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By conservation of energy, the decrease in kinetic energy equals the increase in gravitational potential energy. At height \(h\), the gain in gravitational potential energy is \(mgh = E_k - \frac{1}{3}E_k = \frac{2}{3}E_k\), which gives \(E_k = \frac{3}{2}mgh\). At the maximum height \(H\), the kinetic energy is zero, so the total gravitational potential energy gained is \(mgH = E_k\). Therefore, \(mgH = \frac{3}{2}mgh\), which simplifies to \(H = 1.5 h\).

1 mark for correct option A.

The extension of a wire is given by \(\Delta x = \frac{F L}{A E}\), where \(A = \frac{\pi d^2}{4}\). Thus, \(\Delta x\) is proportional to \(\frac{L}{d^2}\). For wire X, \(\Delta x_X \propto \frac{2L}{(0.5d)^2} = \frac{2L}{0.25d^2} = 8 \left(\frac{L}{d^2}\right)\). Therefore, the ratio of the extensions is 8.

1 mark for correct option C.

Using the grating equation \(d \sin\theta = n\lambda\). For the first scenario, \(d \sin\theta = 3\lambda\). For the second scenario, \(d \sin\theta' = 2(1.5\lambda) = 3\lambda\). Since both expressions equal \(3\lambda\), we have \(d \sin\theta' = d \sin\theta\), which means \(\theta' = \theta\).

1 mark for correct option A.

The total resistance of the network is \(R + \frac{R}{2} = 1.5R\). The total current in the circuit is \(I = \frac{V}{1.5R} = \frac{2V}{3R}\). The current is shared equally between the two parallel resistors, so the current through one parallel resistor is \(I_p = \frac{I}{2} = \frac{V}{3R}\). The power dissipated in this resistor is \(P = I_p^2 R = \left(\frac{V}{3R}\right)^2 R = \frac{V^2}{9R}\

1 mark for correct option A.

Momentum is a vector quantity. Let the initial velocity be \(+v\). After half a revolution, the velocity is in the opposite direction, \(-v\). The change in momentum is \(\Delta p = m(-v) - mv = -2mv\). The magnitude of this change in momentum is \(2mv\).

1 mark for correct option D.

The electric field strength is \(E = \frac{V}{d}\). The force on the electron is \(F = eE = \frac{eV}{d}\). The acceleration is \(a = \frac{F}{m} = \frac{eV}{md\). Using the equation of motion \(s = ut + \frac{1}{2}at^2\) with \(u = 0\) and \(s = d\), we get \(d = \frac{1}{2} \left(\frac{eV}{md}\right) t^2\). Solving for \(t\) gives \(t^2 = \frac{2md^2}{eV}\, which leads to \)t = d \sqrt{\frac{2m}{eV}}\).

1 mark for correct option B.

The mean kinetic energy of gas molecules is directly proportional to the absolute temperature, so \(\frac{1}{2}m\langle c^2 \rangle \propto T\). Thus, doubling the temperature doubles the mean square speed. From the ideal gas law \(pV = nRT\), at constant volume, pressure is directly proportional to temperature, so the pressure also doubles.

1 mark for correct option A.

After three half-lives, the fraction of nuclei remaining is \((1/2)^3 = 1/8\). Thus, the number of remaining nuclei is \(N = \frac{N_0}{8}\). The number of decayed nuclei is \(N_d = N_0 - N = \frac{7}{8}N_0\). The ratio of decayed to remaining nuclei is \(\frac{N_d}{N} = \frac{7/8}{1/8} = 7\).

1 mark for correct option C.

At the highest point of the projectile's trajectory, the vertical component of its velocity is zero. The horizontal component of velocity remains constant and is given by \(v_x = v \cos(60^\circ) = 0.5 v\), where \(v\) is the initial launch speed. Since kinetic energy is proportional to the square of the speed, the kinetic energy at the highest point is \(E_k = \frac{1}{2} m v_x^2 = \frac{1}{2} m (0.5 v)^2 = 0.25 \left(\frac{1}{2} m v^2\right) = 0.25 E\).

1 mark: Correctly identifies that the vertical component of velocity is zero at the highest point, applies the cosine factor to find the horizontal component, and calculates the remaining kinetic energy to be \(0.25 E\).

The Young modulus \(E\) is given by \(E = \frac{F L}{A \Delta x}\), which rearranges to \(\Delta x = \frac{F L}{A E}\). Since both wires are made of the same metal, \(E\) is the same. The tensile force \(F\) is also the same. The cross-sectional area \(A\) is proportional to the square of the diameter, \(d^2\). Thus, extension is proportional to length over diameter squared: \(\Delta x \propto \frac{L}{d^2}\). This gives \(\frac{\Delta x_X}{\Delta x_Y} = \frac{L_X}{L_Y} \times \left(\frac{d_Y}{d_X}\right)^2\). Substituting \(L_X = 2 L_Y\) and \(d_X = 0.5 d_Y\) yields \(\frac{\Delta x_X}{\Delta x_Y} = 2 \times 2^2 = 8\).

1 mark: Correctly relates the extension to the length and diameter squared, applies the given ratios, and determines the correct factor of 8.

According to Einstein's photoelectric equation, \(E_{max} = \frac{hc}{\lambda} - \Phi\), where \(\Phi\) is the work function of the metal. If the wavelength is halved to \(\lambda/2\), the energy of the incident photons is doubled. The new maximum kinetic energy \(E'_{max}\) is given by \(E'_{max} = \frac{hc}{\lambda/2} - \Phi = 2\left(\frac{hc}{\lambda}\right) - \Phi = 2(E_{max} + \Phi) - \Phi = 2 E_{max} + \Phi\). Since the work function \(\Phi\) is a positive quantity, \(E'_{max}\) must be greater than \(2 E_{max}\).

1 mark: Uses Einstein's photoelectric equation to show that halving the wavelength doubles the photon energy, and correctly deduces that the new maximum kinetic energy is greater than twice the original value due to the constant work function.

The total resistance of the circuit is \(R + r\). As the external resistance \(R\) is decreased, the total resistance decreases, so the current \(I = \frac{E}{R+r}\) increases. The terminal potential difference \(V\) is given by \(V = E - Ir\). Since the current \(I\) increases and the internal resistance \(r\) remains constant, the lost volts \(Ir\) increase, which causes the terminal potential difference \(V\) to decrease.

1 mark: Correctly identifies that a decrease in external resistance leads to an increased current and a decreased terminal potential difference.

At the highest point of the circular bridge, the forces acting on the car are its weight \(mg\) acting downwards and the normal contact force \(R\) acting upwards. The resultant force towards the center of the circular path provides the necessary centripetal force: \(mg - R = \frac{m v^2}{r}\). Rearranging this equation for the normal contact force gives \(R = mg - \frac{m v^2}{r} = m\left(g - \frac{v^2}{r}\right)\).

1 mark: Formulates the correct equation of motion for circular motion at the highest point and solves for the normal contact force.

The energy stored in a capacitor is given by \(E = \frac{1}{2} C V^2\). During discharge, the potential difference \(V\) across the capacitor after time \(t\) is given by \(V = V_0 e^{-t/RC\). At \(t = RC\), the potential difference is \(V = V_0 e^{-1}\). Substituting this into the energy formula, the remaining energy is \(E = \frac{1}{2} C \left(V_0 e^{-1}\right)^2 = \left(\frac{1}{2} C V_0^2\right) e^{-2} = E_0 e^{-2}\). Thus, the fraction of the initial energy remaining is \(e^{-2}\).

1 mark: Recognizes that energy is proportional to the square of voltage, relates the voltage decay factor of \(e^{-1}\) after one time constant to an energy decay factor of \(e^{-2}\).

In the annihilation process, the total mass-energy of the proton and the antiproton is converted into electromagnetic radiation. The total initial energy is \(E_{total} = m_p c^2 + m_{\bar{p}} c^2 = 2 m_p c^2\), since the proton and antiproton have identical rest mass. This total energy is shared equally between the two identical photons. Therefore, the energy of each photon is \(E_{photon} = m_p c^2\). Using the relation \(E_{photon} = hf\), we get \(hf = m_p c^2 \implies f = \frac{m_p c^2}{h}\).

1 mark: Equates the total energy of the two particles to the total energy of the two photons, and correctly solves for the frequency of a single photon.

The average kinetic energy of the molecules in an ideal gas is directly proportional to its absolute temperature: \(\frac{1}{2} m \langle c^2 \rangle = \frac{3}{2} k T\), where \(\langle c^2 \rangle\) is the mean square speed. Doubling the temperature \(T\) doubles the average kinetic energy, and thus doubles the mean square speed \(\langle c^2 \rangle\). The root-mean-square (r.m.s.) speed, which is \(\sqrt{\langle c^2 \rangle}\), increases by a factor of \(\sqrt{2}\). The average momentum of the molecules is zero due to isotropic random motion.

1 mark: Identifies that the mean square speed of the molecules is directly proportional to the absolute temperature and therefore doubles.

For the block to move up the slope at constant velocity, the pulling force \(F\) must balance the component of gravity down the slope and the kinetic frictional force. Therefore, \(F = mg\sin\theta + f_k\), where \(f_k = \mu R = \mu mg\cos\theta\). The pulling force is therefore \(F = mg(\sin\theta + \mu\cos\theta)\). The power delivered is \(P = Fv = mgv(\sin\theta + \mu\cos\theta)\).

1 mark for the correct option A. Method: Resolving forces parallel to the incline to find the tension, then multiplying by velocity to obtain power.

The current in the circuit is given by \(I = \frac{\mathcal{E}}{R+r}\). The terminal potential difference is \(V_T = \mathcal{E} - Ir = \mathcal{E}\left(1 - \frac{r}{R+r}\right)\). As \(R\) increases, \(V_T\) increases continuously towards \(\mathcal{E}\). The power dissipated in \(R\) is \(P = I^2 R = \frac{\mathcal{E}^2 R}{(R+r)^2}\). According to the maximum power transfer theorem, this power reaches a maximum when \(R = r\). Therefore, as \(R\) increases, \(P\) first increases to a maximum and then decreases.

1 mark for the correct option A. Method: Relate terminal potential difference to the potential divider equation and analyze the power transfer theorem.

The Young modulus is given by \(E = \frac{F/A}{\Delta x / l} = \frac{F l}{A \Delta x}\). Rearranging for extension gives \(\Delta x = \frac{F l}{A E} = \frac{4 F l}{\pi d^2 E}\). Thus, \(\Delta x \propto \frac{l}{d^2}\). The ratio of extensions is \(\frac{\Delta x_X}{\Delta x_Y} = \frac{l_X}{l_Y} \times \left(\frac{d_Y}{d_X}\right)^2 = \frac{1}{2} \times \left(\frac{1}{2}\right)^2 = \frac{1}{2} \times \frac{1}{4} = \frac{1}{8}\).

1 mark for the correct option A. Method: Express extension in terms of length and diameter using the Young modulus equation, then substitute the ratios.

From Einstein's photoelectric equation, \(E_{\text{max}} = hf - \Phi \implies hf = E_{\text{max}} + \Phi\). When the frequency is doubled, the new maximum kinetic energy is \(E'_\text{max} = h(2f) - \Phi = 2(hf) - \Phi = 2(E_{\text{max}} + \Phi) - \Phi = 2E_{\text{max}} + \Phi\).

1 mark for the correct option B. Method: Apply Einstein's photoelectric equation to both cases and substitute the expression for \(hf\).

The magnetic force provides the centripetal force: \(Bqv = \frac{mv^2}{r} \implies r = \frac{mv}{Bq}\). Therefore, \(r \propto \frac{v}{B}\). Comparing the two states: \(\frac{r_2}{r_1} = \frac{v_2}{v_1} \times \frac{B_1}{B_2} = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4} = 0.25\).

1 mark for the correct option D. Method: Derive the relation for the radius of a charged particle in a magnetic field and compute the ratio based on the changes.

The average kinetic energy of gas molecules is directly proportional to absolute temperature: \(\frac{1}{2}m\langle c^2 \rangle = \frac{3}{2}kT \implies \langle c^2 \rangle \propto T\). Convert temperatures to Kelvin: \(T_1 = 27 + 273.15 = 300.15\text{ K}\) and \(T_2 = 327 + 273.15 = 600.15\text{ K}\). The ratio of the mean square speeds is \(\frac{\langle c^2 \rangle_2}{\langle c^2 \rangle_1} = \frac{T_2}{T_1} = \frac{600.15}{300.15} \approx 2.0\).

1 mark for the correct option B. Method: Relate mean square speed to absolute temperature in Kelvin and calculate the ratio.

The gravitational field strength on a planet's surface is \(g = \frac{GM}{R^2}\). For the hypothetical planet, \(g_{\text{new}} = \frac{G(2M)}{(2R)^2} = \frac{2}{4} \frac{GM}{R^2} = 0.5 g_{\text{Earth}}\). The time period of a simple pendulum is \(T = 2\pi\sqrt{\frac{L}{g}}\). Hence, \(T_{\text{new}} = 2\pi\sqrt{\frac{L}{0.5g_{\text{Earth}}}} = \frac{1}{\sqrt{0.5}} T = \sqrt{2} T \approx 1.41 T\).

1 mark for the correct option B. Method: Determine the ratio of the gravitational field strength of the planet to Earth, then find the effect on the time period of the pendulum.

Let's check the conservation laws for \(n \to p + e^-\): 1) Charge: \(0 \to (+1) + (-1) = 0\) (conserved). 2) Baryon number: \(1 \to 1 + 0 = 1\) (conserved). 3) Lepton number: \(0 \to 0 + 1 = 1\) (violated, since there is no antineutrino to make the total lepton number zero). Thus, lepton number conservation is violated.

1 mark for the correct option B. Method: Analyze conservation laws (charge, baryon number, lepton number) for the reactants and products.

The area under a force-time graph is equal to the impulse, which equals the change in momentum of the body.

\(\text{Impulse} = \text{Area of the triangle} = \frac{1}{2} \times \text{base} \times \text{height}\)

\(\text{Impulse} = \frac{1}{2} \times 6.0\text{ s} \times 10\text{ N} = 30\text{ N s}\)

Since the body starts from rest (\(u = 0\)):

\(\Delta p = m v - m u = m v\)

\(30\text{ N s} = 2.0\text{ kg} \times v\)

\(v = \frac{30}{2.0} = 15\text{ m s}^{-1}\).

Correct answer is C.

1 mark:
- Calculates area of force-time graph to find impulse (\(30\text{ N s}\)).
- Divides impulse by mass to find final velocity (\(15\text{ m s}^{-1}\)).

The Young modulus is given by \(E = \frac{\sigma}{\varepsilon} = \frac{F L}{A \Delta L}\), which rearranged for extension gives \(\Delta L = \frac{F L}{A E}\).

Since both wires are of the same material, \(E\) is identical. They support the same load \(F\).

Thus, \(\Delta L \propto \frac{L}{A}\).

The cross-sectional area of a wire of diameter \(d\) is \(A = \frac{\pi d^2}{4}\), so \(A \propto d^2\).

Therefore, \(\Delta L \propto \frac{L}{d^2}\).

For wire X: \(L_{\text{X}} = 2 L_{\text{Y}}\) and \(d_{\text{X}} = 0.5 d_{\text{Y}}\).

\(\frac{\Delta L_{\text{X}}}{\Delta L_{\text{Y}}} = \frac{L_{\text{X}}}{L_{\text{Y}}} \times \left(\frac{d_{\text{Y}}}{d_{\text{X}}}\right)^2 = 2 \times \left(\frac{1}{0.5}\right)^2 = 2 \times 4 = 8.0\).

Correct answer is D.

1 mark:
- Recalls or derives \(\Delta L \propto \frac{L}{d^2}\).
- Correctly substitutes the ratios to find \(\frac{\Delta L_{\text{X}}}{\Delta L_{\text{Y}}} = 8.0\).

The diffraction grating equation is \(d \sin\theta = n \lambda\).

For the first order (\(n=1\)): \(\sin\theta = \frac{\lambda}{d}\).

Since the slit spacing \(d\) is inversely proportional to the number of lines per millimetre \(N\), we have \(d \propto \frac{1}{N}\), which means \(\sin\theta \propto N \lambda\).

Let \(\sin\theta = k N \lambda\).

For the second setup:

\(\sin\phi = k (1.5N) (1.2\lambda) = 1.8 (k N \lambda) = 1.8 \sin\theta\).

Correct answer is D.

1 mark:
- Relates grating slit spacing to number of lines per millimetre (\(d \propto 1/N\)).
- Uses \(d \sin\theta = \lambda\) to show \(\sin\theta \propto N \lambda\).
- Evaluates \(1.5 \times 1.2 = 1.8\).

