2025 IB DP Physics Practice Paper with Answers

The average translational kinetic energy of molecules in an ideal gas is given by \(E_k = \frac{3}{2} k_B T\), which is directly proportional to the absolute temperature \(T\). Therefore, when the absolute temperature is doubled, the average translational kinetic energy is also doubled. The mean speed, collision frequency, and momentum change are all proportional to \(\sqrt{T}\), so they increase by a factor of \(\sqrt{2}\).

[1 mark] awarded for identifying that the average translational kinetic energy is directly proportional to the absolute temperature, making option B the correct response.

The velocity in simple harmonic motion is given by \(v = \omega \sqrt{A^2 - x^2}\). The maximum speed is \(v_{\text{max}} = \omega A\). When the displacement \(x = \frac{1}{2}A\), we have: \(v = \omega \sqrt{A^2 - \left(\frac{1}{2}A\right)^2} = \omega \sqrt{A^2 - \frac{1}{4}A^2} = \omega \sqrt{\frac{3}{4}A^2} = \omega A \frac{\sqrt{3}}{2} = \frac{\sqrt{3}}{2} v_{\text{max}}\).

[1 mark] for correctly using the relationship between velocity, displacement, and amplitude in simple harmonic motion to calculate the speed as a fraction of maximum speed.

Since the block remains at rest, the forces acting on it must be in equilibrium. In the vertical direction, the downward force of gravity (the weight of the block, \(mg\)) must be balanced by the upward force of static friction \(f_s\). Therefore, \(f_s = mg\). Note that \(\mu_s F\) represents the maximum possible static frictional force, not the actual static frictional force acting on the block in equilibrium.

[1 mark] for recognizing that under static equilibrium, the vertical frictional force must exactly balance the gravitational force \(mg\).

After three half-lives, the fraction of original nuclei remaining is \(\left(\frac{1}{2}\right)^3 = \frac{1}{8}\). Therefore, the number of remaining nuclei is \(N = \frac{1}{8} N_0\). The number of nuclei that have decayed is \(N_0 - N = N_0 - \frac{1}{8} N_0 = \frac{7}{8} N_0\).

[1 mark] for calculating the fraction of nuclei remaining after three half-lives and subtracting it from the initial quantity to find the decayed fraction.

The initial magnetic flux linkage is \(\Phi_{\text{initial}} = N B A\) because the plane of the coil is perpendicular to the magnetic field. The final magnetic flux linkage is \(\Phi_{\text{final}} = 0\) because the plane of the coil is parallel to the magnetic field. The change in flux linkage is \(\Delta \Phi = |0 - N B A| = N B A\). According to Faraday's law of induction, the magnitude of the average induced emf is \(\varepsilon = \frac{\Delta \Phi}{\Delta t} = \frac{NBA}{\Delta t}\).

[1 mark] for determining the change in magnetic flux linkage and applying Faraday's law to obtain the average induced emf.

For a satellite in a stable circular orbit, the gravitational force provides the centripetal acceleration: \(\frac{G M m}{r^2} = \frac{m v^2}{r}\). Multiplying both sides by \(\frac{1}{2}r\) gives the kinetic energy: \(E_k = \frac{1}{2} m v^2 = \frac{G M m}{2r}\). The gravitational potential energy is \(E_p = -\frac{G M m}{r}\). The total mechanical energy \(E\) is the sum of kinetic and potential energy: \(E = E_k + E_p = \frac{G M m}{2r} - \frac{G M m}{r} = -\frac{G M m}{2r}\).

[1 mark] for using the circular orbit condition to find the kinetic energy, adding it to the potential energy, and obtaining the correct total mechanical energy.

Using the conservation of energy, the gravitational potential energy lost is converted into both translational and rotational kinetic energy: \(M g h = \frac{1}{2} M v^2 + \frac{1}{2} I \omega^2\). Since the sphere rolls without slipping, \(\omega = \frac{v}{R}\). Substituting this and the moment of inertia into the equation: \(M g h = \frac{1}{2} M v^2 + \frac{1}{2} \left(\frac{2}{5} M R^2\right) \left(\frac{v}{R}\right)^2 = \frac{1}{2} M v^2 + \frac{1}{5} M v^2 = \frac{7}{10} M v^2\). Solving for \(v\), we get \(v = \sqrt{\frac{10}{7}gh}\).

[1 mark] for equating the initial potential energy to the total kinetic energy (translational + rotational) and solving for linear speed \(v\).

The apparent brightness \(b\) of a star is given by \(b = \frac{L}{4\pi d^2}\), where \(L\) is the luminosity and \(d\) is the distance from Earth. Since both stars have the same apparent brightness, \(b_X = b_Y\). Therefore, \(\frac{L_X}{4\pi d_X^2} = \frac{L_Y}{4\pi d_Y^2}\), which simplifies to \(\frac{L_X}{L_Y} = \left(\frac{d_X}{d_Y}\right)^2\). Since \(d_X = 3 d_Y\), the ratio of their luminosities is \(\left(\frac{3 d_Y}{d_Y}\right)^2 = 3^2 = 9\).

[1 mark] for using the relationship between apparent brightness, luminosity, and distance to show that luminosity is proportional to the square of the distance for stars with equal apparent brightness.

The lifetime \(t\) of a star is proportional to the mass fuel divided by the luminosity: \(t \propto \frac{M}{L}\). Since \(L \propto M^{3.5}\), we have \(t \propto M^{-2.5}\). The ratio of the lifetimes is therefore \(\frac{t_X}{t_Y} = \left(\frac{M_X}{M_Y}\right)^{-2.5} = (3.0)^{-2.5} \approx 0.064\).

Award 1 mark for identifying that lifetime is proportional to \(M^{-2.5}\) and calculating the ratio correctly to obtain 0.064.

Using conservation of linear momentum: \(m v = m v_1 + 3m v_2\), which simplifies to \(v = v_1 + 3v_2\). For a perfectly elastic collision, the relative speed of separation equals the relative speed of approach: \(v_2 - v_1 = v\), which means \(v_1 = v_2 - v\). Substituting \(v_1\) into the momentum equation gives \(v = (v_2 - v) + 3v_2\), which simplifies to \(2v = 4v_2\), and therefore \(v_2 = 0.50 v\).

Award 1 mark for setting up conservation of momentum and energy equations and solving for the final speed of the larger mass to obtain 0.50 v.

The number of remaining nuclei after time \(t\) is \(N = N_0 e^{-\lambda t} = N_0 e^{-\lambda (2/\lambda)} = N_0 e^{-2}\). The number of decayed nuclei is \(N_D = N_0 - N = N_0 (1 - e^{-2})\). The ratio of decayed to remaining nuclei is \(\frac{N_D}{N} = \frac{N_0(1-e^{-2})}{N_0 e^{-2}} = \frac{1-e^{-2}}{e^{-2}} = e^2 - 1\).

Award 1 mark for finding the remaining and decayed populations in terms of exponential decay and dividing them to find the ratio.

