2024 IB DP Physics Practice Paper with Answers

For the first ball thrown upwards: \(y_1 = u t - \frac{1}{2} g t^2\). For the second ball dropped from height \(H\): \(y_2 = H - \frac{1}{2} g t^2\). At the instant they collide, their vertical positions are equal: \(y_1 = y_2\). Therefore, \(u t - \frac{1}{2} g t^2 = H - \frac{1}{2} g t^2\), which simplifies to \(u t = H\). Thus, the time elapsed is \(t = \frac{H}{u}\).

[1 mark] for equating the two displacement equations and solving for \(t\) to get the correct option A.

Since the block is stationary (in static equilibrium), the net force acting on it must be zero. In the vertical direction, the downward force is the weight of the block (\(mg\)) and the upward force is the static frictional force (\(f\)). Therefore, \(f = mg\). Note that \(\mu F\) represents the maximum possible static frictional force, not the actual frictional force in this state.

[1 mark] for recognizing that the block is in static equilibrium, so vertical forces must balance, yielding \(f = mg\).

The temperature of the gas must be converted to Kelvin. The initial temperature is \(T_i = 27 + 273 = 300\text{ K}\) and the final temperature is \(T_f = 54 + 273 = 327\text{ K}\). Since the volume of the container is fixed, Charles's / Gay-Lussac's Law applies: \(P \propto T\). Thus, \(\frac{P_f}{P_i} = \frac{T_f}{T_i} = \frac{327}{300} = 1.09\).

[1 mark] for converting Celsius to Kelvin and applying Gay-Lussac's law to obtain 1.09.

Efficiency \(\eta\) is defined as the ratio of useful power output to total power input. The useful work done by the motor is the gravitational potential energy gained by the box, which is \(W = mgh\). Since this work is done over a time interval \(t\), the useful power output is \(P_{\text{out}} = \frac{mgh}{t}\). The power input is \(P\). Therefore, the efficiency is \(\eta = \frac{P_{\text{out}}}{P_{\text{in}}} = \frac{mgh}{Pt}\).

[1 mark] for calculating useful power output as \(\frac{mgh}{t}\) and dividing by input power \(P\) to get option A.

Let \(N_X\) be the number of remaining \(X\) nuclei and \(N_Y\) be the number of created \(Y\) nuclei. We are given \(\frac{N_Y}{N_X} = 7\), which means \(N_Y = 7 N_X\). The total initial number of \(X\) nuclei was \(N_0 = N_X + N_Y = N_X + 7 N_X = 8 N_X\). This implies that the fraction of \(X\) nuclei remaining is \(\frac{N_X}{N_0} = \frac{1}{8}\). Since \(\frac{1}{8} = \left(\frac{1}{2}\right)^3\), exactly 3 half-lives have elapsed. Therefore, \(t = 3T\).

[1 mark] for determining the remaining fraction of active nuclei is 1/8, corresponding to 3 half-lives, thus \(t=3T\).

The resistance of a wire is given by \(R = \rho \frac{l}{A}\), where \(\rho\) is resistivity, \(l\) is length, and \(A\) is cross-sectional area. The volume \(V = A \cdot l\) is constant, so we can write \(A = \frac{V}{l}\). Substituting this into the resistance formula gives \(R = \rho \frac{l^2}{V}\). Since \(\rho\) and \(V\) are constant for both wires, resistance is directly proportional to the square of the length: \(R \propto l^2\). Since \(l_Q = 2 l_P\), we have \(\frac{R_Q}{R_P} = \left(\frac{l_Q}{l_P}\right)^2 = 2^2 = 4\).

[1 mark] for expressing resistance in terms of volume and length, showing \(R \propto l^2\), and calculating the ratio to be 4.

The orbital speed \(v\) of a satellite in a circular orbit of radius \(r\) around a planet of mass \(M\) is found by equating the gravitational force to the centripetal force: \(\frac{G M m}{r^2} = \frac{m v^2}{r} \implies v = \sqrt{\frac{G M}{r}}\). This shows that the orbital speed is independent of the satellite's mass. For the second satellite, the orbital radius is \(2r\), so its orbital speed is \(v' = \sqrt{\frac{G M}{2r}} = \frac{1}{\sqrt{2}} \sqrt{\frac{G M}{r}} = \frac{v}{\sqrt{2}}\).

[1 mark] for using the orbital speed formula and demonstrating that doubling the radius reduces the speed by a factor of \(\sqrt{2}\).

The total mechanical energy in simple harmonic motion is given by \(E_{\text{total}} = \frac{1}{2} k A^2\). The elastic potential energy at displacement \(x\) is \(E_p = \frac{1}{2} k x^2\). The kinetic energy is \(E_k = E_{\text{total}} - E_p = \frac{1}{2} k (A^2 - x^2)\). Setting \(E_k = E_p\) gives \(\frac{1}{2} k (A^2 - x^2) = \frac{1}{2} k x^2\), which simplifies to \(A^2 - x^2 = x^2 \implies 2x^2 = A^2 \implies x = \frac{A}{\sqrt{2}}\).

[1 mark] for setting kinetic energy equal to potential energy and solving for \(x\) to find \(x = \frac{A}{\sqrt{2}}\).

The forces acting on the falling object of mass \(m\) are its weight \(mg\) downwards and the resistive force \(kv\) upwards. By Newton's second law, \(ma = mg - kv\), which can be written as \(a = g - \frac{k}{m}v\). This is a linear relationship of the form \(y = mx + c\), where the gradient is \(-\frac{k}{m}\) (constant and negative) and the vertical intercept is \(g\). Therefore, the graph of acceleration \(a\) against velocity \(v\) is a straight line with a negative gradient.

Award 1 mark for identifying that the equation of motion is linear with a negative slope, leading to the selection of option A.