The terminal potential difference \(V\) measured by the voltmeter is given by:

\(V = \frac{E R}{R + r}\)

We can write two equations for the two cases:

Case 1: \(6.0 = \frac{2.0 E}{2.0 + r} \implies 12.0 + 6.0 r = 2.0 E \implies E = 6.0 + 3.0 r\)

Case 2: \(9.0 = \frac{6.0 E}{6.0 + r} \implies 54.0 + 9.0 r = 6.0 E \implies E = 9.0 + 1.5 r\)

Equating the two expressions for \(E\):

\(6.0 + 3.0 r = 9.0 + 1.5 r\)

\(1.5 r = 3.0 \implies r = 2.0\ \Omega\).

Correct answer is C.

1 mark:
- Uses the potential divider equation or terminal potential difference formula \(V = E - Ir\) to set up simultaneous equations.
- Solves the simultaneous equations to find \(r = 2.0\ \Omega\).

At the highest point of the bridge, the forces acting on the car are the gravitational force (weight) \(mg\) acting downwards, and the normal contact force \(R\) acting upwards.

The resultant force acting towards the center of the circular path provides the centripetal acceleration:

\(F_{\text{net}} = mg - R = \frac{m v^2}{r}\)

Rearranging for \(R\) gives:

\(R = mg - \frac{m v^2}{r}\).

Correct answer is C.

1 mark:
- Identifies that centripetal force acts towards the center of the circle (downwards at the peak).
- Applies Newton's second law: \(mg - R = \frac{mv^2}{r}\).
- Rearranges to make \(R\) the subject.

The centripetal force is provided by the magnetic force:

\(q v B = \frac{m v^2}{r} \implies r = \frac{m v}{q B}\)

We can relate the momentum \(p = m v\) to kinetic energy \(E_k\):

\(E_k = \frac{p^2}{2m} \implies p = \sqrt{2 m E_k}\)

Thus, the radius of the path is:

\(r = \frac{\sqrt{2 m E_k}}{q B}\)

Since both particles have the same kinetic energy \(E_k\) and enter the same magnetic field \(B\):

\(r \propto \frac{\sqrt{m}}{q}\)

For the proton:

\(R_{\text{proton}} \propto \frac{\sqrt{m}}{e} = R\)

For the alpha particle:

\(R_{\text{alpha}} \propto \frac{\sqrt{4m}}{2e} = \frac{2 \sqrt{m}}{2e} = \frac{\sqrt{m}}{e}\)

Therefore, \(R_{\text{alpha}} = R\).

Correct answer is B.

1 mark:
- Expresses orbital radius in terms of kinetic energy, mass, and charge: \(r = \frac{\sqrt{2 m E_k}}{q B}\).
- Evaluates the ratio of \(\frac{\sqrt{m}}{q}\) for both particles to find that the radii are equal.

The charges of the quarks are:
- Up quark (\(u\)): \(+\frac{2}{3}e\)
- Down quark (\(d\)): \(-\frac{1}{3}e\) \(\implies\) anti-down quark (\(\bar{d}\)): \(+\frac{1}{3}e\)

The total charge of the meson is:

\(Q = +\frac{2}{3}e + \frac{1}{3}e = +1e\)

All quarks have a baryon number of \(+\frac{1}{3}\).
All antiquarks have a baryon number of \(-\frac{1}{3}\).

The total baryon number of the meson is:

\(B = +\frac{1}{3} + \left(-\frac{1}{3}\right) = 0\).

Correct answer is A.

1 mark:
- Determines correct charge: \(+1e\).
- Determines correct baryon number: \(0\).

According to Stefan-Boltzmann's law:

\(L = 4\pi R^2 \sigma T^4\)

Thus, luminosity is proportional to \(R^2 T^4\).

The ratio of the luminosities of star Y and star X is:

\(\frac{L_{\text{Y}}}{L_{\text{X}}} = \left(\frac{R_{\text{Y}}}{R_{\text{X}}}\right)^2 \left(\frac{T_{\text{Y}}}{T_{\text{X}}}\right)^4\)

Substituting the values:

\(\frac{L_{\text{Y}}}{L_{\text{X}}} = (0.5)^2 \times (2)^4 = 0.25 \times 16 = 4\).

Correct answer is C.

1 mark:
- Recalls Stefan-Boltzmann's law: \(L \propto R^2 T^4\).
- Correctly substitutes the ratios and calculates \(\frac{L_{\text{Y}}}{L_{\text{X}}} = 4\).

The car travels up the slope at constant speed, so the net force along the slope is zero. The forces acting down the slope are the component of gravity \( mg \sin\theta \) and the resistive force \( F \). Thus, the forward force \( F_{\text{engine}} \) exerted by the engine is \( F_{\text{engine}} = mg \sin\theta + F \). The rate of work done (power) is \( P = F_{\text{engine}} v = (mg \sin\theta + F)v \).

1 mark for identifying that the engine's forward force must balance the sum of the resistive force and the component of weight parallel to the slope, and correctly multiplying this by speed to obtain the power.

The Young modulus \( E \) is given by \( E = \frac{\text{stress}}{\text{strain}} = \frac{F/A}{\Delta x / L} = \frac{F L}{A \Delta x} \). Rearranging for extension gives \( \Delta x = \frac{F L}{A E} \). Since the cross-sectional area \( A = \frac{\pi d^2}{4} \), the extension is proportional to \( \frac{L}{d^2} \). For wire X: \( \Delta x_X \propto \frac{L}{d^2} \). For wire Y: \( \Delta x_Y \propto \frac{2L}{(2d)^2} = \frac{2L}{4d^2} = \frac{L}{2d^2} \). Thus, \( \frac{\Delta x_X}{\Delta x_Y} = \frac{L/d^2}{L/(2d^2)} = 2 \).

1 mark for using the formula for Young's modulus to establish the relationship between extension, length, and diameter, and correctly calculating the ratio as 2.

According to Einstein's photoelectric equation, the initial maximum kinetic energy is given by \( E_k = hf - \Phi \), where \( \Phi \) is the work function of the metal. If the frequency is doubled, the new maximum kinetic energy is \( E_k' = h(2f) - \Phi = 2hf - \Phi \). Substituting \( hf = E_k + \Phi \) gives \( E_k' = 2(E_k + \Phi) - \Phi = 2E_k + \Phi \). Since the work function \( \Phi \) must be positive for photoelectrons to be emitted, \( E_k' > 2E_k \).

1 mark for applying Einstein's photoelectric equation to show that the new maximum kinetic energy is \( 2E_k + \Phi \), and concluding that since \( \Phi > 0 \), the kinetic energy is greater than \( 2E_k \).

The total current flowing through the parallel combination of \( R_1 \) and \( R_2 \) is the same as the total current flowing through \( R_3 \). Since the potential difference across \( R_3 \) is equal to the potential difference across the parallel combination, the equivalent resistance of the parallel combination \( R_p \) must equal \( R_3 \). The formula for the equivalent resistance of a parallel circuit is \( \frac{1}{R_p} = \frac{1}{R_1} + \frac{1}{R_2} \). Substituting \( R_3 \) for \( R_p \), we get \( \frac{1}{R_3} = \frac{1}{R_1} + \frac{1}{R_2} \).

1 mark for identifying that equal potential differences across series parts implies equal resistances, and setting the parallel equivalent resistance formula equal to \( R_3 \).

The centripetal force acts perpendicular to the instantaneous velocity (direction of motion) of the object at all times. Because the angle between the centripetal force vector and the displacement vector is always \( 90^\circ \), the work done is \( W = F \cdot s \cdot \cos(90^\circ) = 0 \). Therefore, no work is done on the object by the centripetal force.

1 mark for stating or showing that the centripetal force is always perpendicular to the direction of displacement, so the work done is zero.

The centripetal force on a charged particle in a magnetic field is provided by the magnetic force: \( Bqv = \frac{mv^2}{\rho} \implies \rho = \frac{mv}{Bq} \). For the electron, \( r = \frac{m_e v}{Be} \). For the proton, \( R = \frac{m_p v}{Be} \). The ratio is \( \frac{R}{r} = \frac{m_p v / Be}{m_e v / Be} = \frac{m_p}{m_e} \).

1 mark for deriving the formula for orbital radius and finding the correct ratio of the proton mass to the electron mass.

A baryon is made of three quarks. A strangeness of \( -1 \) means the particle must contain exactly one strange quark \( s \) (since \( s \) has strangeness \( -1 \), and \( u \) and \( d \) have strangeness \( 0 \)). The charge of a \( u \) quark is \( +\frac{2}{3}e \), a \( d \) quark is \( -\frac{1}{3}e \), and an \( s \) quark is \( -\frac{1}{3}e \). Let's calculate the total charge for the combination uds: \( Q = (+\frac{2}{3}) + (-\frac{1}{3}) + (-\frac{1}{3}) = 0 \). This matches both requirements.

1 mark for identifying the correct quark combination that satisfies both the strangeness and charge conservation requirements for a baryon.

The kinetic theory equation for gas pressure is \( p = \frac{1}{3} \frac{N m \langle c^2 \rangle}{V} \), where \( \langle c^2 \rangle \) is the mean square speed. Since the root-mean-square speed \( c_{\text{rms}} = \sqrt{\langle c^2 \rangle} \), doubling \( c_{\text{rms}} \) means that \( \langle c^2 \rangle \) increases by a factor of \( 2^2 = 4 \). Because \( N \), \( m \), and \( V \) remain constant, the pressure \( p \) is directly proportional to \( \langle c^2 \rangle \). Thus, the pressure increases to \( 4p \).

1 mark for recognizing that pressure is proportional to the mean square speed, so doubling the r.m.s. speed quadruples the pressure.

Since the speed is constant, the forces along the ramp are balanced. The force pushing up the ramp \( F \) must equal the sum of the component of the weight down the ramp and the frictional force: \( F = mg \sin\theta + f_k \). Frictional force is \( f_k = \mu R = \mu mg \cos\theta \). Thus, \( F = mg(\sin\theta + \mu\cos\theta) \). The work done by the force is \( W = F \times L = mgL(\sin\theta + \mu\cos\theta) \).

A - 1 mark: Correct choice selected.

The original fringe spacing is given by \( x = \frac{\lambda D}{d} \). When submerged in a liquid of refractive index \( n \), the wavelength of light becomes \( \lambda' = \frac{\lambda}{n} \). Since the physical dimensions \( d \) and \( D \) remain unchanged, the new fringe spacing is \( x' = \frac{\lambda' D}{d} = \frac{\lambda D}{n d} = \frac{x}{n} \).

B - 1 mark: Correct choice selected.

The radius of a charged particle's path in a magnetic field is \( R = \frac{mv}{qB} \). Expressing velocity in terms of kinetic energy \( E_k = \frac{1}{2}mv^2 \implies v = \sqrt{\frac{2E_k}{m}} \), we get \( R = \frac{m}{qB}\sqrt{\frac{2E_k}{m}} = \frac{\sqrt{2mE_k}}{qB} \). For the second particle with mass \( 2m \), charge \( 2q \), and same energy \( E_k \), the new radius is \( R' = \frac{\sqrt{2(2m)E_k}}{(2q)B} = \frac{2\sqrt{mE_k}}{2qB} = \frac{\sqrt{mE_k}}{qB} \). Comparing this to the original radius: \( R = \frac{\sqrt{2}\sqrt{mE_k}}{qB} \implies R' = \frac{R}{\sqrt{2}} \).

D - 1 mark: Correct choice selected.

For Configuration 1, the equivalent resistance is \( R_1 = 3R \). The power dissipated is \( P_1 = \frac{V^2}{3R} \). For Configuration 2, the equivalent resistance of the two parallel resistors is \( \frac{R}{2} \), which in series with the third gives \( R_2 = R + \frac{R}{2} = 1.5R = \frac{3}{2}R \). The power dissipated is \( P_2 = \frac{V^2}{1.5R} = \frac{2V^2}{3R} \). The ratio is \( \frac{P_1}{P_2} = \frac{V^2 / 3R}{2V^2 / 3R} = \frac{1}{2} \).

A - 1 mark: Correct choice selected.

Young modulus is \( E = \frac{\text{Stress}}{\text{Strain}} = \frac{F/A}{\Delta x / L} = \frac{FL}{A\Delta x} \implies \Delta x = \frac{FL}{AE} \). The cross-sectional area is \( A = \pi \left(\frac{d}{2}\right)^2 = \frac{\pi d^2}{4} \), so \( \Delta x \propto \frac{L}{d^2} \). For the second wire: \( L' = 2L \) and \( d' = 2d \), so \( \Delta x' \propto \frac{2L}{(2d)^2} = \frac{2L}{4d^2} = \frac{1}{2}\frac{L}{d^2} \). Therefore, the new extension is \( \frac{1}{2}\Delta x \).

C - 1 mark: Correct choice selected.

For an ideal gas, internal energy depends only on temperature. Since the process is isothermal, the temperature remains constant, so \( \Delta U = 0 \). During expansion, the gas does work on its surroundings, meaning the work done by the gas is positive (\( W > 0 \)). According to the first law of thermodynamics, \( \Delta U = Q - W \), where \( W \) is the work done by the gas. Since \( \Delta U = 0 \), we have \( Q = W \), meaning \( Q > 0 \) (heat is supplied to the gas to maintain its temperature).

A - 1 mark: Correct choice selected.

The duration of 36 hours corresponds to \( \frac{36}{12} = 3 \) half-lives. The fraction of active nuclei remaining is \( \left(\frac{1}{2}\right)^3 = \frac{1}{8} \), so the number of remaining active nuclei is \( N = \frac{N_0}{8} \). The number of nuclei that have decayed is \( N_d = N_0 - N = \frac{7N_0}{8} \). The ratio of decayed to remaining nuclei is \( \frac{7N_0/8}{N_0/8} = 7 \).

A - 1 mark: Correct choice selected.

Using Wien's displacement law, \( \lambda_{\text{max}} T = \text{constant} \implies T \propto \frac{1}{\lambda_{\text{max}}} \). Therefore, the temperature ratio is \( \frac{T_X}{T_Y} = \frac{800\text{ nm}}{400\text{ nm}} = 2 \). Using the Stefan-Boltzmann law for luminosity, \( L = 4\pi R^2 \sigma T^4 \implies R \propto \frac{\sqrt{L}}{T^2} \). The ratio of the radii is \( \frac{R_X}{R_Y} = \sqrt{\frac{L_X}{L_Y}} \left(\frac{T_Y}{T_X}\right)^2 = \sqrt{\frac{L}{4L}} \left(\frac{1}{2}\right)^2 = \frac{1}{2} \times \frac{1}{4} = \frac{1}{8} \).

D - 1 mark: Correct choice selected.

Initially, the total resistance of the circuit is \(R_{\text{total}} = r + 3r + 5r = 9r\).

The initial current in the circuit is:
\(I_{\text{initial}} = \frac{E}{9r}\)

The initial potential difference across the variable resistor is:
\(V_{\text{initial}} = I_{\text{initial}} \times 5r = \frac{5}{9}E \approx 0.56E\)

When the variable resistor is decreased to zero, the total resistance of the circuit becomes:
\(R_{\text{total}} = r + 3r = 4r\)

The final current in the circuit is:
\(I_{\text{final}} = \frac{E}{4r}\)

Since \(4r < 9r\), the current in the circuit increases.

The final potential difference across the variable resistor is:
\(V_{\text{final}} = I_{\text{final}} \times 0 = 0\)

Since \(0 < 0.56E\), the potential difference across the variable resistor decreases.

1 mark for the correct option (A).

Award 1 mark for analyzing that reducing circuit resistance increases circuit current, and that decreasing the variable resistance to zero reduces its potential difference to zero.

The relationship between the root-mean-square speed of the gas molecules and the absolute temperature is given by \(v_{\text{rms}} = \sqrt{\frac{3kT}{m}}\), which means \(v_{\text{rms}} \propto \sqrt{T}\) or \(T \propto v_{\text{rms}}^2\).

First, convert the initial temperature from degrees Celsius to Kelvin:
\(T_1 = 27 + 273 = 300\text{ K}\)

Since the r.m.s. speed is doubled, the absolute temperature must increase by a factor of \(2^2 = 4\):
\(T_2 = 4 \times 300\text{ K} = 1200\text{ K}\)

Convert the new absolute temperature back to degrees Celsius:
\(\theta_2 = 1200 - 273 = 927\text{ }^\circ\text{C}\)

1 mark for the correct option (C).

Award 1 mark for recognizing that doubling r.m.s. speed requires quadrupling the absolute temperature, and correctly converting between Celsius and Kelvin scales.

