2026 Edexcel IAL Physics (YPH11) Practice Paper with Answers

The radius \(r\) of a circular path of a particle with charge \(q\) and mass \(m\) moving perpendicular to a uniform magnetic field \(B\) is given by \(r = \frac{p}{qB} = \frac{\sqrt{2m E_k}}{qB}\), where \(E_k\) is the kinetic energy of the particle. Let the proton have mass \(m\) and charge \(e\). Its radius is \(r_p = \frac{\sqrt{2m E_k}}{eB} = r\). An alpha particle has mass \(4m\) and charge \(2e\). Since its kinetic energy is also \(E_k\), its radius is \(r_{\alpha} = \frac{\sqrt{2(4m) E_k}}{(2e)B} = \frac{2\sqrt{2m E_k}}{2eB} = \frac{\sqrt{2m E_k}}{eB} = r\).

B is the correct option (1 mark).

Using Einstein's photoelectric equation, \(E_{\text{max}} = \frac{hc}{\lambda} - \Phi\), which can be rewritten as \(\frac{hc}{\lambda} = E_{\text{max}} + \Phi\). When the wavelength is halved, the new wavelength is \(\lambda' = \frac{\lambda}{2}\). The new maximum kinetic energy \(E'_{\text{max}}\) is given by \(E'_{\text{max}} = \frac{hc}{\lambda'} - \Phi = \frac{2hc}{\lambda} - \Phi\). Substituting the first expression into this equation gives \(E'_{\text{max}} = 2(E_{\text{max}} + \Phi) - \Phi = 2 E_{\text{max}} + \Phi\).

B is the correct option (1 mark).

The power dissipated in the external resistor is given by \(P = I^2 R = \left(\frac{E}{R+r}\right)^2 R\). According to the maximum power transfer theorem, the power delivered to the external load is at its maximum when the load resistance \(R\) equals the internal resistance \(r\). When \(R\) is much smaller than \(r\), \(P\) is approximately proportional to \(R\) and increases. When \(R\) is much larger than \(r\), \(P\) is approximately proportional to \(1/R\) and decreases. Therefore, as \(R\) increases, the power \(P\) increases to a maximum at \(R = r\) and then decreases.

C is the correct option (1 mark).

The rate at which work is done (power) by a force is given by the scalar product of the force and the velocity vector. At the highest point of its trajectory, the vertical component of the velocity is zero, meaning the velocity vector is entirely horizontal. The force of gravity (weight) acts vertically downwards. Since the velocity vector and the force of gravity are perpendicular, their scalar product is zero. Therefore, the rate at which gravity does work on the ball is zero.

A is the correct option (1 mark).

The extension of a wire is given by the formula \(x = \frac{F L}{A E} = \frac{F L}{\pi r^2 E}\), where \(E\) is the Young modulus of the material. For wire X, the extension is \(x = \frac{F L}{\pi r^2 E}\). For wire Y, which is made of the same material, has length \(2L\), and radius \(2r\), the extension is \(x_Y = \frac{F (2L)}{\pi (2r)^2 E} = \frac{2FL}{4\pi r^2 E} = \frac{1}{2}\left(\frac{F L}{\pi r^2 E}\right) = \frac{x}{2}\).

B is the correct option (1 mark).

By conservation of charge, the initial total charge is \((+1) + (0) = +1\), so the final state must have a total charge of \(+1\). Since the kaon \(K^0\) has a charge of \(0\), the baryon must have a charge of \(+1\). By conservation of baryon number, the initial baryon number is \((0) + (1) = 1\), so the final state must have a baryon number of \(1\), which is carried by the baryon. In strong interactions, strangeness must also be conserved. The initial strangeness is \(0\). The kaon \(K^0\) has strangeness \(+1\) due to its \(\bar{s}\) quark. Thus, the baryon must have strangeness \(-1\), requiring one \(s\) quark (charge \(-1/3\)). For a three-quark baryon to have strangeness \(-1\) and a total charge of \(+1\), the remaining two quarks must have a total charge of \(+4/3\), which corresponds to two \(u\) quarks (each with charge \(+2/3\)). Therefore, the quark composition is \(uus\).

A is the correct option (1 mark).

Let \(\theta\) be the angle that the string makes with the vertical. The radius of the horizontal circular path is \(r\), which means \(\sin\theta = \frac{r}{L}\). The horizontal component of the tension provides the centripetal force: \(T \sin\theta = \frac{m v^2}{r}\). Substituting \(\sin\theta = \frac{r}{L}\) into this equation gives \(T \left(\frac{r}{L}\right) = \frac{m v^2}{r}\). Rearranging this formula to solve for \(v^2\) yields \(v^2 = \frac{T r^2}{m L}\).

B is the correct option (1 mark).

The total mechanical energy of a simple harmonic oscillator is given by \(E = \frac{1}{2} m \omega^2 A^2\). The potential energy \(E_p\) at displacement \(x = \frac{1}{2}A\) is \(E_p = \frac{1}{2} m \omega^2 x^2 = \frac{1}{2} m \omega^2 \left(\frac{1}{2}A\right)^2 = \frac{1}{4} \left(\frac{1}{2} m \omega^2 A^2\right) = \frac{1}{4} E\). Since total energy is conserved, the kinetic energy \(E_k\) at this displacement is \(E_k = E - E_p = E - \frac{1}{4}E = \frac{3}{4} E\). The ratio of kinetic energy to potential energy is \(\frac{E_k}{E_p} = \frac{\frac{3}{4} E}{\frac{1}{4} E} = 3\), which represents a ratio of 3 : 1.

D is the correct option (1 mark).

The power dissipated in the load resistor \(R\) is given by \(P = I^2 R = \left(\frac{V_0}{R + r}\right)^2 R\). According to the maximum power transfer theorem, maximum power is transferred to the load when the load resistance equals the internal resistance of the source, \(R = r\). Substituting \(R = r\) into the power formula gives \(P_{\text{max}} = \left(\frac{V_0}{2r}\right)^2 r = \frac{V_0^2}{4r^2} r = \frac{V_0^2}{4r}\).

1 mark: Correctly identifies \(P_{\text{max}} = \frac{V_0^2}{4r}\) (Option C).

The Young modulus is given by \(E = \frac{F L}{A \Delta L}\), so the extension is \(\Delta L = \frac{F L}{A E}\). Since both wires are made of the same material, \(E\) is constant. The force \(F\) is also the same. The cross-sectional area \(A\) is proportional to the square of the diameter \(d^2\). Therefore, \(\Delta L \propto \frac{L}{d^2}\). For wire X, \(L_X = 2 L_Y\) and \(d_X = 0.5 d_Y\). Thus, \(\frac{\Delta L_X}{\Delta L_Y} = \frac{L_X}{L_Y} \times \left(\frac{d_Y}{d_X}\right)^2 = 2 \times \left(\frac{1}{0.5}\right)^2 = 2 \times 4 = 8\).

1 mark: Correctly calculates the ratio of extensions as 8 (Option C).

Einstein's photoelectric equation states that \(E_k = hf - \phi\), where \(\phi\) is the work function. When the frequency is doubled to \(2f\), the new maximum kinetic energy is \(E'_k = h(2f) - \phi = 2hf - \phi\). Since \(hf = E_k + \phi\), we can substitute this into the equation: \(E'_k = (E_k + \phi) + hf - \phi = E_k + hf\).

1 mark: Correctly identifies the new maximum kinetic energy as \(E_k + hf\) (Option B).

Momentum is a vector quantity. Let the initial velocity of the particle be \(\vec{v}_i = v \hat{i}\). After traveling through a semicircle, its direction of motion is reversed, so its final velocity is \(\vec{v}_f = -v \hat{i}\). The change in momentum is \(\Delta \vec{p} = m \vec{v}_f - m \vec{v}_i = -m v \hat{i} - m v \hat{i} = -2 m v \hat{i}\). The magnitude of this change is \(2mv\).

1 mark: Correctly determines the magnitude of the change in momentum as \(2mv\) (Option D).

The initial charge on the first capacitor is \(Q = CV\). When connected to the second capacitor of capacitance \(2C\) in parallel, the total capacitance becomes \(C_{\text{total}} = C + 2C = 3C\). Since the system is isolated, the total charge \(Q\) is conserved. The final energy stored in the system is \(U_f = \frac{Q^2}{2 C_{\text{total}}} = \frac{(CV)^2}{2(3C)} = \frac{C^2 V^2}{6C} = \frac{1}{6} C V^2\).

1 mark: Correctly calculates the final energy stored as \(\frac{1}{6} C V^2\) (Option C).

The mean square speed of ideal gas molecules \(\langle c^2 \rangle\) is directly proportional to the absolute temperature \(T\) in Kelvin, as shown by \(\frac{1}{2} m \langle c^2 \rangle = \frac{3}{2} k_B T\). The initial temperature in Kelvin is \(T_1 = 27 + 273 = 300\text{ K}\). The final temperature in Kelvin is \(T_2 = 327 + 273 = 600\text{ K}\). The ratio of final to initial temperature is \(\frac{T_2}{T_1} = \frac{600}{300} = 2\). Therefore, the ratio of the mean square speeds is also 2.

1 mark: Correctly calculates the ratio of mean square speeds as 2 (Option B).

The period of a simple pendulum is given by \(T = 2\pi\sqrt{\frac{L}{g}}\). On the new planet, the length becomes \(L' = 2L\) and the acceleration due to gravity is \(g' = \frac{g}{2}\). The new period \(T'\) is \(T' = 2\pi\sqrt{\frac{L'}{g'}} = 2\pi\sqrt{\frac{2L}{g/2}} = 2\pi\sqrt{\frac{4L}{g}} = 2 \left(2\pi\sqrt{\frac{L}{g}}\right) = 2T\).

1 mark: Correctly determines the new period as \(2T\) (Option D).

Baryons are hadrons composed of three quarks (and have a baryon number of 1). Protons (\(uud\)), neutrons (\(udd\)), lambdas (\(uds\)), and sigmas (\(uus\)) are all baryons. Pions and kaons are mesons (composed of a quark-antiquark pair). Therefore, only group B (proton, neutron, lambda) consists entirely of baryons.

1 mark: Correctly identifies the group consisting entirely of baryons (Option B).

The resistance of a wire is given by \(R = \rho \frac{l}{A}\), where \(\rho\) is resistivity, \(l\) is length, and \(A\) is cross-sectional area. Since the volume \(V = A \cdot l\) remains constant, the area is given by \(A = \frac{V}{l}\). Substituting this into the resistance formula gives \(R = \rho \frac{l^2}{V}\). If the length is doubled to \(2l\), the new resistance \(R'\) is: \(R' = \rho \frac{(2l)^2}{V} = 4 \left(\rho \frac{l^2}{V}\right) = 4R\).

D - 1 mark: Correctly identifies that resistance increases by a factor of 4.

At maximum height \(H\), the final velocity is zero. Using the equation of motion \(v_f^2 = u^2 + 2as\), we have: \(0 = v^2 - 2gH \implies 2gH = v^2\). At a height of \(h = \frac{3}{4}H\), the velocity \(v_h\) is given by: \(v_h^2 = v^2 - 2gh = v^2 - 2g\left(\frac{3}{4}H\right) = v^2 - \frac{3}{4}(2gH)\). Substituting \(2gH = v^2\) yields: \(v_h^2 = v^2 - \frac{3}{4}v^2 = \frac{1}{4}v^2 \implies v_h = \frac{1}{2}v\).

B - 1 mark: Correctly uses equations of motion to determine that the velocity at three-quarters height is half the initial velocity.

The extension \(\Delta x\) of a wire is given by: \(\Delta x = \frac{FL}{AE} = \frac{4FL}{\pi d^2 E}\), where \(F\) is force, \(L\) is length, \(d\) is diameter, and \(E\) is Young modulus. Since both wires have the same force \(F\) and material \(E\), the extension is proportional to \(\frac{L}{d^2}\). Thus, the ratio of extensions is: \(\frac{\Delta x_X}{\Delta x_Y} = \frac{L_X}{L_Y} \cdot \left(\frac{d_Y}{d_X}\right)^2 = 2 \cdot (2)^2 = 8\).

C - 1 mark: Correctly applies the extension formula to find a ratio of 8.

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 \(N\) acting upwards. The net downward force provides the required centripetal acceleration: \(mg - N = \frac{mv^2}{r}\). Rearranging this equation for \(N\) gives: \(N = mg - \frac{mv^2}{r}\).

D - 1 mark: Correctly identifies the centripetal force equation and rearranges for the normal force.

The magnetic force provides the centripetal force: \(Bqv = \frac{mv^2}{r} \implies r = \frac{mv}{Bq}\). The kinetic energy of the particle is \(E_k = \frac{1}{2}mv^2\), which means the momentum \(p = mv = \sqrt{2mE_k}\). Substituting this expression for momentum into the equation for radius gives: \(r = \frac{\sqrt{2mE_k}}{Bq}\).

A - 1 mark: Correctly relates kinetic energy, momentum, and circular motion in a magnetic field.

Using the grating equation: \(d \sin\theta = n\lambda\). For the initial third-order maximum (\(n=3\)): \(\sin\theta = \frac{3\lambda}{d}\). For the new setup, the new wavelength is \(1.5\lambda\) and the new slit spacing is \(2d\). The equation for the new angle \(\theta'\) is: \(\sin\theta' = \frac{3(1.5\lambda)}{2d} = \frac{4.5\lambda}{2d} = 2.25 \frac{\lambda}{d}\). Since \(\frac{\lambda}{d} = \frac{\sin\theta}{3}\), we can substitute this to get: \(\sin\theta' = 2.25 \left(\frac{\sin\theta}{3}\right) = 0.75 \sin\theta\). Therefore, \(\theta' = \sin^{-1}(0.75 \sin\theta)\).

A - 1 mark: Correctly applies the diffraction grating equation to solve for the new angle.

After three half-lives, the fraction of active nuclei remaining is \(\left(\frac{1}{2}\right)^3 = \frac{1}{8}\). Therefore, the fraction of active nuclei that has decayed is \(1 - \frac{1}{8} = \frac{7}{8}\).

D - 1 mark: Correctly identifies that \(\frac{7}{8}\) of the original nuclei have decayed after 3 half-lives.

According to the ideal gas law, \(pV = nRT\). This can be written as \(\frac{p_1 V_1}{T_1} = \frac{p_2 V_2}{T_2}\). Given \(V_2 = \frac{V_1}{3}\) and \(T_2 = 2T_1\), we substitute these values into the equation: \(\frac{p_1 V_1}{T_1} = \frac{p_2 \left(\frac{V_1}{3}\right)}{2T_1}\). Simplifying gives: \(p_1 = \frac{p_2}{6} \implies p_2 = 6p_1 = 6p\).

C - 1 mark: Correctly uses the ideal gas equation to find that the pressure increases by a factor of 6.

At the maximum height, the vertical component of the velocity is zero. Using the equations of motion for the vertical direction: \(v_y^2 = u_y^2 - 2gH\). Since \(v_y = 0\) and \(u_y = u \sin\theta\), we get \(0 = (u \sin\theta)^2 - 2gH\). Rearranging for \(H\) gives \(H = \frac{u^2 \sin^2\theta}{2g}\).

[1 mark] Correctly identifies Option A as the expression for maximum height.

For a negative temperature coefficient (NTC) thermistor, resistance increases as temperature decreases. Since the thermistor is connected in series with a fixed resistor, an increase in its resistance means it takes a larger proportion of the constant total supply voltage. Therefore, the potential difference across the thermistor increases, causing the voltmeter reading to increase.

[1 mark] Correctly identifies Option A.

The diffraction grating equation is given by \(d \sin\theta = n\lambda\). For the third-order maximum (\(n = 3\)), we have \(d \sin\theta = 3\lambda\), which gives \(\sin\theta = \frac{3\lambda}{d}\). For the new setup, the slit spacing is \(2d\) and the wavelength is \(\frac{\lambda}{2}\). Thus, the equation becomes \(2d \sin\phi = 3\left(\frac{\lambda}{2}\right)\), which simplifies to \(2d \sin\phi = \frac{3\lambda}{2}\), or \(\sin\phi = \frac{3\lambda}{4d}\). Comparing this to the original angle gives \(\sin\phi = \frac{1}{4} \sin\theta\).

[1 mark] Correctly identifies Option D through application of the grating equation.

At the highest point of the vertical loop, both the gravitational force \(mg\) and the normal contact force \(N\) act downwards towards the center of the circle. The centripetal force is the net force towards the center: \(F_c = N + mg = \frac{mv^2}{r}\). Solving for \(N\) yields \(N = \frac{mv^2}{r} - mg\).

[1 mark] Correctly identifies Option B as the normal contact force.

The radius of a charged particle moving in a magnetic field is \(r = \frac{mv}{Bq}\). Since kinetic energy \(E_k = \frac{p^2}{2m} = \frac{(mv)^2}{2m}\), we can write the momentum as \(mv = \sqrt{2mE_k}\). Substituting this into the radius formula gives \(r = \frac{\sqrt{2mE_k}}{Bq}\). Since both particles have the same kinetic energy \(E_k\) and are in the same magnetic field \(B\), the radius is proportional to \(\frac{\sqrt{m}}{q}\). Therefore, the ratio is \(\frac{r_\alpha}{r_e} = \frac{\sqrt{m_\alpha}}{q_\alpha} \cdot \frac{q_e}{\sqrt{m_e}} = \frac{\sqrt{m_\alpha}}{2e} \cdot \frac{e}{\sqrt{m_e}} = \frac{1}{2}\sqrt{\frac{m_\alpha}{m_e}}\).

[1 mark] Correctly derives and identifies Option A.

According to the kinetic theory of gases, the average kinetic energy of the gas molecules is directly proportional to the absolute temperature: \(\frac{1}{2}m\langle c^2 \rangle = \frac{3}{2} k_B T\), where \(\langle c^2 \rangle\) is the mean square speed. Since the mass \(m\) of the molecules is constant, doubling the absolute temperature directly doubles the mean square speed. Root-mean-square speed \(c_{rms}\) is proportional to \(\sqrt{T}\), so it increases by a factor of \(\sqrt{2}\). Average time between collisions decreases by a factor of \(\sqrt{2}\). Pressure of the gas doubles.

[1 mark] Correctly identifies Option A based on kinetic theory principles.

After one half-life, the activity is reduced to half of its initial value: \(\frac{1}{2} A_0\). After two half-lives, it becomes \(\left(\frac{1}{2}\right)^2 A_0 = \frac{1}{4} A_0\). After three half-lives, the remaining activity is \(\left(\frac{1}{2}\right)^3 A_0 = \frac{1}{8} A_0\). The fraction of activity remaining is therefore \(\frac{1}{8}\). The fraction that has decayed is \(1 - \frac{1}{8} = \frac{7}{8}\).

[1 mark] Correctly identifies Option A.

The elastic strain energy stored in a stretched wire is given by \(U = \frac{1}{2} F \Delta L\). Using the definition of Young modulus, \(E = \frac{\text{stress}}{\text{strain}} = \frac{F/A}{\Delta L/L} = \frac{FL}{A \Delta L}\). Rearranging this for force gives \(F = \frac{EA \Delta L}{L}\). Substituting this expression for \(F\) into the strain energy formula yields \(U = \frac{1}{2} \left(\frac{EA \Delta L}{L}\right) \Delta L = \frac{EA(\Delta L)^2}{2L}\).

[1 mark] Correctly derives and identifies Option A.

The terminal potential difference \(V\) is given by \(V = E - Ir\).
Since \(I = \frac{E}{R+r}\), as \(R\) increases, \(I\) decreases.
When \(R \to \infty\), \(I \to 0\), and therefore \(V \to E\). Thus, \(V\) increases towards the limit of \(E\).

The power \(P\) dissipated in the variable resistor is given by:
\[P = I^2 R = \frac{E^2 R}{(R+r)^2}\]

By the maximum power transfer theorem, the power delivered to the load resistance is maximized when the load resistance equals the internal resistance of the source, which is when \(R = r\).

Therefore, \(V\) increases to a limit of \(E\), and \(P\) reaches a maximum value when \(R = r\).

1 mark for the correct option (B).

[1] Correctly identifies that \(V\) increases towards \(E\) and \(P\) peaks when \(R = r\).