The mass of air pushed downwards per unit time is given by \(\frac{\Delta m}{\Delta t} = \rho A v\). Since this air is accelerated from rest to a speed \(v\), the rate of change of momentum of the air is \(\frac{\Delta p}{\Delta t} = \frac{\Delta m}{\Delta t} v = \rho A v^2\). According to Newton's second and third laws, the upward lift force on the helicopter must equal this rate of change of momentum. For the helicopter to hover, this lift force must balance its weight \(Mg\). Thus, \(Mg = \rho A v^2\). Rearranging for \(A\) gives \(A = \frac{Mg}{\rho v^2}\).

Award 1 mark for deriving the relation \(Mg = \rho A v^2\) and correctly solving for the area \(A\).

During the first process (isothermal expansion), the temperature remains constant at \(T\). Since the volume is doubled (\(V_2 = 2V\)), by Boyle's Law, the pressure is halved, so \(P_2 = \frac{P}{2}\). In the second process (isochoric heating), the volume remains constant at \(V_3 = 2V\). The pressure is increased back to \(P_3 = P\). Using the pressure law at constant volume: \(\frac{P_2}{T_2} = \frac{P_3}{T_3}\), which gives \(\frac{P/2}{T} = \frac{P}{T_3}\). Solving for \(T_3\) yields \(T_3 = 2T\).

Award 1 mark for showing that the intermediate pressure is \(P/2\) and applying the constant-volume gas law to find the final temperature of \(2T\).

Let the displacement of the particle be described by \(x(t) = A \sin(\omega t)\), where \(\omega = \frac{2\pi}{T}\). Starting at \(x = 0\) at \(t = 0\), the particle reaches its maximum positive displacement \(x = A\) at \(t = \frac{T}{4}\). The distance travelled during this time is \(A\). To travel a total distance of \(1.5A\), the particle must travel an additional distance of \(0.5A\) in the opposite direction, meaning its displacement decreases from \(x = A\) to \(x = 0.5A\). Setting \(A \sin(\omega t) = 0.5A\) gives \(\sin(\omega t) = 0.5\). The solutions for \(\omega t\) are \(\frac{\pi}{6}\) (on the way up) and \(\frac{5\pi}{6}\) (on the way down). Since we want the time on the way down, we have \(\omega t = \frac{5\pi}{6} \implies \frac{2\pi}{T} t = \frac{5\pi}{6}\), which yields \(t = \frac{5}{12}T\).

Award 1 mark for setting up the equation for the position and solving for the time when the total distance equals \(1.5A\).

The terminal potential difference is given by \(V = I R = \frac{E R}{R + r} = \frac{E}{1 + \frac{r}{R}}\). As \(R\) increases, the fraction \(\frac{r}{R}\) decreases, which means the denominator decreases and \(V\) increases continuously. The power dissipated in the external resistor is given by \(P = I^2 R = \frac{E^2 R}{(R+r)^2}\). According to the maximum power transfer theorem, \(P\) is maximized when \(R = r\). Since \(R\) starts much smaller than \(r\) and ends much larger than \(r\), the power \(P\) first increases, reaches a maximum when \(R = r\), and then decreases.

Award 1 mark for identifying that the terminal voltage increases continuously and the power peaks at \(R=r\).

The number of remaining active nuclei at any time \(t\) is given by \(N(t) = N_0 \left(\frac{1}{2}\right)^{t/T_{1/2}}\). At \(t = T_{1/2}\), the number of remaining nuclei is \(N_1 = \frac{N_0}{2}\). At \(t = 3T_{1/2}\), the number of remaining nuclei is \(N_3 = N_0 \left(\frac{1}{2}\right)^3 = \frac{N_0}{8}\). The number of nuclei that decay during this interval is the difference in active nuclei: \(\Delta N = N_1 - N_3 = \frac{N_0}{2} - \frac{N_0}{8} = \frac{3}{8}N_0\).

Award 1 mark for finding the number of active nuclei at each time boundary and taking their difference to obtain \(\frac{3}{8}N_0\).

The first minimum of the single-slit diffraction envelope occurs at an angle \(\theta\) satisfying \(\sin\theta = \frac{\lambda}{b}\). The central diffraction envelope thus spans the range from \(-\frac{\lambda}{b}\) to \(+\frac{\lambda}{b}\). Interference maxima occur at angles satisfying \(\sin\theta = \frac{m\lambda}{d}\), where \(m\) is an integer. For these maxima to lie within the central envelope, we require \(-\frac{\lambda}{b} < \frac{m\lambda}{d} < \frac{\lambda}{b}\). Since \(d = 5b\), this becomes \(-1 < \frac{m}{5} < 1\), or \(-5 < m < 5\). The integer values of \(m\) that satisfy this are \(-4, -3, -2, -1, 0, 1, 2, 3, 4\). There are exactly 9 such integers, meaning 9 interference maxima lie within the central envelope. (The 5th-order interference maxima at \(m = \pm 5\) are missing because they coincide with the diffraction minima).

Award 1 mark for finding the range of allowed orders \(-5 < m < 5\) and correctly identifying that there are 9 fringes.

The total power radiated by the star is \(P_{\text{star}} = \sigma (4\pi R^2) T^4\). The intensity of this radiation at the distance of the planet is \(I = \frac{P_{\text{star}}}{4\pi d^2} = \frac{\sigma R^2 T^4}{d^2}\). The planet intercepts this radiation with its cross-sectional area \(\pi r^2\), where \(r\) is the planet's radius. The total power absorbed is \(P_{\text{absorbed}} = (1 - \alpha) I (\pi r^2)\). This power is distributed over the planet's entire spherical surface area \(4\pi r^2\). Thus, the average power absorbed per unit area is \(I_{\text{absorbed}} = \frac{P_{\text{absorbed}}}{4\pi r^2} = \frac{(1 - \alpha) I}{4} = \frac{(1 - \alpha) \sigma T^4 R^2}{4 d^2}\).

Award 1 mark for calculating the intensity at distance \(d\), multiplying by \((1-\alpha)\) to account for absorption, and dividing by 4 to account for the ratio of cross-sectional area to surface area.