### Part (a)
1. Identify the forces acting parallel to the slope:
- Component of weight acting down the slope:
\( F_{\text{parallel}} = mg \sin\theta = 75\text{ kg} \times 9.81\text{ m s}^{-2} \times \sin(22^\circ) \approx 275.6\text{ N} \)
- Frictional resistive force acting up the slope:
\( F_{\text{resistive}} = 120\text{ N} \)

2. Calculate the net force parallel to the slope:
\( F_{\text{net}} = 275.6\text{ N} - 120\text{ N} = 155.6\text{ N} \)

3. Apply Newton's Second Law:
\( a = \frac{F_{\text{net}}}{m} = \frac{155.6\text{ N}}{75\text{ kg}} \approx 2.07\text{ m s}^{-2} \)

### Part (b)
Using equations of motion for constant acceleration:
\( v^2 = u^2 + 2as \)
Since \( u = 0 \):
\( v = \sqrt{2 \times 2.07\text{ m s}^{-2} \times 85\text{ m}} \approx 18.8\text{ m s}^{-1} \)

Alternatively, using work-energy conservation:
\( \text{Loss in G.P.E.} = mgh = 75 \times 9.81 \times (85 \sin 22^\circ) \approx 23430\text{ J} \)
\( \text{Work done against friction} = 120\text{ N} \times 85\text{ m} = 10200\text{ J} \)
\( \text{Gain in K.E.} = 23430\text{ J} - 10200\text{ J} = 13230\text{ J} \)
\( \frac{1}{2} m v^2 = 13230\text{ J} \implies v = \sqrt{\frac{2 \times 13230}{75}} \approx 18.8\text{ m s}^{-1} \)

### Part (a) [3 Marks]
- **M1**: Correct expression or calculation for the parallel weight component \( mg \sin(22^\circ) \) [\( 276\text{ N} \)].
- **M2**: Correct subtraction of resistive force from parallel weight component [\( F_{\text{net}} = 156\text{ N} \)].
- **A1**: Correct acceleration in range \( 2.07 \text{ to } 2.1\text{ m s}^{-2} \).

### Part (b) [3 Marks]
- **M1**: Use of \( v^2 = u^2 + 2as \) OR conservation of energy equation \( mgh - W_f = \frac{1}{2}mv^2 \).
- **M2**: Correct substitution of values into chosen kinematic/energy formula.
- **A1**: Correct speed in range \( 18.7 \text{ to } 19.0\text{ m s}^{-1} \).

### Part (a)
1. Volume of the sphere:
\( V = \frac{4}{3}\pi r^3 = \frac{4}{3}\pi (1.50 \times 10^{-3}\text{ m})^3 = 1.414 \times 10^{-8}\text{ m}^3 \)
2. Upthrust is the weight of the displaced oil:
\( U = \rho_f V g = 900\text{ kg m}^{-3} \times 1.414 \times 10^{-8}\text{ m}^3 \times 9.81\text{ m s}^{-2} \approx 1.249 \times 10^{-4}\text{ N} \)
This is approximately \( 1.2 \times 10^{-4}\text{ N} \) (or \( 1.3 \times 10^{-4}\text{ N} \) if rounded up).

### Part (b)
At terminal velocity, the forces are balanced:
\( \text{Weight} = \text{Upthrust} + \text{Viscous Drag Force} \)
\( W = U + F_D \implies F_D = W - U \)

1. Calculate the weight of the steel sphere:
\( W = \rho_s V g = 7800\text{ kg m}^{-3} \times 1.414 \times 10^{-8}\text{ m}^3 \times 9.81\text{ m s}^{-2} = 1.082 \times 10^{-3}\text{ N} \)
2. Calculate the viscous drag force:
\( F_D = 1.082 \times 10^{-3}\text{ N} - 1.249 \times 10^{-4}\text{ N} = 9.571 \times 10^{-4}\text{ N} \)
3. Apply Stokes' Law:
\( F_D = 6\pi \eta r v \)
\( \eta = \frac{F_D}{6\pi r v} = \frac{9.571 \times 10^{-4}}{6\pi \times (1.50 \times 10^{-3}\text{ m}) \times 0.120\text{ m s}^{-1}} \)
\( \eta = \frac{9.571 \times 10^{-4}}{3.393 \times 10^{-3}} \approx 0.282\text{ Pa s} \) (or \( \text{N s m}^{-2} \))

### Part (a) [2 Marks]
- **M1**: Use of volume formula \( V = \frac{4}{3}\pi r^3 \) and upthrust formula \( U = \rho_f V g \).
- **A1**: Correctly calculated intermediate value showing \( 1.25 \times 10^{-4}\text{ N} \) (or \( 1.22 \times 10^{-4}\text{ N} \) if \( g = 9.8\text{ m s}^{-2} \)).

### Part (b) [4 Marks]
- **M1**: Calculate weight of the sphere \( W = 1.08 \times 10^{-3}\text{ N} \).
- **M2**: Correct statement of equilibrium \( F_D = W - U \) and calculation of viscous drag \( F_D \approx 9.6 \times 10^{-4}\text{ N} \).
- **M3**: Rearrangement of Stokes' Law to make viscosity the subject: \( \eta = \frac{F_D}{6\pi r v} \).
- **A1**: Correct value of viscosity in the range \( 0.28 \text{ to } 0.29\text{ Pa s} \) with correct units.

### Part (a)
1. Calculate the energy of the incident photon:
\( E = \frac{hc}{\lambda} = \frac{6.63 \times 10^{-34}\text{ J s} \times 3.00 \times 10^8\text{ m s}^{-1}}{380 \times 10^{-9}\text{ m}} \approx 5.234 \times 10^{-19}\text{ J} \)
2. Convert the work function from eV to Joules:
\( \Phi = 2.28\text{ eV} \times 1.60 \times 10^{-19}\text{ J eV}^{-1} = 3.648 \times 10^{-19}\text{ J} \)
3. Apply Einstein's photoelectric equation:
\( E_{\text{k, max}} = E - \Phi \)
\( E_{\text{k, max}} = 5.234 \times 10^{-19}\text{ J} - 3.648 \times 10^{-19}\text{ J} = 1.586 \times 10^{-19}\text{ J} \approx 1.59 \times 10^{-19}\text{ J} \)

### Part (b)
1. Intensity is proportional to the number of photons incident per unit area per second.
2. Doubling the intensity doubles the number of photons (and hence increases the rate of emission of photoelectrons), but the energy of each individual photon remains the same.
3. Since \( E_{\text{k, max}} \) depends only on the energy of a single photon and the work function, the maximum kinetic energy remains unchanged.

### Part (a) [4 Marks]
- **M1**: Use of \( E = \frac{hc}{\lambda} \).
- **M2**: Correct calculation of photon energy: \( 5.23 \times 10^{-19}\text{ J} \).
- **M3**: Correct conversion of work function: \( 2.28\text{ eV} = 3.65 \times 10^{-19}\text{ J} \).
- **A1**: Correct subtraction yielding \( 1.59 \times 10^{-19}\text{ J} \) (accept \( 1.58 \times 10^{-19}\text{ J} \) to \( 1.60 \times 10^{-19}\text{ J} \)).

### Part (b) [2 Marks]
- **M1**: Identifies that intensity affects only the number of photons per second / rate of electron emission (and not single photon energy).
- **A1**: Explicitly states that maximum kinetic energy remains unchanged.

### Part (a)
Using the potential divider equation:
\( V_{\text{out}} = V_{\text{in}} \times \left( \frac{R_{\text{th}}}{R_{\text{th}} + R_{\text{fixed}}} \right) \)

Substitute known values:
\( 5.4\text{ V} = 9.0\text{ V} \times \left( \frac{R_{\text{th}}}{R_{\text{th}} + 1.2\text{ k}\Omega} \right) \)

Divide both sides by 9.0:
\( 0.6 = \frac{R_{\text{th}}}{R_{\text{th}} + 1200} \)
\( 0.6(R_{\text{th}} + 1200) = R_{\text{th}} \)
\( 0.6R_{\text{th}} + 720 = R_{\text{th}} \)
\( 0.4R_{\text{th}} = 720 \implies R_{\text{th}} = 1800\ \Omega = 1.8\text{ k}\Omega \) (as shown)

### Part (b)
With the new thermistor resistance \( R_{\text{th}} = 650\ \Omega \) and fixed resistor \( R_{\text{fixed}} = 1200\ \Omega \):

\( V_{\text{out}} = V_{\text{in}} \times \left( \frac{R_{\text{th}}}{R_{\text{th}} + R_{\text{fixed}}} \right) \)
\( V_{\text{out}} = 9.0\text{ V} \times \left( \frac{650}{650 + 1200} \right) \)
\( V_{\text{out}} = 9.0\text{ V} \times \left( \frac{650}{1850} \right) \approx 3.16\text{ V} \approx 3.2\text{ V} \)

### Part (a) [2 Marks]
- **M1**: Sets up potential divider formula or uses ratio of voltages: \( \frac{5.4}{3.6} = \frac{R_{\text{th}}}{1.2\text{ k}\Omega} \).
- **A1**: Correct algebraic progression leading directly to \( 1800\ \Omega \) or \( 1.8\text{ k}\Omega \).

### Part (b) [4 Marks]
- **M1**: Recognises total resistance of circuit is now \( 1200\ \Omega + 650\ \Omega = 1850\ \Omega \).
- **M2**: Correct current calculation: \( I = \frac{9.0}{1850} \approx 4.86 \times 10^{-3}\text{ A} \) OR correctly sets up new potential divider fraction.
- **M3**: Multiplies current by thermistor resistance: \( V = 4.86 \times 10^{-3} \times 650 \).
- **A1**: Correct final value of \( 3.2\text{ V} \) (accept \( 3.16\text{ V} \)).

### Part (a)
At the design speed, no lateral frictional force acts. Only two forces act on the car:
1. Weight \( W = mg \) acting vertically downwards.
2. Normal reaction \( N \) (or \( R \)) acting perpendicular to the banked track surface (upwards and angled towards the centre of the circle).

### Part (b)
Resolve forces:
- Vertically: \( N \cos\theta = mg \)
- Horizontally (providing centripetal force): \( N \sin\theta = \frac{mv^2}{r} \)

Divide the two equations:
\( \frac{N \sin\theta}{N \cos\theta} = \frac{mv^2 / r}{mg} \implies \tan\theta = \frac{v^2}{rg} \)

Rearrange to solve for speed \( v \):
\( v = \sqrt{rg \tan\theta} \)
\( v = \sqrt{150\text{ m} \times 9.81\text{ m s}^{-2} \times \tan(18^\circ)} \)
\( v = \sqrt{1471.5 \times 0.3249} \approx \sqrt{478.1} \approx 21.9\text{ m s}^{-1} \) (or \( 22\text{ m s}^{-1} \))

### Part (a) [2 Marks]
- **M1**: Arrow vertically downwards labelled Weight/\( mg \)/Gravity.
- **A1**: Arrow perpendicular to the inclined track labelled Normal reaction / \( R \) / \( N \) (no friction force shown).

### Part (b) [4 Marks]
- **M1**: Resolves normal reaction vertically: \( N \cos(18) = mg \).
- **M2**: Resolves normal reaction horizontally to equate to centripetal force: \( N \sin(18) = \frac{mv^2}{r} \).
- **M3**: Combines equations to give \( v = \sqrt{rg \tan\theta} \).
- **A1**: Correct design speed in range \( 21.8 \text{ to } 22.0\text{ m s}^{-1} \).

### Part (a)
For the electrons to pass through undeflected, the net force must be zero. Therefore, the electric force \( F_E = eE \) must be equal in magnitude and opposite in direction to the magnetic force \( F_B = evB \).

### Part (b)
Equating the forces:
\( eE = evB \implies E = vB \)

We know the relation between uniform electric field and potential difference is:
\( E = \frac{V}{d} \)

Therefore:
\( \frac{V}{d} = vB \implies V = v B d \)

Substitute the given values:
- \( v = 2.5 \times 10^5\text{ m s}^{-1} \)
- \( B = 4.5 \times 10^{-3}\text{ T} \)
- \( d = 8.0 \times 10^{-3}\text{ m} \)

\( V = (2.5 \times 10^5\text{ m s}^{-1}) \times (4.5 \times 10^{-3}\text{ T}) \times (8.0 \times 10^{-3}\text{ m}) \)
\( V = 2.5 \times 10^5 \times 3.60 \times 10^{-5}\text{ V} = 9.0\text{ V} \)

### Part (a) [2 Marks]
- **M1**: Identifies that the electric and magnetic forces must act in opposite directions.
- **A1**: States that the magnitudes of the forces must be equal (or \( eE = evB \)).

### Part (b) [4 Marks]
- **M1**: Recognises and uses \( E = \frac{V}{d} \).
- **M2**: Correctly equates to find \( V = vBd \).
- **M3**: Substitution of values with correct power-of-ten conversions for \( \text{mT} \) and \( \text{mm} \).
- **A1**: Correct final value of \( 9.0\text{ V} \).

### Part (a)
Use the ideal gas equation:
\( pV = N k T \)

Where:
- \( p = 1.5 \times 10^5\text{ Pa} \)
- \( V = 0.025\text{ m}^3 \)
- \( T = 22 + 273.15 = 295.15\text{ K} \)
- \( k = 1.38 \times 10^{-23}\text{ J K}^{-1} \)

Rearrange to solve for \( N \):
\( N = \frac{pV}{kT} = \frac{1.5 \times 10^5 \times 0.025}{1.38 \times 10^{-23} \times 295.15} \)
\( N = \frac{3750}{4.073 \times 10^{-21}} \approx 9.21 \times 10^{23} \text{ atoms} \)

### Part (b)
Since volume is constant, apply Gay-Lussac's Law (Pressure Law):
\( \frac{p_1}{T_1} = \frac{p_2}{T_2} \implies p_2 = p_1 \times \frac{T_2}{T_1} \)

Where:
- \( T_2 = 85 + 273.15 = 358.15\text{ K} \)
- \( T_1 = 295.15\text{ K} \)

Calculate \( p_2 \):
\( p_2 = 1.5 \times 10^5\text{ Pa} \times \frac{358.15\text{ K}}{295.15\text{ K}} \approx 1.82 \times 10^5\text{ Pa} \)

### Part (a) [3 Marks]
- **M1**: Convert Celsius to Kelvin: \( T = 295\text{ K} \).
- **M2**: Rearrange ideal gas equation correctly: \( N = \frac{pV}{kT} \).
- **A1**: Correct number of atoms \( 9.2 \times 10^{23} \) (accept \( 9.20 \times 10^{23} \) to \( 9.23 \times 10^{23} \)).

### Part (b) [3 Marks]
- **M1**: Convert new temperature to Kelvin: \( T_2 = 358\text{ K} \).
- **M2**: Correct application of \( \frac{p_1}{T_1} = \frac{p_2}{T_2} \).
- **A1**: Correct pressure \( 1.8 \times 10^5\text{ Pa} \) (accept \( 1.82 \times 10^5\text{ Pa} \)).

### Part (a)
1. Convert half-life to seconds:
\( t_{1/2} = 6.00\text{ hours} \times 3600\text{ s hour}^{-1} = 21600\text{ s} \)
2. Apply the decay constant relationship:
\( \lambda = \frac{\ln 2}{t_{1/2}} \)
\( \lambda = \frac{0.69315}{21600\text{ s}} \approx 3.209 \times 10^{-5}\text{ s}^{-1} \approx 3.21 \times 10^{-5}\text{ s}^{-1} \)

### Part (b)
Since the elapsed time is \( 24.0\text{ hours} \), which is exactly 4 half-lives (\( 24.0 / 6.00 = 4 \)):
\( A = \frac{A_0}{2^n} = \frac{4.50 \times 10^8\text{ Bq}}{2^4} = \frac{4.50 \times 10^8}{16} = 2.8125 \times 10^7\text{ Bq} \approx 2.81 \times 10^7\text{ Bq} \)

Alternatively, using the exponential decay formula:
\( A = A_0 e^{-\lambda t} \)
Where \( t = 24.0 \times 3600\text{ s} = 86400\text{ s} \):
\( A = (4.50 \times 10^8) \times e^{-(3.209 \times 10^{-5}) \times 86400} \)
\( A = (4.50 \times 10^8) \times e^{-2.773} \approx 2.81 \times 10^7\text{ Bq} \)

### Part (a) [3 Marks]
- **M1**: Convert 6.00 hours to seconds (\( 21600\text{ s} \)).
- **M2**: Use of formula \( \lambda = \frac{\ln 2}{t_{1/2}} \).
- **A1**: Correct decay constant: \( 3.21 \times 10^{-5}\text{ s}^{-1} \) (accept \( 3.2 \times 10^{-5}\text{ s}^{-1} \)).

### Part (b) [3 Marks]
- **M1**: States that 24 hours corresponds to 4 half-lives OR sets up the equation \( A = A_0 e^{-\lambda t} \).
- **M2**: Correct substitution of values into chosen method.
- **A1**: Correct remaining activity: \( 2.81 \times 10^7\text{ Bq} \) (accept range \( 2.8 \times 10^7\text{ Bq} \) to \( 2.82 \times 10^7\text{ Bq} \)).