The induced electromotive force (emf) in the loop as it is pulled out is given by Faraday's Law: \(\varepsilon = B s v\). The electrical power dissipated in the loop is given by \(P = \frac{\varepsilon^2}{R} = \frac{(B s v)^2}{R} = \frac{B^2 s^2 v^2}{R}\).

Award 1 mark for calculating the correct induced emf and substituting it into the power formula to get the correct option.

The total energy is the sum of kinetic energy and potential energy: \(E_T = E_k + E_p\). Given that \(E_k = 3 E_p\), we have \(E_T = 3 E_p + E_p = 4 E_p\). The total energy is \(E_T = \frac{1}{2} m \omega^2 x_0^2\) and the potential energy at displacement \(x\) is \(E_p = \frac{1}{2} m \omega^2 x^2\). Substituting these into the relationship gives \(\frac{1}{2} m \omega^2 x_0^2 = 4 \left(\frac{1}{2} m \omega^2 x^2\right)\), which simplifies to \(x_0^2 = 4 x^2\), so \(x = \pm 0.50 x_0\).

Award 1 mark for relating total energy to potential energy and solving for the displacement in terms of the amplitude.

The maximum height is given by \(H = \frac{v^2 \sin^2(30^\circ)}{2g} = \frac{v^2 (0.5)^2}{2g} = \frac{v^2}{8g}\). The horizontal range is given by \(R = \frac{v^2 \sin(60^\circ)}{g} = \frac{v^2 \sqrt{3}}{2g}\). The ratio of maximum height to horizontal range is \(\frac{H}{R} = \frac{v^2 / (8g)}{v^2 \sqrt{3} / (2g)} = \frac{1}{8} \times \frac{2}{\sqrt{3}} = \frac{1}{4\sqrt{3}}\).

Award 1 mark for expressing height and range using kinematic formulas and determining their ratio algebraically.

The average kinetic energy of the gas molecules is directly proportional to the absolute temperature of the gas: \(\bar{E_k} = \frac{1}{2} m \langle v^2 \rangle = \frac{3}{2} k_B T\). This implies that the mean square speed \(\langle v^2 \rangle\) is directly proportional to the absolute temperature \(T\). Since \(T\) is multiplied by 1.20 (an increase of 20%), the mean square speed \(\langle v^2 \rangle\) also increases by exactly 20%.

Award 1 mark for identifying the proportional relationship between absolute temperature and mean square speed, and concluding that the percentage increase is the same.

For a stable circular orbit, the centripetal force is provided by the gravitational force: \(\frac{m v^2}{r} = \frac{G M m}{r^2}\), which gives the orbital speed \(v = \sqrt{\frac{GM}{r}}\). Therefore, the orbital speed is inversely proportional to the square root of the orbital radius: \(v \propto \frac{1}{\sqrt{r}}\). If the radius is increased to \(4R\), the new speed becomes \(v_{new} = \sqrt{\frac{GM}{4R}} = \frac{1}{2} \sqrt{\frac{GM}{R}} = 0.50 v\).

Award 1 mark for relating orbital speed to orbital radius and calculating the correct speed factor of 0.50.

The luminosity of a star is given by Stefan-Boltzmann's law: \(L = 4\pi R^2 \sigma T^4\). Comparing Star A and Star B, we have \(\frac{L_A}{L_B} = \left(\frac{R_A}{R_B}\right)^2 \left(\frac{T_A}{T_B}\right)^4\). Substituting the given values: \(324 = \left(\frac{R_A}{R_B}\right)^2 (3)^4 \implies 324 = \left(\frac{R_A}{R_B}\right)^2 \times 81 \implies \left(\frac{R_A}{R_B}\right)^2 = 4 \implies \frac{R_A}{R_B} = 2\).

Award 1 mark for the correct answer B. (Method: Use of Stefan-Boltzmann law ratio, solving for the radius ratio: 1 mark for accuracy).

The impulse is equal to the area under the force-time curve. Since the force acts in the opposite direction, the impulse is negative: \(J = -\frac{1}{2} \times \text{base} \times \text{height} = -\frac{1}{2} \times 1.0\text{ s} \times 12\text{ N} = -6.0\text{ N s}\). Using the impulse-momentum theorem: \(\Delta p = J \implies m(v_f - v_i) = J \implies 2.0 \times (v_f - 4.0) = -6.0 \implies v_f - 4.0 = -3.0 \implies v_f = 1.0\text{ m s}^{-1}\) in the original direction.

Award 1 mark for the correct answer B. (Method: Calculate area of triangle as magnitude of impulse = 6 N s, apply change in momentum formula, track direction correctly: 1 mark).

We know that \(0.125 = \frac{1}{8} = \left(\frac{1}{2}\right)^3\). This means that 3 half-lives have elapsed in 12 hours, giving a half-life of \(T_{1/2} = \frac{12}{3} = 4\text{ hours}\). After 16 hours, the number of half-lives elapsed is \(n = \frac{16}{4} = 4\). The remaining activity is \(A = A_0 \left(\frac{1}{2}\right)^4 = \frac{1}{16} A_0 = 0.0625 A_0 \approx 0.063 A_0\).

Award 1 mark for the correct answer B. (Method: Determine half-life as 4 hours, calculate fraction remaining after 4 half-lives as 1/16: 1 mark).

The induced electromotive force (emf) is given by Faraday's law: \(\varepsilon = -\frac{\Delta \Phi}{\Delta t}\). The rate of change of magnetic flux as the loop leaves the field is \(\frac{\Delta \Phi}{\Delta t} = B \frac{\Delta A}{\Delta t} = B s v\). Thus, \(\varepsilon = B s v\). The electrical power dissipated in the resistor is \(P = \frac{\varepsilon^2}{R} = \frac{(B s v)^2}{R} = \frac{B^2 s^2 v^2}{R}\).

Award 1 mark for the correct answer B. (Method: Express induced emf as Bsv, apply electrical power formula P = V^2 / R: 1 mark).

The total energy in simple harmonic motion is \(E = E_k + E_p\). Given that \(E_k = 3 E_p\), we have \(E = 3 E_p + E_p = 4 E_p\). The total energy is \(E = \frac{1}{2} k x_0^2\) and the potential energy at displacement \(x\) is \(E_p = \frac{1}{2} k x^2\). Substituting these: \(\frac{1}{2} k x_0^2 = 4 \left(\frac{1}{2} k x^2\right) \implies x_0^2 = 4 x^2 \implies x = \frac{x_0}{2}\).

Award 1 mark for the correct answer B. (Method: Relate total energy to potential energy, substitute SHM energy formulae and solve for displacement: 1 mark).

According to Gay-Lussac's law for a constant volume, \(\frac{P_1}{T_1} = \frac{P_2}{T_2}\). Temperatures must be in Kelvin. \(T_1 = 27 + 273 = 300\text{ K}\). Since \(P_2 = 2 P_1\), the final temperature is \(T_2 = 2 T_1 = 2 \times 300\text{ K} = 600\text{ K}\). Convert this back to Celsius: \(T_2 = 600 - 273 = 327^\circ\text{C}\).