Taking the upward direction as positive, the displacement \(s\) of the stone when it reaches the bottom of the cliff is \(-h\). Using the kinematic equation: \(s = ut + \frac{1}{2}at^2\) where \(u = +v\), \(a = -g\), and \(t = T\). Substituting these values: \(-h = vT - \frac{1}{2}gT^2\). Rearranging for \(h\): \(h = \frac{1}{2}gT^2 - vT\). Therefore, option C is correct.

Award [1] for the correct answer C. No partial marks.

First, find the common acceleration \(a\) of both blocks: \(a = \frac{F_{\text{net}}}{m_{\text{total}}} = \frac{F}{m + 3m} = \frac{F}{4m}\). Next, consider the block of mass \(3m\). The only horizontal force acting on it is the contact force \(F_{\text{contact}}\) exerted on it by the block of mass \(m\). Using Newton's second law for this block: \(F_{\text{contact}} = (3m) \cdot a = 3m \cdot \frac{F}{4m} = \frac{3}{4}F\). Therefore, the force exerted by the block of mass \(m\) on the block of mass \(3m\) is \(\frac{3}{4}F\).

Award [1] for the correct answer D. No partial marks.

The total mechanical energy \(E\) of the mass-spring system is equal to the initial elastic potential energy when the spring is stretched by \(x_0\): \(E = \frac{1}{2} k x_0^2\). At any position \(x\), the total energy is the sum of the kinetic energy \(E_k\) and the elastic potential energy \(E_p\): \(E = E_k + E_p \implies E_k = E - E_p\). When \(x = \frac{x_0}{2}\), the potential energy is: \(E_p = \frac{1}{2} k \left(\frac{x_0}{2}\right)^2 = \frac{1}{8} k x_0^2\). Thus, the kinetic energy is: \(E_k = \frac{1}{2} k x_0^2 - \frac{1}{8} k x_0^2 = \frac{3}{8} k x_0^2\).

Award [1] for the correct answer C. No partial marks.

The ideal gas equation can be written as: \(P V = N k_B T\), where \(P\) is pressure, \(V\) is volume, \(N\) is the number of molecules, \(k_B\) is the Boltzmann constant, and \(T\) is the absolute temperature. Since the volume \(V\) of the rigid cylinder is constant: \(P \propto N T\). Therefore, the ratio of final pressure \(P_f\) to initial pressure \(P_i\) is: \(\frac{P_f}{P_i} = \frac{N_f T_f}{N_i T_i}\). Given that \(N_f = \frac{1}{2} N_i\) and \(T_f = 2 T_i\): \(\frac{P_f}{P_i} = \left(\frac{1}{2}\right) \cdot (2) = 1\). So, \(P_f = P_i = P\). The pressure remains unchanged.

Award [1] for the correct answer B. No partial marks.

The current \(I\) in the circuit is given by: \(I = \frac{E}{R + r}\). As the resistance \(R\) increases, the total resistance of the circuit increases, which causes the current \(I\) to decrease. The terminal potential difference \(V\) across the cell is: \(V = E - Ir\). Since \(I\) decreases, the 'lost volts' \(Ir\) decrease, and therefore the terminal potential difference \(V\) increases. The power \(P\) dissipated in the internal resistance of the cell is: \(P = I^2 r\). Since the current \(I\) decreases and \(r\) is constant, the power \(P\) must decrease. Thus, the terminal potential difference increases, and the power dissipated in the internal resistance decreases.

Award [1] for the correct answer C. No partial marks.

For a satellite in a circular orbit of radius \(r\) around a planet of mass \(M\), the gravitational force provides the centripetal force: \(\frac{G M m}{r^2} = \frac{m v^2}{r}\). Solving for the orbital speed \(v\): \(v = √\frac{G M}{r}\). If the orbital radius is doubled to \(2r\), the new orbital speed \(v'\) is: \(v' = √\frac{G M}{2r} = \frac{1}{\sqrt{2}}\sqrt{\frac{G M}{r}} = \frac{v}{\sqrt{2}}\). Therefore, option B is correct.

Award [1] for the correct answer B. No partial marks.

Let the initial activity of both isotopes at \(t = 0\) be \(A_0\). For isotope \(X\) with a half-life of \(2\text{ hours}\), the number of half-lives elapsed after \(8\text{ hours}\) is \(n_X = \frac{8}{2} = 4\). The activity of \(X\) is \(A_X = A_0 (1/2)^4 = A_0 / 16\). For isotope \(Y\) with a half-life of \(4\text{ hours}\), the number of half-lives elapsed after \(8\text{ hours}\) is \(n_Y = \frac{8}{4} = 2\). The activity of \(Y\) is \(A_Y = A_0 (1/2)^2 = A_0 / 4\). The ratio of the activities is \(\frac{A_X}{A_Y} = \frac{A_0 / 16}{A_0 / 4} = \frac{4}{16} = \frac{1}{4}\).

Award [1] for the correct answer A. No partial marks.

The displacement of a particle starting from its equilibrium position can be written as: \(x(t) = A \sin(\omega t)\), where \(omega = \frac{2\pi}{T}\). Setting \(x(t) = \frac{A}{2}\), we get: \(\frac{A}{2} = A \sin(\omega t) \implies \sin(\omega t) = \frac{1}{2}\). The minimum positive angle for this is \(\omega t = \frac{\pi}{6}\). Substituting \(\omega = \frac{2\pi}{T}\): \(\frac{2\pi}{T} t = \frac{\pi}{6} \implies t = \frac{T}{12}\).

Award [1] for the correct answer A. No partial marks.