**Part (a)**
The initial kinetic energy of the block is:
\(E_k = \frac{1}{2} m v^2 = \frac{1}{2} \times 0.35\text{ kg} \times (4.5\text{ m s}^{-1})^2 = 3.544\text{ J}\)

The gain in gravitational potential energy is:
\(\Delta E_p = m g h = m g d \sin\theta = 0.35\text{ kg} \times 9.81\text{ m s}^{-2} \times 1.80\text{ m} \times \sin(25^\circ)\)
\(\Delta E_p = 0.35 \times 9.81 \times 1.80 \times 0.4226 = 2.612\text{ J}\)

By conservation of energy, the work done against friction \(W_f\) is:
\(W_f = E_k - \Delta E_p = 3.544\text{ J} - 2.612\text{ J} = 0.932\text{ J}\)
This is approximately \(0.9\text{ J}\).

**Part (b)**
The work done against friction is also given by:
\(W_f = F_f \times d\)
where \(F_f = \mu R\) is the frictional force, and \(R = m g \cos\theta\) is the normal reaction force.

\(R = 0.35\text{ kg} \times 9.81\text{ m s}^{-2} \times \cos(25^\circ) = 3.4335 \times 0.9063 = 3.112\text{ N}\)

Therefore:
\(W_f = \mu R d \implies \mu = \frac{W_f}{R d} = \frac{0.932\text{ J}}{3.112\text{ N} \times 1.80\text{ m}} = 0.166\)

The coefficient of kinetic friction is \(0.17\).

**Part (a) [3 marks]**
* **M1**: Calculates initial kinetic energy: \(E_k = \frac{1}{2} \times 0.35 \times 4.5^2 = 3.54\text{ J}\) (1)
* **M1**: Calculates gain in gravitational potential energy: \(\Delta E_p = 0.35 \times 9.81 \times 1.80 \times \sin(25^\circ) = 2.61\text{ J}\) (1)
* **A1**: Subtracts \(\Delta E_p\) from \(E_k\) to obtain \(0.93\text{ J}\) (must show at least 2 significant figures to justify 'about 0.9 J') (1)

**Part (b) [3 marks]**
* **M1**: Calculates normal reaction force: \(R = m g \cos(25^\circ) = 3.11\text{ N}\) (accept \(3.1\text{ N}\)) (1)
* **M1**: Relates work done to friction and distance: \(W_f = \mu R d\) or \(F_f = \frac{0.93}{1.80} = 0.52\text{ N}\) (1)
* **A1**: Calculates \(\mu = 0.17\) (accept \(0.166\) to \(0.17\); allow ecf from part a) (1)

**Part (a)**
First, calculate the cross-sectional area \(A\) of the wire:
\(A = \frac{\pi d^2}{4} = \frac{\pi \times (0.82 \times 10^{-3}\text{ m})^2}{4} = 5.281 \times 10^{-7}\text{ m}^2\)

The force \(F\) acting on the wire is the weight of the mass:
\(F = m g = 6.5\text{ kg} \times 9.81\text{ m s}^{-2} = 63.77\text{ N}\)

Using the formula for Young modulus \(E = \frac{F L}{A \Delta L}\), we can rearrange for extension \(\Delta L\):
\(\Delta L = \frac{F L}{A E} = \frac{63.77\text{ N} \times 2.40\text{ m}}{5.281 \times 10^{-7}\text{ m}^2 \times 1.2 \times 10^{11}\text{ Pa}} = 2.41 \times 10^{-3}\text{ m}\)

**Part (b)**
The elastic strain energy \(E_{el}\) stored in the wire is:
\(E_{el} = \frac{1}{2} F \Delta L\)
\(E_{el} = \frac{1}{2} \times 63.77\text{ N} \times 2.41 \times 10^{-3}\text{ m} = 0.0768\text{ J}\)
Rounding to two significant figures gives \(0.077\text{ J}\) or \(7.7 \times 10^{-2}\text{ J}\).

**Part (a) [3 marks]**
* **M1**: Calculates cross-sectional area: \(A = 5.28 \times 10^{-7}\text{ m}^2\) (1)
* **M1**: Recalls and rearranges Young modulus equation to make extension the subject: \(\Delta L = \frac{F L}{A E}\) (1)
* **A1**: Calculates extension \(\Delta L = 2.4 \times 10^{-3}\text{ m}\) (accept \(2.41 \times 10^{-3}\text{ m}\)) (1)

**Part (b) [3 marks]**
* **M1**: Recalls formula for elastic strain energy: \(E_{el} = \frac{1}{2} F \Delta L\) (or \(\frac{1}{2} k \Delta L^2\)) (1)
* **M1**: Substitutes values of force (\(63.8\text{ N}\)) and their extension from part (a) (1)
* **A1**: Calculates \(E_{el} = 7.7 \times 10^{-2}\text{ J}\) (accept \(0.077\text{ J}\) or \(0.0768\text{ J}\); allow ecf from part a) (1)

**Part (a)**
The energy of a single photon is:
\(E = \frac{h c}{\lambda} = \frac{6.63 \times 10^{-34}\text{ J s} \times 3.00 \times 10^8\text{ m s}^{-1}}{240 \times 10^{-9}\text{ m}} = 8.288 \times 10^{-19}\text{ J}\)

Convert the work function of sodium into joules:
\(\Phi = 2.36\text{ eV} \times 1.60 \times 10^{-19}\text{ J eV}^{-1} = 3.776 \times 10^{-19}\text{ J}\)

The Einstein photoelectric equation is:
\(h f = \Phi + E_{k,\max} \implies E_{k,\max} = E - \Phi\)
\(E_{k,\max} = 8.288 \times 10^{-19}\text{ J} - 3.776 \times 10^{-19}\text{ J} = 4.512 \times 10^{-19}\text{ J}\)
This is approximately \(4.5 \times 10^{-19}\text{ J}\).

**Part (b)**
The total power incident on the area is:
\(P = I \times A = 1.5\text{ W m}^{-2} \times 2.0 \times 10^{-4}\text{ m}^2 = 3.0 \times 10^{-4}\text{ W}\)

The number of photons incident per second is:
\(N_{\text{photons}} = \frac{P}{E} = \frac{3.0 \times 10^{-4}\text{ W}}{8.288 \times 10^{-19}\text{ J}} = 3.62 \times 10^{14}\text{ s}^{-1}\)

Since only \(1.0\%\) of these photons eject a photoelectron, the rate of photoelectron emission is:
\(R = 0.01 \times N_{\text{photons}} = 0.01 \times 3.62 \times 10^{14}\text{ s}^{-1} = 3.62 \times 10^{12}\text{ s}^{-1}\)

**Part (a) [3 marks]**
* **M1**: Calculates photon energy: \(E = 8.29 \times 10^{-19}\text{ J}\) (1)
* **M1**: Converts work function to J: \(\Phi = 3.78 \times 10^{-19}\text{ J}\) (1)
* **A1**: Subtracts to get \(4.51 \times 10^{-19}\text{ J}\) (must show at least 2 significant figures to justify 'approximately') (1)

**Part (b) [3 marks]**
* **M1**: Calculates total power: \(P = 3.0 \times 10^{-4}\text{ W}\) (1)
* **M1**: Calculates rate of incident photons: \(N = 3.6 \times 10^{14}\text{ s}^{-1}\) (allow ecf from part a) (1)
* **A1**: Multiplies by 0.01 to get \(3.6 \times 10^{12}\text{ s}^{-1}\) (accept range \(3.6 \times 10^{12}\) to \(3.63 \times 10^{12}\)) (1)

**Part (a)**
Using the potential divider equation, the output voltage \(V_{\text{out}}\) across the thermistor is:
\(V_{\text{out}} = V_s \times \frac{R_{\text{th}}}{R_{\text{th}} + R}\)
\(V_{\text{out}} = 9.0\text{ V} \times \frac{2.8\text{ k}\Omega}{2.8\text{ k}\Omega + 1.2\text{ k}\Omega} = 9.0 \times \frac{2.8}{4.0} = 6.3\text{ V}\)

**Part (b)**
As temperature increases, thermal energy is supplied to the semi-conducting material of the thermistor. This liberates charge carriers (electrons), causing them to jump to the conduction band. Thus, the number density \(n\) of charge carriers increases significantly, causing the resistance to decrease.

At \(60^\circ\text{C}\), the thermistor's resistance is \(650\ \Omega = 0.65\text{ k}\Omega\).
The new voltmeter reading is:
\(V_{\text{out}} = 9.0\text{ V} \times \frac{0.65\text{ k}\Omega}{0.65\text{ k}\Omega + 1.2\text{ k}\Omega} = 9.0 \times \frac{0.65}{1.85} = 3.16\text{ V}\)
Rounding to two significant figures, the reading is \(3.2\text{ V}\).

**Part (a) [2 marks]**
* **M1**: Recalls potential divider formula with correct substitution: \(9.0 \times \frac{2.8}{2.8 + 1.2}\) (1)
* **A1**: Obtains \(6.3\text{ V}\) (1)

**Part (b) [4 marks]**
* **M1**: Mentions that thermal energy liberates charge carriers / increases the number density \(n\) of conduction electrons (1)
* **A1**: Connects increased \(n\) to the decrease in resistance (1)
* **M1**: Substitutes new resistance values into potential divider equation: \(9.0 \times \frac{0.65}{0.65 + 1.2}\) (or \(9.0 \times \frac{650}{650 + 1200}\)) (1)
* **A1**: Calculates \(V_{\text{out}} = 3.2\text{ V}\) (accept \(3.16\text{ V}\) to \(3.2\text{ V}\)) (1)

**Part (a)**
The forces acting on the rider are:
1. Weight \(W = m g\) acting vertically downwards from the center of mass.
2. Frictional force \(F\) acting vertically upwards (counteracting gravity).
3. Normal contact force \(R\) (or reaction force) acting horizontally inwards towards the center of rotation (providing the centripetal force).

**Part (b)**
To prevent the rider from sliding down, the upward frictional force must equal the downward gravitational force:
\(F = m g\)

The maximum possible static friction is:
\(F = \mu R\)

The normal contact force provides the centripetal acceleration:
\(R = m \omega^2 r\)

Combining these equations for the minimum speed (where friction is at its maximum limit):
\(m g = \mu (m \omega^2 r)\)
\(g = \mu \omega^2 r\)

Rearranging for the angular velocity \(\omega\):
\(\omega^2 = \frac{g}{\mu r}\)
\(\omega = \sqrt{\frac{9.81\text{ m s}^{-2}}{0.45 \times 3.5\text{ m}}} = \sqrt{\frac{9.81}{1.575}} = \sqrt{6.229} = 2.496\text{ rad s}^{-1}\)
This is approximately \(2.5\text{ rad s}^{-1}\).

**Part (a) [2 marks]**
* **M1**: Two vertical force arrows of roughly equal length: weight pointing down, friction pointing up (1)
* **A1**: One horizontal force arrow pointing inwards, labeled as normal reaction / contact force (1)
*(Reject centrifugal force, reject any outward forces)*

**Part (b) [4 marks]**
* **M1**: Equates vertical forces: \(F = m g\) (1)
* **M1**: Equates horizontal force to centripetal force expression: \(R = m \omega^2 r\) (1)
* **M1**: States or uses the boundary condition \(F = \mu R\) to obtain \(g = \mu \omega^2 r\) (1)
* **A1**: Completes calculation to show \(\omega = 2.50\text{ rad s}^{-1}\) (must show calculation with 3 significant figures: \(2.496\text{ rad s}^{-1}\) or similar) (1)

**Part (a)**
For the protons to pass without deflection:
1. The electric force \(F_E\) and the magnetic force \(F_B\) on each proton must be equal in magnitude and opposite in direction.
2. The net force acting on the protons must be zero.

**Part (b)**
The electric force is:
\(F_E = q E\)

The magnetic force is:
\(F_B = q v B\)

For undeflected protons:
\(q E = q v B \implies v = \frac{E}{B}\)

The electric field strength \(E\) between the plates is:
\(E = \frac{V}{d} = \frac{1.8 \times 10^3\text{ V}}{15 \times 10^{-3}\text{ m}} = 1.20 \times 10^5\text{ V m}^{-1}\)

Substitute this into the expression for velocity:
\(v = \frac{1.20 \times 10^5\text{ V m}^{-1}}{0.12\text{ T}} = 1.00 \times 10^6\text{ m s}^{-1}\)

**Part (a) [2 marks]**
* **M1**: States that the electric force and magnetic force must be equal and opposite (1)
* **A1**: Concludes that the resultant force is zero (1)

**Part (b) [4 marks]**
* **M1**: Equates formulas for electric and magnetic force: \(q E = q v B\) or states \(v = E/B\) (1)
* **M1**: Recalls and uses \(E = V/d\) (1)
* **M1**: Calculates \(E = 1.2 \times 10^5\text{ V m}^{-1}\) (1)
* **A1**: Calculates \(v = 1.0 \times 10^6\text{ m s}^{-1}\) (accept \(1 \times 10^6\text{ m s}^{-1}\)) (1)

**Part (a)**
The energy \(Q\) required to raise the temperature of the water to boiling point is:
\(Q = m c \Delta\theta\)
where \(\Delta\theta = 100^\circ\text{C} - 18^\circ\text{C} = 82\text{ K}\)

\(Q = 1.2\text{ kg} \times 4200\text{ J kg}^{-1}\text{ K}^{-1} \times 82\text{ K} = 4.133 \times 10^5\text{ J}\)

The time \(t\) taken with a power of \(2.2\text{ kW} = 2200\text{ W}\) is:
\(t = \frac{Q}{P} = \frac{4.133 \times 10^5\text{ J}}{2200\text{ W}} = 187.9\text{ s}\)
This rounds to \(188\text{ s}\) (or \(1.9 \times 10^2\text{ s}\)).

**Part (b)**
The energy supplied by the heating element in \(87\text{ s}\) is:
\(Q = P \times t = 2200\text{ W} \times 87\text{ s} = 1.914 \times 10^5\text{ J}\)

The mass of water vaporized is \(m = 85\text{ g} = 0.085\text{ kg}\).

Using \(Q = m L\), where \(L\) is the specific latent heat of vaporization:
\(L = \frac{Q}{m} = \frac{1.914 \times 10^5\text{ J}}{0.085\text{ kg}} = 2.252 \times 10^6\text{ J kg}^{-1}\)
Rounding to two significant figures gives \(2.3 \times 10^6\text{ J kg}^{-1}\).

**Part (a) [3 marks]**
* **M1**: Calculates thermal energy required: \(Q = 1.2 \times 4200 \times 82 = 4.1 \times 10^5\text{ J}\) (1)
* **M1**: Relates energy to power and time: \(t = Q/P\) (1)
* **A1**: Calculates \(t = 190\text{ s}\) or \(188\text{ s}\) (1)

**Part (b) [3 marks]**
* **M1**: Calculates energy supplied: \(Q = 2200 \times 87 = 1.9 \times 10^5\text{ J}\) (1)
* **M1**: Rearranges latent heat equation: \(L = Q/m\) with mass converted to kg (1)
* **A1**: Calculates \(L = 2.3 \times 10^6\text{ J kg}^{-1}\) (accept \(2.25 \times 10^6\text{ J kg}^{-1}\)) (1)

**Part (a)**
The decay constant \(\lambda\) is related to the half-life \(T_{1/2}\) by:
\(\lambda = \frac{\ln 2}{T_{1/2}}\)
\(\lambda = \frac{0.69315}{1.25 \times 10^9\text{ yr}} = 5.545 \times 10^{-10}\text{ yr}^{-1}\)
This is approximately \(5.5 \times 10^{-10}\text{ yr}^{-1}\).

**Part (b)**
Each Argon-40 atom was formed by the decay of a Potassium-40 atom. Therefore, the original number of Potassium-40 atoms \(N_0\) is:
\(N_0 = N_{\text{Potassium}} + N_{\text{Argon}}\)
\(N_0 = 1.50 \times 10^{16} + 4.50 \times 10^{16} = 6.00 \times 10^{16}\text{ atoms}\)

The fraction of Potassium-40 remaining is:
\(\frac{N}{N_0} = \frac{1.50 \times 10^{16}}{6.00 \times 10^{16}} = 0.25 = \frac{1}{4}\)

Since \(\frac{1}{4} = (\frac{1}{2})^2\), exactly two half-lives have elapsed since the rock formed.
Age of the rock \(t\) is:
\(t = 2 \times T_{1/2} = 2 \times 1.25 \times 10^9\text{ years} = 2.50 \times 10^9\text{ years}\)

Alternatively, using the exponential decay equation:
\(N = N_0 e^{-\lambda t}\)
\(0.25 = e^{-\lambda t}\)
\(\ln(0.25) = -\lambda t\)
\(t = \frac{1.3863}{5.545 \times 10^{-10}\text{ yr}^{-1}} = 2.50 \times 10^9\text{ years}\).