The Young modulus \(E\) of a material is defined as:
\[E = \frac{\text{Stress}}{\text{Strain}} = \frac{F / A}{x / L} = \frac{F L}{A x}\]

Rearranging this expression for the extension \(x\):
\[x = \frac{F L}{A E}\]

Since the cross-sectional area of a wire of diameter \(d\) is \(A = \frac{\pi d^2}{4}\), we can write:
\[x = \frac{4 F L}{\pi d^2 E}\]

Because both wires are made of the same material, the Young modulus \(E\) is the same. Since the applied force \(F\) is also the same, we have:
\[x \propto \frac{L}{d^2}\]

Therefore, the ratio of the extensions is:
\[\frac{x_{\text{X}}}{x_{\text{Y}}} = \frac{L_{\text{X}}}{L_{\text{Y}}} \times \left( \frac{d_{\text{Y}}}{d_{\text{X}}} \right)^2\]

We are given that \(L_{\text{X}} = 2 L_{\text{Y}}\) and \(d_{\text{X}} = \frac{1}{2} d_{\text{Y}}\). Substituting these values:
\[\frac{x_{\text{X}}}{x_{\text{Y}}} = 2 \times (2)^2 = 8\]

1 mark for the correct option (A).

[1] Correctly calculates the ratio of the extensions as 8.

Einstein's photoelectric equation is given by:
\[E_{\text{k}} = \frac{hc}{\lambda} - \phi\]
where \(\phi\) is the work function of the metal.

When the wavelength is halved to \(\frac{\lambda}{2}\), the new maximum kinetic energy \(E_{\text{k}}'\) is:
\[E_{\text{k}}' = \frac{hc}{\lambda / 2} - \phi = 2\left( \frac{hc}{\lambda} \right) - \phi\]

Using the original equation, we can substitute \(\frac{hc}{\lambda} = E_{\text{k}} + \phi\):
\[E_{\text{k}}' = 2(E_{\text{k}} + \phi) - \phi = 2 E_{\text{k}} + \phi\]

Since the work function \(\phi\) is a positive constant for any given metal, \(2 E_{\text{k}} + \phi > 2 E_{\text{k}}\).

Therefore, the new maximum kinetic energy is greater than \(2 E_{\text{k}}\).

1 mark for the correct option (B).

[1] Correctly reasons that the photon energy doubles, meaning the excess energy exceeds \(2 E_{\text{k}}\).

The horizontal motion has a constant velocity, so the horizontal displacement in time \(t\) is:
\[x = v t \implies t = \frac{x}{v}\]

Under constant gravitational acceleration, the vertical distance fallen in time \(t\) is:
\[y = \frac{1}{2} g t^2 = \frac{1}{2} g \left( \frac{x}{v} \right)^2 = \frac{g}{2 v^2} x^2\]

By the conservation of energy, the kinetic energy at any point is the sum of the initial kinetic energy and the loss in gravitational potential energy:
\[E_{\text{k}} = E_{\text{k, initial}} + mgy\]

Substituting \(y\) into the equation:
\[E_{\text{k}} = \frac{1}{2} m v^2 + mg \left( \frac{g}{2 v^2} x^2 \right) = \frac{1}{2} m v^2 + \left( \frac{m g^2}{2 v^2} \right) x^2\]

This is a quadratic equation of the form \(E_{\text{k}} = A + B x^2\) where \(A\) and \(B\) are positive constants.

Since \(\frac{\text{d}E_{\text{k}}}{\text{d}x} = 2Bx\), the gradient is positive and increases linearly with \(x\). This corresponds to a curve with an increasing gradient.

1 mark for the correct option (B).

[1] Correctly identifies that \(E_{\text{k}}\) is quadratic with respect to \(x\), resulting in a curve with an increasing gradient.

At the highest point of the vertical circular path, two forces act on the car:
1. The weight \(mg\) directed downwards (towards the center of the circle).
2. The normal contact force \(N\) directed upwards (away from the center of the circle).

The resultant centripetal force \(F_{\text{c}}\) is directed towards the center of the circle, which is downwards:
\[F_{\text{c}} = mg - N\]

Since the centripetal force is also given by \(F_{\text{c}} = \frac{m v^2}{R}\):
\[mg - N = \frac{m v^2}{R}\]

Rearranging for the normal contact force \(N\):
\[N = mg - \frac{m v^2}{R}\]

1 mark for the correct option (A).

[1] Correctly sets up the equation of motion for circular motion at the top of a vertical circle.

The magnetic force on the charged particle acts as the centripetal force:
\[q v B = \frac{m v^2}{r}\]

Rearranging to express the radius \(r\):
\[r = \frac{m v}{q B}\]

Let \(r'\) be the new radius when the magnetic flux density is \(B' = 2B\) and the speed is \(v' = \frac{v}{2}\):
\[r' = \frac{m v'}{q B'} = \frac{m \left(\frac{v}{2}\right)}{q (2B)} = \frac{1}{4} \left( \frac{m v}{q B} \right) = \frac{r}{4}\]

Therefore, the ratio of the new radius to the original radius is \(\frac{1}{4}\).

1 mark for the correct option (D).

[1] Correctly identifies that \(r \propto \frac{v}{B}\) and calculates the new ratio.

A \(\Sigma^-\) is a baryon, which means it must consist of three quarks (not a quark-antiquark pair).

Let's use conservation laws to determine its properties:
1. **Charge Conservation:**
The charge on the right-hand side is \(Q(\text{n}) + Q(\pi^-) = 0 + (-1) = -1\).
Therefore, the charge of the \(\Sigma^-\)
baryon must be \(-1\).
Let's check the charges of the given options:
- \(\text{uds}\): \(+\frac{2}{3} - \frac{1}{3} - \frac{1}{3} = 0\)
- \(\text{dds}\): \(-\frac{1}{3} - \frac{1}{3} - \frac{1}{3} = -1\)
- \(\text{uus}\): \(+\frac{2}{3} + \frac{2}{3} - \frac{1}{3} = +1\)
- \(\text{uss}\): \(+\frac{2}{3} - \frac{1}{3} - \frac{1}{3} = 0\)

Only the quark structure \(\text{dds}\) has a net charge of \(-1\).
Therefore, the correct quark structure is \(\text{dds}\).

1 mark for the correct option (B).

[1] Applies conservation of charge to determine the quark combination of the baryon.

For an ideal gas, the equation of state is:
\[p V = n R T\]

Since the mass of the gas is fixed, the number of moles \(n\) remains constant. Therefore, we can write:
\[\frac{p_1 V_1}{T_1} = \frac{p_2 V_2}{T_2}\]

We are given the following conditions:
- \(p_1 = p\)
- \(T_2 = 2 T_1\)
- \(V_2 = \frac{1}{3} V_1\)

Substituting these values into the equation to find the final pressure \(p_2\):
\[p_2 = p_1 \left( \frac{V_1}{V_2} \right) \left( \frac{T_2}{T_1} \right)\]
\[p_2 = p \left( \frac{V_1}{\frac{1}{3} V_1} \right) \left( \frac{2 T_1}{T_1} \right)\]
\[p_2 = p \times 3 \times 2 = 6p\]

Thus, the final pressure of the gas is \(6p\).

1 mark for the correct option (C).

[1] Correctly applies the ideal gas equation of state to determine the ratio of the pressures.

Since the box moves at a constant speed, the net force parallel to the slope is zero.

The force acting up the slope is \(F\).
The forces acting down the slope are the component of weight parallel to the slope, \(mg \sin \theta\), and the frictional force, \(f\).

Therefore:
\[F = f + mg \sin \theta\]
\[f = F - mg \sin \theta\]

The rate at which work is done against friction (power dissipated due to friction) is given by:
\[P_f = f v = (F - mg \sin \theta)v\]

Hence, the correct option is B.

1 mark for selecting B.

- Reject C because it represents the force of friction, not the rate of doing work (which requires multiplying by velocity).
- Reject A and D as they do not correctly represent the work rate against friction.

The relationship for the extension \(e\) of a wire is derived from the Young modulus formula:
\[E = \frac{\text{Stress}}{\text{Strain}} = \frac{F / A}{e / L} = \frac{F L}{A e}\]
Rearranging for extension:
\[e = \frac{F L}{E A}\]
Since the cross-sectional area of a wire with diameter \(d\) is \(A = \frac{\pi d^2}{4}\), we have:
\[e = \frac{4 F L}{\pi E d^2} \implies e \propto \frac{L}{d^2}\]
For the second wire with length \(2L\) and diameter \(2d\):
\[e_2 \propto \frac{2L}{(2d)^2} = \frac{2L}{4d^2} = \frac{1}{2} \frac{L}{d^2}\]
Therefore, the extension of the second wire is \(\frac{e}{2}\).

Hence, the correct option is B.

1 mark for selecting B.

- Accept B because the ratio of length to cross-sectional area is halved when length is doubled and diameter is doubled.

The diffraction grating equation is given by:
\[d \sin \theta = n \lambda\]
For the third-order maximum, \(n = 3\).

The grating spacing \(d\) is the distance between adjacent lines. Since there are \(N\) lines per millimetre:
\[d = \frac{1 \text{ mm}}{N} = \frac{10^{-3} \text{ m}}{N}\]
Substituting \(d\) into the grating equation:
\[\left(\frac{10^{-3}}{N}\right) \sin \theta = 3 \lambda\]
Rearranging to make \(\sin \theta\) the subject:
\[\sin \theta = \frac{3 N \lambda}{10^{-3}}\]

Hence, the correct option is B.

1 mark for selecting B.

- Accept B as it correctly converts lines per millimetre to line spacing in metres and applies the diffraction grating formula.

The total resistance of the circuit is \(R_{total} = R + r\).
As \(R\) is increased, the total resistance increases, which causes the current in the circuit, \(I = \frac{\mathcal{E}}{R+r}\), to decrease.

1. **Terminal potential difference (\(V\)):**
\[V = \mathcal{E} - I r\]
Since the current \(I\) decreases, the lost volts \(I r\) decrease. Therefore, the terminal potential difference \(V\) increases.

2. **Power dissipated in the internal resistance (\(P_r\)):**
\[P_r = I^2 r\]
Since the current \(I\) decreases and \(r\) is constant, the power dissipated in the internal resistance decreases.

Thus, the terminal potential difference increases, and the power dissipated in the internal resistance decreases.

Hence, the correct option is C.

1 mark for selecting C.

- Reject options with terminal potential difference decreasing because as current decreases, lost volts decrease, increasing terminal potential difference.

The normal contact force \(R_N\) acts perpendicular to the banked surface.
Resolving forces vertically (since there is no vertical acceleration):
\[R_N \cos \theta = mg \implies R_N = \frac{mg}{\cos \theta}\]
Resolving forces horizontally (the horizontal component provides the centripetal force):
\[R_N \sin \theta = \frac{m v^2}{r}\]
Substituting the expression for \(R_N\):
\[\left(\frac{mg}{\cos \theta}\right) \sin \theta = \frac{m v^2}{r}\]
\[mg \tan \theta = \frac{m v^2}{r}\]
Dividing both sides by \(mg\):
\[\tan \theta = \frac{v^2}{rg}\]

Hence, the correct option is A.

1 mark for selecting A.

- Reject B, C, and D as they do not correctly represent the force balance and circular motion condition.

The potential difference across the discharging capacitor at time \(t\) is given by:
\[V = V_0 e^{-\frac{t}{RC}}\]
At \(t = RC\) (one time constant):
\[V = V_0 e^{-1} = \frac{V_0}{e}\]
The energy \(E\) stored in a capacitor is given by:
\[E = \frac{1}{2} C V^2\]
Therefore, the initial energy is \(E_0 = \frac{1}{2} C V_0^2\).

The energy remaining after one time constant is:
\[E = \frac{1}{2} C \left(\frac{V_0}{e}\right)^2 = \frac{1}{2} C V_0^2 \cdot \frac{1}{e^2} = \frac{E_0}{e^2}\]
The fraction of the initial energy remaining is:
\[\frac{E}{E_0} = \frac{1}{e^2}\]

Hence, the correct option is B.

1 mark for selecting B.

- Reject A because \(1/e\) is the fraction of remaining voltage or charge, not energy (energy depends on \(V^2\)).

A baryon consists of three quarks.

1. **Strangeness constraint:**
Each strange quark (\(\text{s}\)) has a strangeness of \(-1\). Since the \(\Xi^0\) baryon has a strangeness of \(-2\), it must contain exactly two strange quarks (\(\text{s}\)). This rules out options C and D.

2. **Charge constraint:**
- The charge of an up quark (\(\text{u}\)) is \(+\frac{2}{3}e\).
- The charge of a down quark (\(\text{d}\)) is \(-\frac{1}{3}e\).
- The charge of a strange quark (\(\text{s}\)) is \(-\frac{1}{3}e\).

Let's calculate the charge of the remaining options:
- For \(\text{u s s}\):
\[Q = \left(+\frac{2}{3}\right) + \left(-\frac{1}{3}\right) + \left(-\frac{1}{3}\right) = 0\]
- For \(\text{d s s}\):
\[Q = \left(-\frac{1}{3}\right) + \left(-\frac{1}{3}\right) + \left(-\frac{1}{3}\right) = -1\]

Since \(\Xi^0\) has a charge of \(0\), the correct quark combination is \(\text{u s s}\).

Hence, the correct option is A.

1 mark for selecting A.

- Reject B as it has a net charge of \(-1\).
- Reject C and D because they do not have strangeness \(-2\).

The mean kinetic energy of the molecules in an ideal gas is directly proportional to its absolute temperature \(T\) in Kelvin:
\[\frac{1}{2} m \langle c^2 \rangle = \frac{3}{2} k_B T \implies \langle c^2 \rangle \propto T\]
First, convert the temperatures from Celsius to Kelvin:
- Initial temperature, \(T_1 = 27 + 273 = 300\text{ K}\)
- Final temperature, \(T_2 = 327 + 273 = 600\text{ K}\)

The ratio of the mean square speeds is equal to the ratio of the absolute temperatures:
\[\frac{\langle c^2 \rangle_2}{\langle c^2 \rangle_1} = \frac{T_2}{T_1} = \frac{600\text{ K}}{300\text{ K}} = 2.0\]

Hence, the correct option is B.

1 mark for selecting B.

- Reject A (1.4) as this is the factor by which the root mean square (r.m.s.) speed increases (i.e. \(\sqrt{2.0} \approx 1.4\)).
- Reject D (12.1) as this is obtained by using temperatures in Celsius (\(327 / 27 = 12.1\)), which is incorrect.

The radius of a circular path of a charged particle in a magnetic field is given by \( R = \frac{mv}{Bq} \).

The kinetic energy of the particle is given by \( E_k = \frac{1}{2}mv^2 \), which can be rearranged to give the momentum \( p = mv = \sqrt{2mE_k} \).

Substituting this into the radius equation gives:
\( R = \frac{\sqrt{2mE_k}}{Bq} \)

Therefore, the radius is proportional to the square root of the kinetic energy, \( R \propto \sqrt{E_k} \).

If the kinetic energy is doubled, the new radius \( R' \) will be:
\( R' = \sqrt{2} R \)

1 mark for identifying the correct relationship \( R \propto \sqrt{E_k} \) and selecting option A.

Using the ideal gas equation:
\( pV = nRT \)

We can write this as:
\( \frac{p_1 V_1}{T_1} = \frac{p_2 V_2}{T_2} \)

Given:
- Initial conditions: \( p_1 = p \), \( V_1 = V \), \( T_1 = T \)
- Final conditions: \( V_2 = \frac{1}{3} V \), \( T_2 = 2T \)

Substituting these values into the equation:
\( \frac{pV}{T} = \frac{p_2 \left(\frac{1}{3} V\right)}{2T} \)

Rearranging to solve for the new pressure \( p_2 \):
\( p_2 = p \times \left(\frac{V}{V_2}\right) \times \left(\frac{T_2}{T_1}\right) \)
\( p_2 = p \times 3 \times 2 = 6p \)

1 mark for using the ideal gas relationship to determine the pressure ratio and selecting option D.

The weight of the rocket is \( W = mg = 0.25\text{ kg} \times 9.81\text{ N kg}^{-1} = 2.45\text{ N} \). The net upward force is \( F_{\text{net}} = \text{Thrust} - W = 6.0\text{ N} - 2.45\text{ N} = 3.55\text{ N} \). The work done on the rocket by this net force over a distance of \( 15\text{ m} \) is equal to its gain in kinetic energy: \( W_{\text{net}} = F_{\text{net}} \times s = 3.55\text{ N} \times 15\text{ m} = 53.25\text{ J} \). Since it starts from rest, \( E_k = \frac{1}{2}mv^2 = 53.25\text{ J} \). Rearranging for speed gives \( v = \sqrt{\frac{2 \times 53.25\text{ J}}{0.25\text{ kg}}} = 20.6\text{ m s}^{-1} \).

1. Calculate net force as \( 3.55\text{ N} \) (or acceleration as \( 14.2\text{ m s}^{-2} \)) [1 Mark].
2. Apply \( v^2 = u^2 + 2as \) or \(\Delta E_k = F_{\text{net}}s\) [1 Mark].
3. Correct speed of \( 21\text{ m s}^{-1} \) or \( 20.6\text{ m s}^{-1} \) with correct unit [1 Mark].

The cross-sectional area of the string is \( A = \frac{\pi d^2}{4} = \frac{\pi \times (0.85 \times 10^{-3}\text{ m})^2}{4} = 5.67 \times 10^{-7}\text{ m}^2 \). The Young modulus is defined as \( E = \frac{\text{Stress}}{\text{Strain}} = \frac{F/A}{\Delta L / L_0} = \frac{F L_0}{A \Delta L} \). Rearranging for extension \( \Delta L \): \( \Delta L = \frac{F L_0}{A E} = \frac{85\text{ N} \times 0.65\text{ m}}{5.67 \times 10^{-7}\text{ m}^2 \times 3.0 \times 10^9\text{ Pa}} = 0.0325\text{ m} \).

1. Correctly calculate cross-sectional area as \( 5.67 \times 10^{-7}\text{ m}^2 \) [1 Mark].
2. Rearrange the Young modulus equation to make extension \( \Delta L \) the subject [1 Mark].
3. Correct value of extension in range \( 3.2 \times 10^{-2}\text{ m} \) to \( 3.3 \times 10^{-2}\text{ m} \) with correct unit [1 Mark].

The potential difference across the thermistor is \( V_{\text{th}} = V_{\text{total}} - V_{\text{fixed}} = 9.0\text{ V} - 5.4\text{ V} = 3.6\text{ V} \). Using the potential divider ratio: \( \frac{R_{\text{th}}}{R_{\text{fixed}}} = \frac{V_{\text{th}}}{V_{\text{fixed}}} \implies R_{\text{th}} = R_{\text{fixed}} \times \frac{V_{\text{th}}}{V_{\text{fixed}}} = 1.5\text{ k}\Omega \times \frac{3.6\text{ V}}{5.4\text{ V}} = 1.0\text{ k}\Omega \). Alternatively, the current in the circuit is \( I = \frac{5.4\text{ V}}{1500\text{ }\Omega} = 3.6 \times 10^{-3}\text{ A} \). Then \( R_{\text{th}} = \frac{3.6\text{ V}}{3.6 \times 10^{-3}\text{ A}} = 1000\text{ }\Omega \).

1. Calculate potential difference across the thermistor as \( 3.6\text{ V} \) [1 Mark].
2. Use of potential divider equation or Ohm's law to express relationship [1 Mark].
3. Correct resistance value of \( 1.0\text{ k}\Omega \) (or \( 1000\text{ }\Omega \)) with unit [1 Mark].

The energy of an incident photon is \( E = \frac{hc}{\lambda} = \frac{6.63 \times 10^{-34}\text{ J s} \times 3.00 \times 10^8\text{ m s}^{-1}}{3.8 \times 10^{-7}\text{ m}} = 5.23 \times 10^{-19}\text{ J} \). According to Einstein's photoelectric equation, \( hf = \Phi + E_{k\text{,max}} \), so the work function is \( \Phi = hf - E_{k\text{,max}} = 5.23 \times 10^{-19}\text{ J} - 1.2 \times 10^{-19}\text{ J} = 4.03 \times 10^{-19}\text{ J} \). Converting this value to electronvolts: \( \Phi = \frac{4.03 \times 10^{-19}\text{ J}}{1.60 \times 10^{-19}\text{ J eV}^{-1}} = 2.52\text{ eV} \).