For a perfectly elastic head-on collision between a moving mass \( m_1 = m \) with velocity \( v_1 = v \) and a stationary mass \( m_2 = 3m \) with \( v_2 = 0 \), the final velocity of the first mass \( v_{1f} \) is given by the formula:

\( v_{1f} = \frac{m_1 - m_2}{m_1 + m_2} v_1 \)

Substituting the given masses:

\( v_{1f} = \frac{m - 3m}{m + 3m} v = \frac{-2m}{4m} v = -\frac{v}{2} \)

The negative sign indicates that the velocity is in the direction opposite to the initial motion. Thus, the velocity is \( \frac{v}{2} \) in the opposite direction.

[1 mark] Award for correct identification of option B.

According to the ideal gas law \( P V = n R T \), since the volume \( V \) is constant, pressure is directly proportional to absolute temperature: \( P \propto T \). Therefore, if the absolute temperature triples, the pressure also triples.

The average kinetic energy of the molecules is directly proportional to the temperature: \( \frac{1}{2} m v_{\text{rms}}^2 = \frac{3}{2} k_B T \). This implies that the root-mean-square (average) speed is proportional to the square root of the temperature: \( v_{\text{rms}} \propto \sqrt{T} \). If temperature triples, the average speed increases by a factor of \( \sqrt{3} \).

[1 mark] Award for correct identification of option B.

The useful mechanical power output of the motor is the rate of doing work against gravity:

\( P_{\text{out}} = F v = m g v \)

Efficiency is defined as:

\( \eta = \frac{P_{\text{out}}}{P_{\text{in}}} = \frac{m g v}{P} \)

Solving for \( v \):

\( v = \frac{\eta P}{m g} \)

[1 mark] Award for correct identification of option A.

The maximum equivalent resistance is obtained when all three resistors are connected in series:

\( R_{\text{max}} = R + R + R = 3R \)

The minimum equivalent resistance is obtained when all three resistors are connected in parallel:

\( \frac{1}{R_{\text{min}}} = \frac{1}{R} + \frac{1}{R} + \frac{1}{R} = \frac{3}{R} \implies R_{\text{min}} = \frac{R}{3} \)

The ratio is:

\( \frac{R_{\text{max}}}{R_{\text{min}}} = \frac{3R}{\frac{R}{3}} = 9 \)

[1 mark] Award for correct identification of option C.

The energy of the emitted photon is equal to the difference in energy between the initial and final states of the transition:

\( E_{\text{photon}} = E_{\text{initial}} - E_{\text{final}} \)

\( E_{\text{photon}} = -1.51\text{ eV} - (-3.40\text{ eV}) = 1.89\text{ eV} \)

[1 mark] Award for correct identification of option B.

The gravitational field strength \( g' \) at a distance \( r \) from the center of a planet is given by:

\( g' = \frac{G M}{r^2} \)

At the surface where \( r = R \), the field strength is \( g = \frac{G M}{R^2} \).

At a height \( h \) above the surface, the distance from the center is \( r = R + h \). We require:

\( g' = \frac{g}{9} \implies \frac{G M}{(R+h)^2} = \frac{1}{9} \frac{G M}{R^2} \)

Taking the square root of both sides:

\( \frac{1}{R+h} = \frac{1}{3R} \implies R + h = 3R \implies h = 2R \)

[1 mark] Award for correct identification of option A.

(a) The relationship between the variables is given by the terminal potential difference equation:
\(V = \varepsilon - Ir\)

(b) Comparing \(V = -rI + \varepsilon\) with the equation of a straight line \(y = mx + c\):
- The y-intercept represents the emf, so \(\varepsilon = 6.0\text{ V}\).
- The gradient of the line is \(-r\), therefore: \(-r = -1.5\ \Omega \implies r = 1.5\ \Omega\).

(c) The total resistance of the series circuit is:
\(R_{\text{total}} = R + r = 3.0 + 1.5 = 4.5\ \Omega\)

The current in the circuit is:
\(I = \frac{\varepsilon}{R_{\text{total}}} = \frac{6.0}{4.5} = 1.33\text{ A}\)

The power dissipated in the variable resistor \(R\) is:
\(P = I^2 R = (1.33)^2 \times 3.0 = 5.33\text{ W} \approx 5.3\text{ W}\)

(a)
[1 mark] for writing the correct equation \(V = \varepsilon - Ir\) or equivalent.

(b)
[1 mark] for correctly identifying \(\varepsilon = 6.0\text{ V}\).
[1 mark] for correctly identifying \(r = 1.5\ \Omega\).

(c)
[1 mark] for calculating total resistance \(R_{\text{total}} = 4.5\ \Omega\).
[1 mark] for calculating circuit current \(I = 1.33\text{ A}\) (accept \(1.3\text{ A}\)).
[1.25 marks] for calculating power as \(5.3\text{ W}\) (accept \(5.33\text{ W}\), allow ECF from earlier parts).

(a)
(i) During an isothermal process, the temperature \(T\) is constant. Since the average kinetic energy of gas molecules is directly proportional to absolute temperature (\(E_k = \frac{3}{2} k_B T\)), the average kinetic energy remains constant.
(ii) In an isobaric cooling, the gas volume decreases at constant pressure, which means temperature must decrease (Charles's Law, \(V \propto T\)). Therefore, the average kinetic energy of the molecules decreases.

(b) For an isothermal process of a fixed mass of gas, Boyle's Law applies:
\(P_1 V_1 = P_2 V_2\)
\((3.0 \times 10^5\text{ Pa}) \times (2.0 \times 10^{-3}\text{ m}^3) = P_2 \times (4.0 \times 10^{-3}\text{ m}^3)\)
\(P_2 = \frac{6.0 \times 10^2}{4.0 \times 10^{-3}} = 1.5 \times 10^5\text{ Pa}\)

(c) For an isobaric process, Charles's Law applies:
\(\frac{V_2}{T_2} = \frac{V_3}{T_3}\)
Where \(V_2 = 4.0 \times 10^{-3}\text{ m}^3\), \(T_2 = 400\text{ K}\), and \(V_3 = 2.0 \times 10^{-3}\text{ m}^3\).
\(\frac{4.0 \times 10^{-3}}{400} = \frac{2.0 \times 10^{-3}}{T_3}\)
\(T_3 = 200\text{ K}\)

(a)
[1 mark] for stating average kinetic energy is constant because temperature is constant.
[1 mark] for stating average kinetic energy decreases because temperature decreases.