**Part (a) [2 marks]**
* **M1**: Recalls \(\lambda = \frac{\ln 2}{T_{1/2}}\) (1)
* **A1**: Shows calculation resulting in \(5.5 \times 10^{-10}\text{ yr}^{-1}\) (must show at least 3 sig figs: \(5.54 \times 10^{-10}\)) (1)

**Part (b) [4 marks]**
* **M1**: Identifies original number of atoms: \(N_0 = 1.50 \times 10^{16} + 4.50 \times 10^{16} = 6.00 \times 10^{16}\) (1)
* **M1**: Calculates the ratio \(N/N_0 = 0.25\) (1)
* **M1**: Deduces that 2 half-lives have passed (or uses decay equation \(t = \ln(N_0/N)/\lambda\)) (1)
* **A1**: Calculates age as \(2.50 \times 10^9\text{ years}\) (accept \(2.5 \times 10^9\text{ years}\)) (1)

(a) Under vertical acceleration due to gravity, the vertical displacement is given by:
\(s = ut + \frac{1}{2}at^2\)
\(12.5 = 0 + \frac{1}{2}(9.81)t^2\)
\(t = \sqrt{\frac{2 \times 12.5}{9.81}} = 1.596\text{ s}\), which is approximately \(1.6\text{ s}\).

(b) The horizontal velocity remains constant throughout the flight:
\(v = \frac{s}{t} = \frac{22.0\text{ m}}{1.596\text{ s}} = 13.78\text{ m s}^{-1} \approx 13.8\text{ m s}^{-1}\)
(If using \(1.6\text{ s}\), \(v = 13.75\text{ m s}^{-1} \approx 13.8\text{ m s}^{-1}\))

(c) The vertical velocity component just before landing is:
\(v_y = gt = 9.81 \times 1.596 = 15.66\text{ m s}^{-1}\)
(Alternatively, using \(v_y^2 = 2gh = 2 \times 9.81 \times 12.5 \implies v_y = 15.66\text{ m s}^{-1}\))

The final speed is the magnitude of the combined components:
\(v_{\text{final}} = \sqrt{v_x^2 + v_y^2} = \sqrt{13.78^2 + 15.66^2} = \sqrt{189.89 + 245.25} = 20.86\text{ m s}^{-1} \approx 20.9\text{ m s}^{-1}\)

(a) (2 marks):
- Clear use of \(s = \frac{1}{2}gt^2\) or equivalent equation of motion (1)
- Correct calculation showing \(t \approx 1.596\text{ s}\) before rounding to \(1.6\text{ s}\) (1)

(b) (2 marks):
- Correct use of \(v = d/t\) with either \(1.6\text{ s}\) or \(1.596\text{ s}\) (1)
- Correct value of \(13.8\text{ m s}^{-1}\) (or \(13.75\text{ m s}^{-1}\)) with correct units (1)

(c) (2 marks):
- Correct calculation of vertical velocity component \(v_y = 15.7\text{ m s}^{-1}\) (1)
- Correct calculation of final speed using Pythagoras, yielding \(20.9\text{ m s}^{-1}\) (accept \(20.8 - 21.0\text{ m s}^{-1}\)) (1)

(a) First, calculate the tension force acting on the wire:
\(F = mg = 6.5 \times 9.81 = 63.77\text{ N}\)

Using the Young modulus formula:
\(E = \frac{\text{Stress}}{\text{Strain}} = \frac{F/A}{\Delta L/L}\)
\(\Delta L = \frac{F L}{A E}\)
\(\Delta L = \frac{63.77 \times 2.4}{1.5 \times 10^{-7} \times 1.8 \times 10^{11}} = \frac{153.05}{27000} = 5.67 \times 10^{-3}\text{ m}\) (or \(5.7 \times 10^{-3}\text{ m}\))

(b) The elastic strain energy is the area under the force-extension graph:
\(E_{\text{el}} = \frac{1}{2} F \Delta L\)
\(E_{\text{el}} = 0.5 \times 63.77 \times 5.67 \times 10^{-3} = 0.181\text{ J} \approx 0.18\text{ J}\)

(a) (3 marks):
- Correct calculation of the tension force (\(63.8\text{ N}\) or \(65\text{ N}\) if \(g=10\text{ N kg}^{-1}\)) (1)
- Correct rearrangement of Young modulus formula for extension \(\Delta L\) (1)
- Correct final answer for extension: \(5.7 \times 10^{-3}\text{ m}\) (accept \(5.6 \times 10^{-3}\text{ m}\) to \(5.8 \times 10^{-3}\text{ m}\)) (1)

(b) (3 marks):
- Use of \(E_{\text{el}} = \frac{1}{2} F \Delta L\) (or \(E_{\text{el}} = \frac{1}{2} k(\Delta L)^2\)) (1)
- Correct substitution of calculated force and extension (1)
- Correct final answer with units: \(0.18\text{ J}\) (allow ecf from (a)) (1)

(a) Calculate the grating spacing \(d\):
\(d = \frac{1}{500 \times 10^3\text{ m}^{-1}} = 2.0 \times 10^{-6}\text{ m}\)

Using the grating equation \(d \sin\theta = n\lambda\):
\(2.0 \times 10^{-6} \times \sin(38.5^\circ) = 2 \lambda\)
\(\sin(38.5^\circ) = 0.6225\)
\(\lambda = \frac{2.0 \times 10^{-6} \times 0.6225}{2} = 6.23 \times 10^{-7}\text{ m}\)
This is approximately \(6.2 \times 10^{-7}\text{ m}\).

(b) Replacing the grating with one that has 300 lines per mm instead of 500 lines per mm increases the slit separation \(d\) (since \(d = 1/N\)).
From \(\sin\theta = \frac{n\lambda}{d}\), because \(d\) increases while \(n\) and \(\lambda\) remain constant, \(\sin\theta\) must decrease.
Therefore, the diffraction angle \(\theta\) decreases, and the second-order maximum occurs closer to the central zero-order maximum.

(a) (3 marks):
- Correct calculation of grating spacing \(d = 2.0 \times 10^{-6}\text{ m}\) (1)
- Correct substitution into the grating formula \(d \sin\theta = n\lambda\) (1)
- Shows intermediate calculation leading to \(6.23 \times 10^{-7}\text{ m}\) (1)

(b) (3 marks):
- States that fewer lines per mm means the grating spacing \(d\) is larger (1)
- Refers to \(d \sin\theta = n\lambda\) (or \(\sin\theta \propto 1/d\)) to show \(\sin\theta\) must decrease (1)
- Concludes that the second-order maximum is observed at a smaller angle (or closer to the normal) (1)

(a) The current in each case can be found using the terminal potential difference \(V\) across the external resistor \(R\):
\(I_1 = \frac{V_1}{R_1} = \frac{1.35\text{ V}}{4.5\ \Omega} = 0.30\text{ A}\)
\(I_2 = \frac{V_2}{R_2} = \frac{1.44\text{ V}}{9.0\ \Omega} = 0.16\text{ A}\)

(b) Use the relation \(\varepsilon = V + Ir\):
Case 1: \(\varepsilon = 1.35 + 0.30r\)
Case 2: \(\varepsilon = 1.44 + 0.16r\)

Equating the two expressions for \(\varepsilon\):
\(1.35 + 0.30r = 1.44 + 0.16r\)
\(0.14r = 0.09\)
\(r = \frac{0.09}{0.14} \approx 0.643\ \Omega \approx 0.64\ \Omega\)

(c) Substitute \(r\) back into one of the equations:
\(\varepsilon = 1.35 + (0.30 \times 0.643) = 1.35 + 0.193 = 1.543\text{ V} \approx 1.54\text{ V}\)

(a) (2 marks):
- Calculates current for Case 1 correctly, \(I_1 = 0.30\text{ A}\) (1)
- Calculates current for Case 2 correctly, \(I_2 = 0.16\text{ A}\) (1)

(b) (2 marks):
- Equates two terminal voltage and internal resistance expressions (1)
- Correct calculation of \(r = 0.64\ \Omega\) (accept \(0.64\ \Omega\) to \(0.65\ \Omega\)) (1)

(c) (2 marks):
- Substitutes their values of \(I\) and \(r\) into \(\varepsilon = V + Ir\) (1)
- Correct calculation of \(\varepsilon = 1.54\text{ V}\) (accept range \(1.50 - 1.55\text{ V}\) depending on rounding) (1)

(a) At the top of the circular hill, the forces acting on the car are gravity \(mg\) (downwards) and normal contact force \(N\) (upwards). The resultant force acts towards the center of the circle:
\(mg - N = \frac{mv^2}{r}\)

To avoid losing contact with the track, we require \(N \ge 0\). The maximum speed occurs when the car is on the verge of flying off, where \(N = 0\):
\(mg = \frac{mv^2}{r} \implies v_{\text{max}} = \sqrt{gr}\)
\(v_{\text{max}} = \sqrt{9.81 \times 18} = 13.28\text{ m s}^{-1} \approx 13.3\text{ m s}^{-1}\)

(b) At the bottom of the dip, the net centripetal force acts upwards (towards the center):
\(N - mg = \frac{mv^2}{r}\)
\(N = m\left(g + \frac{v^2}{r}\right)\)
\(N = 450 \left(9.81 + \frac{12^2}{18}\right)\)
\(N = 450 (9.81 + 8.0) = 450 \times 17.81 = 8014.5\text{ N} \approx 8.0 \times 10^3\text{ N}\) (or \(8010\text{ N}\))

(a) (3 marks):
- Identifies that contact is lost when \(N = 0\) / gravity provides all of the centripetal force (1)
- Uses \(v = \sqrt{gr}\) with correct numbers (1)
- Obtains \(13.3\text{ m s}^{-1}\) (accept \(13\text{ m s}^{-1}\) if using \(g = 10\text{ m s}^{-2}\)) (1)

(b) (3 marks):
- Identifies correct equation for circular motion at the bottom: \(N - mg = \frac{mv^2}{r}\) (1)
- Correct substitution of mass, speed, and radius into this equation (1)
- Calculates normal contact force as \(8.0 \times 10^3\text{ N}\) (accept \(8000\text{ N}\) to \(8100\text{ N}\)) (1)

(a) Kinetic energy gained by the electron is equal to the work done by the accelerating potential difference:
\(\frac{1}{2} m_e v^2 = eV\)
\(v = \sqrt{\frac{2eV}{m_e}} = \sqrt{\frac{2 \times 1.60 \times 10^{-19}\text{ C} \times 2500\text{ V}}{9.11 \times 10^{-31}\text{ kg}}}\)
\(v = \sqrt{\frac{8.00 \times 10^{-16}}{9.11 \times 10^{-31}}} = \sqrt{8.782 \times 10^{14}} = 2.96 \times 10^7\text{ m s}^{-1}\)
This is approximately \(3.0 \times 10^7\text{ m s}^{-1}\).

(b) When entering the perpendicular magnetic field, the magnetic force acts as the centripetal force:
\(B e v = \frac{m_e v^2}{r}\)
\(r = \frac{m_e v}{B e}\)
Using the unrounded value of \(v = 2.96 \times 10^7\text{ m s}^{-1}\):
\(r = \frac{9.11 \times 10^{-31}\text{ kg} \times 2.96 \times 10^7\text{ m s}^{-1}}{4.2 \times 10^{-3}\text{ T} \times 1.60 \times 10^{-19}\text{ C}} = \frac{2.697 \times 10^{-23}}{6.72 \times 10^{-22}} = 4.01 \times 10^{-2}\text{ m}\) (or \(4.0\text{ cm}\))

(If using the given show-that value \(3.0 \times 10^7\text{ m s}^{-1}\)):
\(r = \frac{9.11 \times 10^{-31} \times 3.0 \times 10^7}{4.2 \times 10^{-3} \times 1.60 \times 10^{-19}} = 4.07 \times 10^{-2}\text{ m}\) (or \(4.1\text{ cm}\))

(a) (3 marks):
- Equates electrical work to kinetic energy, \(eV = \frac{1}{2}mv^2\) (1)
- Converts \(2.5\text{ kV}\) to \(2500\text{ V}\) and substitutes values correctly (1)
- Calculates the speed showing \(2.96 \times 10^7\text{ m s}^{-1}\) (1)

(b) (3 marks):
- Equates magnetic force to centripetal force, \(Bev = \frac{mv^2}{r}\) (1)
- Substitutes values into rearranged equation for \(r\) (1)
- Obtains a radius of \(4.0 \times 10^{-2}\text{ m}\) or \(4.1 \times 10^{-2}\text{ m}\) with correct unit (1)

(a) A pion is a meson, consisting of a quark-antiquark pair (such as \(u\bar{u}\) or \(d\bar{d}\)).

(b) Momentum must be conserved. Since the initial pion is stationary, the total momentum of the system before the decay is zero. Therefore, the total momentum of the two photons after the decay must also be zero. This requires the two photons to travel in opposite directions with equal and opposite momentum (\(p_1 = -p_2\)). Since the energy of a photon is related to its momentum by \(E = pc\), having equal momentum magnitudes means they must have equal energy.

(c) The rest mass-energy of the pion is completely converted into the energy of the two photons.
Total energy = \(135\text{ MeV}\).
Since there are two identical photons of equal energy, the energy of each photon is:
\(E = \frac{135\text{ MeV}}{2} = 67.5\text{ MeV}\)

Convert this energy to joules:
\(E = 67.5 \times 10^6\text{ eV} \times 1.60 \times 10^{-19}\text{ J eV}^{-1} = 1.08 \times 10^{-11}\text{ J}\)

(a) (1 mark):
- Identifies quark-antiquark structure (accept meson) (1)

(b) (2 marks):
- States that initial momentum is zero, so final total momentum must be zero to conserve momentum (1)
- Explains that equal momentum means equal energy because \(E = pc\) (1)

(c) (3 marks):
- Determines energy of one photon in MeV is \(67.5\text{ MeV\) (1)
- Correct conversion factor from MeV to J (multiplies by \(10^6\) and \(1.60 \times 10^{-19}\)) (1)
- Correct calculation to yield \(1.08 \times 10^{-11}\text{ J}\) (1)

(a) Convert the temperature to kelvin:
\(T_1 = 27 + 273.15 = 300.15\text{ K}\) (using \(300\text{ K}\) is acceptable)

Use the ideal gas equation:
\(pV = N k T\)
\(N = \frac{pV}{kT} = \frac{1.2 \times 10^5\text{ Pa} \times 3.5 \times 10^{-2}\text{ m}^3}{1.38 \times 10^{-23}\text{ J K}^{-1} \times 300.15\text{ K}} = \frac{4200}{4.142 \times 10^{-21}} = 1.014 \times 10^{24}\text{ atoms}\)

(b) For a gas of constant volume and constant mass, the pressure is directly proportional to absolute temperature:
\(\frac{p_1}{T_1} = \frac{p_2}{T_2}\)

Convert the new temperature to kelvin:
\(T_2 = 85 + 273.15 = 358.15\text{ K}\)

Calculate the new pressure:
\(p_2 = p_1 \times \frac{T_2}{T_1} = 1.2 \times 10^5\text{ Pa} \times \frac{358.15\text{ K}}{300.15\text{ K}} = 1.432 \times 10^5\text{ Pa} \approx 1.43 \times 10^5\text{ Pa}\)

(a) (3 marks):
- Converts temperature from Celsius to Kelvin (\(300\text{ K}\)) (1)
- Uses \(pV = NkT\) or \(pV = nRT\) with correct substitution (1)
- Calculates number of atoms correctly as \(1.0 \times 10^{24}\) (or \(1.01 \times 10^{24}\)) (1)

(b) (3 marks):
- Converts \(85^\circ\text{C}\) to Kelvin (\(358\text{ K}\)) (1)
- Uses the pressure law relation \(\frac{p_1}{T_1} = \frac{p_2}{T_2}\) (1)
- Obtains a pressure of \(1.43 \times 10^5\text{ Pa}\) (accept \(1.4 \times 10^5\text{ Pa}\) to \(1.44 \times 10^5\text{ Pa}\)) (1)

First, find the component of weight down the slope: \(W_{\parallel} = mg \sin(25^{\circ}) = 75 \times 9.81 \times \sin(25^{\circ}) = 310.9\text{ N}\). Next, find the net force: \(F_{\text{net}} = 310.9\text{ N} - 120\text{ N} = 190.9\text{ N}\). Calculate acceleration: \(a = F_{\text{net}} / m = 190.9 / 75 = 2.55\text{ m s}^{-2}\). Use \(v^2 = u^2 + 2as\) with \(u = 0\): \(v = \sqrt{2 \times 2.55 \times 80} = 20.2\text{ m s}^{-1}\). Use \(v = u + at\) to find the time: \(t = v/a = 20.2 / 2.55 = 7.92\text{ s}\).