1. Use of \( E = \frac{hc}{\lambda} \) to calculate photon energy in joules [1 Mark].
2. Correct application of photoelectric equation to calculate work function in joules [1 Mark].
3. Correct conversion to \(\text{eV}\) yielding \( 2.5\text{ eV} \) (accept range \( 2.5\text{ eV} \) to \( 2.52\text{ eV} \)) [1 Mark].

The force of static friction provides the necessary centripetal force for circular motion: \( F_f = F_c \implies \mu mg = m\omega^2r \). Dividing both sides by mass \( m \) gives \( \mu g = \omega^2r \). Rearranging for the angular velocity: \( \omega = \sqrt{\frac{\mu g}{r}} = \sqrt{\frac{0.35 \times 9.81\text{ m s}^{-2}}{0.14\text{ m}}} = \sqrt{24.525} = 4.95\text{ rad s}^{-1} \).

1. State that friction provides centripetal force, \( \mu mg = m\omega^2 r \) [1 Mark].
2. Rearrange the equation for angular velocity \( \omega \) [1 Mark].
3. Correct final value of \( 5.0\text{ rad s}^{-1} \) (or \( 4.95\text{ rad s}^{-1} \)) with unit [1 Mark].

The electric field strength between the parallel plates is \( E = \frac{V}{d} = \frac{1800\text{ V}}{12 \times 10^{-3}\text{ m}} = 1.5 \times 10^5\text{ V m}^{-1} \). The charge of an alpha particle is \( q = 2e = 2 \times 1.6 \times 10^{-19}\text{ C} = 3.2 \times 10^{-19}\text{ C} \). The electrostatic force acting on it is \( F = Eq = 1.5 \times 10^5\text{ V m}^{-1} \times 3.2 \times 10^{-19}\text{ C} = 4.8 \times 10^{-14}\text{ N} \).

1. Calculate electric field strength \( E = 1.5 \times 10^5\text{ V m}^{-1} \) [1 Mark].
2. Use of \( F = Eq \) with the charge of an alpha particle as \( 3.2 \times 10^{-19}\text{ C} \) [1 Mark].
3. Correct force value of \( 4.8 \times 10^{-14}\text{ N} \) with unit [1 Mark].

The de Broglie wavelength is \( \lambda = \frac{h}{p} \), so the final momentum of the electron is \( p = \frac{h}{\lambda} = \frac{6.63 \times 10^{-34}\text{ J s}}{1.2 \times 10^{-10}\text{ m}} = 5.53 \times 10^{-24}\text{ kg m s}^{-1} \). The kinetic energy of the electron is \( E_k = \frac{p^2}{2m_e} = \frac{(5.53 \times 10^{-24}\text{ kg m s}^{-1})^2}{2 \times 9.11 \times 10^{-31}\text{ kg}} = 1.68 \times 10^{-17}\text{ J} \). Since the electron is accelerated from rest by potential difference \( V \), the work done is \( eV = E_k \), giving \( V = \frac{E_k}{e} = \frac{1.68 \times 10^{-17}\text{ J}}{1.60 \times 10^{-19}\text{ C}} = 105\text{ V} \).

1. Correctly determine electron momentum using \( p = h / \lambda \) [1 Mark].
2. Relate potential difference \( V \) to kinetic energy or momentum via \( eV = \frac{p^2}{2m} \) [1 Mark].
3. Correct potential difference value of \( 105\text{ V} \) (or \( 1.0 \times 10^2\text{ V} \) to 2 s.f.) with unit [1 Mark].

Using the ideal gas equation \( pV = nRT \), the final absolute temperature \( T \) is given by: \( T = \frac{pV}{nR} = \frac{1.8 \times 10^5\text{ Pa} \times 0.025\text{ m}^3}{1.2\text{ mol} \times 8.31\text{ J mol}^{-1}\text{ K}^{-1}} = \frac{4500}{9.972} = 451.3\text{ K} \). To convert this temperature to degrees Celsius: \( \theta = T - 273.15 = 451.3 - 273.15 = 178.1\text{ }^\circ\text{C} \). Rounded to 2 significant figures, this is \( 180\text{ }^\circ\text{C} \).

1. State or use \( pV = nRT \) [1 Mark].
2. Correctly calculate the final temperature in kelvin as \( 451\text{ K} \) [1 Mark].
3. Correct conversion to degrees Celsius with unit, yielding \( 178\text{ }^\circ\text{C} \) or \( 180\text{ }^\circ\text{C} \) [1 Mark].

First, calculate the vertical displacement \(s\) of the drone during the first \(4.0 \text{ s}\) using \(s = ut + \frac{1}{2}at^2\):
\(s = 0 + \frac{1}{2} \times 1.8 \text{ m s}^{-2} \times (4.0 \text{ s})^2 = 14.4 \text{ m}\).

Next, calculate the upward force \(F\) exerted by the drone's motors. The net force is \(F - mg = ma\), so:
\(F = m(g + a) = 1.4 \text{ kg} \times (9.81 \text{ m s}^{-2} + 1.8 \text{ m s}^{-2}) = 1.4 \times 11.61 = 16.25 \text{ N}\).

Finally, calculate the useful work done \(W\):
\(W = F \times s = 16.25 \text{ N} \times 14.4 \text{ m} = 234 \text{ J}\).

Alternatively, using energy conservation:
\(\Delta E_p = mgh = 1.4 \times 9.81 \times 14.4 = 197.8 \text{ J}\)
\(v = at = 1.8 \times 4.0 = 7.2 \text{ m s}^{-1}\)
\(\Delta E_k = \frac{1}{2}mv^2 = 0.5 \times 1.4 \times 7.2^2 = 36.3 \text{ J}\)
\(W_{\text{total}} = \Delta E_p + \Delta E_k = 197.8 + 36.3 = 234.1 \text{ J}\).

To 2 significant figures, this is \(230 \text{ J}\).

1 Mark: Calculation of vertical distance \(s = 14.4 \text{ m}\) (or final velocity \(v = 7.2 \text{ m s}^{-1}\)).
1 Mark: Correct substitution into the force equation \(F = m(g+a)\) to find \(F = 16.3 \text{ N}\) OR correct determination of both gravitational potential energy gain (\(198 \text{ J}\)) and kinetic energy gain (\(36 \text{ J}\)).
1 Mark: Final work calculated to give \(230 \text{ J}\) (or \(234 \text{ J}\)) with appropriate units.

First, find the tension force \(F\) using the Young Modulus equation \(E = \frac{F L}{A \Delta x}\):
\(F = \frac{E A \Delta x}{L} = \frac{1.2 \times 10^{11} \text{ Pa} \times 4.5 \times 10^{-7} \text{ m}^2 \times 1.5 \times 10^{-3} \text{ m}}{2.4 \text{ m}} = 33.75 \text{ N}\).

Now, calculate the stored elastic strain energy \(E_{el}\):
\(E_{el} = \frac{1}{2} F \Delta x = \frac{1}{2} \times 33.75 \text{ N} \times 1.5 \times 10^{-3} \text{ m} = 0.0253 \text{ J}\).

To 2 significant figures, the stored energy is \(0.025 \text{ J}\) (or \(2.5 \times 10^{-2} \text{ J}\)).

1 Mark: Use of \(E = \frac{\text{stress}}{\text{strain}}\) to express tension \(F\) or direct formulation of energy \(E_{el} = \frac{E A (\Delta x)^2}{2L}\).
1 Mark: Correct calculation of force \(F = 33.8 \text{ N}\) or correct substitution of values into the combined energy expression.
1 Mark: Correct final value for stored energy of \(0.025 \text{ J}\) (or \(2.5 \times 10^{-2} \text{ J}\)) with appropriate unit.

We set up two simultaneous equations using the equation \(\mathcal{E} = I(R + r)\).

For the first case:
\(\mathcal{E} = 1.2 \times (4.0 + r) = 4.8 + 1.2r\) (Equation 1)

For the second case:
\(\mathcal{E} = 0.60 \times (9.0 + r) = 5.4 + 0.60r\) (Equation 2)

Equating Equation 1 and Equation 2:
\(4.8 + 1.2r = 5.4 + 0.60r\)
\(0.60r = 0.6\)
\(r = 1.0\ \Omega\).

Substitute \(r = 1.0\ \Omega\) back into Equation 1 to find \(\mathcal{E}\):
\(\mathcal{E} = 1.2 \times (4.0 + 1.0) = 6.0 \text{ V}\).

1 Mark: Set up two correct equations based on \(\mathcal{E} = I(R + r)\).
1 Mark: Solve for internal resistance to find \(r = 1.0\ \Omega\).
1 Mark: Solve for e.m.f. to find \(\mathcal{E} = 6.0 \text{ V}\).

First, find the grating spacing \(d\):
\(d = \frac{1.0 \times 10^{-3} \text{ m}}{600} = 1.667 \times 10^{-6} \text{ m}\).

The maximum angle of diffraction is \(90^\circ\), so \(\sin\theta \le 1\).
Using the grating equation \(d \sin\theta = n \lambda\):
\(n \le \frac{d}{\lambda} = \frac{1.667 \times 10^{-6} \text{ m}}{589 \times 10^{-9} \text{ m}} = 2.83\).

Since the order \(n\) must be an integer, the highest order maximum that can be observed is \(n = 2\).

The total number of observed maxima includes the central maximum (\(n = 0\)), and the first and second orders on both sides:
\(\text{Total} = 2n + 1 = 2(2) + 1 = 5\).

1 Mark: Correct calculation of grating spacing \(d = 1.67 \times 10^{-6} \text{ m}\).
1 Mark: Using \(d \sin\theta = n \lambda\) with \(\sin\theta \le 1\) to determine maximum possible order \(n = 2\).
1 Mark: Realizing total maxima count includes the zero order and both sides to give \(5\).

Let \(N\) be the normal contact force and \(\theta = 25^\circ\) be the angle to the vertical.
Resolving vertically (no vertical acceleration):
\(N \cos\theta = mg\) (Equation 1)

Resolving horizontally (the horizontal component provides the centripetal force):
\(N \sin\theta = \frac{mv^2}{r}\) (Equation 2)

Dividing Equation 2 by Equation 1:
\(\tan\theta = \frac{v^2}{rg}\)

Solve for speed \(v\):
\(v = \sqrt{rg \tan\theta} = \sqrt{0.80 \text{ m} \times 9.81 \text{ m s}^{-2} \times \tan(25^\circ)}\)
\(v = \sqrt{7.848 \times 0.4663} = \sqrt{3.66} = 1.91 \text{ m s}^{-1}\).

To 2 significant figures, \(v = 1.9 \text{ m s}^{-1}\).

1 Mark: Identification of vertical force balance \(N \cos 25^\circ = mg\).
1 Mark: Linking horizontal force to centripetal expression \(N \sin 25^\circ = \frac{mv^2}{r}\) to obtain \(\tan 25^\circ = \frac{v^2}{rg}\).
1 Mark: Correct speed calculation yielding \(1.9 \text{ m s}^{-1}\) (or \(1.91 \text{ m s}^{-1}\)).

The charge of the alpha particle is \(q = 2e = 2 \times 1.60 \times 10^{-19} \text{ C} = 3.20 \times 10^{-19} \text{ C}\).

Using the balance between magnetic and centripetal force, \(q v B = \frac{m v^2}{r}\):
\(v = \frac{q B r}{m} = \frac{3.20 \times 10^{-19} \text{ C} \times 0.80 \text{ T} \times 0.45 \text{ m}}{6.64 \times 10^{-27} \text{ kg}} = 1.735 \times 10^7 \text{ m s}^{-1}\).

Now calculate kinetic energy in Joules:
\(E_k = \frac{1}{2} m v^2 = \frac{1}{2} \times 6.64 \times 10^{-27} \text{ kg} \times (1.735 \times 10^7 \text{ m s}^{-1})^2 = 1.00 \times 10^{-12} \text{ J}\).

Convert this energy to MeV:
\(E_k = \frac{1.00 \times 10^{-12} \text{ J}}{1.60 \times 10^{-13} \text{ J MeV}^{-1}} = 6.25 \text{ MeV}\).

Rounding to 2 significant figures gives \(6.3 \text{ MeV}\).

1 Mark: Correct calculation of speed of the alpha particle, \(v = 1.7 \times 10^7 \text{ m s}^{-1}\).
1 Mark: Correct calculation of kinetic energy in Joules, \(E_k = 1.0 \times 10^{-12} \text{ J}\).
1 Mark: Correct conversion factor applied to find energy as \(6.3 \text{ MeV}\) (accept range \(6.2\) to \(6.3 \text{ MeV}\)).

First, calculate the decay constant \(\lambda\) in \(\text{s}^{-1}\):
\(t_{1/2} = 5730 \times 365.25 \times 24 \times 3600 = 1.808 \times 10^{11} \text{ s}\)
\(\lambda = \frac{\ln 2}{t_{1/2}} = \frac{0.69315}{1.808 \times 10^{11} \text{ s}} = 3.83 \times 10^{-12} \text{ s}^{-1}\).

Next, find the activity \(A\) at \(t = 1.0 \times 10^4 \text{ years} = 3.156 \times 10^{11} \text{ s}\):
\(A = A_0 e^{-\lambda t} = 350 \times e^{-(3.83 \times 10^{-12} \times 3.156 \times 10^{11})} = 350 \times e^{-1.209} = 104.5 \text{ Bq}\).

Now, find the remaining number of nuclei \(N\) using \(A = \lambda N\):
\(N = \frac{A}{\lambda} = \frac{104.5}{3.83 \times 10^{-12}} = 2.73 \times 10^{13}\) nuclei.

Finally, calculate the mass of Carbon-14 remaining:
Each Carbon-14 nucleus has a mass of approximately \(14 \times 1.66 \times 10^{-27} \text{ kg} = 2.324 \times 10^{-26} \text{ kg}\).
\(m = 2.73 \times 10^{13} \times 2.324 \times 10^{-26} \text{ kg} = 6.34 \times 10^{-13} \text{ kg}\).

1 Mark: Determination of decay constant \(\lambda = 3.83 \times 10^{-12} \text{ s}^{-1}\) (or \(1.21 \times 10^{-4} \text{ yr}^{-1}\)).
1 Mark: Calculation of the remaining number of nuclei \(N = 2.7 \times 10^{13}\) (or remaining activity \(104 \text{ Bq}\)).
1 Mark: Correct conversion to mass in kilograms to obtain \(6.3 \times 10^{-13} \text{ kg}\) (accept \(6.3 \times 10^{-10} \text{ g}\)).

Since the container has a constant volume, no work is done on or by the gas. Thus, the thermal energy transferred \(Q\) equals the increase in internal energy \(\Delta U\) of the helium gas.

For a monatomic ideal gas, the internal energy is \(U = \frac{3}{2}pV\). Therefore:
\(\Delta U = \frac{3}{2} V \Delta p\)

Calculate the change in pressure \(\Delta p\):
\(\Delta p = 1.8 \times 10^5 \text{ Pa} - 1.2 \times 10^5 \text{ Pa} = 6.0 \times 10^4 \text{ Pa}\).

Substitute the values to find the energy:
\(Q = 1.5 \times 0.045 \text{ m}^3 \times 6.0 \times 10^4 \text{ Pa} = 4050 \text{ J}\).

Alternatively, calculate the number of molecules \(N = \frac{p_1 V}{kT_1} = 1.35 \times 10^{24}\), the final temperature \(T_2 = 435 \text{ K}\), and use \(\Delta U = \frac{3}{2}N k \Delta T = 4050 \text{ J}\).

This rounds to \(4100 \text{ J}\) or \(4.1 \text{ kJ}\).

1 Mark: Correct identification of temperature change \(\Delta T = 145 \text{ K}\) OR pressure change \(\Delta p = 6.0 \times 10^4 \text{ Pa}\).
1 Mark: Use of a valid thermal energy equation, such as \(Q = \frac{3}{2} V \Delta p\) or \(Q = \frac{3}{2} N k \Delta T\).
1 Mark: Correct calculation of thermal energy to give \(4.1 \text{ kJ}\) (or \(4050 \text{ J}\)).

First, calculate the cross-sectional area of the wire: \( A = \pi r^2 = \pi \left( \frac{0.45 \times 10^{-3}}{2} \right)^2 \approx 1.59 \times 10^{-7}\text{ m}^2 \). Next, determine the tensile force: \( F = mg = 4.0\text{ kg} \times 9.81\text{ N kg}^{-1} = 39.24\text{ N} \). Using the Young Modulus equation \( E = \frac{F L}{A \Delta x} \), rearrange for extension \( \Delta x \): \( \Delta x = \frac{F L}{A E} = \frac{39.24 \times 1.8}{1.59 \times 10^{-7} \times 2.0 \times 10^{11}} = 2.22 \times 10^{-3}\text{ m} \).

1 mark: Correct calculation of cross-sectional area \( A = 1.59 \times 10^{-7}\text{ m}^2 \).
1 mark: Rearranging Young Modulus equation for extension \( \Delta x = \frac{FL}{AE} \) and substituting correct force value.
1 mark: Correct final answer with unit \( (2.2 \text{ or } 2.22) \times 10^{-3}\text{ m} \).

The current in the circuit is calculated using the terminal potential difference and the external resistance: \( I = \frac{V}{R} = \frac{4.8\text{ V}}{8.0\ \Omega} = 0.60\text{ A} \). Using the formula for e.m.f. \( E = V + Ir \), substitute the known values: \( 6.0 = 4.8 + 0.60 r \). Solving for the internal resistance gives: \( 1.2 = 0.60 r \Rightarrow r = 2.0\ \Omega \).

1 mark: Calculate correct circuit current of \( 0.60\text{ A} \).
1 mark: Use of \( E = V + Ir \) or equivalent potential divider formula.
1 mark: Correct internal resistance of \( 2.0\ \Omega \).

The energy of the incident photon is given by: \( E = h f = 6.63 \times 10^{-34}\text{ J s} \times 7.5 \times 10^{14}\text{ Hz} = 4.97 \times 10^{-19}\text{ J} \). Using Einstein's photoelectric equation: \( h f = \phi + E_{\text{k,max}} \Rightarrow \phi = h f - E_{\text{k,max}} = 4.97 \times 10^{-19}\text{ J} - 1.20 \times 10^{-19}\text{ J} = 3.77 \times 10^{-19}\text{ J} \). To convert this energy to electronvolts: \( \phi = \frac{3.77 \times 10^{-19}\text{ J}}{1.60 \times 10^{-19}\text{ J eV}^{-1}} \approx 2.36\text{ eV} \).

1 mark: Correct calculation of incident photon energy \( 4.97 \times 10^{-19}\text{ J} \).
1 mark: Correct subtraction to find work function in Joules \( 3.77 \times 10^{-19}\text{ J} \).
1 mark: Correct conversion of work function to eV giving \( 2.4\text{ eV} \) (accept \( 2.36\text{ eV} \)).

The centripetal force required to keep the car in circular motion is provided by friction: \( F = \frac{m v^2}{r} \). Substituting the maximum frictional force for \( F \) gives: \( 1.8 = \frac{0.25 v^2}{0.80} \). Solving for the velocity: \( v^2 = \frac{1.8 \times 0.80}{0.25} = 5.76 \Rightarrow v = 2.4\text{ m s}^{-1} \).

1 mark: Recognise that maximum frictional force equals centripetal force \( F = \frac{m v^2}{r} \).
1 mark: Correct rearrangement of formula to find \( v = \sqrt{\frac{F r}{m}} \).
1 mark: Correct calculation of maximum speed \( 2.4\text{ m s}^{-1} \) with units.

The magnetic force on the proton provides the centripetal acceleration: \( B q v = \frac{m v^2}{r} \). Simplifying this gives: \( v = \frac{B q r}{m} \). Substitute the given values (converting \( 2.4\text{ cm} \) to \( 2.4 \times 10^{-2}\text{ m} \)): \( v = \frac{0.15 \times 1.60 \times 10^{-19} \times 2.4 \times 10^{-2}}{1.67 \times 10^{-27}} = 3.45 \times 10^5\text{ m s}^{-1} \).