Award 1 mark for the correct answer C. (Method: Convert temperature to Kelvin, apply pressure-temperature relationship, convert back to Celsius: 1 mark).

The total energy of a satellite in a circular orbit of radius \(r\) is \(E_{\text{orb}} = -\frac{GMm}{2r}\). For the satellite to just escape to infinity, its total energy must be at least zero (\(E_{\text{escape}} = 0\)). The additional energy required is \(\Delta E = E_{\text{escape}} - E_{\text{orb}} = 0 - \left(-\frac{GMm}{2r}\right) = \frac{GMm}{2r}\).

Award 1 mark for the correct answer A. (Method: State orbital total energy as negative half G M m / r, determine escape threshold as 0, calculate the difference: 1 mark).

The translational kinetic energy is \(E_{\text{trans}} = \frac{1}{2} M v^2\). Since it rolls without slipping, \(\omega = \frac{v}{R}\). The rotational kinetic energy is \(E_{\text{rot}} = \frac{1}{2} I \omega^2 = \frac{1}{2} \left(\frac{2}{5} M R^2\right) \left(\frac{v}{R}\right)^2 = \frac{1}{5} M v^2\). The total kinetic energy is \(E_{\text{total}} = E_{\text{trans}} + E_{\text{rot}} = \frac{1}{2} M v^2 + \frac{1}{5} M v^2 = \frac{7}{10} M v^2\). The fraction is \(\frac{E_{\text{rot}}}{E_{\text{total}}} = \frac{\frac{1}{5}}{\frac{7}{10}} = \frac{2}{7}\).

Award 1 mark for the correct answer B. (Method: Express translational and rotational kinetic energies, sum them to find total energy, compute fraction: 1 mark).

The lifetime \(T\) of a star is proportional to the fuel available (which is proportional to its mass \(M\)) divided by the rate at which it burns fuel (which is the luminosity \(L\)). Therefore:

\(T \propto \frac{M}{L}\)

Since \(L \propto M^{3.5}\), we have:

\(T \propto \frac{M}{M^{3.5}} = M^{-2.5}\)

The ratio of the lifetime of star X to star Y is:

\(\frac{T_X}{T_Y} = \left(\frac{M_X}{M_Y}\right)^{-2.5} = \left(\frac{2}{6}\right)^{-2.5} = 3^{2.5} = 9\sqrt{3} \approx 15.6\)

[1 mark] for establishing that \(T \propto M^{-2.5}\) and calculating the ratio to be 15.6.

The block begins to slide when the applied horizontal force \(F\) overcomes the maximum static friction force:

\(F = f_{s,\text{max}} = \mu_s m g = 0.40 m g\)

Once the block is in motion, the friction force opposing the motion becomes kinetic friction:

\(f_k = \mu_k m g = 0.30 m g\)

Since the applied force remains constant at \(F = 0.40 m g\), the net horizontal force acting on the block immediately after it starts to slide is:

\(F_{\text{net}} = F - f_k = 0.40 m g - 0.30 m g = 0.10 m g\)

Using Newton's second law, the acceleration \(a\) is:

\(a = \frac{F_{\text{net}}}{m} = 0.10 g = 0.10 \times 10\text{ m s}^{-2} = 1.0\text{ m s}^{-2}\)

[1 mark] for identifying the applied force as \(\mu_s m g\) and the kinetic friction as \(\mu_k m g\), then calculating the net acceleration of \(1.0\text{ m s}^{-2}\).

The activity of isotope A as a function of time \(t\) is:

\(R_A(t) = A_0 \left(\frac{1}{2}\right)^{\frac{t}{6.0}}\)

The activity of isotope B is:

\(R_B(t) = 2 A_0 \left(\frac{1}{2}\right)^{\frac{t}{3.0}}\)

Setting the two activities equal:

\(A_0 \left(\frac{1}{2}\right)^{\frac{t}{6.0}} = 2 A_0 \left(\frac{1}{2}\right)^{\frac{t}{3.0}}\)

\(\left(\frac{1}{2}\right)^{\frac{t}{6.0}} = 2 \left(\frac{1}{2}\right)^{\frac{t}{3.0}}\)

Let \(x = \frac{t}{6.0}\), so that \(\frac{t}{3.0} = 2x\):

\(\left(\frac{1}{2}\right)^x = 2 \left(\frac{1}{2}\right)^{2x}\)

\(\frac{\left(1/2\right)^x}{\left(1/2\right)^{2x}} = 2 \implies \left(\frac{1}{2}\right)^{-x} = 2 \implies 2^x = 2^1\)

Thus, \(x = 1\), which gives:

\(\frac{t}{6.0} = 1 \implies t = 6.0\text{ hours}\)

[1 mark] for equating the two activity functions or using inspection to find the time when both activities are equal to \(A_0 / 2\).

As the loop enters the magnetic field, the magnetic flux through the loop increases. According to Faraday's law of induction, this induces an electromotive force (emf) in the loop:

\(\varepsilon = B L v\)

This induced emf causes a current \(I\) to flow through the loop of resistance \(R\):

\(I = \frac{\varepsilon}{R} = \frac{B L v}{R}\)

According to Lenz's law, the induced current and the magnetic field will produce a magnetic force that opposes the motion of the loop (i.e., directed upward). Only the bottom side of the loop is within the magnetic field, experiencing an upward magnetic force:

\(F = I L B = \left(\frac{B L v}{R}\right) L B = \frac{B^2 L^2 v}{R}\)

Thus, the force has magnitude \(\frac{B^2 L^2 v}{R}\) and is directed upward.

[1 mark] for correctly identifying the induced emf and current, and applying Lenz's law to find that the force is upward with a magnitude of \(\frac{B^2 L^2 v}{R}\).

The total energy \(E_T\) in simple harmonic motion is given by:

\(E_T = \frac{1}{2} k A^2\)

The potential energy \(E_p\) at displacement \(x\) is:

\(E_p = \frac{1}{2} k x^2\)

The kinetic energy \(E_k\) is the difference between total energy and potential energy:

\(E_k = E_T - E_p = \frac{1}{2} k (A^2 - x^2)\)

We are given that \(E_k = 3 E_p\):

\(\frac{1}{2} k (A^2 - x^2) = 3 \left(\frac{1}{2} k x^2\right)\)

Simplifying this equation:

\(A^2 - x^2 = 3 x^2 \implies A^2 = 4 x^2\)

Solving for \(x\):

\(x = \pm \frac{A}{2}\)

[1 mark] for equating \(E_k = 3 E_p\) in terms of displacement and amplitude, and solving to find \(x = \pm A/2\).

Since there is no air resistance, the horizontal component of velocity remains constant:

\(v_x = v\)

When the projectile lands, its velocity vector makes an angle of \(45^\circ\) with the horizontal, which means the magnitudes of the horizontal and vertical velocity components are equal:

\(|v_y| = |v_x| = v\)

For a vertical fall from rest over a height \(h\), the vertical speed \(v_y\) upon impact is related to \(h\) by:

\(v_y^2 = 2 g h\)

Substituting \(v_y = v\):

\(v^2 = 2 g h \implies h = \frac{v^2}{2g}\)

[1 mark] for equating vertical and horizontal speeds due to the landing angle, and using kinematics to determine the correct height expression.

