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The change in momentum of the ball is: \(\Delta p = m v_f - m v_i = 0.20 \times (-10) - 0.20 \times 15 = -2.0 - 3.0 = -5.0\text{ N s}\). The magnitude of the change in momentum is \(5.0\text{ N s}\). Using Newton's second law, the average force is given by \(F = \frac{\Delta p}{\Delta t} = \frac{5.0}{0.050} = 100\text{ N}\).

Award [1] for the correct calculation of force: 100 N.

The electrical energy supplied is \(Q = P \times t = 50\text{ W} \times (2.0 \times 60\text{ s}) = 6000\text{ J}\). Since there are no heat losses, this energy equals the thermal energy gained by the liquid: \(Q = m c \Delta T\). Thus, \(6000 = 0.25 \times c \times 10\), which simplifies to \(6000 = 2.5 c\). Solving for \(c\) gives \(c = 2400\text{ J kg}^{-1}\text{ K}^{-1}\).

Award [1] for the correct calculation of specific heat capacity: 2400 J kg^-1 K^-1.

The activity decreases from \(800\text{ Bq}\) to \(100\text{ Bq}\), which is a factor of \(\frac{100}{800} = \frac{1}{8}\). Since \(\frac{1}{8} = \left(\frac{1}{2}\right)^3\), exactly three half-lives have elapsed. Therefore, the half-life \(T_{1/2}\) is \(\frac{12\text{ hours}}{3} = 4.0\text{ hours}\).

Award [1] for the correct identification of the half-life as 4.0 hours.

The gravitational field strength at the surface of a spherical body of mass \(M\) and radius \(R\) is given by \(g = \frac{GM}{R^2}\). For the planet, the mass is \(2M\) and the radius is \(2R\). Therefore, its gravitational field strength is: \(g_{\text{planet}} = \frac{G(2M)}{(2R)^2} = \frac{2GM}{4R^2} = 0.50 g\).

Award [1] for the correct determination of the gravitational field strength: 0.50g.

For single-slit diffraction, the angle \(\theta\) of the first minimum is given by the single-slit diffraction formula \(\theta \approx \frac{\lambda}{b}\). Under the new conditions, the new angle \(\theta'\) is: \(\theta' = \frac{2\lambda}{0.5b} = 4\frac{\lambda}{b} = 4\theta\).

Award [1] for the correct relationship leading to 4 theta.

The total energy in simple harmonic motion is proportional to the square of the amplitude, and is equal to \(E_{\text{max}}\). The potential energy at displacement \(x\) is given by \(E_p = E_{\text{max}} \left(\frac{x}{A}\right)^2\). At \(x = \frac{A}{2}\), the potential energy is \(E_p = E_{\text{max}} \left(\frac{A/2}{A}\right)^2 = 0.25 E_{\text{max}}\). Since total energy is conserved, the kinetic energy is \(E_k = E_{\text{max}} - E_p = E_{\text{max}} - 0.25 E_{\text{max}} = 0.75 E_{\text{max}}\).

Award [1] for the correct calculation of kinetic energy: 0.75 E_max.

The radius of the circular path of a charged particle in a magnetic field is given by \(R = \frac{mv}{qB}\). For the second particle, the radius is \(R' = \frac{(2m)(2v)}{(2q)B} = 2 \frac{mv}{qB} = 2R\).

Award [1] for the correct calculation of the radius: 2R.

The energy of the emitted photon is proportional to its frequency, \(\Delta E = hf\). For the transition from \(n = 2\) to \(n = 1\): \(hf = E_2 - E_1 = -\frac{E_0}{4} - \left(-\frac{E_0}{1}\right) = \frac{3}{4} E_0\). For the transition from \(n = 3\) to \(n = 2\): \(hf' = E_3 - E_2 = -\frac{E_0}{9} - \left(-\frac{E_0}{4}\right) = \frac{5}{36} E_0\). Dividing the two expressions: \(\frac{f'}{f} = \frac{5/36}{3/4} = \frac{5}{36} \times \frac{4}{3} = \frac{5}{27}\). Thus, \(f' = \frac{5}{27} f\).

Award [1] for the correct ratio of frequencies: 5/27 f.

The initial momentum of the block is \(p_i = mv\). Defining the direction of rebound as negative, the final momentum is \(p_f = -\frac{1}{3}mv\). The change in momentum is \(\Delta p = p_f - p_i = -\frac{1}{3}mv - mv = -\frac{4}{3}mv\). The magnitude of the average force is \(F_{\text{avg}} = \frac{|\Delta p|}{\Delta t} = \frac{4mv}{3\Delta t}\).

Award [1] for the correct choice of C. Award [0] for incorrect choices.

From Wien's displacement law, the temperature \(T\) is inversely proportional to the peak wavelength: \(T \propto \frac{1}{\lambda}\). Thus, the temperature of Star Y is \(T_Y = \frac{1}{2} T_X\). According to the Stefan-Boltzmann law, luminosity is given by \(L = 4\pi R^2 \sigma T^4\). Comparing the luminosities: \(\frac{L_Y}{L_X} = \left(\frac{R_Y}{R_X}\right)^2 \left(\frac{T_Y}{T_X}\right)^4 = (2)^2 \times \left(\frac{1}{2}\right)^4 = 4 \times \frac{1}{16} = \frac{1}{4}\).