1 mark: Equate magnetic force to centripetal force: \( Bqv = \frac{mv^2}{r} \) or \( r = \frac{mv}{Bq} \).
1 mark: Correct substitution of values with radius in meters.
1 mark: Correct final velocity \( 3.5 \times 10^5\text{ m s}^{-1} \) (accept \( 3.4 \times 10^5\text{ m s}^{-1} \) to \( 3.45 \times 10^5\text{ m s}^{-1} \)).

First, convert the initial temperature to Kelvin: \( T_1 = 20 + 273 = 293\text{ K} \). Since volume is constant, we apply Pressure Law: \( \frac{P_1}{T_1} = \frac{P_2}{T_2} \). Rearranging for \( T_2 \) gives: \( T_2 = T_1 \times \frac{P_2}{P_1} = 293\text{ K} \times \frac{1.8 \times 10^5\text{ Pa}}{1.2 \times 10^5\text{ Pa}} = 439.5\text{ K} \point\). Convert back to Celsius: \( T_2 = 439.5 - 273 = 166.5\text{ }^\circ\text{C} \).

1 mark: Correct conversion of initial temperature to Kelvin (\( 293\text{ K} \)).
1 mark: Use of pressure law ratio \( \frac{P_1}{T_1} = \frac{P_2}{T_2} \) to find new temperature in Kelvin (\( 440\text{ K} \)).
1 mark: Correct final Celsius temperature \( 167\text{ }^\circ\text{C} \) (accept \( 166.5\text{ }^\circ\text{C} \) to \( 167\text{ }^\circ\text{C} \)).

First, determine the number of half-lives that have elapsed: \( \frac{105}{840} = 0.125 = \left(\frac{1}{2}\right)^3 \), so 3 half-lives have elapsed. Therefore, the half-life \( t_{1/2} \) is: \( t_{1/2} = \frac{12.0\text{ hours}}{3} = 4.0\text{ hours} \). Convert the half-life into seconds: \( t_{1/2} = 4.0 \times 3600\text{ s} = 14400\text{ s} \). Finally, calculate the decay constant: \( \lambda = \frac{\ln(2)}{t_{1/2}} = \frac{0.693}{14400\text{ s}} \approx 4.81 \times 10^{-5}\text{ s}^{-1} \).

1 mark: Calculate the half-life to be \( 4.0\text{ hours} \).
1 mark: Convert half-life to seconds (\( 14400\text{ s} \)).
1 mark: Use \( \lambda = \frac{\ln 2}{t_{1/2}} \) to obtain correct value \( 4.8 \times 10^{-5}\text{ s}^{-1} \) (accept \( 4.81 \times 10^{-5}\text{ s}^{-1} \)).

First, find the surface temperature \( T \) of the star using Wien's displacement law: \( \lambda_{\text{max}} T = 2.90 \times 10^{-3} \Rightarrow T = \frac{2.90 \times 10^{-3}}{480 \times 10^{-9}} = 6041.7\text{ K} \). Next, use Stefan-Boltzmann law for luminosity: \( L = A \sigma T^4 = 4 \pi r^2 \sigma T^4 \). Substituting the values: \( L = 4 \pi (6.2 \times 10^8)^2 \times 5.67 \times 10^{-8} \times (6041.7)^4 \approx 3.64 \times 10^{26}\text{ W} \).

1 mark: Correct calculation of temperature using Wien's law (\( 6040\text{ K} \)).
1 mark: Recall and substitute into \( L = 4 \pi r^2 \sigma T^4 \).
1 mark: Correct value for luminosity \( 3.6 \times 10^{26}\text{ W} \) (accept range \( 3.6 \times 10^{26}\text{ W} \) to \( 3.7 \times 10^{26}\text{ W} \)).

First, calculate the cross-sectional area of the wire: \( A = \pi r^2 = \pi (0.40 \times 10^{-3}\text{ m})^2 = 5.03 \times 10^{-7}\text{ m}^2 \).

Next, use the drift velocity formula: \( I = nAve \).

Rearranging for drift velocity \( v \):
\( v = \frac{I}{nAe} = \frac{3.2\text{ A}}{8.5 \times 10^{28}\text{ m}^{-3} \times 5.03 \times 10^{-7}\text{ m}^2 \times 1.60 \times 10^{-19}\text{ C}} \)
\( v = \frac{3.2}{6.84 \times 10^3} = 4.68 \times 10^{-4}\text{ m s}^{-1} \), which rounds to \( 4.7 \times 10^{-4}\text{ m s}^{-1} \).

1. Calculate cross-sectional area \( A \) using correct radius (\( 5.0 \times 10^{-7}\text{ m}^2 \)) (1 mark)
2. Correct rearrangement of \( I = nAve \) (1 mark)
3. Correct final answer of \( 4.7 \times 10^{-4}\text{ m s}^{-1} \) (accept \( 4.68 \times 10^{-4}\text{ m s}^{-1} \)) with appropriate unit (1 mark)

Calculate the cross-sectional area: \( A = \pi r^2 = \pi (5.5 \times 10^{-3}\text{ m})^2 = 9.50
\times 10^{-5}\text{ m}^2 \).

Calculate the force (weight of the climber): \( F = mg = 75.0\text{ kg} \times 9.81\text{ N kg}^{-1} = 736\text{ N} \).

Use the Young modulus equation: \( E = \frac{\text{stress}}{\text{strain}} = \frac{F / A}{\Delta L / L} \implies \Delta L = \frac{FL}{AE} \).

Substitute the values:
\(
\Delta L = \frac{736\text{ N} \times 15.0\text{ m}}{9.50 \times 10^{-5}\text{ m}^2 \times 3.0 \times 10^9\text{ Pa}} = 0.0387\text{ m} \approx 3.9 \times 10^{-2}\text{ m} \text{ (or } 3.9\text{ cm)} \).

1. Correct calculation of weight (\( 736\text{ N} \)) and cross-sectional area (\( 9.50 \times 10^{-5}\text{ m}^2 \)) (1 mark)
2. Recall and correct rearrangement of the Young modulus formula for extension \( \Delta L \) (1 mark)
3. Correct final answer of \( 0.039\text{ m} \) or \( 3.9\text{ cm} \) with unit (1 mark)

First, analyze the vertical motion to find the time of flight \( t \).
Using \( s = ut + \frac{1}{2}at^2 \) with \( u_y = 0 \):
\( s = \frac{1}{2}gt^2 \implies 45\text{ m} = \frac{1}{2} \times 9.81\text{ m s}^{-2} \times t^2 \)
\( t = \sqrt{\frac{2 \times 45}{9.81}} = 3.03\text{ s} \).

Since horizontal velocity is constant:
\( v_x = \frac{\text{horizontal distance}}{t} = \frac{54\text{ m}}{3.03\text{ s}} = 17.8\text{ m s}^{-1} \approx 18\text{ m s}^{-1} \).

1. Use of vertical kinematic equation \( s = \frac{1}{2}gt^2 \) to express time (1 mark)
2. Correct time of flight \( t = 3.0\text{ s} \) or \( 3.03\text{ s} \) (1 mark)
3. Correct initial horizontal velocity with unit (\( 18\text{ m s}^{-1} \)) (1 mark)

The horizontal centripetal force is provided by the static friction between the tires and the track.
Using the centripetal force formula:
\( F = \frac{mv^2}{r} \)

Substitute the given values:
\( F = \frac{0.25\text{ kg} \times (3.5\text{ m s}^{-1})^2}{1.2\text{ m}} \)
\( F = \frac{0.25 \times 12.25}{1.2} = 2.55\text{ N} \), which rounds to \( 2.6\text{ N} \).

1. Recall of the centripetal force formula \( F = \frac{mv^2}{r} \) (1 mark)
2. Correct substitution of mass, velocity, and radius (1 mark)
3. Correct calculation of force to 2 significant figures with unit (\( 2.6\text{ N} \)) (1 mark)

The magnetic force provides the centripetal force acting on the electron:
\( Bqv = \frac{mv^2}{r} \implies r = \frac{mv}{Bq} \).

Substitute the mass of an electron \( m_e = 9.11 \times 10^{-31}\text{ kg} \) and the elementary charge \( q = 1.60 \times 10^{-19}\text{ C} \):
\( r = \frac{9.11 \times 10^{-31}\text{ kg} \times 4.2 \times 10^6\text{ m s}^{-1}}{0.15\text{ T} \times 1.60 \times 10^{-19}\text{ C}} \)
\( r = \frac{3.826 \times 10^{-24}}{2.4 \times 10^{-20}} = 1.59 \times 10^{-4}\text{ m} \approx 1.6 \times 10^{-4}\text{ m} \).

1. Equate magnetic force to centripetal force to obtain \( r = \frac{mv}{Bq} \) (1 mark)
2. Substitution of correct values, including standard values for \( m_e \) and \( e \) (1 mark)
3. Correct calculation of radius with unit (\( 1.6 \times 10^{-4}\text{ m} \)) (1 mark)

First, calculate the kinetic energy \( E_k \) gained by the alpha particle. Since its charge is \( q = 2e = 2 \times 1.60 \times 10^{-19}\text{ C} \):
\( E_k = qV = 2 \times 1.60 \times 10^{-19}\text{ C} \times 12000\text{ V} = 3.84 \times 10^{-15}\text{ J} \).

Now, calculate the momentum \( p \):
\( p = \sqrt{2mE_k} = \sqrt{2 \times 6.64 \times 10^{-27}\text{ kg} \times 3.84 \times 10^{-15}\text{ J}} = 7.14 \times 10^{-21}\text{ kg m s}^{-1} \).

Finally, calculate the de Broglie wavelength \( \lambda \):
\( \lambda = \frac{h}{p} = \frac{6.63 \times 10^{-34}\text{ J s}}{7.14 \times 10^{-21}\text{ kg m s}^{-1}} = 9.29 \times 10^{-14}\text{ m} \approx 9.3 \times 10^{-14}\text{ m} \).

1. Calculate kinetic energy using \( E_k = 2eV \) (\( 3.84 \times 10^{-15}\text{ J} \)) (1 mark)
2. Calculate momentum \( p \) or velocity (\( 1.08 \times 10^6\text{ m s}^{-1} \)) (1 mark)
3. Calculate de Broglie wavelength with correct unit (\( 9.3 \times 10^{-14}\text{ m} \)) (1 mark)

Use the diffraction grating equation:
\( d \sin\theta = n\lambda \)

For \( n = 2 \):
\( d \sin(34.5^\circ) = 2 \times 589 \times 10^{-9}\text{ m} \)
\( d = \frac{1.178 \times 10^{-6}\text{ m}}{\sin(34.5^\circ)} = 2.08 \times 10^{-6}\text{ m} \).

The number of lines per metre \( N_{\text{metre}} = \frac{1}{d} = \frac{1}{2.08 \times 10^{-6}} = 4.81 \times 10^5\text{ m}^{-1} \).

To convert this to lines per millimetre, divide by 1000:
\( N_{\text{mm}} = \frac{4.81 \times 10^5}{1000} = 481\text{ mm}^{-1} \approx 4.8 \times 10^2\text{ lines mm}^{-1} \).

1. Use \( d \sin\theta = n\lambda \) with correct substitution to find grating spacing \( d \) (1 mark)
2. Realize that number of lines per unit length is \( 1/d \) (1 mark)
3. Correct calculation of lines per millimetre with unit (\( 480\text{ lines mm}^{-1} \)) (1 mark)

Since the volume \( V \) and number of moles \( n \) remain constant, the ideal gas law (\( PV = nRT \)) dictates that pressure is directly proportional to the absolute temperature:
\( P \propto T \).

First, convert the initial temperature to Kelvin:
\( T_1 = 20^\circ\text{C} + 273.15 = 293.15\text{ K} \).

Since pressure doubles (\( P_2 = 2P_1 \)), the absolute temperature must also double:
\( T_2 = 2 \times 293.15\text{ K} = 586.3\text{ K} \).

Convert back to Celsius:
\( \theta_2 = 586.3 - 273.15 = 313.15^\circ\text{C} \approx 313^\circ\text{C} \).

1. Convert the initial Celsius temperature to Kelvin (\( 293\text{ K} \)) (1 mark)
2. Correctly deduce that absolute temperature must double (\( 586\text{ K} \)) because pressure doubles at constant volume (1 mark)
3. Convert temperature back to Celsius and state the correct value and unit (\( 313^\circ\text{C} \)) (1 mark)

Using conservation of energy: Initial Kinetic Energy = Gain in Gravitational Potential Energy + Work Done against Friction. \(E_k = \frac{1}{2}mv^2\). \(\Delta E_p = mgh = mgd \sin\theta\). \(W_f = f \cdot d\) where friction \(f = \mu R = \mu mg \cos\theta\). Thus, \(\frac{1}{2}mv^2 = mgd \sin\theta + \mu mgd \cos\theta\). Dividing by \(m\): \(\frac{1}{2}v^2 = gd(\sin\theta + \mu \cos\theta)\). \(d = \frac{v^2}{2g(\sin\theta + \mu \cos\theta)}\). Substituting values: \(v = 4.2\text{ m s}^{-1}\), \(g = 9.81\text{ m s}^{-2}\), \(\theta = 30^\circ\), \(\mu = 0.25\). \(d = \frac{4.2^2}{2 \times 9.81 \times (\sin 30^\circ + 0.25 \cos 30^\circ)}\). \(d = \frac{17.64}{19.62 \times (0.5 + 0.2165)} = \frac{17.64}{14.058} \approx 1.25\text{ m}\) which rounds to \(1.3\text{ m}\).

1. Equate initial kinetic energy to work done against gravity plus work done against friction: \(\frac{1}{2}mv^2 = mgh + W_f\) (1 mark) 2. Express work done against friction as \(\mu mgd \cos\theta\) (1 mark) 3. Derive the expression: \(d = \frac{v^2}{2g(\sin\theta + \mu \cos\theta)}\) (1 mark) 4. Substitute correct values into the expression (1 mark) 5. Correct final value of \(d = 1.3\text{ m}\) (accept \(1.25\text{ m}\) to \(1.3\text{ m}\)) with unit (1 mark)

First, calculate the tension force \(F = mg = 12 \times 9.81 = 117.72\text{ N}\). The Young modulus is given by \(E = \frac{F L}{A \Delta L}\). Rearranging for extension: \(\Delta L = \frac{F L}{A E} = \frac{117.72 \times 2.4}{1.5 \times 10^{-6} \times 2.0 \times 10^{11}} = 9.4176 \times 10^{-4}\text{ m}\). The elastic strain energy stored is \(E_{el} = \frac{1}{2} F \Delta L = 0.5 \times 117.72 \times 9.4176 \times 10^{-4} = 0.0554\text{ J}\), which is \(0.055\text{ J}\) to two significant figures.

1. Calculate tension force using \(F = mg\) (1 mark) 2. Recall or use Young modulus formula to express extension (1 mark) 3. Correct value for extension \(\Delta L = 9.4 \times 10^{-4}\text{ m}\) (1 mark) 4. Use elastic strain energy formula \(E = \frac{1}{2} F \Delta L\) (1 mark) 5. Calculate correct final energy \(0.055\text{ J}\) (accept \(0.055\text{ J}\) to \(0.056\text{ J}\)) (1 mark)

The total resistance of the circuit is \(R_{total} = R + r\). The current \(I\) in the circuit is \(I = \frac{E}{R+r}\). The power dissipated in the variable resistor is \(P = I^2 R\). Substituting the current gives \(P = \left(\frac{E}{R+r}\right)^2 R = \frac{E^2 R}{(R+r)^2}\). Maximum power transfer occurs when the external resistance equals the internal resistance: \(R = r = 3.5\ \Omega\). Substituting this into the power equation: \(P_{max} = \frac{E^2 r}{(2r)^2} = \frac{E^2}{4r} = \frac{9.0^2}{4 \times 3.5} = \frac{81}{14} = 5.79\text{ W}\), which is \(5.8\text{ W}\) to 2 significant figures.

1. State equation for total current \(I = \frac{E}{R+r}\) (1 mark) 2. Substitute into \(P = I^2 R\) and algebraically show the derived equation (1 mark) 3. State that maximum power occurs when \(R = r\) (hence \(R = 3.5\ \Omega\)) (1 mark) 4. Substitute values into the power equation (1 mark) 5. Obtain \(5.8\text{ W}\) (accept \(5.79\text{ W}\)) (1 mark)

Grating spacing \(d = \frac{1 \times 10^{-3}\text{ m}}{600} = 1.667 \times 10^{-6}\text{ m}\). Using the diffraction grating equation \(d \sin\theta = n \lambda\) for \(n = 1\): \(\lambda = 1.667 \times 10^{-6} \times \sin(18.4^\circ) = 5.26 \times 10^{-7}\text{ m} \approx 5.3 \times 10^{-7}\text{ m}\). For the highest order, the maximum possible angle \(\theta\) is \(90^\circ\) where \(\sin\theta = 1\). So, \(n_{max} \le \frac{d}{\lambda} = \frac{1.667 \times 10^{-6}}{5.26 \times 10^{-7}} = 3.17\). Since order must be an integer, the highest order is \(3\).

1. Calculate grating spacing \(d = 1.67 \times 10^{-6}\text{ m}\) (1 mark) 2. Use \(d \sin\theta = n\lambda\) with \(n=1\) (1 mark) 3. Calculate \(\lambda = 5.3 \times 10^{-7}\text{ m}\) (1 mark) 4. Set \(\sin\theta \le 1\) to find maximum order (1 mark) 5. State maximum integer order \(n = 3\) (1 mark)

For a conical pendulum, resolving forces: Vertically: \(T \cos\theta = mg\). Horizontally: \(T \sin\theta = \frac{mv^2}{r}\). Dividing the equations gives \(\tan\theta = \frac{v^2}{rg}\). The radius is \(r = L \sin\theta\). Thus, \(v^2 = L g \sin\theta \tan\theta\), giving \(v = \sqrt{L g \sin\theta \tan\theta}\). Substituting the values: \(v = \sqrt{0.85 \times 9.81 \times \sin 25^\circ \times \tan 25^\circ} = \sqrt{0.85 \times 9.81 \times 0.4226 \times 0.4663} = \sqrt{1.6432} = 1.28\text{ m s}^{-1}\), which is \(1.3\text{ m s}^{-1}\) to 2 significant figures.

1. State horizontal and vertical force equations (1 mark) 2. Identify radius relationship \(r = L \sin\theta\) (1 mark) 3. Correctly derive \(v = \sqrt{L g \sin\theta \tan\theta}\) (1 mark) 4. Substitute given values into the equation (1 mark) 5. Correctly calculate speed \(1.3\text{ m s}^{-1}\) (accept \(1.28\text{ m s}^{-1}\)) (1 mark)

First, equating work done by the electrical field to the gain in kinetic energy: \(eV = \frac{1}{2} m_e v^2\), which gives \(v = \sqrt{\frac{2eV}{m_e}}\). Second, magnetic force provides the centripetal force: \(evB = \frac{m_e v^2}{R}\), which simplifies to \(R = \frac{m_e v}{eB}\). Substituting \(v\) into the radius equation: \(R = \frac{m_e}{eB} \sqrt{\frac{2eV}{m_e}} = \frac{1}{B} \sqrt{\frac{2 m_e V}{e}}\). Plugging in the given numbers: \(R = \frac{1}{3.4 \times 10^{-3}} \sqrt{\frac{2 \times 9.11 \times 10^{-31} \times 120}{1.60 \times 10^{-19}}} = \frac{3.697 \times 10^{-5}}{3.4 \times 10^{-3}} = 0.01087\text{ m}\), which is \(1.1 \times 10^{-2}\text{ m}\) (or \(1.1\text{ cm}\)) to two significant figures.