Let the initial state parameters be \(P_1 = P\), \(V_1 = V\), and \(T_1 = T\).

Step 1: Isothermal halving of volume.
During an isothermal process, \(P V = \text{constant}\). Since the volume is halved (\(V_2 = \frac{V}{2}\)), the pressure must double:

\(P_2 = 2 P\)

Step 2: Isochoric doubling of temperature.
During an isochoric (constant volume) process, \(\frac{P}{T} = \text{constant}\). Since the absolute temperature is doubled (\(T_3 = 2 T_2\)), the pressure must double:

\(P_3 = 2 P_2 = 2 (2 P) = 4 P\)

[1 mark] for applying the gas laws to find the intermediate pressure as \(2P\) and the final pressure as \(4P\).

The total energy \(E\) of a satellite of mass \(m\) in a circular orbit of radius \(r\) around a planet of mass \(M\) is given by:

\(E = -\frac{G M m}{2r}\)

The initial total energy of the satellite is:

\(E_i = -\frac{G M m}{2r}\)

The final total energy of the satellite in the new orbit of radius \(2r\) is:

\(E_f = -\frac{G M m}{2(2r)} = -\frac{G M m}{4r}\)

The energy required to be supplied is the difference between the final and initial energies:

\(\Delta E = E_f - E_i = -\frac{G M m}{4r} - \left(-\frac{G M m}{2r}\right) = \frac{G M m}{4r}\)

[1 mark] for utilizing the total orbital energy formula and calculating the correct positive difference \(\Delta E = \frac{G M m}{4r}\).

The Stefan-Boltzmann law states that the luminosity of a star is given by \(L = 4\pi R^2 \sigma T^4\). Comparing the two stars: \(\frac{L_Y}{L_X} = \left(\frac{R_Y}{R_X}\right)^2 \left(\frac{T_Y}{T_X}\right)^4 = 3^2 \times 2^4 = 9 \times 16 = 144\).

Award 1 mark for the correct answer C. No partial marks are awarded for multiple-choice questions.

Using the impulse-momentum theorem: \(\Delta p = m\Delta v = m(v_f - v_i)\). If the initial direction is positive, \(v_i = 20\text{ m s}^{-1}\) and \(v_f = -15\text{ m s}^{-1}\). Therefore, \(\Delta p = 0.15 \times (-15 - 20) = -5.25\text{ N s}\). The magnitude of the change in momentum is \(5.25\text{ N s}\). The average force is \(F = \frac{\Delta p}{\Delta t} = \frac{5.25}{0.050} = 105\text{ N}\).

Award 1 mark for the correct answer C.

After three half-lives, the remaining fraction of the radioactive nuclei is \(N = N_0 \left(\frac{1}{2}\right)^3 = \frac{N_0}{8}\). The number of nuclei that have decayed is \(N_{\text{decayed}} = N_0 - \frac{N_0}{8} = \frac{7N_0}{8}\). The ratio of decayed to remaining nuclei is \(\frac{7N_0/8}{N_0/8} = 7\).

Award 1 mark for the correct answer C.

The magnetic flux linkage through the coil at any time \(t\) is \(N\Phi = N B A \cos(\omega t)\). According to Faraday's law of induction, the induced electromotive force (emf) is \(\varepsilon = -\frac{d(N\Phi)}{dt} = N B A \omega \sin(\omega t)\). The maximum value of this function occurs when \(\sin(\omega t) = 1\), giving \(\varepsilon_{\max} = N B A \omega\).

Award 1 mark for the correct answer B.

The total energy in SHM is constant and given by \(E_T = \frac{1}{2} k x_0^2\). The potential energy at displacement \(x\) is \(E_P = \frac{1}{2} k x^2\). The kinetic energy is \(E_K = E_T - E_P = \frac{1}{2} k (x_0^2 - x^2)\). Setting \(E_K = 3 E_P\) gives \(\frac{1}{2} k (x_0^2 - x^2) = 3 \left(\frac{1}{2} k x^2\right)\), which simplifies to \(x_0^2 - x^2 = 3 x^2 \implies x_0^2 = 4 x^2 \implies x = \pm \frac{x_0}{2}\).

Award 1 mark for the correct answer B.

During projectile motion, the acceleration is constant and equals \(g\) downwards at all times, including at the maximum height. At the highest point, the vertical component of the velocity is zero, and only the horizontal component remains, which is constant throughout the motion and equal to \(u \cos\theta\).

Award 1 mark for the correct answer C.

The root-mean-square speed of an ideal gas molecule is given by \(v_{\text{rms}} = \sqrt{\frac{3 k_B T}{m}}\), where \(T\) is the absolute temperature. Therefore, \(v_{\text{rms}} \propto \sqrt{T}\). When the absolute temperature is tripled, the rms speed increases by a factor of \(\sqrt{3}\).

Award 1 mark for the correct answer B.

By conservation of angular momentum, \(I_1 \omega_1 = I_{\text{total}} \omega_2\). The initial moment of inertia is \(I_1 = I\). Since the cylinders are identical, the final moment of inertia is \(I_{\text{total}} = 2I\). Therefore, \(I \omega_1 = 2I \omega_2 \implies \omega_2 = \frac{1}{2} \omega_1\), so the ratio \(\frac{\omega_2}{\omega_1} = 0.50\).

Award 1 mark for the correct answer B.

(a) The period \(T\) is the total time divided by the number of oscillations:\
\(T = \frac{35.8\text{ s}}{20} = 1.79\text{ s}\).\
\
The absolute uncertainty in the period is:\
\(\Delta T = \frac{\Delta t}{20} = \frac{0.2\text{ s}}{20} = 0.01\text{ s}\).\
\
So, \(T = 1.79 \pm 0.01\text{ s}\).\
\
(b) Rearranging \(T = 2\pi\sqrt{\frac{L}{g}}\):\
\(T^2 = \frac{4\pi^2 L}{g} \implies g = \frac{4\pi^2 L}{T^2}\).\
\
Substituting the values:\
\(g = \frac{4\pi^2 \times 0.800}{(1.79)^2} \approx 9.86\text{ m s}^{-2}\).\
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(c) The fractional uncertainty in \(g\) is given by:\
\(\frac{\Delta g}{g} = \frac{\Delta L}{L} + 2\frac{\Delta T}{T}\).\
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Calculating each term:\
\(\frac{\Delta L}{L} = \frac{0.005}{0.800} = 0.00625\) (or \(0.625\\%\))\
\
\(\frac{\Delta T}{T} = \frac{0.01}{1.79} \approx 0.00559\) (or \(0.559\\%\))\
\
\(\frac{\Delta g}{g} = 0.00625 + 2(0.00559) = 0.01743\).\
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Percentage uncertainty in \(g = 0.01743 \times 100\\% \approx 1.7\\%\) (or \(1.74\\%\)).