(b)
[1 mark] for recognizing the use of Boyle's Law \(P_1 V_1 = P_2 V_2\) and substituting values.
[1 mark] for the correct final answer \(1.5 \times 10^5\text{ Pa}\).

(c)
[1 mark] for recognizing the use of Charles's Law \(\frac{V_2}{T_2} = \frac{V_3}{T_3}\) and substituting values.
[1.25 marks] for the correct final temperature of \(200\text{ K}\).

(a) Using the double-slit interference formula:
\(s = \frac{\lambda D}{d}\)
Substitute the given values into the formula:
\(\lambda = 632.8 \times 10^{-9}\text{ m}\)
\(d = 0.120 \times 10^{-3}\text{ m}\)
\(D = 1.50\text{ m}\)

\(s = \frac{(632.8 \times 10^{-9}) \times 1.50}{0.120 \times 10^{-3}} = 7.91 \times 10^{-3}\text{ m} = 7.91\text{ mm}\)

(b) When immersed in water, the speed of light decreases according to \(v = \frac{c}{n}\).
Since the frequency \(f\) of the light remains constant, the wavelength in water is given by:
\(\lambda_{\text{water}} = \frac{\lambda}{n} = \frac{632.8\text{ nm}}{1.33} \approx 476\text{ nm}\)
Since fringe separation is directly proportional to wavelength (\(s \propto \lambda\)), the fringe separation will decrease:
\(s_{\text{water}} = \frac{s}{n} = \frac{7.91\text{ mm}}{1.33} \approx 5.95\text{ mm}\)

(a)
[1 mark] for stating/recalling the formula \(s = \frac{\lambda D}{d}\).
[1 mark] for correct substitution of values (showing consistent units, e.g., meters).
[1.25 marks] for the correct final answer of \(7.91\text{ mm}\) (or \(7.91 \times 10^{-3}\text{ m}\)).

(b)
[1 mark] for stating that the fringe separation decreases.
[1 mark] for explaining that the wavelength of light decreases in a medium with a higher refractive index (\(\lambda_{\text{medium}} = \frac{\lambda}{n}\)).
[1 mark] for linking the smaller wavelength to a smaller fringe separation using \(s \propto \lambda\).

(a) The total linear momentum of a system remains constant, provided that no net external force acts on the system (it is an isolated system).

(b) Let the direction to the right be positive.
Initial momentum of glider A: \(p_A = m_A v_A = 0.40 \times 2.5 = 1.00\text{ kg m s}^{-1}\)
Initial momentum of glider B: \(p_B = m_B v_B = 0.60 \times (-1.5) = -0.90\text{ kg m s}^{-1}\)

Using conservation of momentum:
\(p_{\text{initial}} = p_{\text{final}}\)
\(m_A v_A + m_B v_B = (m_A + m_B) v_f\)
\(1.00 - 0.90 = (0.40 + 0.60) v_f\)
\(0.10 = 1.00 v_f \implies v_f = +0.10\text{ m s}^{-1}\)

So the final velocity is \(0.10\text{ m s}^{-1}\) to the right.

(c) Initial kinetic energy:
\(E_{k,i} = \frac{1}{2} m_A v_A^2 + \frac{1}{2} m_B v_B^2\)
\(E_{k,i} = \frac{1}{2} (0.40) (2.5)^2 + \frac{1}{2} (0.60) (-1.5)^2 = 1.25 + 0.675 = 1.925\text{ J}\)

Final kinetic energy:
\(E_{k,f} = \frac{1}{2} (m_A + m_B) v_f^2 = \frac{1}{2} (1.00) (0.10)^2 = 0.005\text{ J}\)

Kinetic energy lost:
\(\Delta E_k = E_{k,i} - E_{k,f} = 1.925 - 0.005 = 1.92\text{ J}\)

(a)
[1 mark] for stating that total linear momentum is conserved.
[1 mark] for stating the condition: no net external force / isolated system.

(b)
[1 mark] for setting up the conservation of momentum equation with correct signs.
[1.25 marks] for the final velocity of \(0.10\text{ m s}^{-1}\) (must specify the value; sign or direction "to the right" is required for full credit).

(c)
[1 mark] for calculating both initial kinetic energy (\(1.925\text{ J}\)) and final kinetic energy (\(0.005\text{ J}\)).
[1 mark] for finding the difference to get \(1.92\text{ J}\).

(a) Considering the vertical motion, with initial vertical velocity \(u_y = 0\):
\(s_y = u_y t + \frac{1}{2} g t^2\)
\(45 = 0 + \frac{1}{2} (9.81) t^2\)
\(t^2 = \frac{90}{9.81} \approx 9.174\)
\(t = \sqrt{9.174} \approx 3.03\text{ s}\), which is approximately \(3.0\text{ s}\).

(b) The horizontal velocity is constant because air resistance is negligible:
\(s_x = u_x t\)
Using the calculated time \(t = 3.03\text{ s}\):
\(s_x = 12 \times 3.03 = 36.4\text{ m}\) (or \(36\text{ m}\) if using \(t = 3.0\text{ s}\)).

(c) Just before impact:
- The horizontal component of velocity is \(v_x = 12\text{ m s}^{-1}\).
- The vertical component of velocity is \(v_y = u_y + g t = 0 + 9.81 \times 3.03 = 29.7\text{ m s}^{-1}\).

The magnitude of the velocity is:
\(v = \sqrt{v_x^2 + v_y^2} = \sqrt{12^2 + 29.7^2} = \sqrt{144 + 882.1} = \sqrt{1026.1} \approx 32\text{ m s}^{-1}\).

