times 10^7\text{ N C}^{-1}\)).
* **[1 mark]** for recognizing that both fields point in the same direction and adding them.
* **[1 mark]** for the correct final answer: \(3.3 \times 10^7\text{ N C}^{-1}\) (accept \(3.25 \times 10^7\text{ N C}^{-1}\)) AND stating the direction is to the right / towards \(q_2\).

### Part (b) [3 marks]
* **[1 mark]** for identifying that the zero-field point must lie to the left of \(q_1\) (or set up the equations with correct signs/regions).
* **[1 mark]** for setting up the correct equation: \(\frac{4.0}{d^2} = \frac{9.0}{(d+0.12)^2}\) (or equivalent).
* **[1 mark]** for the correct final answer of \(0.24\text{ m}\) or \(24\text{ cm}\).

### Part (c) [3 marks]
* **[1 mark]** for stating that the proton moves/accelerates to the right (towards \(q_2\)) / its velocity increases.
* **[1 mark]** for stating that the acceleration increases.
* **[1 mark]** for explaining that the net electric field (or electric force) increases as the proton moves closer to the charge of larger magnitude (\(q_2\)).

(a) Coherent waves have a constant phase difference over time and share the same frequency.

(b) The path difference for destructive interference (dark fringes) is given by \((m + \frac{1}{2})\lambda\). For the second-order dark fringe (the second minimum from the center, corresponding to \(m = 1\)), the path difference is:
\text{Path difference} = 1.5\lambda = 1.5 \times 532 \text{ nm} = 798 \text{ nm} \text{ (or } 7.98 \times 10^{-7} \text{ m)}

(c) The distance between adjacent bright fringes (fringe spacing \(s\)) is:
\(s = \frac{\lambda D}{d}\)
\(s = \frac{532 \times 10^{-9} \text{ m} \times 1.50 \text{ m}}{0.120 \times 10^{-3} \text{ m}} = 6.65 \times 10^{-3} \text{ m}\)

The distance to the second-order bright fringe (where \(m = 2\)) from the central maximum is:
\(y = 2s = 2 \times 6.65 \times 10^{-3} \text{ m} = 1.33 \times 10^{-2} \text{ m}\) (or \(13.3 \text{ mm}\)).

(a)
- Constant phase difference (relationship) between the light waves. [1]
*(Do not accept "in phase" on its own without mentioning constancy, but accept "constant phase relationship".)*

(b)
- Path difference of \(1.5\lambda\) or \(1.5 \times 532 \text{ nm}\) or \(798 \text{ nm}\) (or \(7.98 \times 10^{-7} \text{ m}\)). [1]

(c)
- Correct formula for fringe spacing: \(s = \frac{\lambda D}{d}\) OR use of \(d \sin\theta = m\lambda\) with small angle approximation \(\theta \approx \frac{y}{D}\). [1]
- Correct substitution resulting in fringe spacing of \(6.65 \times 10^{-3} \text{ m}\) OR substitution of \(y_2 = \frac{2 \lambda D}{d}\). [1]
- Final answer of \(1.33 \times 10^{-2} \text{ m}\) (or \(13.3 \text{ mm}\) or \(1.33 \text{ cm}\)). [1]
*(Award [3] marks for a correct final answer with working. Allow ECF if wrong fringe spacing formula is used consistently.)*

Part (a)
(i) High temperatures are required so that the hydrogen nuclei (protons) have sufficient kinetic energy to overcome the mutual electrostatic repulsion (the Coulomb barrier) between their positive charges. High pressure (and density) ensures a high frequency of collisions, keeping the nuclei close enough (within \(\approx 10^{-15}\text{ m}\)) for the short-range strong nuclear force to bind them together.
(ii) The overall reaction is: \(4 \,^1_1\text{H} \rightarrow \,^4_2\text{He} + 2 \,^0_1\text{e}^+ + 2 \nu_e + 2 \gamma\) (alternatively, \(4\text{p} \rightarrow \text{He}-4 + 2\text{e}^+ + 2\nu_e\)).