Award marks as follows:\
(a) [2 marks]\
- \(T = 1.79\text{ s}\) [1]\
- \(\Delta T = 0.01\text{ s}\) [1]\
\
(b) [2 marks]\
- Correct rearrangement to show \(g = \frac{4\pi^2 L}{T^2}\) [1]\
- \(g = 9.86\text{ m s}^{-2}\) (accept 9.9 or 9.85) [1]\
\
(c) [2.66 marks]\
- Correct fractional uncertainty equation: \(\frac{\Delta g}{g} = \frac{\Delta L}{L} + 2\frac{\Delta T}{T}\) [1]\
- Correct substitution of values [1]\
- Final answer of \(1.7\\%\) (accept \(1.74\\%\) or \(1.8\\%\) if using rounded intermediate values) [0.66]

(a) Since the actual volume is \(V_{\text{total}} = V_{\text{read}} + V_{\text{tubing}}\\, the recorded volume is systematically smaller than the actual volume (\)V_{\\text{read}} < V_{\\text{total}}\)). Therefore, \(\frac{1}{V_{\text{read}}}\) will be systematically larger than \(\frac{1}{V_{\text{total}}}\). This shift means that a graph of \(P\) vs \(\frac{1}{V_{\text{read}}}\) will not be a straight line through the origin, but will instead curve or display a negative horizontal intercept.\
\
(b) The total volume is:\
\(V_{\text{total}} = 20.0\text{ cm}^3 + 2.0\text{ cm}^3 = 22.0\text{ cm}^3\).\
\
The absolute uncertainty in \(V_{\text{total}}\) remains the same as that of \(V_{\text{read}}\) since \(V_{\text{tubing}}\) is treated as a constant with no uncertainty: \(\Delta V_{\text{total}} = 0.5\text{ cm}^3\).\
\
The percentage uncertainty is:\
\(\frac{0.5}{22.0} \times 100\\% \approx 2.27\\% \approx 2.3\\%\).\
\
(c) Repeating and averaging measurements reduces the effect of random errors because random variations above and below the true value tend to cancel out. However, it has no effect on systematic errors because the systematic offset (neglecting the tubing volume) remains constant in every trial.

Award marks as follows:\
(a) [2 marks]\
- Identifies that \(V_{\text{read}}\) is systematically smaller than the actual volume (or \(\frac{1}{V_{\text{read}}}\) is systematically larger) [1]\
- Concludes that the graph of \(P\) vs \(\frac{1}{V_{\text{read}}}\) will not be a straight line through the origin / will show systematic deviation or curve [1]\
\
(b) [2 marks]\
- \(V_{\text{total}} = 22.0\text{ cm}^3\) [1]\
- Percentage uncertainty \(= 2.3\\%\) (accept \(2.27\\%\)) [1]\
\
(c) [2.66 marks]\
- States that repeating and averaging reduces random error [1]\
- Explains that random variations cancel out/average out [1]\
- States that systematic error is unaffected because it is a constant bias in all trials [0.66]

(a) The resistance of a wire is given by \(R = \rho \frac{L}{A}\\, where \)A\) is the cross-sectional area. Rearranging gives:\
\(\frac{R}{L} = \frac{\rho}{A}\).\
\
Since the gradient of the graph of \(R\) against \(L\) is \(m = \frac{R}{L}\\, we have:\n\)m = \\frac{\\rho}{A} \\implies \\rho = m A\).\
\
Since the cross-section is circular, \(A = \frac{\pi d^2}{4}\). Substituting this yields:\
\(\rho = m \frac{\pi d^2}{4}\).\
\
(b) Convert the diameter to meters: \(d = 0.38 \times 10^{-3}\text{ m}\).\
\(\rho = 9.20 \times \frac{\pi (0.38 \times 10^{-3})^2}{4} = 9.20 \times 1.1341 \times 10^{-7} \approx 1.043 \times 10^{-6}\text{ \Omega\text{ m}}\).\
\
(c) The fractional uncertainty in \(\rho\) is given by:\
\(\frac{\Delta\rho}{\rho} = \frac{\Delta m}{m} + 2\frac{\Delta d}{d}\).\
\
Substituting the values:\
\(\frac{\Delta\rho}{\rho} = \frac{0.15}{9.20} + 2 \times \frac{0.02}{0.38} = 0.0163 + 0.1053 = 0.1216\).\
\
Therefore, the absolute uncertainty is:\
\(\Delta\rho = 0.1216 \times 1.043 \times 10^{-6}\text{ \Omega\text{ m}} \approx 0.127 \times 10^{-6}\text{ \Omega\text{ m}}\).\
\
Using two significant figures for the uncertainty (matching the input data), this is \(0.13 \times 10^{-6}\text{ \Omega\text{ m}}\).

Award marks as follows:\
(a) [2 marks]\
- Relates resistance formula \(R = \rho \frac{L}{A}\) to gradient \(m = \frac{R}{L}\) [1]\
- Correctly substitutes circular area \(A = \frac{\pi d^2}{4}\) to obtain the given expression [1]\
\
(b) [2 marks]\
- Correct substitution of values with diameter converted to meters [1]\
- Correct final value of \(1.04 \times 10^{-6}\text{ \Omega\text{ m}}\) (accept \(1.0 \times 10^{-6}\) or \(1.043 \times 10^{-6}\)) [1]\
\
(c) [2.66 marks]\
- Correct expression for fractional uncertainty (with a factor of 2 for diameter) [1]\
- Calculates fractional uncertainty as \(12\\%\) or \(12.2\\%\) (or \(0.122\)) [1]\
- Correct absolute uncertainty of \(0.13 \times 10^{-6}\text{ \Omega\text{ m}}\) (accept \(0.12 \times 10^{-6}\) to \(0.13 \times 10^{-6}\)) [0.66]

(a) High temperatures are needed to provide the nuclei with sufficient kinetic energy to overcome the electrostatic coulomb repulsion between the positively charged protons. High pressures are needed to ensure a high density of nuclei, which increases the probability of collision.

(b)
- Initial mass: \(2 \times 3.016029\text{ u} = 6.032058\text{ u}\)
- Final mass: \(4.002603\text{ u} + (2 \times 1.007276\text{ u}) = 6.017155\text{ u}\)
- Mass defect \(\Delta m = 6.032058\text{ u} - 6.017155\text{ u} = 0.014903\text{ u}\)
- Energy released: \(0.014903\text{ u} \times 931.5\text{ MeV u}^{-1} = 13.882\text{ MeV} \approx 13.9\text{ MeV}\)

(c)
- Energy per reaction: \(26.7\text{ MeV} = 26.7 \times 1.60 \times 10^{-13}\text{ J} = 4.272 \times 10^{-12}\text{ J}\)
- Number of reactions per second: \(N = \frac{3.8 \times 10^{26}\text{ W}}{4.272 \times 10^{-12}\text{ J}} = 8.895 \times 10^{37}\text{ s}^{-1}\)
- Since 4 protons are consumed per reaction, the number of hydrogen nuclei used per second is \(4 \times 8.895 \times 10^{37} = 3.558 \times 10^{38}\text{ s}^{-1}\)
- Total mass of hydrogen converted per second: \(3.558 \times 10^{38} \times 1.67 \times 10^{-27}\text{ kg} = 5.94 \times 10^{11}\text{ kg s}^{-1} \approx 5.9 \times 10^{11}\text{ kg s}^{-1}\)

(d) After the main sequence, hydrogen fusion in the core ceases and the star expands to become a red giant as helium fusion begins. Once helium in the core is depleted, the outer layers are ejected as a planetary nebula. The remaining hot core collapses under gravity to form a white dwarf. This collapse is halted by electron degeneracy pressure, which arises because the Pauli exclusion principle prevents electrons from occupying the same quantum states.