1. Equate kinetic energy to electric work done: \(eV = \frac{1}{2}m_ev^2\) (1 mark) 2. Equate magnetic force to centripetal force: \(evB = \frac{m_ev^2}{R}\) (1 mark) 3. Derive the expression \(R = \frac{1}{B}\sqrt{\frac{2 m_e V}{e}}\) (1 mark) 4. Correct substitution of all physical constants and given values (1 mark) 5. Calculate correct final radius \(1.1 \times 10^{-2}\text{ m}\) (or \(1.1\text{ cm}\)) (1 mark)

First, calculate the half-life in seconds: \(T_{1/2} = 1600 \times 3.16 \times 10^7\text{ s} = 5.056 \times 10^{10}\text{ s}\). Use this to calculate the decay constant: \(\lambda = \frac{\ln 2}{T_{1/2}} = \frac{0.69315}{5.056 \times 10^{10}} = 1.37 \times 10^{-11}\text{ s}^{-1}\). Next, find the number of nuclei using the activity formula \(A = \lambda N\): \(N = \frac{A}{\lambda} = \frac{3.7 \times 10^{10}}{1.37 \times 10^{-11}} = 2.70 \times 10^{21}\) nuclei. Convert the number of nuclei to moles: \(n = \frac{N}{N_A} = \frac{2.70 \times 10^{21}}{6.02 \times 10^{23}} = 4.485 \times 10^{-3}\text{ mol}\). Finally, calculate the mass: \(m = n \times 226\text{ g mol}^{-1} = 4.485 \times 10^{-3} \times 226 = 1.01\text{ g}\), which is \(1.0\text{ g}\) to 2 significant figures.

1. Convert half-life to seconds and calculate \(\lambda = 1.37 \times 10^{-11}\text{ s}^{-1}\) (1 mark) 2. Use \(A = \lambda N\) to find number of nuclei \(N\) (1 mark) 3. Correct value for \(N = 2.7 \times 10^{21}\) (1 mark) 4. Divide by Avogadro constant to find moles (1 mark) 5. Calculate mass \(1.0\text{ g}\) (accept range \(1.0\text{ g}\) to \(1.02\text{ g}\)) (1 mark)

First, convert temperature to Kelvin: \(T = 20 + 273 = 293\text{ K}\). Use the ideal gas equation \(PV = Nk_BT\) to solve for \(N\): \(N = \frac{PV}{k_BT} = \frac{1.5 \times 10^5 \times 0.045}{1.38 \times 10^{-23} \times 293} = \frac{6750}{4.043 \times 10^{-21}} = 1.67 \times 10^{24}\) atoms, which is \(1.7 \times 10^{24}\) to two significant figures. The average kinetic energy of one atom is \(\frac{3}{2}k_BT\). The total kinetic energy of \(N\) atoms is \(E_{total} = N \left(\frac{3}{2}k_BT\right) = \frac{3}{2}PV = 1.5 \times 1.5 \times 10^5 \times 0.045 = 10125\text{ J}\), which is \(1.0 \times 10^4\text{ J}\) (or \(10\text{ kJ}\)) to two significant figures.

1. Convert temperature to Kelvin (293 K) (1 mark) 2. Use \(PV = Nk_BT\) to express and solve for \(N\) (1 mark) 3. Correct value of \(N = 1.7 \times 10^{24}\) (accept \(1.67 \times 10^{24}\)) (1 mark) 4. State or use \(E_{total} = \frac{3}{2}Nk_BT\) or \(\frac{3}{2}PV\) (1 mark) 5. Correct total kinetic energy of \(1.0 \times 10^4\text{ J}\) (accept \(1.01 \times 10^4\text{ J}\) or \(10\text{ kJ}\)) (1 mark)

First, calculate the vertical component of the velocity \(v_y\) just before impact using conservation of energy or equations of motion: \(v_y^2 = u_y^2 + 2gy\). Since the helicopter flies horizontally, \(u_y = 0\): \(v_y^2 = 0 + 2 \times 9.81\text{ m s}^{-2} \times 120\text{ m} = 2354.4\text{ m}^2\text{ s}^{-2}\), which gives \(v_y = 48.52\text{ m s}^{-1}\). Combine the horizontal and vertical components to find the impact speed: \(v = \sqrt{v_x^2 + v_y^2} = \sqrt{45^2 + 2354.4} = \sqrt{2025 + 2354.4} = 66.18\text{ m s}^{-1}\). This is approximately \(66\text{ m s}^{-1}\). Now, calculate the angle \(\theta\) to the horizontal at impact: \(\tan \theta = \frac{v_y}{v_x} = \frac{48.52}{45} = 1.078\). Therefore, \(\theta = \arctan(1.078) = 47.15^\circ\), which is approximately \(47.2^\circ\).

1. [M1] Uses \(v_y^2 = u_y^2 + 2gy\) or \(\frac{1}{2}mv^2 = \frac{1}{2}mu^2 + mgh\) to relate initial and final velocities. 2. [A1] Correct vertical speed component \(v_y = 48.5\text{ m s}^{-1}\). 3. [A1] Correct substitution to show total speed \(v = 66.2\text{ m s}^{-1}\). 4. [M1] Uses \(\tan\theta = \frac{v_y}{v_x}\) to find the direction. 5. [A1] Correct angle \(\theta = 47^\circ\) to \(47.2^\circ\).

Using Young modulus formula: \(E = \frac{F L}{A \Delta x}\), which rearranges to \(\Delta x = \frac{F L}{A E}\). Substituting the given values: \(\Delta x = \frac{4.5 \times 10^4\text{ N} \times 3.5\text{ m}}{2.5 \times 10^{-4}\text{ m}^2 \times 2.0 \times 10^{11}\text{ Pa}} = 3.15 \times 10^{-3}\text{ m}\) (or \(3.15\text{ mm}\)). For the elastic energy stored: \(E_{\text{el}} = \frac{1}{2} F \Delta x = \frac{1}{2} \times (4.5 \times 10^4\text{ N}) \times (3.15 \times 10^{-3}\text{ m}) = 70.88\text{ J}\), which rounds to \(70.9\text{ J}\).

1. [M1] Recalls or uses the Young modulus equation rearranged for extension. 2. [A1] Correct substitution of all variables in SI units. 3. [A1] Correct extension value \(3.15\text{ mm}\). 4. [M1] Uses formula for elastic energy: \(E = \frac{1}{2}F\Delta x\). 5. [A1] Correct elastic energy \(71\text{ J}\) (accept \(70.9\text{ J}\) to \(71\text{ J}\), allow ECF from incorrect extension).

First, express the total current \(I\) in the circuit using Ohm's law: \(I = \frac{\mathcal{E}}{R+r}\). The power dissipated in the external resistor is given by: \(P = I^2 R = \left(\frac{\mathcal{E}}{R+r}\right)^2 R = \frac{\mathcal{E}^2 R}{(R+r)^2}\). To find the maximum power, differentiate \(P\) with respect to \(R\) using the quotient rule: \(\frac{\mathrm{d}P}{\mathrm{d}R} = \mathcal{E}^2 \frac{(R+r)^2 \cdot 1 - R \cdot 2(R+r)}{(R+r)^4} = \mathcal{E}^2 \frac{(R+r)[(R+r) - 2R]}{(R+r)^4} = \mathcal{E}^2 \frac{r-R}{(R+r)^3}\). Setting the derivative to zero for maximum power: \(r - R = 0 \implies R = r\). Thus, the power is maximized when the external resistance equals the internal resistance.

1. [M1] Expresses total current as \(I = \frac{\mathcal{E}}{R+r}\) and substitutes this expression into \(P = I^2 R\). 2. [A1] Clearly derives the given expression for power \(P\). 3. [M1] Correctly applies the quotient rule or product rule to differentiate \(P\) with respect to \(R\). 4. [A1] Shows the correct derivative \(\frac{\mathrm{d}P}{\mathrm{d}R} = \frac{\mathcal{E}^2(r-R)}{(R+r)^3}\). 5. [A1] Sets derivative to zero and concludes that \(R = r\).

First, calculate the energy \(E\) of the incident photons: \(E = \frac{hc}{\lambda} = \frac{6.63 \times 10^{-34}\text{ J s} \times 3.00 \times 10^8\text{ m s}^{-1}}{365 \times 10^{-9}\text{ m}} = 5.449 \times 10^{-19}\text{ J}\). Convert the work function \(\Phi\) of sodium to Joules: \(\Phi = 2.28\text{ eV} \times 1.60 \times 10^{-19}\text{ J eV}^{-1} = 3.648 \times 10^{-19}\text{ J}\). Use Einstein's photoelectric equation to find \(E_{k,\text{max}}\): \(E_{k,\text{max}} = E - \Phi = 5.449 \times 10^{-19}\text{ J} - 3.648 \times 10^{-19}\text{ J} = 1.801 \times 10^{-19}\text{ J}\). Finally, calculate the stopping potential \(V_s\) using \(E_{k,\text{max}} = e V_s\): \(V_s = \frac{E_{k,\text{max}}}{e} = \frac{1.801 \times 10^{-19}\text{ J}}{1.60 \times 10^{-19}\text{ C}} = 1.13\text{ V}\).

1. [M1] Uses \(E = \frac{hc}{\lambda}\) to determine photon energy in J or eV. 2. [M1] Converts the work function to Joules (or photon energy to eV) correctly. 3. [A1] Correctly calculates \(E_{k,\text{max}} = 1.80 \times 10^{-19}\text{ J}\) (accept range \(1.79 \times 10^{-19}\text{ J}\) to \(1.81 \times 10^{-19}\text{ J}\)). 4. [M1] Uses stopping potential formula \(V_s = \frac{E_{k,\text{max}}}{e}\). 5. [A1] Correct stopping potential \(1.13\text{ V}\) (accept range \(1.11\text{ V}\) to \(1.14\text{ V}\)).

Let \(\theta\) be the angle between the string and the vertical. Resolving forces vertically, where there is no acceleration: \(T \cos\theta = mg\). Therefore, the tension \(T\) is expressed as: \(T = \frac{mg}{\cos\theta}\). To find the angle \(\theta\), use the geometric properties of the conical pendulum: \(\sin\theta = \frac{\text{radius}}{\text{length}} = \frac{0.50\text{ m}}{1.2\text{ m}} = 0.4167\). This gives \(\theta = \arcsin(0.4167) = 24.62^\circ\). Now, find the cosine of this angle: \(\cos(24.62^\circ) = 0.9091\). Finally, substitute the values to calculate the tension \(T\): \(T = \frac{0.15\text{ kg} \times 9.81\text{ N kg}^{-1}}{0.9091} = 1.618\text{ N} \approx 1.62\text{ N}\).

1. [M1] Resolves forces vertically to establish \(T \cos\theta = mg\). 2. [A1] Correctly rearranges the equation to show \(T = \frac{mg}{\cos\theta}\). 3. [M1] Uses trigonometry to find \(\theta = 24.6^\circ\) or \(\cos\theta = 0.909\). 4. [M1] Correct substitution of mass and \(g\) into their tension equation. 5. [A1] Obtains a correct final value for tension: \(1.62\text{ N}\) (accept range \(1.60\text{ N}\) to \(1.65\text{ N}\)).

(a) For an ion to pass through the region undeflected, the magnetic force \(F_B\) and the electric force \(F_E\) acting on the ion must balance each other: \(F_B = F_E \implies q v B = q E\). Dividing both sides of the equation by \(q\) gives: \(v = \frac{E}{B}\). (b) Rearranging the formula to solve for the electric field strength: \(E = v B\). Substituting the given values: \(E = 2.4 \times 10^5\text{ m s}^{-1} \times 0.15\text{ T} = 3.6 \times 10^4\text{ V m}^{-1}\) (or \(\text{N C}^{-1}\)).

1. [M1] Equates electric and magnetic force expressions: \(qE = qvB\). 2. [A1] Shows clearly that speed is independent of charge: \(v = \frac{E}{B}\). 3. [M1] Rearranges the equation to express electric field: \(E = vB\). 4. [A1] Calculates the correct numerical value: \(3.6 \times 10^4\). 5. [A1] States the correct unit of electric field strength (\(\text{V m}^{-1}\) or \(\text{N C}^{-1}\)).

(a) Convert both temperatures from Celsius to Kelvin: \(T_1 = 27.0 + 273.15 = 300.15\text{ K}\) and \(T_2 = 127.0 + 273.15 = 400.15\text{ K}\). Since the root-mean-square speed is proportional to the square root of the absolute temperature: \(v_{\text{rms}} = \sqrt{\frac{3k_B T}{m}} \implies v_{\text{rms}} \propto \sqrt{T}\). Therefore, the ratio is: \(\frac{v_{\text{rms,2}}}{v_{\text{rms,1}}} = \sqrt{\frac{T_2}{T_1}} = \sqrt{\frac{400.15}{300.15}} = \sqrt{1.333} \approx 1.155\), which is approximately \(1.15\). (b) The mean kinetic energy of a molecule is given by: \(E_k = \frac{3}{2} k_B T\). At \(127.0^\circ\text{C}\) (\(400.15\text{ K}\)): \(E_k = \frac{3}{2} \times (1.38 \times 10^{-23}\text{ J K}^{-1}) \times 400.15\text{ K} = 8.283 \times 10^{-21}\text{ J} \approx 8.28 \times 10^{-21}\text{ J}\).

1. [M1] Converts both temperatures to Kelvin (accept \(300\text{ K}\) and \(400\text{ K}\)). 2. [M1] Uses \(v_{\text{rms}} \propto \sqrt{T}\) to set up the ratio calculation. 3. [A1] Correctly calculates ratio as \(1.15\) (or \(1.155\)). 4. [M1] Recalls and uses \(E_k = \frac{3}{2} k_B T\) for kinetic energy. 5. [A1] Correctly calculates \(E_k = 8.28 \times 10^{-21}\text{ J}\) (accept \(8.3 \times 10^{-21}\text{ J}\)).

First, find the decay constant \(\lambda\) in \(\text{s}^{-1}\): \(\lambda = \frac{\ln 2}{T_{1/2}} = \frac{0.693}{5.27 \times 3.16 \times 10^7\text{ s}} = 4.162 \times 10^{-9}\text{ s}^{-1}\). Now, calculate the remaining activity \(A\) after \(12.0\text{ years}\): \(t = 12.0 \times 3.16 \times 10^7\text{ s} = 3.792 \times 10^8\text{ s}\). \(A = A_0 e^{-\lambda t} = (3.7 \times 10^5\text{ Bq}) \times e^{-(4.162 \times 10^{-9} \times 3.792 \times 10^8)} = 3.7 \times 10^5 \times e^{-1.578} = 7.64 \times 10^4\text{ Bq}\). Finally, use the formula \(A = \lambda N\) to find the remaining number of nuclei \(N\): \(N = \frac{A}{\lambda} = \frac{7.64 \times 10^4\text{ Bq}}{4.162 \times 10^{-9}\text{ s}^{-1}} = 1.836 \times 10^{13}\) nuclei.

1. [M1] Correctly calculates decay constant \(\lambda = 4.16 \times 10^{-9}\text{ s}^{-1}\) or equivalent in \(\text{year}^{-1}\). 2. [M1] Uses exponential decay formula with consistent time units. 3. [A1] Correctly calculates remaining activity \(7.6 \times 10^4\text{ Bq}\). 4. [M1] Uses \(A = \lambda N\) to find remaining number of nuclei. 5. [A1] Correctly calculates \(N = 1.8 \times 10^{13}\) (accept range \(1.75 \times 10^{13}\) to \(1.90 \times 10^{13}\)).

First, we derive the formula: 1) From the conservation of charge, the total current \(I\) from the source is equal to the sum of the currents through the individual parallel branches: \(I = I_1 + I_2\). 2) According to Ohm's law, the current through each resistor is given by \(I = V / R\). Since the resistors are connected in parallel, they both experience the same potential difference \(V\) across them. Thus, \(I_1 = V / R_1\) and \(I_2 = V / R_2\). 3) Substituting these into the current equation gives \(V / R_p = V / R_1 + V / R_2\). Dividing both sides by \(V\) yields: \(1/R_p = 1/R_1 + 1/R_2\). Second, we perform the calculation: 1) Find the equivalent parallel resistance \(R_p\): \(1/R_p = 1/6.0 + 1/12.0 = 3/12.0 = 1/4.0\ \Omega^{-1}\), so \(R_p = 4.0\ \Omega\). 2) Find the total resistance of the circuit, which includes the internal resistance \(r\): \(R_{total} = R_p + r = 4.0 + 1.5 = 5.5\ \Omega\). 3) Find the total current \(I\) using the e.m.f. \(E\): \(I = E / R_{total} = 12.0 / 5.5 = 2.18\text{ A}\).

[1] States conservation of charge in terms of current: \(I = I_1 + I_2\). [1] Expresses current using Ohm's law and concludes that potential difference across parallel branches is equal, leading to \(1/R_p = 1/R_1 + 1/R_2\). [1] Calculates equivalent parallel resistance \(R_p = 4.0\ \Omega\). [1] Adds internal resistance to obtain total resistance \(R_{total} = 5.5\ \Omega\). [1] Calculates total current to be \(2.18\text{ A}\) (accept \(2.2\text{ A}\)).

1) For vertical equilibrium, the vertical component of tension balances the weight: \(T \cos\theta = mg\), which rearranges to \(T = \frac{mg}{\cos\theta}\). 2) The horizontal component of tension provides the necessary centripetal force for circular motion: \(T \sin\theta = \frac{mv^2}{r}\). 3) Substituting the expression for \(T\) into the centripetal force equation: \(\left(\frac{mg}{\cos\theta}\right) \sin\theta = \frac{mv^2}{r}\). This simplifies to \(mg \tan\theta = \frac{mv^2}{r}\), and solving for speed gives \(v = \sqrt{rg\tan\theta}\). 4) Determine the radius \(r\) of the circular path using trigonometry: \(r = L \sin\theta = 0.85 \times \sin 35^\circ = 0.4875\text{ m}\). 5) Substitute the values to calculate \(v\): \(v = \sqrt{0.4875 \times 9.81 \times \tan 35^\circ} = \sqrt{0.4875 \times 9.81 \times 0.7002} = \sqrt{3.349} = 1.83\text{ m s}^{-1}\).

[1] Sets up vertical equilibrium equation \(T \cos\theta = mg\) and derives \(T = \frac{mg}{\cos\theta}\). [1] Identifies centripetal force equation \(T \sin\theta = \frac{mv^2}{r}\). [1] Correctly substitutes \(T\) into the centripetal equation to derive \(v = \sqrt{rg\tan\theta}\). [1] Calculates radius \(r = 0.488\text{ m}\) using \(r = L \sin\theta\). [1] Calculates final speed \(v = 1.83\text{ m s}^{-1}\) (accept \(1.8\text{ m s}^{-1}\)).

1) From Einstein's photoelectric equation, the maximum kinetic energy of the emitted photoelectrons is given by \(E_{k,\max} = hf - \Phi\). 2) The stopping potential \(V_s\) is the potential difference required to stop the fastest-moving photoelectrons, so the work done by the electric field on these electrons is equal to their maximum kinetic energy: \(E_{k,\max} = eV_s\). 3) Substituting this into the equation yields \(eV_s = hf - \Phi\), which divides by \(e\) to give \(V_s = \frac{hf}{e} - \frac{\Phi}{e}\). 4) Convert the wavelength of ultraviolet light to find the energy of the incident photons in eV: \(E = \frac{hc}{\lambda} = \frac{6.63 \times 10^{-34} \times 3.00 \times 10^8}{260 \times 10^{-9}} = 7.65 \times 10^{-19}\text{ J}\). In electronvolts: \(E = \frac{7.65 \times 10^{-19}}{1.60 \times 10^{-19}} = 4.78\text{ eV}\). 5) Calculate \(V_s\): \(V_s = 4.78\text{ eV} - 4.3\text{ eV} = 0.48\text{ V}\).

[1] Recalls Einstein's photoelectric equation \(E_{k,\max} = hf - \Phi\). [1] Relates maximum kinetic energy to stopping potential via \(E_{k,\max} = eV_s\) and completes the derivation. [1] Calculates the energy of the incident photon in J or eV: \(E = 7.65 \times 10^{-19}\text{ J}\) or \(4.78\text{ eV}\). [1] Converts work function to J (\(6.88 \times 10^{-19}\text{ J}\)) or keeps photon energy in eV for direct subtraction. [1] Calculates stopping potential \(V_s = 0.48\text{ V}\) (accept range \(0.47\text{ V}\) to \(0.50\text{ V}\)).