1 mark: Calculate parallel weight component of 311 N. 1 mark: Deduce net accelerating force of 191 N. 1 mark: Calculate acceleration as 2.55 m s^-2. 1 mark: Use correct kinematic equation to find velocity. 1 mark: Obtain velocity of 20.2 m s^-1. 1 mark: Obtain time of 7.92 s.

Calculate cross-sectional area: \(A = \pi d^2 / 4 = \pi \times (0.80 \times 10^{-3})^2 / 4 = 5.03 \times 10^{-7}\text{ m}^2\). Calculate tension: \(F = mg = 12 \times 9.81 = 117.7\text{ N}\). Use Young modulus equation to find extension: \(\Delta L = FL / (AE) = (117.7 \times 2.5) / (5.03 \times 10^{-7} \times 2.0 \times 10^{11}) = 2.93 \times 10^{-3}\text{ m}\) (or 2.93 mm). Calculate stored elastic strain energy: \(E = 0.5 \times F \times \Delta L = 0.5 \times 117.7 \times 2.93 \times 10^{-3} = 0.172\text{ J}\).

1 mark: Calculate cross-sectional area of wire as 5.03 x 10^-7 m^2. 1 mark: Calculate tension as 118 N. 1 mark: Correctly rearrange Young modulus formula. 1 mark: Obtain extension of 2.93 mm. 1 mark: Use elastic strain energy formula 1/2 Fx. 1 mark: Obtain energy of 0.172 J.

Calculate grating spacing: \(d = 1 \times 10^{-3} / 600 = 1.67 \times 10^{-6}\text{ m}\). Use diffraction formula: \(d \sin\theta = n\lambda\). For \(n = 2\), \(\lambda = (1.67 \times 10^{-6} \sin 44.0^{\circ}) / 2 = 5.80 \times 10^{-7}\text{ m}\). Calculate photon energy: \(E = hc/\lambda = (6.63 \times 10^{-34} \times 3.00 \times 10^8) / 5.80 \times 10^{-7} = 3.43 \times 10^{-19}\text{ J}\). Convert to eV: \(E = 3.43 \times 10^{-19} / 1.60 \times 10^{-19} = 2.14\text{ eV}\). Calculate max kinetic energy: \(K_{\text{max}} = E - \Phi = 2.14\text{ eV} - 1.85\text{ eV} = 0.29\text{ eV}\).

1 mark: Calculate grating spacing d = 1.67 x 10^-6 m. 1 mark: Use d sin(theta) = n lambda. 1 mark: Obtain wavelength of 5.80 x 10^-7 m. 1 mark: Use E = hc/lambda to calculate photon energy in joules or eV. 1 mark: Convert photon energy to 2.14 eV. 1 mark: Obtain maximum kinetic energy of 0.29 eV.

Set up simultaneous equations using \(\varepsilon = I(R+r)\). Equation 1: \(\varepsilon = 1.20(4.0 + r) = 4.80 + 1.20r\). Equation 2: \(\varepsilon = 0.55(10.0 + r) = 5.50 + 0.55r\). Equating the expressions: \(4.80 + 1.20r = 5.50 + 0.55r\), which gives \(0.65r = 0.70\), so \(r = 1.08\ \Omega\). Substitute back: \(\varepsilon = 1.20(4.0 + 1.08) = 6.10\text{ V}\). Graphically: Measure terminal potential difference \(V = IR\) and current \(I\). Plot \(V\) against \(I\). The equation is \(V = \varepsilon - Ir\). The y-intercept gives the emf \(\varepsilon\), and the negative of the gradient gives the internal resistance \(r\).

1 mark: Set up both simultaneous equations. 1 mark: Solve for internal resistance to get 1.08 ohms. 1 mark: Solve for emf to get 6.10 V. 1 mark: State that terminal PD V should be plotted against current I. 1 mark: Identify that the y-intercept is the emf. 1 mark: Identify that the gradient is equal to negative internal resistance.

Resolve forces vertically: \(N \cos\theta = mg\). Resolve forces horizontally: \(N \sin\theta = m v^2 / r\). Divide the horizontal equation by the vertical equation: \(\tan\theta = v^2 / (rg)\). Rearrange for speed: \(v = \sqrt{rg \tan\theta}\). Substitute the values: \(v = \sqrt{1.50 \times 9.81 \times \tan(20.0^{\circ})} = \sqrt{14.715 \times 0.3640} = 2.31\text{ m s}^{-1}\).

1 mark: Write vertical balance equation N cos(theta) = mg. 1 mark: Write horizontal centripetal force equation N sin(theta) = m v^2 / r. 1 mark: Combine equations to show tan(theta) = v^2 / rg. 1 mark: Rearrange correctly to solve for v. 1 mark: Substitute correct values. 1 mark: Obtain final speed of 2.31 m s^-1.

Equate kinetic energy to electrical potential energy: \(0.5 m v^2 = eV\). Rearrange for speed: \(v = \sqrt{2eV/m} = \sqrt{2 \times 1.60 \times 10^{-19} \times 2500 / 9.11 \times 10^{-31}} = 2.96 \times 10^7\text{ m s}^{-1}\). For circular path, magnetic force equals centripetal force: \(Bev = m v^2 / r\). Rearrange for radius: \(r = mv / (Be) = (9.11 \times 10^{-31} \times 2.96 \times 10^7) / (4.20 \times 10^{-3} \times 1.60 \times 10^{-19}) = 0.0402\text{ m}\) (or 4.02 cm).

1 mark: Use conservation of energy equation 1/2 m v^2 = eV. 1 mark: Perform correct substitutions for speed calculation. 1 mark: Obtain speed of 2.96 x 10^7 m s^-1. 1 mark: Equate magnetic force to centripetal force. 1 mark: Rearrange for radius r = mv / Be. 1 mark: Obtain radius of 0.0402 m (or 4.02 cm).

Use ideal gas law: \(pV = nRT\). Solve for moles: \(n = pV / RT = (1.05 \times 10^5 \times 0.0250) / (8.31 \times 293) = 1.08\text{ moles}\). Since volume is constant, \(p_1 / T_1 = p_2 / T_2\). Solve for final temperature: \(T_2 = T_1 \times (p_2 / p_1) = 293 \times (1.85 \times 10^5 / 1.05 \times 10^5) = 516\text{ K}\). The mean kinetic energy of a molecule is \(E = 1.5 k_B T\). Change in mean kinetic energy: \(\Delta E = 1.5 k_B \Delta T = 1.5 \times 1.38 \times 10^{-23} \times (516 - 293) = 4.62 \times 10^{-21}\text{ J}\).

1 mark: Use pV = nRT to find moles. 1 mark: Obtain 1.08 moles. 1 mark: Use p1/T1 = p2/T2 for constant volume. 1 mark: Obtain final temperature of 516 K. 1 mark: Use 3/2 kT for mean kinetic energy. 1 mark: Obtain change in mean kinetic energy of 4.62 x 10^-21 J.

Use Wien's displacement law: \(\lambda_{\text{max}} T = 2.898 \times 10^{-3}\text{ m K}\). Calculate temperature: \(T = 2.898 \times 10^{-3} / 4.10 \times 10^{-7} = 7068\text{ K}\) (rounded to 7070 K). Use Stefan's law: \(L = 4 \pi r^2 \sigma T^4\). Rearrange for radius: \(r = \sqrt{L / (4 \pi \sigma T^4)}\). Calculate denominator term: \(4 \pi \sigma T^4 = 4 \pi \times 5.67 \times 10^{-8} \times (7068)^4 = 1.778 \times 10^9\text{ W m}^{-2}\). Solve for radius: \(r = \sqrt{3.20 \times 10^{27} / 1.778 \times 10^9} = 1.34 \times 10^9\text{ m}\).

1 mark: Use Wien's displacement law. 1 mark: Calculate surface temperature as 7070 K. 1 mark: State Stefan's law. 1 mark: Rearrange Stefan's law for radius r. 1 mark: Calculate T^4 correctly. 1 mark: Obtain final radius of 1.34 x 10^9 m.

1. Find the vertical displacement to drop: \(s_y = 3.2\text{ m} - 1.8\text{ m} = 1.4\text{ m}\).
2. Using \(s_y = u_y t + \frac{1}{2}g t^2\) with \(u_y = 0\):
\(1.4 = 0.5 \times 9.81 \times t^2 \implies t^2 = 0.2854\text{ s}^2 \implies t = 0.534\text{ s}\).
3. Horizontal motion: \(v = \frac{s_x}{t} = \frac{8.5\text{ m}}{0.534\text{ s}} = 15.9\text{ m s}^{-1} \approx 16\text{ m s}^{-1}\).

- Calculated vertical drop displacement of 1.4 m (1 mark)
- Uses vertical equation with u = 0 (1 mark)
- Correct calculation of time of flight t = 0.53 s (1 mark)
- Uses horizontal velocity equation v = s / t (1 mark)
- Substitutes values of s_x and t correctly (1 mark)
- Final answer of 16 m s^-1 (or 15.9 m s^-1) with appropriate units (1 mark)

Both materials experience the same extension \(\Delta L\).
Total tension is the sum of tensions: \(F_{\text{total}} = F_{\text{steel}} + F_{\text{alu}}\).
Using \(F = \frac{E A \Delta L}{L}\):
\(F_{\text{total}} = \left( \frac{E_{\text{steel}} A_{\text{steel}}}{L} + \frac{E_{\text{alu}} A_{\text{alu}}}{L} \right) \Delta L\)
Substitute values:
\(650 = \frac{\Delta L}{2.5} \times \left( (2.0 \times 10^{11} \times 2.0 \times 10^{-6}) + (7.0 \times 10^{10} \times 4.5 \times 10^{-6}) \right)\)
\(650 = \frac{\Delta L}{2.5} \times (4.0 \times 10^5 + 3.15 \times 10^5)\)
\(1625 = 7.15 \times 10^5 \Delta L\)
\(\Delta L = 2.27 \times 10^{-3}\text{ m} \approx 2.3 \times 10^{-3}\text{ m}\).

- Recalls or uses E = stress / strain or F = E A delta L / L (1 mark)
- Identifies that total force is the sum of the forces in steel and aluminium (1 mark)
- Expresses total force in terms of a single extension delta L (1 mark)
- Substitutes area and Young modulus values correctly into the expression (1 mark)
- Rearranges to solve for delta L (1 mark)
- Final answer of 2.3 x 10^-3 m (or 2.27 x 10^-3 m) (1 mark)

1. Grating spacing: \(d = \frac{1\text{ mm}}{450} = 2.222 \times 10^{-6}\text{ m}\).
2. The angle from the central maximum to a second-order maximum is half the total angle: \(\theta_2 = 15.6^\circ\).
3. Using \(d \sin\theta = n \lambda\) for \(n=2\):
\(\lambda = \frac{d \sin\theta_2}{2} = \frac{2.222 \times 10^{-6} \times \sin(15.6^\circ)}{2} = 2.99 \times 10^{-7}\text{ m}\).
4. For \(n = 4\): \(\sin\theta_4 = \frac{4\lambda}{d} = \frac{4 \times 2.99 \times 10^{-7}}{2.222 \times 10^{-6}} = 0.538\).
5. Since \(\sin\theta_4 < 1\), the fourth-order maximum is geometrically possible and can be observed.

- Correct calculation of d = 2.22 x 10^-6 m (1 mark)
- Identifies half-angle theta = 15.6 degrees (1 mark)
- Uses d sin(theta) = n lambda with n = 2 (1 mark)
- Obtains wavelength lambda = 2.99 x 10^-7 m (1 mark)
- Calculates sin(theta_4) or theoretical maximum order (1 mark)
- Concludes that n=4 is possible since sin(theta_4) < 1 (1 mark)

1. Use \(\varepsilon = I(R + r)\).
2. First scenario: \(\varepsilon = 0.45(8.0 + r)\)
3. Second scenario: \(\varepsilon = 0.22(18.0 + r)\)
4. Equate the two: \(0.45(8.0 + r) = 0.22(18.0 + r)\)
5. Solve for \(r\): \(3.6 + 0.45r = 3.96 + 0.22r \implies 0.23r = 0.36 \implies r = 1.57\ \Omega \approx 1.6\ \Omega\).
6. Substitute back to find \(\varepsilon\): \(\varepsilon = 0.45(8.0 + 1.57) = 4.3\text{ V}\).

- Recalls or uses emf = I(R + r) (1 mark)
- Sets up first simultaneous equation (1 mark)
- Sets up second simultaneous equation (1 mark)
- Equates or uses substitution to eliminate emf (1 mark)
- Correct calculation of internal resistance r = 1.6 ohms (accept 1.57 ohms) (1 mark)
- Correct calculation of emf = 4.3 V (accept 4.31 V) (1 mark)

Forces acting on the car are the downward weight \(mg\) and the normal contact force \(N\) directed inward and upward perpendicular to the curve (at \(35^\circ\) above horizontal).
Resolving vertically: \(N \sin(35^\circ) = mg\)
Resolving horizontally: \(N \cos(35^\circ) = \frac{m v^2}{r}\)
Dividing the two equations: \(\frac{N \sin(35^\circ)}{N \cos(35^\circ)} = \frac{mg}{m v^2 / r} \implies \tan(35^\circ) = \frac{rg}{v^2}\)
Rearranging for \(v\): \(v = \sqrt{\frac{rg}{\tan(35^\circ)}} = \sqrt{\frac{0.45 \times 9.81}{\tan(35^\circ)}} = \sqrt{\frac{4.4145}{0.7002}} = 2.51\text{ m s}^{-1}\).

- Vector diagram showing Weight downwards and Normal force at 35 degrees to horizontal (1 mark)
- Formulates vertical equilibrium equation N sin(35) = mg (1 mark)
- Formulates horizontal centripetal force equation N cos(35) = m v^2 / r (1 mark)
- Combines both equations to eliminate mass and normal force (1 mark)
- Rearranges to solve for v (1 mark)
- Obtains final speed v = 2.5 m s^-1 (1 mark)

1. Faraday's law states that the induced e.m.f. is proportional to the rate of change of magnetic flux linkage.
2. Area of the coil: \(A = (0.040\text{ m})^2 = 1.6 \times 10^{-3}\text{ m}^2\).
3. Magnetic flux linkage: \(N\Phi = NBA = 150 \times 0.18 \times 1.6 \times 10^{-3} = 0.0432\text{ Wb-turns}\).
4. Average induced e.m.f.: \(\varepsilon = \frac{\Delta(N\Phi)}{\Delta t}\).
5. \(\varepsilon = \frac{0.0432\text{ Wb-turns}}{0.25\text{ s}} = 0.1728\text{ V}\).
Rounding to 2 significant figures gives \(0.17\text{ V}\).

- Statement of Faraday's Law in terms of rate of change of magnetic flux linkage (1 mark)
- Calculates area of the square coil A = 1.6 x 10^-3 m^2 (1 mark)
- Calculates magnetic flux linkage N*Phi = 0.0432 Wb (1 mark)
- Recalls or uses epsilon = delta(N*Phi) / delta(t) (1 mark)
- Substitutes values into the e.m.f. equation (1 mark)
- Obtains final answer of 0.17 V (or 0.173 V) (1 mark)

1. For constant pressure: \(\frac{V_1}{T_1} = \frac{V_2}{T_2}\).
2. \(T_2 = T_1 \times \frac{V_2}{V_1} = 290\text{ K} \times \frac{3.6 \times 10^{-3}\text{ m}^3}{2.4 \times 10^{-3}\text{ m}^3} = 435\text{ K}\).
3. Work done at constant pressure: \(W = P \Delta V\).
4. Volume change: \(\Delta V = (3.6 - 2.4) \times 10^{-3} = 1.2 \times 10^{-3}\text{ m}^3\).
5. Work done: \(W = 1.1 \times 10^5\text{ Pa} \times 1.2 \times 10^{-3}\text{ m}^3 = 132\text{ J}\).

- Uses V_1 / T_1 = V_2 / T_2 or PV = NkT (1 mark)
- Calculates final temperature T_2 = 435 K (or 440 K) (1 mark)
- Recalls or uses W = P * delta(V) (1 mark)
- Calculates the change in volume delta(V) = 1.2 x 10^-3 m^3 (1 mark)
- Correct substitution of pressure and change in volume (1 mark)
- Final work done of 132 J (or 130 J) with correct unit (1 mark)

1. State Stefan-Boltzmann law: \(L = 4 \pi r^2 \sigma T^4\).
2. Rearrange for radius: \(r = \sqrt{\frac{L}{4 \pi \sigma T^4}}\).
3. Calculate \(T^4 = (8200)^4 = 4.521 \times 10^{15}\text{ K}^4\).
4. Substitute values: \(r = \sqrt{\frac{4.6 \times 10^{27}}{4 \times \pi \times 5.67 \times 10^{-8} \times 4.521 \times 10^{15}}}\).
5. Calculate denominator: \(3.221 \times 10^9\text{ W m}^{-2}\).
6. Solve for \(r\): \(r = \sqrt{1.428 \times 10^{18}} = 1.19 \times 10^9\text{ m}\). This is approximately \(1.2 \times 10^9\text{ m}\).