Part (b)
(i) From Wien's displacement law: \(\lambda_{\max} T = 2.90 \times 10^{-3}\text{ m K}\). Substituting the temperature: \(\lambda_{\max} = \frac{2.90 \times 10^{-3}\text{ m K}}{4500\text{ K}} = 6.44 \times 10^{-7}\text{ m}\) (or \(644\text{ nm}\)).
(ii) Using the Stefan-Boltzmann law for luminosity: \(L = \sigma A T^4 = 5.67 \times 10^{-8} \times 4\pi R^2 T^4\). Substituting the given values: \(L = 5.67 \times 10^{-8} \times 4\pi \times (1.2 \times 10^9)^2 \times (4500)^4 = 5.67 \times 10^{-8} \times 1.81 \times 10^{19} \times 4.10 \times 10^{14} = 4.2 \times 10^{26}\text{ W}\).

Part (c)
(i) Stable equilibrium in a main-sequence star is called hydrostatic equilibrium. It is maintained by a balance between the inward gravitational force (which pulls the stellar material toward the core) and the outward thermal gas pressure and radiation pressure (which are generated by nuclear fusion reactions in the core).
(ii) As core hydrogen is depleted, fusion ceases and the core contracts. This heats up surrounding hydrogen shells, causing the outer envelope to expand and cool, evolving Star X into a Red Giant. Eventually, outer layers are ejected to form a planetary nebula. Since the remaining core mass is less than the Chandrasekhar limit (\(1.4 M_{\odot}\)), electron degeneracy pressure prevents further collapse, leaving behind a stable White Dwarf.

Part (d)
(i) The distance in parsecs is \(d = \frac{1}{p} = \frac{1}{0.075} = 13.3\text{ pc}\). Converting to meters: \(d = 13.33 \times (3.09 \times 10^{16}\text{ m/pc}) = 4.12 \times 10^{17}\text{ m}\) (accept \(4.1 \times 10^{17}\text{ m}\)).

Part (a)
(i)
M1: mention of electrostatic / Coulomb repulsion between positively charged protons [1]
M2: explanation that high temperature corresponds to high kinetic energy allowing protons to overcome this barrier [1]
M3: explanation that high pressure increases collision frequency / density, bringing protons within range of the strong nuclear force [1]
(ii)
M1: correct reactants (\(4 \text{p}\) or \(4 \,^1_1\text{H}\)) and products (\(\,^4_2\text{He}\)) [1]
M2: correct conservation of leptons and charge showing \(2\text{e}^+\) (positrons) and \(2\nu_e\) (neutrinos) [1]

Part (b)
(i)
M1: substitution of values into Wien's law: \(\lambda_{\max} = \frac{2.90 \times 10^{-3}}{4500}\) [1]
A1: \(\lambda_{\max} = 6.4 \times 10^{-7}\text{ m}\) (or \(6.44 \times 10^{-7}\text{ m}\)) [1]
(ii)
M1: recall/use of \(L = \sigma 4\pi R^2 T^4\) [1]
M2: correct substitution: \(L = 5.67 \times 10^{-8} \times 4\pi \times (1.2 \times 10^9)^2 \times 4500^4\) [1]
A1: \(L = 4.2 \times 10^{26}\text{ W}\) (accept range \(4.2 \times 10^{26}\text{ W}\) to \(4.21 \times 10^{26}\text{ W}\)) [1]

Part (c)
(i)
M1: identifies the state as hydrostatic equilibrium [1]
M2: describes inward gravitational force [1]
M3: describes outward radiation / gas pressure resulting from fusion [1]
(ii)
M1: describes expansion into a Red Giant after core hydrogen depletion [1]
M2: describes ejection of outer layers as a planetary nebula [1]
M3: identifies the final remnant as a White Dwarf [1]
M4: explains that stability is due to electron degeneracy pressure and relates to core mass being below the Chandrasekhar limit of \(1.4 M_{\odot}\) [1]

Part (d)
(i)
M1: correct use of \(d = \frac{1}{p}\) to find \(d = 13.3\text{ pc}\) [1]
M2: correct conversion factor from parsecs to meters (\(1\text{ pc} \approx 3.09 \times 10^{16}\text{ m}\)) [1]
A1: \(d = 4.1 \times 10^{17}\text{ m}\) (accept range \(4.1 \times 10^{17}\text{ m}\) to \(4.12 \times 10^{17}\text{ m}\)) [1]