1) The magnetic force on the electron acts as the centripetal force: \(Bev = \frac{mv^2}{r}\). 2) Rearrange this expression to find the radius of the orbit: \(r = \frac{mv}{Be}\). 3) The time \(T\) for one full orbit is the circumference divided by the speed: \(T = \frac{2\pi r}{v}\). Substituting the expression for \(r\) gives \(T = \frac{2\pi}{v} \left(\frac{mv}{Be}\right) = \frac{2\pi m}{Be}\). 4) The frequency \(f\) of the motion is the reciprocal of the time period: \(f = \frac{1}{T} = \frac{Be}{2\pi m}\). 5) Substitute the known values to calculate \(f\): \(f = \frac{2.5 \times 10^{-3} \times 1.60 \times 10^{-19}}{2\pi \times 9.11 \times 10^{-31}} = 6.99 \times 10^7\text{ Hz}\).

[1] Equates magnetic force to centripetal force: \(Bev = \frac{mv^2}{r}\). [1] Eliminates \(v\) using \(v = \frac{2\pi r}{T}\) to show \(T = \frac{2\pi m}{Be}\). [1] Identifies relation between frequency and period: \(f = \frac{1}{T}\) or \(f = \frac{Be}{2\pi m}\). [1] Substitutes correct values for \(e = 1.60 \times 10^{-19}\text{ C}\) and \(m = 9.11 \times 10^{-31}\text{ kg}\). [1] Calculates frequency \(f = 7.0 \times 10^7\text{ Hz}\) (accept range \(6.9 \times 10^7\text{ Hz}\) to \(7.0 \times 10^7\text{ Hz}\)).

1) Work done (elastic strain energy) is \(E_{el} = \frac{1}{2} F \Delta x\). Since stress \(\sigma = \frac{F}{A}\) and strain \(\varepsilon = \frac{\Delta x}{L}\), we can write \(F = \sigma A\) and \(\Delta x = \varepsilon L\). 2) Substituting these gives \(E_{el} = \frac{1}{2} (\sigma A) (\varepsilon L) = \frac{1}{2} \sigma \varepsilon (A L)\). Since the volume of the wire is \(V = A L\), we get \(E_{el} = \frac{1}{2} \sigma \varepsilon V\). 3) Using the definition of Young Modulus \(E = \frac{\sigma}{\varepsilon}\), we substitute \(\sigma = E \varepsilon\) into the energy equation to obtain \(E_{el} = \frac{1}{2} E \varepsilon^2 V\). 4) First calculate the cross-sectional area \(A\) of the wire: \(A = \pi r^2 = \pi (0.40 \times 10^{-3})^2 = 5.03 \times 10^{-7}\text{ m}^2\). 5) Calculate the extension: \(\Delta x = \frac{F L}{A E} = \frac{150 \times 2.2}{5.03 \times 10^{-7} \times 2.0 \times 10^{11}} = 3.28 \times 10^{-3}\text{ m}\). The energy stored is \(E_{el} = \frac{1}{2} F \Delta x = 0.5 \times 150 \times 3.28 \times 10^{-3} = 0.246\text{ J}\) (or \(0.25\text{ J}\)).

[1] Connects force and extension to stress and strain: \(F = \sigma A\) and \(\Delta x = \varepsilon L\). [1] Expresses work done in terms of stress, strain, and volume: \(E_{el} = \frac{1}{2} \sigma \varepsilon V\). [1] Replaces stress using \(\sigma = E \varepsilon\) to complete the derivation of \(E_{el} = \frac{1}{2} E \varepsilon^2 V\). [1] Calculates cross-sectional area \(A = 5.03 \times 10^{-7}\text{ m}^2\) or volume \(V = 1.11 \times 10^{-6}\text{ m}^3\). [1] Obtains the final energy value \(0.25\text{ J}\) (accept range \(0.24\text{ J}\) to \(0.25\text{ J}\)).

1) The mean kinetic energy of a single molecule of mass \(m\) is \(E_k = \frac{1}{2} m v_{rms}^2\). 2) Equating the two expressions for kinetic energy: \(\frac{1}{2} m v_{rms}^2 = \frac{3}{2} k T\), which simplifies to \(v_{rms} = \sqrt{\frac{3 k T}{m}}\). 3) Since \(k = \frac{R}{N_A}\) and the mass of one molecule is \(m = \frac{M}{N_A}\), substituting these gives: \(v_{rms} = \sqrt{\frac{3 (R / N_A) T}{M / N_A}} = \sqrt{\frac{3 R T}{M}}\). 4) Convert the temperature to kelvin: \(T = 20 + 273.15 = 293.15\text{ K}\). 5) Calculate the root-mean-square speed: \(v_{rms} = \sqrt{\frac{3 \times 8.31 \times 293.15}{0.028}} = \sqrt{\frac{7308.2}{0.028}} = \sqrt{261007} = 511\text{ m s}^{-1}\).

[1] Equates kinetic energy of a molecule to thermal energy: \(\frac{1}{2} m v_{rms}^2 = \frac{3}{2} k T\). [1] Expresses Boltzmann constant \(k\) and molecular mass \(m\) in terms of molar quantities to derive the given formula. [1] Converts temperature to Kelvin: \(293\text{ K}\) or \(293.15\text{ K}\). [1] Substitutes molar mass and temperature correctly into the derived formula. [1] Calculates speed \(v_{rms} = 511\text{ m s}^{-1}\) (accept range \(508\text{ m s}^{-1}\) to \(512\text{ m s}^{-1}\)).

1) By definition, at \(t = t_{1/2}\), the number of remaining nuclei is half the initial value, so \(N = \frac{1}{2} N_0\). 2) Substituting into the decay equation gives \(\frac{1}{2} N_0 = N_0 e^{-\lambda t_{1/2}}\). Dividing by \(N_0\) gives \(0.5 = e^{-\lambda t_{1/2}}\). Taking the natural logarithm of both sides: \(\ln(0.5) = -\lambda t_{1/2}\), which simplifies to \(t_{1/2} = \frac{\ln 2}{\lambda}\). 3) Calculate the decay constant \(\lambda\) for Carbon-14: \(\lambda = \frac{\ln 2}{5730\text{ years}} = 1.21 \times 10^{-4}\text{ year}^{-1}\). 4) Since activity \(A\) is proportional to the number of nuclei \(N\), we can write the decay law as \(A = A_0 e^{-\lambda t}\). Substitute the activities: \(4.2 = 15.5 e^{-\lambda t}\), which gives \(\ln(4.2 / 15.5) = -\lambda t\). 5) Calculate the age: \(t = \frac{\ln(15.5 / 4.2)}{1.21 \times 10^{-4}\text{ year}^{-1}} = \frac{1.3056}{1.21 \times 10^{-4}} = 1.08 \times 10^4\text{ years}\) (or \(11000\text{ years}\)).

[1] States condition for half-life: \(N = \frac{1}{2} N_0\) at \(t = t_{1/2}\). [1] Employs natural logarithms correctly to show \(t_{1/2} = \frac{\ln 2}{\lambda}\). [1] Calculates decay constant \(\lambda = 1.21 \times 10^{-4}\text{ year}^{-1}\) (or in seconds: \(3.83 \times 10^{-12}\text{ s}^{-1}\)). [1] Uses \(A = A_0 e^{-\lambda t}\) with correct activities substituted. [1] Calculates age \(1.08 \times 10^4\text{ years}\) (accept range \(1.07 \times 10^4\text{ years}\) to \(1.10 \times 10^4\text{ years}\)).

1) According to Newton's second law, the net force on the object is \(F = ma\). Equating this to the restoring force: \(ma = -kx\). 2) Rearranging for acceleration gives \(a = -\left(\frac{k}{m}\right)x\). Comparing this with the defining equation for simple harmonic motion \(a = -\omega^2 x\) yields \(\omega^2 = \frac{k}{m}\), so \(\omega = \sqrt{\frac{k}{m}}\). 3) The maximum speed \(v_{\max}\) of an object in simple harmonic motion is given by \(v_{\max} = \omega A\), where \(A\) is the amplitude. 4) Calculate the angular frequency: \(\omega = \sqrt{\frac{25.0}{0.350}} = 8.452\text{ rad s}^{-1}\). 5) Substitute the amplitude \(A = 0.060\text{ m}\) to calculate maximum speed: \(v_{\max} = 8.452 \times 0.060 = 0.507\text{ m s}^{-1}\) (or \(0.51\text{ m s}^{-1}\)).

[1] Equates restoring force to mass times acceleration: \(ma = -kx\). [1] Equates coefficient of \(-x\) to \(\omega^2\) to derive \(\omega = \sqrt{\frac{k}{m}}\). [1] States the expression for maximum speed: \(v_{\max} = \omega A\). [1] Calculates angular frequency \(\omega = 8.45\text{ rad s}^{-1}\). [1] Calculates maximum speed \(0.51\text{ m s}^{-1}\) (accept range \(0.50\text{ m s}^{-1}\) to \(0.51\text{ m s}^{-1}\)).

1. Set up the conservation of energy equation: \(E_{\text{initial}} = E_{\text{final}}\).

2. Express total energy: \(\frac{1}{2} m v_i^2 + mgh = \frac{1}{2} m v_f^2\).

3. Cancel the mass \(m\) from each term: \(\frac{1}{2} v_i^2 + gh = \frac{1}{2} v_f^2\).

4. Rearrange for \(v_f\): \(v_f = \sqrt{v_i^2 + 2gh}\).

5. Substitute the given values: \(v_f = \sqrt{18.0^2 + 2 \times 9.81 \times 25.0}\).

6. Calculate the result: \(v_f = \sqrt{324 + 490.5} = \sqrt{814.5} = 28.54\text{ m s}^{-1}\).

7. Standard rounding to 3 significant figures gives \(28.5\text{ m s}^{-1}\).

- State energy conservation equation or \(\frac{1}{2} m v_i^2 + mgh = \frac{1}{2} m v_f^2\) (1 mark)
- Show that mass cancels out to give \(v_f = \sqrt{v_i^2 + 2gh}\) (1 mark)
- Correct substitution of values into equation (1 mark)
- Correct calculation of \(v_f^2 = 814.5\text{ m}^2\text{ s}^{-2}\) (1 mark)
- Correct final speed of \(28.5\text{ m s}^{-1}\) (or \(28.6\text{ m s}^{-1}\)) with unit (1 mark)

1. Recall the formula for Young modulus: \(E = \frac{\text{Stress}}{\text{Strain}} = \frac{F L}{A \Delta L}\).

2. Substitute the values: \(E = \frac{75.0 \times 2.40}{4.50 \times 10^{-7} \times 1.60 \times 10^{-3}} = \frac{180.0}{7.20 \times 10^{-10}} = 2.50 \times 10^{11}\text{ Pa}\).

3. Recall the formula for elastic strain energy: \(E_{\text{elastic}} = \frac{1}{2} F \Delta L\).

4. Substitute the values: \(E_{\text{elastic}} = 0.5 \times 75.0 \times 1.60 \times 10^{-3} = 0.060\text{ J}\) (or \(6.00 \times 10^{-2}\text{ J}\)).

- State \(E = \frac{FL}{A \Delta L}\) or equivalent stress and strain equations (1 mark)
- Substitute values and calculate Young modulus \(E = 2.50 \times 10^{11}\) (1 mark)
- Correct unit for Young modulus (\(\text{Pa}\) or \(\text{N m}^{-2}\)) (1 mark)
- State or use \(E_{\text{elastic}} = \frac{1}{2} F \Delta L\) (1 mark)
- Calculate elastic strain energy as \(0.060\text{ J}\) (or \(6.00 \times 10^{-2}\text{ J}\)) (1 mark)

1. The total resistance of the series circuit is \(R_{\text{total}} = R + r\).

2. The current \(I\) flowing through the circuit is \(I = \frac{\mathcal{E}}{R+r}\).

3. The power \(P\) dissipated in the variable resistor \(R\) is given by \(P = I^2 R\).

4. Substitute \(I\) into the power formula: \(P = \left(\frac{\mathcal{E}}{R+r}\right)^2 R = \frac{\mathcal{E}^2 R}{(R+r)^2}\).

5. For maximum power transfer, we are given \(R = r = 1.50\ \Omega\).

6. Substitute the values: \(P_{\text{max}} = \frac{6.00^2 \times 1.50}{(1.50 + 1.50)^2} = \frac{36.0 \times 1.50}{3.00^2} = \frac{54.0}{9.00} = 6.00\text{ W}\).

- Write current equation \(I = \frac{\mathcal{E}}{R+r}\) (1 mark)
- Substitute \(I\) into \(P = I^2 R\) to derive \(P = \frac{\mathcal{E}^2 R}{(R+r)^2}\) (1 mark)
- State that for maximum power, \(R = r = 1.50\ \Omega\) (1 mark)
- Substitute values correctly into the power equation (1 mark)
- Obtain final answer \(6.00\text{ W}\) (or \(6\text{ W}\)) (1 mark)

1. Energy of the incident photon: \(E = \frac{hc}{\lambda} = \frac{6.63 \times 10^{-34} \times 3.00 \times 10^8}{240 \times 10^{-9}} = 8.2875 \times 10^{-19}\text{ J}\).

2. Convert maximum kinetic energy to joules: \(E_{k,\text{max}} = 1.85 \times 1.60 \times 10^{-19}\text{ J} = 2.96 \times 10^{-19}\text{ J}\).

3. Apply Einstein's photoelectric equation: \(E = \phi + E_{k,\text{max}} \implies \phi = E - E_{k,\text{max}}\).

4. Calculate the work function: \(\phi = 8.2875 \times 10^{-19} - 2.96 \times 10^{-19} = 5.3275 \times 10^{-19}\text{ J}\), which rounds to \(5.33 \times 10^{-19}\text{ J}\).

5. Calculate threshold frequency using \(\phi = h f_0\): \(f_0 = \frac{\phi}{h} = \frac{5.3275 \times 10^{-19}}{6.63 \times 10^{-34}} = 8.04 \times 10^{14}\text{ Hz}\).

- Calculate photon energy: \(E = 8.29 \times 10^{-19}\text{ J}\) (1 mark)
- Convert max kinetic energy to Joules: \(E_{k,\text{max}} = 2.96 \times 10^{-19}\text{ J}\) (1 mark)
- Use photoelectric equation to find work function \(\phi = 5.33 \times 10^{-19}\text{ J}\) (1 mark)
- Use \(\phi = h f_0\) to express threshold frequency (1 mark)
- Calculate threshold frequency \(f_0 = 8.04 \times 10^{14}\text{ Hz}\) (accept range \(8.0 \times 10^{14}\) to \(8.1 \times 10^{14}\)) (1 mark)

1. Analyze the forces acting on the car of mass \(m\): the normal reaction force \(N\) and the weight \(mg\).

2. Resolve the normal reaction force vertically: \(N \cos\theta = mg\).

3. Resolve the normal reaction force horizontally, where the horizontal component provides the centripetal force: \(N \sin\theta = \frac{m v^2}{r}\).

4. Divide the horizontal equation by the vertical equation: \(\frac{N \sin\theta}{N \cos\theta} = \frac{m v^2 / r}{mg} \implies \tan\theta = \frac{v^2}{rg}\).

5. Rearrange for \(v\): \(v = \sqrt{rg\tan\theta}\).

6. Substitute the given values: \(v = \sqrt{120 \times 9.81 \times \tan(22.0^\circ)}\).

7. Calculate \(v\): \(v = \sqrt{1177.2 \times 0.4040} = \sqrt{475.6} = 21.8\text{ m s}^{-1}\).

- State vertical equilibrium equation: \(N \cos\theta = mg\) (1 mark)
- State horizontal centripetal force equation: \(N \sin\theta = \frac{m v^2}{r}\) (1 mark)
- Combine equations to derive \(\tan\theta = \frac{v^2}{rg}\) and hence \(v = \sqrt{rg\tan\theta}\) (1 mark)
- Substitute given values correctly into the formula (1 mark)
- Calculate final speed \(21.8\text{ m s}^{-1}\) (or \(22\text{ m s}^{-1}\)) with correct unit (1 mark)

1. Equate electrical work done to kinetic energy gained: \(e V = \frac{1}{2} m v^2\).

2. Rearrange for speed: \(v = \sqrt{\frac{2 e V}{m}}\).

3. Substitute the given values: \(v = \sqrt{\frac{2 \times 1.60 \times 10^{-19} \times 1200}{9.11 \times 10^{-31}}} = \sqrt{4.215 \times 10^{14}} = 2.053 \times 10^7\text{ m s}^{-1}\).

4. Equate magnetic force to centripetal force: \(B e v = \frac{m v^2}{r}\).

5. Rearrange for radius: \(r = \frac{m v}{B e}\).

6. Substitute the values: \(r = \frac{9.11 \times 10^{-31} \times 2.053 \times 10^7}{2.50 \times 10^{-3} \times 1.60 \times 10^{-19}} = \frac{1.870 \times 10^{-23}}{4.00 \times 10^{-22}} = 0.04675\text{ m}\) (or \(4.68\text{ cm}\)).

- Recall and use \(eV = \frac{1}{2}mv^2\) (1 mark)
- Obtain entry speed \(v = 2.05 \times 10^7\text{ m s}^{-1}\) (1 mark)
- Equate magnetic force and centripetal force: \(Bev = \frac{mv^2}{r}\) (1 mark)
- Rearrange to get \(r = \frac{mv}{Be}\) (1 mark)
- Calculate correct radius \(0.0467\text{ m}\) (or \(4.67\text{ cm}\) to \(4.68\text{ cm}\)) (1 mark)

1. Calculate the energy needed to heat the ice from \(-10.0^\circ\text{C}\) to \(0.0^\circ\text{C}\):
\(Q_1 = m c \Delta T = 0.350 \times 2100 \times 10.0 = 7350\text{ J}\).

2. Calculate the energy needed to melt the ice at \(0.0^\circ\text{C}\):
\(Q_2 = m L = 0.350 \times 3.34 \times 10^5 = 116,900\text{ J}\).

3. Calculate the total energy required:
\(Q_{\text{total}} = Q_1 + Q_2 = 7350 + 116,900 = 124,250\text{ J}\).

4. Calculate the time taken using power: \(P = \frac{Q_{\text{total}}}{t} \implies t = \frac{Q_{\text{total}}}{P}\).

5. Substitute the values: \(t = \frac{124,250}{85.0} = 1461.76\text{ s}\), which is \(1460\text{ s}\) to 3 significant figures.

- Calculate thermal energy to heat ice: \(Q_1 = 7350\text{ J}\) (1 mark)
- Calculate latent heat to melt ice: \(Q_2 = 116,900\text{ J}\) (1 mark)
- Sum the energies to find total energy \(Q_{\text{total}} = 124,250\text{ J}\) (1 mark)
- Recall and use \(t = \frac{Q}{P}\) (1 mark)
- Calculate correct final time: \(1460\text{ s}\) (or \(24.3 - 24.4\text{ minutes}\)) (1 mark)

1. Calculate the angular frequency \(\omega\):
\(\omega = \frac{2\pi}{T} = \frac{2\pi}{1.20} = 5.236\text{ rad s}^{-1}\).

2. Calculate the maximum acceleration \(a_{\text{max}}\) using \(a_{\text{max}} = \omega^2 A\):
\(a_{\text{max}} = (5.236)^2 \times 0.0600 = 27.42 \times 0.0600 = 1.65\text{ m s}^{-2}\).

3. Calculate the maximum velocity \(v_{\text{max}}\) using \(v_{\text{max}} = \omega A\):
\(v_{\text{max}} = 5.236 \times 0.0600 = 0.3142\text{ m s}^{-1}\).

4. Calculate the total mechanical energy \(E\) (which is equal to maximum kinetic energy):
\(E = \frac{1}{2} m v_{\text{max}}^2 = 0.5 \times 0.250 \times (0.3142)^2 = 0.0123\text{ J}\) (or \(1.23 \times 10^{-2}\text{ J}\)).