- Recalls or uses Stefan-Boltzmann Law: L = A * sigma * T^4 (1 mark)
- Identifies surface area of a sphere A = 4 * pi * r^2 (1 mark)
- Correctly rearranges equation for r or r^2 (1 mark)
- Calculates temperature to the fourth power correctly (1 mark)
- Substitutes values correctly into the expression (1 mark)
- Evaluates final answer as 1.19 x 10^9 m, demonstrating it is approximately 1.2 x 10^9 m (1 mark)

According to classical wave theory:
1. Light waves carry continuous energy. If light of any frequency is incident, energy should accumulate over time until electrons gain enough energy to escape. Thus, there should be no minimum (threshold) frequency.
2. Higher intensity means larger amplitude waves carrying more energy per second, which should increase the maximum kinetic energy of the emitted electrons.

Experimental observations of the photoelectric effect show:
1. There is a threshold frequency \(f_0\) below which no photoelectrons are emitted, regardless of the intensity of the light.
2. Photoelectric emission is instantaneous (no measurable time delay), even at extremely low intensities.
3. The maximum kinetic energy of emitted photoelectrons depends solely on the frequency of the incident light, not its intensity. Increasing the intensity only increases the number of photoelectrons emitted per second (current).

These observations support the photon (particulate) model of light:
1. Light consists of discrete packets of energy called photons, each with energy \(E = hf\).
2. One photon interacts with exactly one electron in a one-to-one collision, transferring its energy instantaneously.
3. If the photon energy is less than the work function \(\Phi\) of the metal (i.e., \(hf < \Phi\)), no emission occurs, explaining the threshold frequency \(f_0 = \Phi / h\).
4. Einstein's photoelectric equation \(hf = \Phi + E_{k(\text{max})}\) shows that \(E_{k(\text{max})}\) depends linearly on frequency \(f\).

Award marks based on the following criteria (max 7 marks for physics content, 0.4 marks for quality of written communication):

- **Wave Theory Predictions (Max 2 marks):**
- Wave theory predicts energy accumulates over time, so emission should occur at any frequency given time. (1)
- Wave theory predicts that increasing intensity should increase the maximum kinetic energy of the emitted electrons. (1)

- **Experimental Observations (Max 2 marks):**
- Observation of a threshold frequency below which no emission occurs. (1)
- Emission is instantaneous/there is no time delay, and maximum kinetic energy is independent of intensity. (1)

- **Photon Model Explanations (Max 3 marks):**
- Description of light as discrete packets/photons with energy \(E = hf\). (1)
- Explanation of the 1-to-1 interaction between a photon and an electron. (1)
- Use of the equation \(hf = \Phi + E_{k(\text{max})}\) to show why \(E_{k(\text{max})}\) depends only on frequency, or why threshold frequency exists when \(hf = \Phi\). (1)

- **QWC (0.4 marks):**
- The explanation is logical, uses appropriate scientific terminology (e.g., work function, threshold frequency, intensity), and shows a clear structure. (0.4)

1. **Role of the Magnetic Field:** A uniform magnetic field acts perpendicular to the plane of the D-shaped electrodes (Dees). It exerts a magnetic force \(F_B = Bqv\) perpendicular to the direction of motion of the protons. This acts as a centripetal force: \(Bqv = \frac{mv^2}{r}\). This forces the protons into circular paths within the Dees without changing their speed.
2. **Role of the Electric Field:** A high-frequency alternating potential difference (and hence electric field) is applied across the narrow gap between the Dees. This electric field accelerates the protons as they cross the gap, increasing their kinetic energy.
3. **Increasing Radius:** Each time the protons are accelerated across the gap, their speed \(v\) increases. Rearranging the centripetal force equation gives the radius \(r = \frac{mv}{Bq}\). As \(v\) increases, the radius \(r\) of their circular path increases, causing them to spiral outwards.
4. **Constant Frequency of the AC Supply:** The time spent inside one Dee is half of the period of rotation: \(t = \frac{T}{2} = \frac{\pi r}{v}\). Substituting \(r = \frac{mv}{Bq}\) yields \(t = \frac{\pi m}{Bq}\). Since the mass \(m\), magnetic flux density \(B\), and charge \(q\) are constant, the time \(t\) spent in each Dee is constant, independent of the proton's speed and path radius. Therefore, the frequency of the alternating potential difference must remain constant at \(f = \frac{1}{T} = \frac{Bq}{2\pi m}\) to maintain synchronization with the proton crossings.

Award marks based on the following criteria (max 7 marks for physics content, 0.4 marks for quality of written communication):

- **Magnetic Field Role (Max 2 marks):**
- Magnetic field is perpendicular to the motion/Dees. (1)
- Magnetic force acts as a centripetal force: \(Bqv = \frac{mv^2}{r}\) to maintain circular motion inside the Dees. (1)

- **Electric Field Role (Max 2 marks):**
- Electric field exists in the gap between the Dees. (1)
- Electric field accelerates protons/increases kinetic energy only when crossing the gap. (1)

- **Frequency Derivation & Explanation (Max 3 marks):**
- Shows that speed and radius increase proportionally, leading to \(r = \frac{mv}{Bq}\). (1)
- Derives or states the time in a Dee: \(t = \frac{\pi m}{Bq}\) or period \(T = \frac{2\pi m}{Bq}\). (1)
- Explains that because \(t\) is independent of speed/radius, the frequency of the alternating voltage can and must remain constant. (1)

- **QWC (0.4 marks):**
- Clear, logical description of the physical process from start to finish with appropriate vocabulary. (0.4)

1. **At the start (instant of jump, \(t = 0\)):** The velocity of the skydiver is zero. Therefore, there is zero air resistance (drag \(D = 0\)). The only force acting on the skydiver is weight (\(W = mg\)) acting downwards. The resultant force is equal to \(mg\), so the initial acceleration is equal to \(g = 9.81\text{ m s}^{-2}\).
2. **During the fall (as velocity increases):** As the skydiver accelerates and their downward velocity increases, the air resistance \(D\) acting upwards increases. The net downward force on the skydiver is given by \(F_{\text{net}} = W - D = ma\). Since \(D\) is increasing and \(W\) remains constant, the net force \(F_{\text{net}}\) decreases, which causes the acceleration \(a\) of the skydiver to decrease. The velocity continues to increase, but at a decreasing rate.
3. **Reaching terminal velocity:** The drag force eventually increases until it is equal in magnitude to the weight of the skydiver (\(D = W\)). At this point, the net force is zero (\(F_{\text{net}} = 0\)), and according to Newton's First Law, the acceleration becomes zero (\(a = 0\)). The skydiver ceases to accelerate and continues to fall at a constant velocity, known as the terminal velocity.

Award marks based on the following criteria (max 7 marks for physics content, 0.4 marks for quality of written communication):

- **Initial State (Max 2 marks):**
- At \(t=0\), velocity is zero so drag/air resistance is zero. (1)
- The only force is weight, resulting in an initial acceleration of \(g\). (1)

- **Intermediate Phase (Max 3 marks):**
- As velocity increases, drag/air resistance increases. (1)
- State the equation \(F_{\text{net}} = W - D = ma\) or describe that net force decreases. (1)
- Explanation that because net force decreases, acceleration decreases (but velocity still increases). (1)

- **Terminal Velocity Phase (Max 2 marks):**
- Drag force eventually equals weight (\(D = W\)). (1)
- Resultant force is zero, acceleration becomes zero, and velocity remains constant. (1)

- **QWC (0.4 marks):**
- The response must be structured sequentially in terms of chronological phases (initial, intermediate, final) with accurate force-acceleration relationships. (0.4)

1. **Elastic Deformation:** Both materials experience elastic deformation at low stresses, where strain is proportional to stress (Hooke's Law is obeyed), and the material returns to its original dimensions when the stress is removed. At this stage, interatomic bonds are stretched but not broken.
2. **Plastic Deformation (Copper):** Beyond its elastic limit, copper undergoes significant plastic (permanent) deformation. In copper, metal ions are arranged in a regular lattice surrounded by a sea of delocalised electrons. The bonding is non-directional. This structure contains defects called dislocations. Under stress, these dislocations can slide through the lattice, allowing planes of atoms to slip past each other without breaking the metallic bond. This makes the material ductile.
3. **Brittle Behavior (Glass):** Glass does not undergo plastic deformation. It has an amorphous or highly ordered network structure with strong covalent or ionic directional bonds. There are no mobile dislocations or slip planes. If a high stress is applied, the bonds cannot slide; instead, they stretch until they break.
4. **Fracture Mechanisms:** Copper undergoes 'necking' where the cross-sectional area decreases at a specific point, and it eventually undergoes a ductile fracture after absorbing a large amount of energy. Glass fractures suddenly and catastrophically. Tiny surface cracks concentrate stress at their tips, causing bonds at the crack tip to snap sequentially, propagating the crack across the material at the speed of sound with little energy absorption.

Award marks based on the following criteria (max 7 marks for physics content, 0.4 marks for quality of written communication):

- **Macroscopic Behavior Comparison (Max 2 marks):**
- Both undergo elastic deformation initially. (1)
- Copper undergoes plastic deformation before breaking, whereas glass breaks suddenly with no/negligible plastic deformation. (1)

- **Microscopic Structure of Ductile Copper (Max 3 marks):**
- Copper has a metallic lattice with non-directional bonds / sea of electrons. (1)
- Presence of dislocations that can move / slide along slip planes under stress. (1)
- Atoms slide into new stable positions, resulting in permanent shape change (ductility) and necking. (1)

- **Microscopic Structure of Brittle Glass (Max 2 marks):**
- Glass has strong, directional covalent/ionic bonds that cannot slide. (1)
- Stress concentrates at microscopic surface cracks, leading to rapid crack propagation and clean, sudden fracture. (1)

- **QWC (0.4 marks):**
- Clear comparative structure ('both', 'whereas', 'on the other hand') and correct terminology (dislocations, plastic deformation, stress concentration). (0.4)

1. **How tracks are formed:** Charged particles pass through the superheated liquid hydrogen, ionizing the atoms. This triggers the formation of tiny vapor bubbles along their path, creating visible tracks.
2. **Determining Charge:** The bubble chamber is situated within a uniform magnetic field. Charged particles experience a magnetic force \(F_B = Bqv\) perpendicular to their velocity, causing their tracks to curve. Using Fleming's Left Hand Rule (or the right-hand rule for positive charges), the direction of curvature (clockwise vs. counter-clockwise) reveals the sign of the charge (positive or negative).
3. **Determining Momentum:** The radius of curvature \(r\) of a track is related to the particle's momentum \(p\) by the formula: \(Bqv = \frac{mv^2}{r} \implies r = \frac{mv}{Bq} = \frac{p}{Bq}\). Since \(B\) and the magnitude of charge \(q\) (usually \(\pm e\)) are known, measuring \(r\) allows the momentum \(p\) to be calculated. Tracks with a larger radius of curvature have higher momentum, and spiral inward as they lose kinetic energy and momentum through collisions with hydrogen atoms.
4. **Identifying Neutral Particles:** Neutral particles do not ionize the hydrogen and therefore leave no tracks in the bubble chamber. Their presence is detected by observing a gap in a reaction (e.g., a neutral particle decaying into two charged particles, creating a 'V' shape track that starts from empty space) or by applying the law of conservation of momentum. If the total momentum of the visible outgoing particles does not equal the momentum of the incoming particles, a neutral particle must have been produced to conserve momentum.

Award marks based on the following criteria (max 7 marks for physics content, 0.4 marks for quality of written communication):

- **Track Formation (Max 1 mark):**
- Charged particles ionize the liquid hydrogen, forming bubble tracks. (1)

- **Charge Identification (Max 2 marks):**
- Magnetic field causes charged particles to curve. (1)
- Direction of curvature indicates charge sign using Fleming's Left Hand Rule. (1)

- **Momentum Calculation (Max 2 marks):**
- Relates magnetic force to centripetal force: \(Bqv = \frac{mv^2}{r}\) or states \(r = \frac{p}{Bq}\). (1)
- States that radius \(r\) is proportional to momentum \(p\) (larger radius means higher momentum), and tracks spiral in as energy/momentum is lost. (1)

- **Neutral Particles (Max 2 marks):**
- Neutral particles do not ionize and thus leave no tracks. (1)
- Their presence is inferred via conservation of momentum/charge at vertices, or via 'V' tracks appearing from 'nothing'. (1)

- **QWC (0.4 marks):**
- Structured explanation with precise physics vocabulary (ionization, centripetal force, conservation laws). (0.4)

1. **Key Assumptions of the Kinetic Theory:**
- The gas contains a very large number of molecules in rapid, continuous, random motion.
- The volume of the gas molecules is negligible compared to the total volume of the container.
- There are no intermolecular forces acting between the molecules except during collisions (the molecules travel in straight lines between collisions).
- All collisions between molecules and with the walls of the container are perfectly elastic (kinetic energy is conserved).
- The duration of a collision is negligible compared to the time between collisions.

2. **Microscopic Explanation of Pressure:**
- Consider a molecule of mass \(m\) moving with velocity \(v\) perpendicular to a container wall.
- When it collides elastically with the wall, it rebounds with velocity \(-v\).
- The change in momentum of the molecule is \(\Delta p = p_{\text{final}} - p_{\text{initial}} = -mv - mv = -2mv\).
- By Newton's Third Law, the wall exerts a force on the molecule, and the molecule exerts an equal and opposite force on the wall.
- According to Newton's Second Law, this force is equal to the rate of change of momentum of the molecule: \(F = \frac{\Delta p}{\Delta t} = \frac{2mv}{\Delta t}\).
- There are a vast number of molecules colliding with the walls constantly. The sum of these individual collision forces gives a total average force \(F_{\text{total}}\) on the wall.
- Pressure \(P\) is defined as this total force per unit area (\(P = \frac{F_{\text{total}}}{A}\)). Since the collisions are numerous and frequent, this results in a constant, macroscopic pressure.

Award marks based on the following criteria (max 7 marks for physics content, 0.4 marks for quality of written communication):

- **Assumptions (Max 2 marks):**
- Identifies at least two valid assumptions (e.g., elastic collisions, negligible molecular volume, no intermolecular forces, random motion). (2) [1 mark for one assumption, 2 marks for two or more]

- **Momentum Change (Max 2 marks):**
- Molecules collide with the wall and reverse direction/rebound. (1)
- States that change in momentum is \(\Delta p = -2mv\) (or equivalent vector change). (1)

- **Force and Newton's Laws (Max 2 marks):**
- Relates force to rate of change of momentum (Newton's Second Law: \(F = \frac{\Delta p}{\Delta t}\)). (1)
- Identifies that the wall exerts a force on the molecule, and the molecule exerts an equal and opposite force on the wall (Newton's Third Law). (1)

- **Pressure Definition (Max 1 mark):**
- Explains that pressure is total force divided by area (\(P = \frac{F}{A}\)) over a large number of collisions. (1)

- **QWC (0.4 marks):**
- Clear, logical sequence transitioning from microscopic assumptions to macroscopic pressure with correct algebraic representation of momentum. (0.4)

1. **Forced Oscillations:** Forced oscillations occur when an external periodic force (the driver) is applied to an oscillating system. In this case, the regular, rhythmic footsteps of the walking pedestrians act as the periodic driving force. The bridge is forced to oscillate at the frequency of this driving force, rather than its own natural frequency.
2. **Resonance:** Resonance occurs when the frequency of the periodic driving force matches the natural frequency of the bridge. When this happens, there is a maximum transfer of energy from the pedestrians to the bridge. This causes the amplitude of the bridge's oscillations to increase rapidly to a maximum value, which can damage the structure.
3. **Role of Damping:** Damping is the process where energy is removed from an oscillating system by dissipative forces (such as friction or viscous drag).
4. **Mechanism of Damping Systems:** Damping systems (like tuned mass dampers or viscous dampers) are designed to exert forces opposite to the direction of motion of the bridge. These damping forces do work against the motion, converting the mechanical kinetic energy of the oscillating bridge into thermal energy (heat). This significantly reduces the maximum amplitude of the oscillations at resonance and prevents the bridge from reaching dangerous levels of motion. Additionally, heavier damping broadens the resonance curve, making the bridge less sensitive to driving frequencies close to its natural frequency.