The direction is given by the angle \(\theta\) below the horizontal:
\(\tan \theta = \frac{v_y}{v_x} = \frac{29.7}{12} = 2.475 \implies \theta = \arctan(2.475) \approx 68^\circ\) below the horizontal.

(a)
[1 mark] for selecting the correct kinematic equation for vertical motion \(s = ut + \frac{1}{2}gt^2\) with \(u_y = 0\).
[1.25 marks] for correct calculation showing \(t = 3.03\text{ s}\), hence validating \(\approx 3.0\text{ s}\).

(b)
[1 mark] for using \(s_x = u_x t\).
[1 mark] for the correct answer: \(36.4\text{ m}\) (accept \(36\text{ m}\) if using \(3.0\text{ s}\)).

(c)
[1 mark] for finding both velocity components (\(v_x = 12\text{ m s}^{-1}\) and \(v_y = 29.7\text{ m s}^{-1}\) or \(29.4\text{ m s}^{-1}\)).
[1 mark] for finding the magnitude \(32\text{ m s}^{-1}\) and the direction angle \(68^\circ\) (accept \(68^\circ\) to \(68.2^\circ\) below the horizontal).

(a) Albedo is the ratio of the total reflected radiation power from a planet to the total incident radiation power.

(b) The total incident solar power is intercepted by a disc of area \(\pi R^2\), but the Earth's total surface area is \(4\pi R^2\).
Therefore, the average solar intensity incident on the surface is:
\(I_{\text{average}} = \frac{S}{4} = \frac{1360}{4} = 340\text{ W m}^{-2}\)

Accounting for the albedo \(\alpha = 0.30\), the fraction of radiation absorbed is \((1 - \alpha)\):
\(I_{\text{absorbed}} = I_{\text{average}} \times (1 - \alpha) = 340 \times (1 - 0.30) = 340 \times 0.70 = 238\text{ W m}^{-2}\)

(c) The Earth's surface absorbs short-wavelength (visible/UV) solar radiation and re-emits longer-wavelength infrared (IR) radiation. Greenhouse gases (such as \(\text{CO}_2\) and water vapor) have molecular bonds with vibrational energy transitions that correspond to the energy of IR photons. Because of this match, these molecules undergo resonance and absorb the IR radiation. They then re-radiate this energy in all directions, with a significant fraction directed back towards the Earth's surface, resulting in an elevated surface temperature.

(a)
[1 mark] for stating that albedo is the ratio of reflected power to incident power (or equivalent definition in terms of intensity/energy).

(b)
[1 mark] for dividing the solar constant by 4 to find the average incident intensity on the spherical surface (\(340\text{ W m}^{-2}\)).
[1.25 marks] for applying the \((1 - 0.30)\) factor to get the final answer of \(238\text{ W m}^{-2}\).

(c)
[1 mark] for stating that the Earth re-emits longer wavelength infrared radiation.
[1 mark] for explaining that greenhouse gas molecules absorb IR due to resonance (vibrational frequencies matching the IR frequency).
[1 mark] for stating that these molecules re-radiate IR in all directions, including back to the Earth's surface, keeping it warmer.

(a) Gravitational potential at a point is the work done per unit mass in bringing a small test mass from infinity to that point.

(b) The orbital radius \(r\) is the sum of the Earth's radius and the altitude:
\(r = R + H = R + R = 2R = 2 \times 6.4 \times 10^6 = 1.28 \times 10^7\text{ m}\)

The gravitational potential energy \(E_p\) is:
\(E_p = -\frac{GMm}{r} = -\frac{(6.67 \times 10^{-11}) \times (6.0 \times 10^{24}) \times 250}{1.28 \times 10^7}\)
\(E_p = -\frac{1.0005 \times 10^{17}}{1.28 \times 10^7} = -7.816 \times 10^9\text{ J} \approx -7.82 \times 10^9\text{ J}\)

(c) The gravitational force acts as the centripetal force:
\(\frac{m v^2}{r} = \frac{GMm}{r^2}\)
\(v = \sqrt{\frac{GM}{r}}\)

Using \(r = 1.28 \times 10^7\text{ m}\):
\(v = \sqrt{\frac{(6.67 \times 10^{-11}) \times (6.0 \times 10^{24})}{1.28 \times 10^7}} = \sqrt{\frac{4.002 \times 10^{14}}{1.28 \times 10^7}} = \sqrt{3.1266 \times 10^7} \approx 5.59 \times 10^3\text{ m s}^{-1}\)

(a)
[1 mark] for defining gravitational potential as work done per unit mass in moving a test mass from infinity to the point.

(b)
[1 mark] for recognizing that \(r = 2R = 1.28 \times 10^7\text{ m}\).
[1.25 marks] for the correct calculation of \(-7.82 \times 10^9\text{ J}\) (negative sign required for full marks).

(c)
[1 mark] for equating gravitational force to centripetal force \(\frac{m v^2}{r} = \frac{GMm}{r^2}\).
[1 mark] for rearranging to solve for velocity \(v =
\sqrt{\frac{GM}{r}}\).
[1 mark] for the correct calculation of \(5.59 \times 10^3\text{ m s}^{-1}\) (accept \(5.6 \times 10^3\text{ m s}^{-1}\)).

(a) The half-life of a radioactive isotope is the time taken for half of the radioactive nuclei in a sample to decay (or the time taken for the activity of the sample to halve).