- Calculate angular frequency \(\omega = 5.24\text{ rad s}^{-1}\) (1 mark)
- Calculate maximum acceleration \(a_{\text{max}} = 1.65\text{ m s}^{-2}\) (1 mark)
- Recall total mechanical energy formula \(E = \frac{1}{2} m \omega^2 A^2\) or \(E = \frac{1}{2} m v_{\text{max}}^2\) (1 mark)
- Calculate maximum velocity \(v_{\text{max}} = 0.314\text{ m s}^{-1}\) (or substitute directly) (1 mark)
- Calculate total mechanical energy \(E = 0.0123\text{ J}\) (accept range \(0.012\text{ J}\) to \(0.013\text{ J}\)) (1 mark)

1. Identify the forces acting on the block parallel to the ramp: the component of weight pulling down the slope \( F_g = mg\sin\theta \) and the frictional force opposing motion \( F_f = \mu R = \mu mg\cos\theta \).
2. The total decelerating force is \( F = mg\sin\theta + \mu mg\cos\theta = mg(\sin\theta + \mu\cos\theta) \).
3. Using the work-energy theorem, the initial kinetic energy is fully converted into gravitational potential energy and work done against friction:
\( \frac{1}{2}mv^2 = Fd = mgd(\sin\theta + \mu\cos\theta) \).
4. Cancelling mass \( m \) and multiplying by 2 yields:
\( v^2 = 2gd(\sin\theta + \mu\cos\theta) \implies v = \sqrt{2gd(\sin\theta + \mu\cos\theta)} \).
5. Substituting the values:
\( v = \sqrt{2 \times 9.81 \times 3.2 \times (\sin 30^\circ + 0.15\cos 30^\circ)} \)
\( v = \sqrt{62.784 \times (0.5 + 0.130)} = \sqrt{39.55} \approx 6.3\text{ m s}^{-1} \).

MP1: State the frictional force equation \( F_f = \mu mg\cos\theta \) and gravity component \( mg\sin\theta \). (1)
MP2: Set up work-energy balance \( \frac{1}{2}mv^2 = mgd\sin\theta + \mu mgd\cos\theta \) or use Newton\'s second law for deceleration. (1)
MP3: Correctly show algebraic steps to arrive at the derived equation for \( v \). (1)
MP4: Substitute correct numerical values into the derived equation. (1)
MP5: Final answer of \( 6.3\text{ m s}^{-1} \) (accept \( 6.28\text{ m s}^{-1} \) to \( 6.3\text{ m s}^{-1} \)) with correct unit. (1)

1. Work done in stretching a wire is \( W = \frac{1}{2} F \Delta x \) since the force is proportional to extension.
2. Energy stored per unit volume is \( \frac{W}{V} = \frac{\frac{1}{2} F \Delta x}{A L} = \frac{1}{2} \left(\frac{F}{A}\right) \left(\frac{\Delta x}{L}\right) \).
3. Since \( \text{stress} = \frac{F}{A} \) and \( \text{strain} = \frac{\Delta x}{L} \), this simplifies to \( \frac{1}{2} \times \text{stress} \times \text{strain} \).
4. Find stress: \( \text{stress} = \frac{180}{1.5 \times 10^{-6}} = 1.2 \times 10^8\text{ Pa} \).
5. Find strain: \( \text{strain} = \frac{\text{stress}}{E} = \frac{1.2 \times 10^8}{2.0 \times 10^{11}} = 6.0 \times 10^{-4} \).
6. Total energy stored \( W = \text{energy density} \times V = \left(\frac{1}{2} \times 1.2 \times 10^8 \times 6.0 \times 10^{-4}\right) \times (1.5 \times 10^{-6} \times 2.0) = 3.6 \times 10^4 \times 3.0 \times 10^{-6} = 0.108\text{ J} \approx 0.11\text{ J} \).

MP1: State work done formula \( W = \frac{1}{2}F\Delta x \). (1)
MP2: Show division of work done by volume \( AL \) to derive the energy density formula. (1)
MP3: Calculate the stress or the extension \( \Delta x = 1.2 \times 10^{-3}\text{ m} \). (1)
MP4: Substitute values to find the energy density or total energy equation. (1)
MP5: Correct final value of \( 0.11\text{ J} \) (accept \( 0.108\text{ J} \)) with unit. (1)

1. The grating equation is \( d\sin\theta = n\lambda \). For the second-order maximum, \( n = 2 \), so \( d\sin\theta = 2\lambda \).
2. The line spacing \( d \) is related to the number of lines per metre \( N_{\text{metre}} \) by \( d = \frac{1}{N_{\text{metre}}} \).
3. Since \( N \) is the number of lines per millimetre, \( N_{\text{metre}} = N \times 10^3 \). Hence, \( d = \frac{1}{N \times 10^3} \).
4. Substitute this into the grating equation: \( \frac{\sin\theta}{N \times 10^3} = 2\lambda \implies N = \frac{\sin\theta}{2\lambda \times 10^3} \).
5. Calculate \( N \):
\( N = \frac{\sin 34.5^\circ}{2 \times 589 \times 10^{-9} \times 10^3} = \frac{0.5664}{1.178 \times 10^{-3}} \approx 480.81 \approx 480\text{ lines mm}^{-1} \).

MP1: Use \( d\sin\theta = n\lambda \) with \( n = 2 \). (1)
MP2: State relationship between line spacing \( d \) and lines per mm: \( d = \frac{1}{10^3 N} \). (1)
MP3: Correctly show algebraic steps to derive the given expression for \( N \). (1)
MP4: Substitute wavelength \( 589 \times 10^{-9}\text{ m} \) and angle \( 34.5^\circ \) correctly. (1)
MP5: Correct calculation of \( 480\text{ lines mm}^{-1} \) (accept \( 481\text{ lines mm}^{-1} \)). (1)

1. The total resistance of the series circuit is \( R_{\text{total}} = R + R_T \).
2. The current in the circuit is \( I = \frac{E}{R_{\text{total}}} = \frac{E}{R + R_T} \).
3. The output voltage is measured across the thermistor, so \( V_{\text{out}} = I R_T = \frac{E R_T}{R + R_T} \).
4. Rearrange the formula to solve for \( R_T \):
\( V_{\text{out}}(R + R_T) = E R_T \implies V_{\text{out}} R = R_T (E - V_{\text{out}}) \implies R_T = \frac{V_{\text{out}} R}{E - V_{\text{out}}} \).
5. Substitute values:
\( R_T = \frac{3.6 \times 1200}{9.0 - 3.6} = \frac{4320}{5.4} = 800\text{ }\Omega \).

MP1: Express current in terms of e.m.f. and total resistance: \( I = \frac{E}{R + R_T} \). (1)
MP2: Show derivation of \( V_{\text{out}} = I R_T = \frac{E R_T}{R + R_T} \). (1)
MP3: Rearrange equation to make \( R_T \) the subject. (1)
MP4: Substitute correct numerical values (using \( R = 1200\text{ }\Omega \)). (1)
MP5: Correct final calculation of \( 800\text{ }\Omega \) (or \( 0.80\text{ k}\Omega \)). (1)

1. Resolve the forces acting on the mass: vertically, tension \( T_s \) balances weight: \( T_s\cos\theta = mg \).
2. Horizontally, the component of tension provides the centripetal force: \( T_s\sin\theta = m\omega^2 r \).
3. Divide the horizontal equation by the vertical equation: \( \tan\theta = \frac{\omega^2 r}{g} \).
4. Since \( r = L\sin\theta \), substitute this into the equation: \( \frac{\sin\theta}{\cos\theta} = \frac{\omega^2 L\sin\theta}{g} \implies \omega^2 = \frac{g}{L\cos\theta} \).
5. Use the relationship \( T = \frac{2\pi}{\omega} \) to get: \( T = 2\pi\sqrt{\frac{L\cos\theta}{g}} \).
6. Substitute the given values:
\( T = 2\pi\sqrt{\frac{0.85\cos 25^\circ}{9.81}} = 2\pi\sqrt{\frac{0.7704}{9.81}} = 2\pi\sqrt{0.0785} \approx 1.76\text{ s} \approx 1.8\text{ s} \).

MP1: Resolve tension force vertically \( T_s\cos\theta = mg \) and horizontally \( T_s\sin\theta = m\omega^2 r \). (1)
MP2: Use the geometry relationship \( r = L\sin\theta \) to eliminate \( r \). (1)
MP3: Substitute \( \omega = \frac{2\pi}{T} \) and complete the algebraic derivation of the given period formula. (1)
MP4: Substitute values into the derived formula correctly. (1)
MP5: Correct calculation of \( 1.8\text{ s} \) (accept \( 1.76\text{ s} \)). (1)

1. The magnetic force provides the centripetal force: \( Bqv = \frac{mv^2}{r} \).
2. Rearranging this gives the radius of the orbit: \( r = \frac{mv}{Bq} \).
3. The time taken for one orbit is \( t = \frac{2\pi r}{v} \).
4. Substituting \( r \) into the period expression gives \( t = \frac{2\pi \left(\frac{mv}{Bq}\right)}{v} = \frac{2\pi m}{Bq} \), showing that \( t \) is independent of velocity \( v \).
5. To find \( B \), rearrange \( Bqv = \frac{mv^2}{r} \) to get \( B = \frac{mv}{qr} \).
6. Substitute the values:
\( B = \frac{1.67 \times 10^{-27} \times 4.2 \times 10^6}{1.60 \times 10^{-19} \times 0.15} = \frac{7.014 \times 10^{-21}}{2.40 \times 10^{-20}} \approx 0.292\text{ T} \approx 0.29\text{ T} \).

MP1: State the force balance equation: \( Bqv = \frac{mv^2}{r} \). (1)
MP2: Express orbital time period as \( t = \frac{2\pi r}{v} \). (1)
MP3: Substitute \( r \) into \( t \) to derive the formula \( t = \frac{2\pi m}{Bq} \). (1)
MP4: Rearrange the force equation to solve for \( B \) and substitute known values correctly. (1)
MP5: Correct calculation of \( 0.29\text{ T} \) (accept \( 0.292\text{ T} \)) with appropriate unit. (1)

1. Total energy needed consists of two stages: raising the temperature of ice from \( -10^\circ\text{C} \) to \( 0^\circ\text{C} \) (\( Q_1 = mc_{\text{ice}}\Delta T \)), and melting the ice at \( 0^\circ\text{C} \) (\( Q_2 = mL_f \)).
2. Summing these: \( Q = Q_1 + Q_2 = mc_{\text{ice}}\Delta T + mL_f = m(c_{\text{ice}}\Delta T + L_f) \).
3. Substitute the values into this formula:
\( Q = 0.45 \times (2100 \times 10 + 3.3 \times 10^5) = 0.45 \times (21000 + 330000) = 0.45 \times 351000 = 1.58 \times 10^5\text{ J} \).
4. The power of the heater is given by \( P = \frac{Q}{t} \).
5. Calculate \( P \):
\( P = \frac{1.58 \times 10^5}{350} \approx 451\text{ W} \approx 450\text{ W} \).

MP1: Identify the two phases of energy absorption (temperature increase and phase change). (1)
MP2: Algebraically combine expressions to derive \( Q = m(c_{\text{ice}}\Delta T + L_f) \). (1)
MP3: Correctly calculate the total energy absorbed as \( 1.58 \times 10^5\text{ J} \). (1)
MP4: State and use the power-energy formula \( P = \frac{Q}{t} \). (1)
MP5: Correct calculation of power as \( 450\text{ W} \) (accept \( 451\text{ W} \)) with unit. (1)

1. The radioactive decay law is \( N = N_0 e^{-\lambda t} \).
2. By definition, at \( t = t_{1/2} \), the number of remaining nuclei is \( N = \frac{N_0}{2} \).
3. Substitute this in: \( \frac{1}{2} = e^{-\lambda t_{1/2}} \implies 2 = e^{\lambda t_{1/2}} \). Taking natural logs: \( \ln 2 = \lambda t_{1/2} \implies \lambda = \frac{\ln 2}{t_{1/2}} \).
4. Convert half-life to seconds: \( t_{1/2} = 8.0 \times 24 \times 3600 = 691200\text{ s} \).
5. Calculate \( \lambda \):
\( \lambda = \frac{\ln 2}{691200} \approx 1.00 \times 10^{-6}\text{ s}^{-1} \).
6. Fraction remaining after 30 days is given by \( \frac{A}{A_0} = e^{-\lambda t} \) or \( (0.5)^{t/t_{1/2}} \):
\( \text{Fraction} = (0.5)^{\frac{30}{8}} = (0.5)^{3.75} \approx 0.0743 \approx 0.074 \) (or \( 7.4\% \)).

MP1: Use \( N = N_0 e^{-\lambda t} \) and definition of half-life to set up \( 0.5 = e^{-\lambda t_{1/2}} \). (1)
MP2: Correctly show algebraic derivation to arrive at \( \lambda = \frac{\ln 2}{t_{1/2}} \). (1)
MP3: Convert half-life into seconds correctly (\( 6.91 \times 10^5\text{ s} \)). (1)
MP4: Calculate \( \lambda \) correctly as \( 1.0 \times 10^{-6}\text{ s}^{-1} \). (1)
MP5: Calculate fraction remaining as \( 0.074 \) (or \( 7.4\% \)). (1)

1. Express current \(I\) in the circuit:
\(I = \frac{E}{R+r}\)

2. Derive the power \(P\) dissipated in the variable resistor:
\(P = I^2 R = \left(\frac{E}{R+r}\right)^2 R = \frac{E^2 R}{(R+r)^2}\)

3. Use the maximum power theorem (or differentiate and equate to zero) which states that maximum power is transferred when the load resistance equals the internal resistance, \(R = r\). Substitute this back to find maximum power \(P_{\text{max}}\):
\(P_{\text{max}} = \frac{E^2 r}{(r+r)^2} = \frac{E^2 r}{4r^2} = \frac{E^2}{4r}\)

4. Substitute the given values \(P_{\text{max}} = 4.5\text{ W}\) and \(E = 6.0\text{ V}\) to find \(r\):
\(4.5 = \frac{6.0^2}{4r} = \frac{36}{4r} = \frac{9}{r}\)
\(r = \frac{9}{4.5} = 2.0\ \Omega\)

5. Assumption: The electromotive force \(E\) and internal resistance \(r\) of the cell remain constant as the current varies.

MP1: State expression for current: \(I = \frac{E}{R+r}\) [1]
MP2: Substitute current into \(P = I^2 R\) to derive the expression \(P = \frac{E^2 R}{(R+r)^2}\) [1]
MP3: State or derive that maximum power occurs when \(R = r\), yielding \(P_{\text{max}} = \frac{E^2}{4r}\) [1]
MP4: Correct calculation to find \(r = 2.0\ \Omega\) (allow ECF from incorrect power derivation) [1]
MP5: State that the internal resistance or emf of the cell is assumed to remain constant [1]

1. Identify and resolve the forces acting on the car. Let \(N\) be the normal contact force and \(mg\) be the weight of the car.
- Vertically, with no vertical acceleration: \(N \cos\theta = mg\)
- Horizontally, the component of the normal force provides the centripetal force: \(N \sin\theta = \frac{mv^2}{r}\)

2. Divide the horizontal equation by the vertical equation to eliminate \(N\) and \(m\):
\(\frac{N \sin\theta}{N \cos\theta} = \frac{mv^2 / r}{mg} \implies \tan\theta = \frac{v^2}{rg}\)

3. Rearrange the formula to solve for the target speed \(v\):
\(v = \sqrt{rg \tan\theta}\)

4. Substitute the values \(r = 120\text{ m}\), \(g = 9.81\text{ m s}^{-2}\), and \(\theta = 18^\circ\):
\(v = \sqrt{120 \times 9.81 \times \tan(18^\circ)} = \sqrt{1177.2 \times 0.3249} = \sqrt{382.5} \approx 19.6\text{ m s}^{-1}\)

MP1: Resolve vertical forces correctly to find \(N \cos\theta = mg\) [1]
MP2: Relate horizontal centripetal force correctly to find \(N \sin\theta = \frac{mv^2}{r}\) [1]
MP3: Combine equations to derive \(v = \sqrt{rg \tan\theta}\) [1]
MP4: Substitute numerical values correctly into the derived formula [1]
MP5: Calculate the final speed to be \(19.6\text{ m s}^{-1}\) (Accept \(20\text{ m s}^{-1}\) for 2 sig figs, or \(19.5\text{ m s}^{-1}\) if \(g = 9.8\text{ m s}^{-2}\) is used) [1]

1. Write the radioactive decay law:
\(N = N_0 e^{-\lambda t}\)

2. By definition of half-life, when \(t = T_{1/2}\), \(N = \frac{N_0}{2}\). Substitute these into the equation:
\(\frac{1}{2} = e^{-\lambda T_{1/2}}\)
Take the natural logarithm of both sides:
\(\ln(0.5) = -\lambda T_{1/2} \implies -\ln(2) = -\lambda T_{1/2} \implies \lambda = \frac{\ln 2}{T_{1/2}}\)

3. Calculate the decay constant \(\lambda\):
\(\lambda = \frac{\ln 2}{8.0\text{ days}} \approx 0.08664\text{ day}^{-1}\)

4. Use the activity decay law \(A = A_0 e^{-\lambda t}\) where \(t = 20\text{ days}\) and \(A_0 = 3.6 \times 10^7\text{ Bq}\):
\(A = (3.6 \times 10^7) \times e^{-0.08664 \times 20} = (3.6 \times 10^7) \times e^{-1.733}\)
\(A = (3.6 \times 10^7) \times 0.1768 \approx 6.36 \times 10^6\text{ Bq}\)

Alternatively, using the half-life ratio method:
\(A = A_0 \times (0.5)^{t/T_{1/2}} = 3.6 \times 10^7 \times (0.5)^{20/8} = 3.6 \times 10^7 \times (0.5)^{2.5} \approx 6.36 \times 10^6\text{ Bq}\)

MP1: Substitute \(N = N_0 / 2\) and \(t = T_{1/2}\) into the exponential decay equation [1]
MP2: Complete algebraic steps to show clearly that \(\lambda = \frac{\ln 2}{T_{1/2}}\) [1]
MP3: Calculate \(\lambda = 0.0866\text{ day}^{-1}\) (or \(1.00 \times 10^{-6}\text{ s}^{-1}\)) OR deduce that the time period corresponds to \(2.5\) half-lives [1]
MP4: Substitute values into the decay equation correctly [1]
MP5: Obtain the final activity of \(6.4 \times 10^6\text{ Bq}\) (Accept answers in range \(6.3 \times 10^6\text{ Bq}\) to \(6.4 \times 10^6\text{ Bq}\) with correct units) [1]

1. For a wire obeying Hooke's law, the work done (which is the elastic strain energy stored) is represented by the area under the force-extension graph:
\(E_{\text{el}} = \frac{1}{2} F \Delta x\)

2. From the definition of Young modulus \(E = \frac{\text{Stress}}{\text{Strain}} = \frac{F / A}{\Delta x / L}\):
\(F = \frac{E A \Delta x}{L}\)

3. Substitute the expression for \(F\) into the energy equation:
\(E_{\text{el}} = \frac{1}{2} \left(\frac{E A \Delta x}{L}\right) \Delta x = \frac{E A (\Delta x)^2}{2L}\)

4. Calculate the cross-sectional area \(A\) of the wire using its diameter \(d = 0.80\text{ mm}\):
\(A = \pi r^2 = \pi (0.40 \times 10^{-3}\text{ m})^2 = 5.03 \times 10^{-7}\text{ m}^2\)

5. Calculate the extension \(\Delta x\) of the wire:
\(\Delta x = \frac{F L}{E A} = \frac{150 \times 2.5}{2.0 \times 10^{11} \times 5.03 \times 10^{-7}} = 3.73 \times 10^{-3}\text{ m}\)

6. Substitute to find \(E_{\text{el}}\):
\(E_{\text{el}} = \frac{1}{2} F \Delta x = 0.5 \times 150 \times 3.73 \times 10^{-3} = 0.28\text{ J}\)

MP1: State the equation for work done as \(E_{\text{el}} = \frac{1}{2} F \Delta x\) [1]
MP2: State \(E = \frac{FL}{A\Delta x}\) and substitute to show derivation of \(E_{\text{el}} = \frac{E A (\Delta x)^2}{2L}\) [1]
MP3: Calculate cross-sectional area \(A = 5.0 \times 10^{-7}\text{ m}^2\) [1]
MP4: Calculate extension \(\Delta x = 3.7 \times 10^{-3}\text{ m}\) [1]
MP5: Calculate elastic strain energy as \(0.28\text{ J}\) (Accept answers from \(0.27\text{ J}\) to \(0.29\text{ J}\)) [1]

1. A maximum is formed when waves from adjacent slits meet in phase and interfere constructively. This occurs when the path difference between adjacent slits is an integer number of wavelengths, \(n\lambda\).