(a)
- [1M] for mentioning that high kinetic energy/temperature is needed to overcome electrostatic/Coulomb repulsion.
- [1M] for mentioning that high pressure/density ensures high frequency of collisions.

(b)
- [1M] for calculating mass of reactants and products correctly.
- [1M] for finding the correct mass defect (\(0.014903\text{ u}\)).
- [1M] for the final energy value with correct unit (\(13.9\text{ MeV}\)).

(c)
- [1M] for converting the reaction energy from \(\text{MeV}\) to Joules (\(4.27 \times 10^{-12}\text{ J}\)).
- [1M] for calculating the rate of reactions (\(8.9 \times 10^{37}\text{ s}^{-1}\)) and accounting for the factor of 4 hydrogen nuclei.
- [1M] for obtaining the correct final mass converted per second (accept \(5.9 \times 10^{11}\text{ kg s}^{-1}\) to \(6.0 \times 10^{11}\text{ kg s}^{-1}\)).

(d)
- [1M] for detailing the red giant and planetary nebula phases.
- [1M] for stating that the remaining core becomes a white dwarf.
- [1.25M] for explaining that electron degeneracy pressure balances the gravitational collapse.

(a) The law of conservation of momentum states that the total momentum of a closed/isolated system remains constant. According to Newton's third law, the force exerted by object A on B is equal and opposite to the force exerted by B on A (\(F_{AB} = -F_{BA}\)). Since the time of contact \(\Delta t\) is the same for both, the impulses are equal and opposite (\(F_{AB}\Delta t = -F_{BA}\Delta t\)). This means the change in momentum of A is equal and opposite to the change in momentum of B, keeping the total momentum of the system constant.

(b)
(i) Conservation of linear momentum:
\(m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2\)
\((0.50 \times 2.4) + 0 = (0.50 \times 0.60) + (0.30 \times v_2)\)
\(1.20 = 0.30 + 0.30 v_2\)
\(0.90 = 0.30 v_2 \implies v_2 = 3.0\text{ m s}^{-1}\) (to the right)

(ii)
- Initial kinetic energy: \(E_{k,i} = \frac{1}{2}(0.50)(2.4)^2 = 1.44\text{ J}\)
- Final kinetic energy: \(E_{k,f} = \frac{1}{2}(0.50)(0.60)^2 + \frac{1}{2}(0.30)(3.0)^2 = 0.09 + 1.35 = 1.44\text{ J}\)
Since \(E_{k,i} = E_{k,f}\), the collision is perfectly elastic.

(c)
(i) Force on the \(0.30\text{ kg}\) glider:
\(F = \frac{\Delta p}{\Delta t} = \frac{m_2 v_2 - m_2 u_2}{\Delta t} = \frac{0.30 \times 3.0 - 0}{0.080} = 11.25\text{ N} \approx 11\text{ N}\)

(ii) By Newton's third law, action and reaction are equal and opposite. The force exerted by the \(0.30\text{ kg}\) glider on the \(0.50\text{ kg}\) glider must be equal in magnitude (\(11.25\text{ N}\)) but opposite in direction. Since the force on the second glider is to the right, the force on the first glider is directed to the left.

(a)
- [1M] for defining conservation of momentum under closed system conditions.
- [1M] for setting up the Newton's third law action-reaction force pair relationship.
- [1M] for linking equal and opposite impulses to conservation of total momentum.

(b)
- (i) [1M] for applying conservation of momentum equation, [1M] for \(3.0\text{ m s}^{-1}\).
- (ii) [1M] for calculating initial and final kinetic energies, [1M] for showing they are equal and concluding 'elastic'.

(c)
- (i) [1M] for applying the impulse-momentum equation, [1M] for \(11\text{ N}\) (or \(11.3\text{ N}\)).
- (ii) [1.25M] for stating the magnitude is equal and direction is to the left, and referencing Newton's third law.

(a)
(i) Activity equation: \(A = A_0 e^{-\lambda t}\)
\(1.0 \times 10^5 = 8.0 \times 10^5 e^{-\lambda (15.0 \times 3600)}\)
\(\ln(1/8) = -54000 \lambda \implies -2.0794 = -54000 \lambda\)
\(\lambda = 3.851 \times 10^{-5}\text{ s}^{-1} \approx 3.85 \times 10^{-5}\text{ s}^{-1}\)

(ii) Half-life: \(T_{1/2} = \frac{\ln 2}{\lambda} = \frac{0.69315}{3.851 \times 10^{-5}} \approx 18000\text{ s} = 5.0\text{ hours}\)
(Alternatively: \(8.0 \rightarrow 4.0 \rightarrow 2.0 \rightarrow 1.0\) represents exactly 3 half-lives. Thus \(3 T_{1/2} = 15.0\text{ hours} \implies T_{1/2} = 5.0\text{ hours}\)).

(b) Initial activity: \(A_0 = \lambda N_0\)
\(8.0 \times 10^5 = (3.851 \times 10^{-5}) \times N_0\)
\(N_0 = \frac{8.0 \times 10^5}{3.851 \times 10^{-5}} = 2.077 \times 10^{10} \approx 2.08 \times 10^{10}\) nuclei.

(c)
(i) In beta-minus decay, a neutron converts into a proton, an electron, and an electron antineutrino (\(\text{n} \rightarrow \text{p} + \text{e}^- + \bar{
u}_e\)). Since it is a three-body decay, the total decay energy (Q-value) is shared continuously between the electron and the antineutrino in varying proportions, leading to a continuous spectrum of kinetic energies for the beta particle.

(ii) Alpha particles are highly ionizing but have extremely low penetrating power and are stopped by a thin sheet of paper or a few centimeters of air. Beta particles have higher penetrating power and can easily pass through paper, but are fully stopped by a few millimeters of aluminum or a thin sheet of lead. Placing paper in front of the source blocks alpha but not beta; placing a lead sheet blocks both.

(a)
- (i) [1M] for setting up the exponential decay equation correctly with conversion of hours to seconds.
- [1M] for finding \(\lambda \approx 3.85 \times 10^{-5}\text{ s}^{-1}\).
- (ii) [1M] for finding the half-life of \(5.0\text{ hours}\) (or \(1.8 \times 10^4\text{ s}\)).