Award marks based on the following criteria (max 7 marks for physics content, 0.4 marks for quality of written communication):

- **Forced Oscillations (Max 2 marks):**
- Explains that forced oscillations are driven by an external periodic force. (1)
- Identifies pedestrian footsteps as the periodic driving force, driving the bridge at the footstep frequency. (1)

- **Resonance (Max 2 marks):**
- Explains that resonance occurs when the driving frequency equals the natural frequency of the bridge. (1)
- Identifies that resonance results in maximum energy transfer and maximum amplitude of oscillation. (1)

- **Damping and Energy Transfer (Max 3 marks):**
- Defines damping as the dissipation of energy due to frictional/viscous forces. (1)
- Explains that dampers do work against the motion, converting kinetic energy to thermal energy. (1)
- States that damping reduces the peak amplitude at resonance and broadens the resonance peak. (1)

- **QWC (0.4 marks):**
- Precise use of oscillation terms (natural frequency, driving frequency, amplitude, dissipation, thermal energy) in a structured explanation. (0.4)

1. **Main Sequence Phase:** The massive star fuses hydrogen into helium in its core. It is in hydrostatic equilibrium, where the outward radiation pressure from fusion balances the inward gravitational collapse.
2. **Transition to Red Supergiant:** When core hydrogen is depleted, fusion ceases in the core. The core contracts under gravity and heats up. This heating ignites a hydrogen-fusing shell around the core, and eventually, the core temperature is high enough to fuse helium into carbon and oxygen. The increased radiation pressure causes the outer layers of the star to expand vastly and cool, forming a red supergiant.
3. **Advanced Shell Fusion:** Unlike lower-mass stars, a massive star has enough gravitational energy to compress and heat the core repeatedly. This allows successive fusion of heavier elements (carbon, neon, oxygen, silicon) in concentric shells ('onion skin' structure) until an iron core is formed.
4. **Supernova Explosion:** Iron fusion is endothermic (requires energy rather than releasing it), so fusion stops in the core once iron is reached. The outward radiation pressure drops to zero. Gravity causes a sudden, catastrophic collapse of the core. The outer layers crash down onto the collapsing core and rebound in a massive shockwave, resulting in a supernova explosion that ejects most of the star's mass into space.
5. **Remnant (Neutron Star or Black Hole):** The remaining core is compressed. If the remaining core mass is between approximately 1.4 and 3 solar masses (the Chandrasekhar and TOV limits), electron/neutron degeneracy pressure halts gravity, forming a dense neutron star. If the remaining core mass exceeds about 3 solar masses, gravity overcomes neutron degeneracy pressure, and the core collapses completely to form a black hole.

Award marks based on the following criteria (max 7 marks for physics content, 0.4 marks for quality of written communication):

- **Main Sequence (Max 1 mark):**
- Identifies hydrogen-to-helium fusion and the balance between radiation pressure and gravitational collapse (hydrostatic equilibrium). (1)

- **Red Supergiant & Shell Fusion (Max 2 marks):**
- Depletion of hydrogen leads to core contraction, heating, and shell/core helium fusion. (1)
- Describes further fusion of heavier elements up to iron in a massive star, causing outer layers to expand. (1)

- **Supernova Collapse (Max 2 marks):**
- Explains that iron fusion is endothermic/cannot release energy, leading to a sudden loss of radiation pressure. (1)
- Gravitational collapse of the core causes a violent rebound shockwave / supernova explosion that ejects outer layers. (1)

- **Remnant Formation (Max 2 marks):**
- Identifies that the core remnant becomes either a neutron star or a black hole. (1)
- Explains that the outcome depends on the mass of the remaining core (role of degeneracy pressure vs gravity). (1)

- **QWC (0.4 marks):**
- Sequential and logical scientific narrative of stellar evolution with accurate terminology (hydrostatic equilibrium, endothermic fusion, supernova, degeneracy pressure). (0.4)

Wave theory predicts that electromagnetic waves deliver energy continuously to a metal surface. Thus, given enough time, electrons should eventually absorb enough energy to escape, regardless of the frequency of the light. However, observations show that if the frequency of the incident radiation is below a specific value called the threshold frequency \(f_0\), no electrons are emitted, no matter how intense the light is.

In the photon model, light is composed of discrete packets of energy called photons, where each photon has energy \(E = hf\). An electron can only interact with a single photon in a one-to-one interaction. If the photon's energy is less than the work function \(\Phi\) of the metal (i.e., \(hf < \Phi\)), no electron can escape. This explains the threshold frequency.

Furthermore, wave theory predicts that increasing the intensity increases the wave amplitude and therefore the rate of energy delivery, which should increase the maximum kinetic energy of the emitted electrons. In practice, increasing the intensity only increases the number of photoelectrons emitted per second (the photocurrent) but has no effect on their maximum kinetic energy. The maximum kinetic energy is determined solely by the frequency of the incident radiation according to Einstein's photoelectric equation: \(E_{\text{k,max}} = hf - \Phi\).

Finally, wave theory predicts a measurable time delay for low-intensity light while electrons accumulate energy. In reality, electron emission is instantaneous once the threshold frequency is exceeded. This is explained by the instantaneous energy transfer during the one-to-one collision between a single photon and a single electron.

• IC1: Wave theory prediction of continuous energy delivery, which should lead to emission at any frequency over time. (1 mark)
• IC2: Explains threshold frequency in terms of the photon model: light is composed of discrete packets of energy \(E=hf\), and a one-to-one interaction occurs. (1 mark)
• IC3: Explains that if \(hf < \Phi\), no emission occurs, which justifies the existence of a threshold frequency. (1 mark)
• IC4: Contrast with wave theory prediction for intensity: wave theory predicts higher intensity increases maximum kinetic energy, but in reality, it only increases the rate of emission (photocurrent). (1 mark)
• IC5: States that maximum kinetic energy depends only on frequency, as shown by \(E_{\text{k,max}} = hf - \Phi\). (1 mark)
• IC6: Explains the lack of time delay: in the photon model, the energy transfer is instantaneous during the one-to-one collision. (1 mark)
• QWC: Clarity, logical structure, and correct physics terminology (photon, work function, threshold frequency, one-to-one interaction). (1.4 marks)

On a flat, horizontal circular track, the only vertical forces are the weight of the car \(mg\) acting downwards and the normal contact force \(N\) acting vertically upwards. These balance each other: \(N = mg\). The centripetal force required for circular motion, \(F_{\text{c}} = \frac{mv^2}{r}\), must be provided entirely by the static friction \(f\) between the tires and the road surface. Because friction has a maximum limit given by \(f_{\text{max}} = \mu N = \mu mg\), there is a strict maximum safe speed \(v_{\text{max}} = \sqrt{\mu g r}\) above which the car will slip.

On a banked track inclined at an angle \(\theta\) to the horizontal, the normal contact force \(N\) acts perpendicular to the track's surface and is therefore tilted at an angle \(\theta\) to the vertical. This tilted normal force has a vertical component \(N \cos\theta\) and a horizontal component \(N \sin\theta\) directed toward the center of the circular path.

Even in the absence of friction, the vertical force balance \(N \cos\theta = mg\) and the horizontal centripetal force requirement \(N \sin\theta = \frac{mv^2}{r}\) allow a design speed \(v = \sqrt{gr \tan\theta}\) where no friction is required at all. When friction is present, the component of friction acting parallel to the slope also contributes to both vertical balance and centripetal force. Therefore, the combination of the horizontal component of the normal force and the horizontal component of friction allows a much greater centripetal force to be achieved, enabling a significantly higher maximum speed before the car slips outward.

• IC1: States that on a flat track, the centripetal force is provided solely by the frictional force between the tires and the road. (1 mark)
• IC2: Identifies that on a flat track, maximum speed is limited because the maximum friction is limited by the normal contact force (\(f_{\text{max}} = \mu mg\)). (1 mark)
• IC3: Explains that on a banked track, the normal contact force \(N\) is perpendicular to the track, meaning it has both vertical and horizontal components. (1 mark)
• IC4: Identifies that the horizontal component of the normal force (\(N \sin\theta\)) acts towards the center of the circle, contributing directly to the centripetal force. (1 mark)
• IC5: States that because the normal force contributes to the centripetal force, less friction is required to negotiate the turn at a given speed. (1 mark)
• IC6: Concludes that the total available centripetal force is much larger on a banked track, allowing the car to travel at a significantly higher speed without slipping. (1 mark)
• QWC: Clear and logical progression of ideas, appropriate use of force components, and clear differentiation between horizontal and vertical force balances. (1.4 marks)

A cyclotron consists of two hollow D-shaped electrodes called 'Dees' inside a vacuum chamber, placed within a uniform magnetic field that acts perpendicular to the plane of the Dees.

Inside the hollow Dees, there is no electric field due to electrostatic shielding. Thus, only the uniform magnetic field acts on the charged particles. The magnetic force \(F = Bqv\) acts perpendicular to the velocity of the protons, providing the centripetal force \(\frac{mv^2}{r} = Bqv\). This causes the protons to move in circular paths with a radius of \(r = \frac{mv}{Bq}\).

An alternating potential difference is applied across the narrow gap between the Dees, generating an electric field in this region. Each time the protons cross the gap, they are accelerated by this electric field, increasing their speed \(v\). Because the radius is directly proportional to the speed (\(r \propto v\)), the protons move into a larger circular path upon re-entering the next Dee.

The time spent by a proton in completing a semi-circular path in one Dee is half of the orbital period: \(t = \frac{\pi r}{v} = \frac{\pi m}{Bq}\). Since mass \(m\), magnetic flux density \(B\), and charge \(q\) are constant (for non-relativistic speeds), this time \(t\) is independent of the speed \(v\) and the radius \(r\). Therefore, the protons always take the exact same time to complete each semi-circle, requiring an alternating potential difference of a constant frequency \(f = \frac{Bq}{2\pi m}\) to ensure that the electric field reverses direction in synchronization with each crossing of the gap.

• IC1: States that the magnetic field is perpendicular to the motion of the particles and provides the centripetal force (\(Bqv = \frac{mv^2}{r}\)), leading to circular motion. (1 mark)
• IC2: States that the electric field exists only in the gap between the Dees and accelerates the particles (increases kinetic energy). (1 mark)
• IC3: Explains that as the speed increases in the gap, the radius of the path inside the Dees increases according to \(r = \frac{mv}{Bq}\). (1 mark)
• IC4: States or derives the equation for the time spent in a Dee: \(t = \frac{\pi m}{Bq}\) (or period \(T = \frac{2\pi m}{Bq}\)). (1 mark)
• IC5: Explains that because this time/period is independent of both radius and speed, the particles take the same duration to complete each semi-circle. (1 mark)
• IC6: Explains that the frequency of the alternating potential difference must remain constant to maintain synchronization with the particle crossings. (1 mark)
• QWC: Logical structure, clear explanations of field interactions, and precise physical terminology (centripetal force, alternating potential difference, synchronization). (1.4 marks)

An alpha (\(\alpha\)) particle is a helium nucleus containing two protons and two neutrons, meaning it is relatively massive and has a high charge of \(+2e\). Because of this high charge and mass, alpha particles are strongly ionizing, readily stripping electrons from surrounding atoms and molecules. However, because they ionize so frequently, they lose their kinetic energy over a very short distance, resulting in a low penetrating power. They can only travel a few centimeters in air and are stopped by a thin sheet of paper or the dead outer layer of human skin.

Consequently, an external alpha source is of low risk because the radiation cannot penetrate the dead outer layer of skin to reach living tissue. However, if an alpha emitter is ingested or inhaled, it comes into direct contact with delicate internal living cells (e.g., in the lungs or stomach) that lack a protective dead layer. The highly ionizing alpha particles cause extensive cellular damage and DNA mutations, leading to a high risk of cancer.

In contrast, gamma (\(\gamma\)) radiation consists of high-energy, uncharged, and massless electromagnetic photons. Because they lack charge, they have a low ionizing power, interacting relatively infrequently with matter. This low ionization rate gives them highly penetrating properties; they can travel great distances in air and pass easily through human body tissues, requiring thick lead or concrete to absorb them. Therefore, an external gamma emitter poses a major threat because the radiation can easily penetrate the body from a distance to damage vital internal organs.

• IC1: Identifies the physical nature of alpha particles (high charge, massive) and gamma photons (uncharged, massless). (1 mark)
• IC2: Explains that alpha particles have very high ionizing power, while gamma rays have low ionizing power. (1 mark)
• IC3: Explains that alpha particles have very low penetration (stopped by skin/paper) because they lose energy quickly through frequent ionizations. (1 mark)
• IC4: Explains that gamma rays have high penetrating power (can pass through tissue) due to their low ionization probability. (1 mark)
• IC5: Explains why an external alpha source is harmless (stopped by the dead outer layer of skin) but highly dangerous internally (direct contact with living cells causing mutations). (1 mark)
• IC6: Explains why a gamma source is dangerous externally (can penetrate from a distance through skin and tissue to damage internal organs). (1 mark)
• QWC: Well-structured comparison, clearly distinguishing between external and internal exposure, and using appropriate terminology (ionization, penetration, cellular damage). (1.4 marks)

The suspension bridge possesses a natural frequency \(f_0\) of oscillation, which is the frequency at which it would vibrate if disturbed and allowed to oscillate freely (free oscillations).

When wind blows across the bridge, it can generate periodic aerodynamic forces (such as those from vortex shedding). This acts as an external periodic driving force, driving the bridge to perform forced oscillations at the driving frequency \(f\) of the wind forces.

When the driving frequency of the wind forces matches or is close to the natural frequency of the bridge (\(f \approx f_0\)), the system undergoes resonance. Under resonance, there is a maximum transfer of energy from the driving force to the bridge, causing the amplitude of the oscillations to increase to a very large value, potentially leading to structural failure.

Dampers mitigate this issue by introducing resistive forces that act in the direction opposite to the bridge's motion. These forces do work against the oscillation, removing kinetic energy from the bridge and dissipating it as thermal energy (heat). This continuous energy dissipation prevents the accumulation of energy in the structure, drastically reducing the peak amplitude of oscillation at resonance. In addition, increased damping broadens the resonance curve, making the bridge less sensitive to a narrow range of wind speeds.

• IC1: Identifies that the bridge has a natural frequency (\(f_0\)) of oscillation associated with free oscillations. (1 mark)
• IC2: Identifies that the wind provides a periodic driving force, causing forced oscillations at a driving frequency \(f\). (1 mark)
• IC3: Explains resonance: when the driving frequency matches the natural frequency (\(f \approx f_0\)), maximum energy transfer occurs, leading to peak amplitude. (1 mark)
• IC4: States that dampers introduce dissipative forces (such as viscous forces) that oppose the motion of the bridge. (1 mark)
• IC5: Explains that dampers remove energy from the oscillating bridge and convert it into thermal energy (heat). (1 mark)
• IC6: Describes the effect of damping on the resonance curve: it reduces the peak amplitude significantly and broadens the response curve, reducing the risk of failure. (1 mark)
• QWC: Clear logical structure, precise differentiation between free, forced, and resonant states, and appropriate terminology (natural frequency, driving frequency, dissipation). (1.4 marks)

The absolute temperature \(T\) of an ideal gas is directly proportional to the mean kinetic energy of its molecules: \(\frac{1}{2}m\langle c^2 \rangle = \frac{3}{2}kT\). Therefore, when the temperature of the gas increases, the mean kinetic energy of the gas molecules increases, meaning they have a higher root-mean-square speed \(c_{\text{rms}}\).

Pressure \(p\) is defined macroscopically as the average force exerted per unit area of the container walls (\(p = F/A\)). Microscopic collisions between the gas molecules and the container walls produce this force. According to Newton's Second Law, the average force exerted by a molecule during a collision is equal to its rate of change of momentum (\(F = \frac{\Delta p}{\Delta t}\)).

When the molecules move faster due to higher temperatures, two key effects occur:
1. Each collision of a molecule with a wall involves a greater change in momentum \(\Delta p\), because the velocity of the molecule is higher (momentum change is \(2mv\)).
2. The molecules travel across the container in less time, causing them to collide with the walls of the fixed-volume container more frequently (a higher collision rate).

Since both the momentum change per collision increases and the frequency of collisions increases, the total rate of change of momentum of the gas molecules colliding with the walls is significantly higher. This increases the total average force exerted on the walls, and since the surface area of the rigid container is constant, the gas pressure increases.

• IC1: States that absolute temperature is directly proportional to the mean kinetic energy of the gas molecules (\(\frac{3}{2}kT\)). (1 mark)
• IC2: Explains that heating the gas increases the average speed of the molecules. (1 mark)
• IC3: Defines pressure as force per unit area and links force to the rate of change of momentum during collisions. (1 mark)
• IC4: Explains that faster-moving molecules experience a greater change in momentum (\(\Delta p\)) when they rebound off the container walls. (1 mark)
• IC5: Explains that because the volume is fixed and molecules move faster, they collide with the walls more frequently. (1 mark)
• IC6: Concludes that the combination of greater momentum change per collision and higher collision frequency results in a larger total rate of change of momentum, hence a larger average force and higher pressure. (1 mark)
• QWC: Well-reasoned explanation from microscopic motion to macroscopic pressure, with precise use of physics terms (mean kinetic energy, momentum, collision rate). (1.4 marks)