(b) Let the half-life be \(T_{1/2}\). The relationship between initial activity \(A_0\) and activity \(A\) at time \(t\) is:
\(A = A_0 \left(\frac{1}{2}\right)^{\frac{t}{T_{1/2}}}\)

Substitute the given values:
\(1.0 \times 10^5 = 8.0 \times 10^5 \left(\frac{1}{2}\right)^{\frac{60}{T_{1/2}}}\)
\(\frac{1}{8} = \left(\frac{1}{2}\right)^{\frac{60}{T_{1/2}}}\)

Since \(\frac{1}{8} = \left(\frac{1}{2}\right)^3\), we equate the exponents:
\(3 = \frac{60}{T_{1/2}}\)
\(T_{1/2} = 20\text{ minutes}\)

(c) First, calculate the decay constant \(\lambda\) in \(\text{s}^{-1}\):
\(T_{1/2} = 20\text{ minutes} = 20 \times 60 = 1200\text{ s}\)
\(\lambda = \frac{\ln 2}{T_{1/2}} = \frac{0.6931}{1200} \approx 5.776 \times 10^{-4}\text{ s}^{-1}\)

The activity is related to the number of nuclei by \(A = \lambda N\). At \(60\text{ minutes}\), the activity is \(A = 1.0 \times 10^5\text{ Bq}\):
\(N = \frac{A}{\lambda} = \frac{1.0 \times 10^5}{5.776 \times 10^{-4}} \approx 1.73 \times 10^8\text{ nuclei}\)

(a)
[1 mark] for stating that it is the time taken for half the nuclei to decay (or activity to halve).

(b)
[1 mark] for setting up the decay relation, showing that the activity has decreased by a factor of 8.
[1.25 marks] for equating the number of half-lives to 3 (since \(2^3 = 8\)) and solving for \(T_{1/2} = 20\text{ minutes}\).

(c)
[1 mark] for converting the half-life to seconds (\(1200\text{ s}\)) and calculating the decay constant \(\lambda \approx 5.78 \times 10^{-4}\text{ s}^{-1}\).
[1 mark] for using the equation \(A = \lambda N\).
[1 mark] for the final answer of \(1.73 \times 10^8\text{ nuclei}\) (accept \(1.7 \times 10^8\)).

Part (a): Independent variable is the volume of gas in the syringe \(V\). Dependent variable is the pressure \(p\) (or \(\frac{1}{p}\)). Part (b): Rearranging Boyle's law: \(p(V + V_0) = C\) gives \(V + V_0 = \frac{C}{p}\), which can be written as \(\frac{1}{p} = \frac{1}{C}V + \frac{V_0}{C}\). This is a linear equation of the form \(y = mx + c\) where \(y = \frac{1}{p}\) and \(x = V\). Since \(C\) and \(V_0\) are constants, the gradient \(m = \frac{1}{C}\) and intercept \(c = \frac{V_0}{C}\) are constant, giving a straight line. Part (c)(i): The volume of the connection tube is \(V_0 = \frac{c}{m} = \frac{2.50 \times 10^{-4}}{1.25 \times 10^{-5}} = 20.0\text{ cm}^3\). The fractional uncertainty in \(c\) is \(\frac{0.15}{2.50} = 0.06\) (6%) and in \(m\) is \(\frac{0.05}{1.25} = 0.04\) (4%). The sum of these fractional uncertainties is \(0.06 + 0.04 = 0.10\) (10%). Therefore, the absolute uncertainty in \(V_0\) is \(0.10 \times 20.0 = 2.0\text{ cm}^3\). The final value is \(V_0 = 20 \pm 2\text{ cm}^3\). Part (c)(ii): If the plunger is compressed too rapidly, the gas temperature increases because there is insufficient time for thermal energy to transfer to the surroundings. Since pressure is directly proportional to temperature at constant volume, this temperature rise will systematically increase the pressure, introducing a systematic error.

Part (a): Volume \(V\) as independent and Pressure \(p\) (or \(\frac{1}{p}\)) as dependent. [1 mark] Part (b): Correct algebraic rearrangement to express \(\frac{1}{p}\) as a function of \(V\). [1 mark] Explains that gradient and intercept are constants because \(C\) and \(V_0\) are constants. [1 mark] Part (c)(i): Correct calculation of \(V_0 = 20\text{ cm}^3\). [1 mark] Correct addition of fractional uncertainties (0.10 or 10%). [1 mark] Correct absolute uncertainty of \(2\text{ cm}^3\) given with appropriate significant figures: \(20 \pm 2\text{ cm}^3\). [1 mark] Part (c)(ii): Explains that rapid compression causes temperature of gas to increase. [1 mark] Explains that this causes pressure readings to be systematically higher than expected under isothermal conditions. [1 mark]

Part (a): If the mass of the spring is ignored, the calculated spring constant \(k\) will be systematically lower than the actual value. This is because the actual oscillating mass is larger than \(M\), leading to a longer period \(T\) and thus a lower value for the ratio \(\frac{4\pi^2 M}{T^2}\). Part (b): At the x-intercept, \(T^2 = 0\). Substituting this into the equation: \(0 = \frac{4\pi^2}{k} M + \frac{4\pi^2 m_s}{3k}\). Dividing both sides by the non-zero factor \(\frac{4\pi^2}{k}\) gives \(0 = M + \frac{m_s}{3}\), which simplifies to \(M = -\frac{m_s}{3}\). Therefore, the x-intercept on the mass axis is \(-\frac{m_s}{3}\). Part (c)(i): The gradient of the graph is \(m = \frac{4\pi^2}{k} = 1.62\text{ s}^2\text{ kg}^{-1}\). Solving for \(k\): \(k = \frac{4\pi^2}{1.62} \approx 24.37\text{ N m}^{-1}\) (or \(24.4\text{ N m}^{-1}\) to three significant figures). Part (c)(ii): The y-intercept is \(c = \frac{4\pi^2 m_s}{3k} = 0.054\text{ s}^2\). Since \(\frac{4\pi^2}{k}\) is equal to the gradient \(1.62\), this simplifies to \(c = 1.62 \times \frac{m_s}{3} = 0.54 m_s\). Thus, \(0.054 = 0.54 m_s\), which gives \(m_s = \frac{0.054}{0.54} = 0.10\text{ kg}\) (or \(100\text{ g}\)).