(b)
- [2M] for recognizing that \(A_0 = \lambda N_0\).
- [2M] for calculating the correct number of nuclei \(2.08 \times 10^{10}\) (accept answers consistent with rounding of \(\lambda\)).

(c)
- (i) [1M] for mentioning the simultaneous emission of an antineutrino.
- [1.25M] for explaining that the decay energy is shared continuously between the electron and the antineutrino.
- (ii) [1M] for stating that alpha is stopped by paper/skin whereas beta passes through.
- [1M] for stating that a lead sheet (or aluminum sheet) is needed to block beta radiation.

(a) Faraday's law states that the induced electromotive force (emf) is directly proportional to the rate of change of magnetic flux linkage. Lenz's law states that the direction of the induced current is such that it opposes the change in magnetic flux that produces it. If Lenz's law were not true, the induced field would assist the change, creating an ever-increasing magnetic field and electrical energy without the input of external work, which violates the law of conservation of energy.

(b)
(i) Magnetic flux linkage: \(\Phi = N B A\)
Change in flux linkage: \(\Delta \Phi = N \Delta B A = 150 \times (0.10 - 0.60) \times 2.5 \times 10^{-3} = -0.1875\text{ Wb}\)
Magnitude of induced emf: \(\epsilon = \left| -\frac{\Delta \Phi}{\Delta t} \right| = \frac{0.1875}{0.15} = 1.25\text{ V}\)

(ii) Average current: \(I = \frac{\epsilon}{R} = \frac{1.25\text{ V}}{12\,\Omega} \approx 0.104\text{ A}\)

(c) If the time interval \(\Delta t\) is doubled from \(0.15\text{ s}\) to \(0.30\text{ s}\), the rate of change of magnetic flux linkage is halved because the total change in flux remains constant (\(-0.1875\text{ Wb}\)). By Faraday's law, the induced emf is halved to \(\epsilon = 0.625\text{ V}\). Consequently, by Ohm's law, the induced current is also halved to \(I = \frac{0.625}{12} \approx 0.052\text{ A}\).

(a)
- [1M] for stating Faraday's law correctly.
- [1M] for stating Lenz's law correctly.
- [1M] for explaining the link between Lenz's law and energy conservation (preventing infinite energy loops).

(b)
- (i) [1.25M] for correctly computing the change in flux linkage \(\Delta \Phi = -0.1875\text{ Wb}\).
- [1M] for calculating the emf as \(1.25\text{ V}\).
- (ii) [1M] for using Ohm's law, [1M] for the final current of \(0.104\text{ A}\).

(c)
- [1M] for noting that doubling the time halves the rate of change of flux.
- [1M] for stating that the induced emf is halved.
- [1M] for stating that the induced current is halved.
- [1M] for showing calculations supporting the new values (\(0.625\text{ V}\) and \(0.052\text{ A}\)).

(a)
(i) Simple harmonic motion is defined as periodic motion where the acceleration of the object is directly proportional to its displacement from equilibrium and is always directed opposite to the displacement (\(a \propto -x\)).

(ii) Angular frequency: \(\omega = \frac{2\pi}{T} = \frac{2\pi}{1.2} = 5.236\text{ rad s}^{-1}\)
Maximum acceleration: \(a_{\text{max}} = \omega^2 x_0 = (5.236)^2 \times 0.080 = 27.415 \times 0.080 = 2.193\text{ m s}^{-2} \approx 2.2\text{ m s}^{-2}\)

(b) Total mechanical energy: \(E_T = \frac{1}{2} m \omega^2 x_0^2 = \frac{1}{2} (0.25) \times (5.236)^2 \times (0.080)^2 = 0.125 \times 27.415 \times 0.0064 = 0.0219\text{ J} \approx 0.022\text{ J}\)

(c)
(i) Velocity at displacement \(x\):
\(v = \pm \omega \sqrt{x_0^2 - x^2}\)
\(v = \pm 5.236 \sqrt{0.080^2 - 0.040^2} = \pm 5.236 \sqrt{0.0064 - 0.0016} = \pm 5.236 \sqrt{0.0048} = \pm 5.236 \times 0.06928 \approx \pm 0.36\text{ m s}^{-1}\)

(ii) The graph of kinetic energy versus displacement is a downward-opening parabola. At the equilibrium position (\(x = 0\)), the kinetic energy is maximum and equal to the total mechanical energy (\(0.022\text{ J}\)). At the maximum displacements (\(x = \pm 8.0\text{ cm}\)), the kinetic energy is zero.

(a)
- (i) [1.5M] for defining acceleration proportional to displacement and opposite in direction (\(a = -\omega^2 x\)).
- (ii) [0.5M] for calculating \(\omega = 5.24\text{ rad s}^{-1}\), [1M] for showing substitution to get \(2.19\text{ m s}^{-2}\).

(b)
- [2M] for stating the formula \(E_T = \frac{1}{2} m \omega^2 x_0^2\).
- [2M] for calculating the value of \(0.022\text{ J}\) (or \(0.0219\text{ J}\)) with correct units.

(c)
- (i) [1M] for using \(v = \pm \omega \sqrt{x_0^2 - x^2}\).
- [1.25M] for the correct answer of \(\pm 0.36\text{ m s}^{-1}\).
- (ii) [1M] for describing the shape as a downward-opening parabola.
- [1M] for correctly identifying maximum energy at \(x=0\) and zero energy at \(x=\pm 8.0\text{ cm}\).

(a)
(i) Vertical motion: \(s_y = u_y t + \frac{1}{2} g t^2\)
With \(s_y = 45\text{ m}\), \(u_y = 0\text{ m s}^{-1}\), and \(g = 9.81\text{ m s}^{-2}\):
\(45 = 0 + \frac{1}{2} (9.81) t^2 \implies t^2 = \frac{90}{9.81} \approx 9.174\)
\(t = 3.03\text{ s}\)

(ii) Horizontal distance: \(s_x = v_x t\)
\(s_x = 15\text{ m s}^{-1} \times 3.029\text{ s} = 45.44\text{ m} \approx 45\text{ m}\) (to 2 s.f., or \(45.4\text{ m}\))

(b) Just before impact:
- Horizontal component of velocity: \(v_x = 15\text{ m s}^{-1}\)
- Vertical component of velocity: \(v_y = u_y + g t = 0 + (9.81 \times 3.029) = 29.71\text{ m s}^{-1}\)
- Magnitude of velocity: \(v = \sqrt{v_x^2 + v_y^2} = \sqrt{15^2 + 29.71^2} = \sqrt{225 + 882.7} = \sqrt{1107.7} = 33.28\text{ m s}^{-1} \approx 33\text{ m s}^{-1}\)
- Direction: \(\theta = \tan^{-1}\left(\frac{v_y}{v_x}\right) = \tan^{-1}\left(\frac{29.71}{15}\right) = \tan^{-1}(1.98) \approx 63.2^\circ\) below the horizontal.