2. From the geometry of the grating, the path difference between light diffracted at an angle \(\theta\) from adjacent slits spaced \(d\) apart is given by \(d \sin\theta\). Therefore, for constructive interference:
\(d \sin\theta = n\lambda\)

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

4. Find the maximum possible order \(n\) by setting the maximum diffraction angle \(\theta = 90^\circ\), so \(\sin\theta \le 1\):
\(n \le \frac{d}{\lambda} = \frac{2.0 \times 10^{-6}\text{ m}}{632.8 \times 10^{-9}\text{ m}} = 3.16\)
Since \(n\) must be an integer, the maximum observable order is \(n = 3\).

5. The total number of maxima is the central maximum (order 0) plus the orders 1, 2, 3 on each side:
\text{Total} = 2n + 1 = 2(3) + 1 = 7

MP1: State that constructive interference occurs when the path difference is an integer number of wavelengths, \(n\lambda\) [1]
MP2: Use geometric reasoning to show that the path difference between adjacent slits is \(d \sin\theta\), leading to \(d \sin\theta = n\lambda\) [1]
MP3: Calculate the slit spacing \(d = 2.0 \times 10^{-6}\text{ m}\) [1]
MP4: Set \(\sin\theta \le 1\) and calculate the maximum integer order \(n = 3\) [1]
MP5: Calculate the total number of maxima to be \(7\) (must be an integer and account for both sides and the central maximum) [1]

Copper is a ductile material with a metallic structure consisting of a regular lattice of positive ions surrounded by a sea of delocalised electrons. This structure contains dislocations (faults in the lattice). When stress is applied, these dislocations can move easily, allowing layers of atoms to slide over one another (plastic deformation). This leads to necking and permanent deformation before fracture. In contrast, glass is a brittle material with an amorphous covalent structure containing strong, highly directional bonds and no mobile dislocations. Under stress, it undergoes only elastic deformation. High stress concentrations build up at microscopic surface cracks, causing them to propagate rapidly through the material, leading to sudden, catastrophic fracture without plastic deformation.

1. Identifies copper as ductile and glass as brittle. (1) 2. Describes copper's metallic structure (lattice of ions in sea of electrons) or glass's amorphous/covalent structure. (1) 3. Explains that copper contains dislocations which can move under stress, leading to slip planes. (1) 4. Explains that this dislocation movement allows copper to deform plastically (permanent deformation) before breaking. (1) 5. Explains that glass lacks mobile dislocations due to strong covalent bonds. (1) 6. Explains that stress concentrations at surface defects in glass lead to sudden crack propagation and brittle failure. (1)

Light passing through each slit diffracts, spreading out and overlapping on the screen. The two slits act as coherent sources, emitting waves with a constant phase relationship and the same frequency. Where the two waves overlap, superposition occurs. At positions where the path difference is an integer number of wavelengths, \(n\lambda\), the waves arrive in phase, resulting in constructive interference and a bright fringe. At positions where the path difference is an odd number of half-wavelengths, \((n+0.5)\lambda\), the waves arrive in antiphase, resulting in destructive interference and a dark fringe. Newton's corpuscular theory suggests light consists of streams of particles, which would travel in straight lines and form only two bright bands directly behind the slits. The continuous alternating fringe pattern can only be explained by wave properties.

1. Stating that light diffracts as it passes through the slits and the waves overlap. (1) 2. Explaining that the slits act as coherent sources (same frequency and constant phase relationship). (1) 3. Explaining that bright fringes are caused by constructive interference where waves arrive in phase / path difference is \(n\lambda\). (1) 4. Explaining that dark fringes are caused by destructive interference where waves arrive in antiphase / path difference is \((n+0.5)\lambda\). (1) 5. Explaining that Newton's corpuscular theory predicts only two bright bands behind the slits. (1) 6. Concluding that the multi-fringe interference pattern is evidence for the wave nature of light. (1)

A uniform magnetic field is directed perpendicular to the plane of the dees. This magnetic field exerts a magnetic force \(F = Bqv\) perpendicular to the velocity of the protons, acting as a centripetal force and keeping them in a circular path. An alternating electric field is established across the gap between the dees. When protons cross this gap, they are accelerated by the electric field, increasing their kinetic energy and speed. As their speed increases, they move in a circular path of larger radius within the next dee. The centripetal force equation is \(Bqv = \frac{mv^2}{r}\), which gives \(r = \frac{mv}{Bq}\). The time spent in one dee is half the orbital period, \(t = \frac{\pi r}{v}\). Substituting \(r\) into the time equation gives \(t = \frac{\pi m}{Bq}\). The total orbital period is \(T = \frac{2\pi m}{Bq}\), and the frequency is \(f = \frac{Bq}{2\pi m}\). Since mass \(m\), magnetic flux density \(B\), and charge \(q\) are constant, the frequency is independent of the proton's speed and orbital radius, and therefore must remain constant.

1. States that the magnetic field is perpendicular to motion and provides centripetal force to keep protons in a circular path. (1) 2. States that the electric field in the gap accelerates the protons, increasing their kinetic energy. (1) 3. Equates centripetal force to magnetic force: \(Bqv = \frac{mv^2}{r}\) and solves for radius or velocity. (1) 4. Expresses the time for half an orbit as \(t = \frac{\pi r}{v}\). (1) 5. Shows that \(t = \frac{\pi m}{Bq}\) (or \(T = \frac{2\pi m}{Bq}\)) is independent of velocity and radius. (1) 6. Concludes that the frequency \(f = \frac{Bq}{2\pi m}\) is constant because all parameters in the equation are constant. (1)

From a thermodynamic perspective, because the compression is rapid, there is negligible time for heat exchange with the surroundings, making the process adiabatic (\(\Delta Q = 0\)). Work is done on the gas as the piston is pushed inward (\(W > 0\) as energy transferred to the gas). According to the First Law of Thermodynamics, \(\Delta U = \Delta Q + W\). Since \(\Delta Q = 0\), the internal energy \(\Delta U\) increases by the amount of work done. For an ideal gas, the internal energy is directly proportional to the absolute temperature as it consists purely of molecular translational kinetic energy. Thus, an increase in internal energy results in an increase in temperature. From a microscopic perspective, when gas molecules collide with the piston while it is moving inwards, they rebound with a higher speed relative to the cylinder walls, gaining kinetic energy. The mean kinetic energy of the molecules increases, which corresponds directly to an increase in the macroscopically measured temperature.

1. Identifies the rapid compression as adiabatic / negligible heat transfer (\(\Delta Q = 0\)). (1) 2. States that work is done on the gas during compression. (1) 3. Applies the First Law of Thermodynamics (\(\Delta U = \Delta Q + W\)) to show internal energy increases. (1) 4. Relates internal energy of an ideal gas directly to its temperature. (1) 5. Explains microscopically that molecules collide with the moving piston and rebound with greater speed/kinetic energy. (1) 6. Links the increase in mean molecular kinetic energy to the increase in temperature. (1)

The incoming gamma-ray photon is neutral and does not ionise the chamber medium, so it leaves no track. When pair production occurs, an electron and a positron are created. These charged particles ionise the medium, leaving visible tracks. Due to the perpendicular magnetic field, they experience a magnetic force \(F = Bqv\) acting as a centripetal force. Because they have opposite charges, they curve in opposite directions. The tracks are spirals with decreasing radii because the particles continuously lose kinetic energy and momentum as they ionise the chamber medium, and since \(r = \frac{p}{Bq}\), a decrease in momentum \(p\) decreases the radius \(r\). Conservation laws are satisfied as follows: Charge is conserved because the net charge before is 0 and after is \((+1e) + (-1e) = 0\). Lepton number is conserved because the net lepton number before is 0 and after is \((+1) + (-1) = 0\). Mass-energy is conserved because the photon energy \(hf\) must be at least the sum of the rest energies of the pair, \(2m_e c^2\), with any excess photon energy converted into the kinetic energies of the electron and positron.

1. Explains that the photon leaves no track because it is neutral. (1) 2. Explains that the electron and positron curve in opposite directions because they have opposite charges. (1) 3. Explains that the tracks spiral inwards (radius decreases) because the particles lose energy/momentum, referencing \(r = \frac{p}{Bq}\). (1) 4. Explains conservation of charge: initial charge is 0 and final total charge is 0. (1) 5. Explains conservation of lepton number: initial lepton number is 0 and final total lepton number is 0. (1) 6. Explains conservation of mass-energy: photon energy must be at least the rest mass-energy of the electron-positron pair (\(hf \ge 2m_e c^2\)). (1)

During the main sequence phase, the star is in hydrostatic equilibrium, where outward radiation and thermal pressure from hydrogen fusion in the core balances the inward gravitational pull. When hydrogen in the core is depleted, the core contracts under gravity and heats up, allowing helium fusion to begin. This heating causes outer shells to expand and cool, forming a red supergiant. The star undergoes successive stages of shell fusion, creating heavier elements up to iron. Fusion stops at iron because fusing iron is endothermic (requires energy input). Once the core is iron, fusion ceases, and the outward radiation pressure drops. The core undergoes sudden, catastrophic gravitational collapse. The outer layers of the star bounce off the collapsing core, triggering a massive shockwave known as a supernova explosion, ejecting most of the stellar material. The remaining core collapses further. If its mass is between the Chandrasekhar limit and the Tolman-Oppenheimer-Volkoff (TOV) limit, it becomes a neutron star supported by neutron degeneracy pressure. If its mass exceeds the TOV limit, gravity overcomes all degeneracy pressures, and it collapses into a black hole.

1. Explains hydrostatic equilibrium during the main sequence (radiation pressure balancing gravity). (1) 2. Describes the core contraction and shell fusion of heavier elements leading to expansion into a red supergiant. (1) 3. States that fusion stops at iron because iron fusion is endothermic / iron has highest binding energy per nucleon. (1) 4. Explains that the loss of radiation pressure leads to core collapse, resulting in a supernova explosion. (1) 5. Identifies that the remaining core becomes a neutron star if supported by neutron degeneracy pressure. (1) 6. Identifies that the core collapses into a black hole if its mass exceeds the TOV limit (or is extremely high). (1)

At very low driving frequencies, the system oscillates with a small amplitude, practically equal to the displacement of the driver, and is in phase with the driver (phase difference of 0). As the driving frequency approaches the natural frequency of the spring-mass system, the system absorbs energy highly efficiently from the driver, leading to resonance. The amplitude of oscillation reaches a maximum, and the phase difference between the driver and the system becomes \(\frac{\pi}{2}\) radians (or 90 degrees), meaning the driver leads the displacement. At very high driving frequencies, the inertia of the mass prevents it from keeping up with the rapid changes of the driver, so the amplitude of oscillation decreases towards zero, and the system oscillates in antiphase with the driver (phase difference of \(\pi\) radians or 180 degrees). Damping is the loss of energy from the system due to resistive forces. Introducing damping reduces the maximum amplitude at resonance, broadens the resonance peak, and slightly shifts the resonant frequency to a value below the natural frequency.

1. Describes the behaviour at low frequencies: small amplitude and in phase (phase difference 0). (1) 2. Describes resonance: maximum amplitude when driving frequency equals natural frequency. (1) 3. States that at resonance, the phase difference is \(\frac{\pi}{2}\) rad (or 90 degrees). (1) 4. Describes high-frequency behaviour: amplitude approaches zero and oscillations are in antiphase (phase difference \(\pi\) rad or 180 degrees). (1) 5. Explains that damping reduces the peak amplitude at resonance. (1) 6. Explains that damping broadens the resonance curve and slightly lowers the resonant frequency. (1)

Electromotive force (e.m.f., \(\varepsilon\)) is the total electrical energy transferred per unit charge by the chemical cell. Every real cell has an internal resistance, \(r\), due to the materials of the cell resisting the flow of charge. When a current \(I\) flows through the cell, some energy is dissipated as thermal energy inside the cell itself. The potential difference across this internal resistance is given by \(Ir\), often referred to as 'lost volts'. According to the principle of conservation of energy, the total energy supplied per unit charge (e.m.f.) must equal the sum of the energy delivered to the external circuit per unit charge (terminal potential difference, \(V\)) and the energy dissipated internally per unit charge (lost volts). This is expressed as \(\varepsilon = V + Ir\), which rearranges to \(V = \varepsilon - Ir\). Since the e.m.f. \(\varepsilon\) and internal resistance \(r\) are constant characteristics of the cell, as the current \(I\) increases, the product \(Ir\) increases. Consequently, the terminal potential difference \(V\) must decrease because a greater portion of the cell's energy is lost internally as heat.

1. Defines e.m.f. (\(\varepsilon\)) as total energy supplied per unit charge or internal resistance (\(r\)) as resistance within the cell. (1) 2. States the relationship between e.m.f., terminal potential difference, and internal resistance: \(V = \varepsilon - Ir\). (1) 3. Explains that \(Ir\) represents the potential difference across the internal resistance ('lost volts'). (1) 4. States that as current \(I\) increases, the 'lost volts' \(Ir\) also increases. (1) 5. Concludes that because \(\varepsilon\) is constant, an increase in \(Ir\) results in a decrease in the terminal potential difference \(V\). (1) 6. Explains this in terms of conservation of energy: more electrical energy is converted to thermal energy within the cell, leaving less available to the external circuit. (1)

In a cyclotron, a uniform magnetic field is applied perpendicular to the plane of the hollow D-shaped electrodes (dees). This magnetic field provides a centripetal force perpendicular to the velocity of the protons, causing them to move in circular paths of radius \(r\), given by \(Bqv = \frac{mv^2}{r}\). The magnetic field does no work on the protons and does not change their speed. An alternating electric field is applied across the narrow gap between the dees. When protons cross this gap, they are accelerated by the electric field, which increases their kinetic energy and speed. As their speed increases, they enter the next dee and travel in a larger semi-circular path. To ensure the electric field always accelerates the protons in the correct direction when they cross the gap, the field must alternate. The time spent in one dee is \(t = \frac{\pi r}{v}\). Substituting \(\frac{r}{v} = \frac{m}{Bq}\), we get \(t = \frac{\pi m}{Bq}\). Since \(m\), \(B\), and \(q\) are constant, the time \(t\) is independent of the radius and speed of the protons. Therefore, the frequency of the alternating potential difference, given by \(f = \frac{1}{2t} = \frac{Bq}{2\pi m}\), must remain constant.

1 mark: State that the magnetic field is perpendicular to the motion and provides the centripetal force, \(Bqv = \frac{mv^2}{r}\). 1 mark: State that the magnetic field keeps the protons in circular paths but does not increase their speed. 1 mark: State that the electric field in the gap accelerates the protons, increasing their speed/kinetic energy. 1 mark: State that the electric field must alternate in direction so that it always accelerates the protons as they arrive at the gap. 1 mark: Derive or state that the time spent in each dee is \(t = \frac{\pi m}{Bq}\), which is independent of the speed or radius of the path. 1 mark: Conclude that since the time for half an orbit is constant, the frequency of the alternating potential difference must be kept constant.

Copper is a ductile material. When a tensile stress is applied, it initially undergoes elastic deformation, where stress is directly proportional to strain. Beyond its elastic limit, copper undergoes significant plastic deformation before reaching its breaking point. Glass is a brittle material. Under tensile stress, glass undergoes only elastic deformation up to its breaking point, where it fractures suddenly without any plastic deformation. At the microscopic level, copper has a metallic structure consisting of positive metal ions in a regular lattice surrounded by a sea of delocalised electrons. When stress is applied beyond the elastic limit, planes of ions can slide past each other (due to the movement of dislocations) without breaking the non-directional metallic bonds, resulting in permanent plastic deformation. Glass has an amorphous giant covalent network structure with strong, highly directional covalent bonds. In glass, there are no dislocations that can move to allow sliding of atomic planes. When stressed, stress concentrates at microscopic surface cracks, causing the covalent bonds at the crack tips to break, which propagates the cracks rapidly and results in a sudden, brittle fracture.

1 mark: Identify copper as ductile (undergoes elastic and plastic deformation) and glass as brittle (undergoes only elastic deformation before fracture). 1 mark: State that copper exhibits permanent deformation beyond its elastic limit, whereas glass obeys Hooke's law up to its breaking point. 1 mark: Describe the microscopic structure of copper as a lattice of positive ions in a sea of delocalised electrons. 1 mark: Explain plastic deformation in copper as the sliding of planes of ions (movement of dislocations) without breaking the non-directional metallic bonds. 1 mark: Describe the microscopic structure of glass as a giant covalent network with strong, directional bonds. 1 mark: Explain brittle fracture in glass as the propagation of cracks due to high stress concentration, with no dislocation movement possible.

According to Huygens' principle, every point on a wavefront acts as a source of secondary spherical wavelets. When the laser light passes through the two narrow slits, each slit behaves as a source of coherent secondary wavelets that spread out (diffract) into the region beyond the slits. The waves from the two slits are coherent because they originate from the same laser source, meaning they have a constant phase difference. These diffracted waves overlap in the space between the slits and the screen, and they superpose. At any point on the screen, the resultant displacement is the vector sum of the individual displacements of the waves. Due to the geometry, waves travelling from each slit to a specific point on the screen travel different distances, resulting in a path difference. At points on the screen where the path difference is an integer number of wavelengths \(n\lambda\), the waves arrive in phase (phase difference is a multiple of \(2\pi\) rad). This leads to constructive interference, producing a bright fringe. At points on the screen where the path difference is an odd number of half-wavelengths \((n + 0.5)\lambda\), the waves arrive in antiphase (phase difference is an odd multiple of \(\pi\) rad). This leads to destructive interference, producing a dark fringe.

1 mark: Explain that according to Huygens' principle, each slit acts as a source of secondary wavelets that diffract/spread out. 1 mark: State that the waves from the two slits are coherent (constant phase relationship) because they originate from the same source. 1 mark: Explain that the waves overlap and superpose (resultant displacement is the vector sum of individual displacements). 1 mark: State that a path difference is created because waves travel different distances from each slit to different points on the screen. 1 mark: Link bright fringes to constructive interference, occurring when path difference is \(n\lambda\) (or waves arrive in phase). 1 mark: Link dark fringes to destructive interference, occurring when path difference is \((n + 0.5)\lambda\) (or waves arrive in antiphase).

Trigonometric parallax is used for relatively nearby stars. As the Earth orbits the Sun, the position of a nearby star is observed at two opposite points in the orbit, six months apart. The nearby star appears to shift its position relative to very distant background stars. The parallax angle \(\theta\) (or \(p\)) is half of this total angular shift. Using trigonometry, the distance \(d\) to the star is calculated using \(d = \frac{r}{\theta}\), where \(r\) is the radius of the Earth's orbit (1 AU), or \(d = \frac{1}{p}\) if using parsecs and arcseconds. This method cannot be used for very distant stars because the parallax angle becomes extremely small. At very large distances, the angular shift is too small to be measured with sufficient precision due to limits in the angular resolution of telescopes. For more distant objects, astronomers use standard candles, which are objects of known luminosity \(L\). Cepheid variables are a type of standard candle. The period of pulsation of a Cepheid variable is directly related to its average luminosity. By measuring the pulsation period, astronomers can determine its luminosity \(L\) from the period-luminosity relationship. They then measure the radiation flux \(F\) (apparent brightness) of the Cepheid on Earth. Using the inverse square law for radiation, \(F = \frac{L}{4\pi d^2}\), they calculate the distance \(d\) to the star or its host galaxy.

1 mark: Explain that trigonometric parallax involves measuring the apparent angular shift of a nearby star against distant background stars from two opposite positions in Earth's orbit. 1 mark: State that the distance is calculated using the parallax angle \(\theta\) (or \(p\)) via \(d = 1/p\) or \(d = r/\theta\). 1 mark: Explain that for very distant stars, the parallax angle is too small to measure accurately due to limits in telescope resolution. 1 mark: Define a standard candle as an astronomical object of known luminosity. 1 mark: Explain that for Cepheid variables, the luminosity is determined by measuring its period of pulsation (using the period-luminosity relationship). 1 mark: Explain that the distance is calculated using the measured flux \(F\) on Earth and the inverse square law formula \(F = \frac{L}{4\pi d^2}\).