Part (a): States that the calculated spring constant is systematically smaller than the actual value. [1 mark] Part (b): Sets \(T^2 = 0\) at the x-intercept. [1 mark] Shows correct algebraic steps to arrive at \(M = -\frac{m_s}{3}\). [1 mark] Part (c)(i): Equates gradient \(1.62\) to \(\frac{4\pi^2}{k}\). [1 mark] Calculates \(k = 24.4\text{ N m}^{-1}\) (accept range 24.0 to 24.4). [1 mark] Part (c)(ii): Equates y-intercept \(0.054\) to \(\frac{4\pi^2 m_s}{3k}\) or \(\text{gradient} \times \frac{m_s}{3}\). [1 mark] Calculates \(m_s = 0.10\text{ kg}\) (or \(100\text{ g}\)). [1 mark]

a) Luminosity is the total power radiated by a star (at all wavelengths), whereas apparent brightness is the power received per unit area at Earth.

b) i) Using Stefan-Boltzmann law:
\(L = 4\pi R^2 \sigma T^4\) where \(\sigma = 5.67 \times 10^{-8}\text{ W m}^{-2}\text{ K}^{-4}\).
\(L = 4\pi \times (3.5 \times 10^9\text{ m})^2 \times (5.67 \times 10^{-8}\text{ W m}^{-2}\text{ K}^{-4}) \times (4500\text{ K})^4 = 3.58 \times 10^{27}\text{ W} \approx 3.6 \times 10^{27}\text{ W}\).

ii) From apparent brightness formula:
\(b = \frac{L}{4\pi d^2}\)
Rearranging for distance:
\(d = \sqrt{\frac{L}{4\pi b}} = \sqrt{\frac{3.58 \times 10^{27}}{4\pi \times 1.8 \times 10^{-11}}} \approx 3.98 \times 10^{18}\text{ m}\).
Since \(1\text{ ly} = 9.46 \times 10^{15}\text{ m}\):
\(d = \frac{3.98 \times 10^{18}\text{ m}}{9.46 \times 10^{15}\text{ m ly}^{-1}} \approx 420\text{ ly}\) (accept values between 410 and 430 ly due to rounding).

c) According to Wien's displacement law, \(\lambda_{\text{max}} T = \text{constant}\). Since the surface temperature \(T\) increases, the peak wavelength \(\lambda_{\text{max}}\) must decrease to maintain a constant product.

a)
- Luminosity: total power radiated by a star [1]
- Apparent brightness: power received per unit area at an observer / Earth [1]

b) i)
- Uses \(L = 4\pi R^2 \sigma T^4\) [1]
- Correctly substitutes values and gets \(3.58 \times 10^{27}\text{ W}\) (must show at least 3 sig figs to justify "show that") [1]

b) ii)
- Recalls / uses \(d = \sqrt{\frac{L}{4\pi b}}\) [1]
- Calculates \(d \approx 3.98 \times 10^{18}\text{ m}\) (or \(4.0 \times 10^{18}\text{ m}\) if using \(3.6 \times 10^{27}\text{ W}\)) [1]
- Converts correctly to light years to yield \(420\text{ ly}\) (accept \(410 - 430\text{ ly}\)) [1]

c)
- Peak wavelength decreases [1]
- Identifies Wien's displacement law (\(\lambda_{\text{max}} T = \text{constant}\)) [1]
- Explains that because temperature increases, peak wavelength must decrease to keep the product constant [1]

a) A constellation is a pattern of stars as projected onto the sky from Earth that may not be physically close or gravitationally bound to one another. A stellar cluster is a group of stars that are physically close in space and held together by gravity.

b) i) Redshift:
\(z = \frac{\Delta \lambda}{\lambda_0} = \frac{682.1\text{ nm} - 656.3\text{ nm}}{656.3\text{ nm}} = \frac{25.8}{656.3} \approx 0.0393\).

ii) Since \(z \ll 1\), the non-relativistic Doppler redshift formula is valid:
\(v = z c = 0.0393 \times 3.00 \times 10^8\text{ m s}^{-1} \approx 1.18 \times 10^7\text{ m s}^{-1}\) (or \(1.18 \times 10^4\text{ km s}^{-1}\)).

iii) Using Hubble's law:
\(v = H_0 d \Rightarrow d = \frac{v}{H_0}\)
With \(v = 1.18 \times 10^4\text{ km s}^{-1}\) and \(H_0 = 70\text{ km s}^{-1}\text{ Mpc}^{-1}\):
\(d = \frac{1.18 \times 10^4}{70} \approx 169\text{ Mpc}\) (accept \(170\text{ Mpc}\)).

c) There is uncertainty in the Hubble constant because measuring precise astronomical distances to distant galaxies (e.g., standard candles) is highly challenging and prone to systematic errors. Additionally, different observational methods (such as Cosmic Microwave Background measurements versus local distance ladders) yield slightly different values, known as the Hubble tension.

a)
- Constellation: stars projected close together from Earth but not gravitationally bound / physically close [1]
- Stellar cluster: stars physically close and held together by gravity [1]

b) i)
- Uses \(z = \frac{\Delta \lambda}{\lambda_0}\) [1]
- Obtains \(z = 0.0393\) (accept \(0.039\)) [1]

b) ii)
- Uses \(v = z c\) [1]
- Obtains \(v = 1.18 \times 10^7\text{ m s}^{-1}\) or \(1.18 \times 10^4\text{ km s}^{-1}\) (accept \(1.2 \times 10^7\text{ m s}^{-1}\) or \(1.2 \times 10^4\text{ km s}^{-1}\)) [1]

b) iii)
- Uses \(d = \frac{v}{H_0}\) [1]
- Obtains \(d = 169\text{ Mpc}\) (accept \(160 - 175\text{ Mpc}\) depending on rounding) [1]

c)
- Award [1] mark per valid reason up to [2] marks:
- Difficulty in measuring accurate distances to far galaxies / standard candles [1]
- Systematic errors in cosmic distance ladder calibration [1]
- Hubble tension / different values obtained from CMB vs local methods [1]