(c) If air resistance is not negligible:
- A resistive drag force acts on the stone opposing its motion.
- The horizontal acceleration is no longer zero (it is negative), which causes the horizontal velocity component to decrease throughout the flight.
- The vertical acceleration is less than \(g\) due to upward drag, resulting in a lower final vertical velocity.
- Consequently, the projectile range is shorter, the trajectory becomes asymmetric (steeper near the end), and the final impact angle is steeper (larger angle below the horizontal).

(a)
- (i) [1M] for setting up the vertical kinematic equation correctly, [1M] for \(3.03\text{ s}\).
- (ii) [1M] for setting up the horizontal kinematic equation, [1M] for \(45\text{ m}\) (or \(45.4\text{ m}\)).

(b)
- [1M] for state horizontal velocity component is constant (\(15\text{ m s}^{-1}\)).
- [1M] for calculating final vertical velocity component (\(29.7\text{ m s}^{-1}\)).
- [1.25M] for calculating the magnitude of velocity (\(33\text{ m s}^{-1}\) or \(33.3\text{ m s}^{-1}\)).
- [1M] for calculating the direction as \(63^\circ\) (or \(63.2^\circ\)) below the horizontal.

(c)
- [1M] for mentioning that horizontal drag reduces horizontal velocity component.
- [1M] for mentioning that vertical drag reduces vertical acceleration and terminal vertical velocity.
- [1M] for concluding that the overall range is reduced and the landing trajectory is steeper.

(a) Basic assumptions of the kinetic theory of gases:
1. The molecules are identical, hard, negligible-volume spheres.
2. The collisions between molecules and with the walls of the container are perfectly elastic.
3. There are no intermolecular forces except during collisions.
4. The molecules are in continuous, random motion.
5. The duration of collisions is negligible compared to the time between collisions.

(b)
(i) Temperature: \(T = 27 + 273.15 = 300.15\text{ K} \approx 300\text{ K}\)
Using the ideal gas equation: \(P V = N k_B T\)
\(P = \frac{N k_B T}{V} = \frac{1.8 \times 10^{24} \times 1.38 \times 10^{-23} \times 300}{0.040}\)
\(P = \frac{7.452}{0.040} = 1.863 \times 10^5\text{ Pa} \approx 1.9 \times 10^5\text{ Pa}\)

(ii) Average kinetic energy of a molecule:
\(E_k = \frac{3}{2} k_B T = 1.5 \times (1.38 \times 10^{-23}) \times 300 = 6.21 \times 10^{-21}\text{ J}\)

(c) When the volume of the container is halved at constant temperature:
- Since the temperature is constant, the average speed and kinetic energy of the gas molecules remain unchanged.
- The number of molecules per unit volume (density) is doubled.
- Because the density is doubled and molecular speeds are constant, molecules collide with the container walls twice as frequently.
- The average change of momentum per collision with the wall remains the same because the molecular speed is constant.
- Therefore, the total rate of change of momentum (force) on the walls doubles, leading to a doubling of the pressure (Boyles's Law: \(P \propto 1/V\)), which increases pressure to \(3.73 \times 10^5\text{ Pa}\).

(a)
- [3M] for any three correct assumptions from the kinetic theory of gases.

(b)
- (i) [0.25M] for temperature conversion to \(300\text{ K}\).
- [1M] for selecting the formula \(P = \frac{N k_B T}{V}\).
- [1M] for correctly calculating \(1.86 \times 10^5\text{ Pa}\).
- (ii) [1M] for using the equation \(E_k = \frac{3}{2} k_B T\).
- [1M] for obtaining \(6.21 \times 10^{-21}\text{ J}\).

(c)
- [1M] for stating that pressure doubles.
- [1M] for explaining that constant temperature means average molecular speed is constant.
- [1M] for stating that halving the volume doubles the density/concentration of molecules.
- [1.25M] for linking this to double the frequency of collisions on the walls, resulting in twice the average force.

(a) Gravitational field strength is defined as the gravitational force per unit mass exerted on a small test mass placed at that point. Gravitational potential at a point is the work done per unit mass in bringing a small test mass from infinity to that point. Because the gravitational force is attractive, work is done by the field (or negative work is done by an external force) as the mass moves from infinity. Since the potential at infinity is defined as zero, all potential values closer to the mass must be negative.

(b)
(i) Equating gravitational force to centripetal force:
\(\frac{G M m}{r^2} = \frac{m v^2}{r} \implies v = \sqrt{\frac{G M}{r}}\)
\(v = \sqrt{\frac{6.67 \times 10^{-11} \times 6.0 \times 10^{24}}{8.0 \times 10^6}} = \sqrt{\frac{4.002 \times 10^{14}}{8.0 \times 10^6}} = \sqrt{5.0025 \times 10^7} = 7073\text{ m s}^{-1} \approx 7.1\text{ km s}^{-1}\)

(ii) Total mechanical energy \(E_T\) is the sum of kinetic and potential energies:
\(E_T = E_k + E_p = \frac{1}{2} m v^2 - \frac{G M m}{r} = -\frac{G M m}{2r}\)
\(E_T = -\frac{6.67 \times 10^{-11} \times 6.0 \times 10^{24} \times 1500}{2 \times 8.0 \times 10^6} = -3.75 \times 10^{10}\text{ J}\)

(c) The escape speed \(v_{\text{esc}}\) is reached when the kinetic energy from the surface is equal to the magnitude of the potential energy at the surface (total energy is zero):
\(\frac{1}{2} m v_{\text{esc}}^2 - \frac{G M m}{R} = 0 \implies v_{\text{esc}} = \sqrt{\frac{2 G M}{R}}\)
\(v_{\text{esc}} = \sqrt{\frac{2 \times 6.67 \times 10^{-11} \times 6.0 \times 10^{24}}{6.4 \times 10^6}} = \sqrt{\frac{8.004 \times 10^{14}}{6.4 \times 10^6}} = \sqrt{1.250625 \times 10^8} = 11183\text{ m s}^{-1} \approx 11.2\text{ km s}^{-1}\) (or \(11\text{ km s}^{-1}\) to 2 s.f.).

(a)
- [1M] for defining gravitational field strength (force per unit mass).
- [1M] for defining gravitational potential with reference to infinity as zero potential.
- [1M] for explaining that gravity is attractive, so work is done by the field, making potential negative.

(b)
- (i) [1.25M] for showing derivation from equating gravitational force to centripetal force.
- [1M] for substitute values and obtaining \(7.07\text{ km s}^{-1}\) (approx \(7.1\text{ km s}^{-1}\)).
- (ii) [1M] for stating formula \(E_T = -\frac{G M m}{2r}\) (or sum of kinetic and potential energies).
- [1M] for calculating value as \(-3.75 \times 10^{10}\text{ J}\) with correct negative sign.

(c)
- [2.25M] for deriving the escape velocity expression \(v = \sqrt{\frac{2GM}{R}}\).
- [1M] for substituting the numerical values correctly.
- [1M] for the final speed of \(11.2\text{ km s}^{-1}\) (or \(11\text{ km s}^{-1}\)).