Award [1] for the correct choice of C. Award [0] for incorrect choices.

The total energy of a satellite in a circular orbit of radius \(r\) is \(E = -\frac{GMm}{2r}\). The initial total energy is \(E_i = -\frac{GMm}{2r}\), and the final total energy is \(E_f = -\frac{GMm}{2(2r)} = -\frac{GMm}{4r}\). The change in total mechanical energy is \(\Delta E = E_f - E_i = -\frac{GMm}{4r} - \left(-\frac{GMm}{2r}\right) = +\frac{GMm}{4r}\).

Award [1] for the correct choice of B. Award [0] for incorrect choices.

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

Award [1] for the correct choice of C. Award [0] for incorrect choices.

After passing through the first vertical polarizer, the unpolarized light becomes linearly polarized with intensity \(I_1 = \frac{1}{2} I_0\). Using Malus's Law for the second polarizer: \(I_2 = I_1 \cos^2(60^\circ)\). Since \(\cos(60^\circ) = 0.5\), \(I_2 = \frac{1}{2} I_0 \times (0.5)^2 = \frac{1}{8} I_0\).

Award [1] for the correct choice of C. Award [0] for incorrect choices.

The total energy of the simple harmonic oscillator is given by \(E_T = \frac{1}{2}m\omega^2 A^2\). The potential energy when the displacement is \(x = \frac{1}{2}A\) is \(E_p = \frac{1}{2}m\omega^2 x^2 = \frac{1}{2}m\omega^2 \left(\frac{1}{2}A\right)^2 = \frac{1}{8}m\omega^2 A^2\). The kinetic energy is \(E_k = E_T - E_p = \frac{1}{2}m\omega^2 A^2 - \frac{1}{8}m\omega^2 A^2 = \frac{3}{8}m\omega^2 A^2\).

Award [1] for the correct choice of C. Award [0] for incorrect choices.

The energy difference between the two states is \(\Delta E = -1.5\text{ eV} - (-3.4\text{ eV}) = 1.9\text{ eV}\). Converting this energy to Joules: \(\Delta E = 1.9 \times 1.60 \times 10^{-19}\text{ J} = 3.04 \times 10^{-19}\text{ J}\). The wavelength \(\lambda\) of the emitted photon is calculated using \(\lambda = \frac{hc}{\Delta E} = \frac{6.63 \times 10^{-34} \times 3.00 \times 10^8}{3.04 \times 10^{-19}} \approx 6.54 \times 10^{-7}\text{ m} = 654\text{ nm}\), which is closest to \(650\text{ nm}\).

Award [1] for the correct choice of C. Award [0] for incorrect choices.

For the droplet to remain stationary, the upward electric force must equal the downward gravitational force: \(F_e = F_g \implies qE = mg\). The electric field \(E\) between the parallel plates is related to the potential difference \(V\) and separation \(d\) by \(E = \frac{V}{d}\). Substituting this in gives \(q \left(\frac{V}{d}\right) = mg\), which simplifies to \(V = \frac{mgd}{q}\).

Award [1] for the correct choice of C. Award [0] for incorrect choices.

The change in momentum (impulse) of the block is given by \(\Delta p = m(v - u)\). Taking the direction of the initial velocity as positive: \(u = 4.0\text{ m s}^{-1}\) and \(v = -2.0\text{ m s}^{-1}\). Thus, \(\Delta p = 0.50 \times (-2.0 - 4.0) = -3.0\text{ N s}\). The magnitude of the impulse is \(3.0\text{ N s}\). The area under the force-time graph represents the impulse. For a triangular force profile: \(\text{Impulse} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 0.080 \times F_{\text{max}} = 0.040 F_{\text{max}}\). Equating the impulse magnitudes: \(0.040 F_{\text{max}} = 3.0\text{ N s}\), which gives \(F_{\text{max}} = 75\text{ N}\).

Award 1 mark for the correct calculation of impulse of 3.0 N s and equating it to the area under the curve to find the correct value of 75 N.

First, calculate the energy required to melt all of the ice: \(Q_{\text{melt}} = m_{\text{ice}} L_f = 0.20 \times 3.3 \times 10^5 = 6.6 \times 10^4\text{ J}\). Next, calculate the maximum thermal energy that the water can release when cooling from \(20^\circ\text{C}\) to \(0^\circ\text{C}\): \(Q_{\text{water}} = m_{\text{water}} c_w \Delta T = 0.50 \times (4.2 \times 10^3) \times 20 = 4.2 \times 10^4\text{ J}\). Since \(Q_{\text{water}} < Q_{\text{melt}}\), there is not enough energy to melt all the ice. The temperature of the mixture will remain at \(0^\circ\text{C}\). The mass of ice that actually melts is \(m_{\text{melt}} = \frac{Q_{\text{water}}}{L_f} = \frac{4.2 \times 10^4}{3.3 \times 10^5} \approx 0.127\text{ kg}\). Therefore, the remaining mass of ice is \(0.20\text{ kg} - 0.127\text{ kg} = 0.073\text{ kg} \approx 0.07\text{ kg}\).

Award 1 mark for calculating the energy needed to melt the ice and the energy released by the water, determining that not all ice melts, and correctly calculating the remaining mass of ice to be 0.07 kg.

The gravitational potential energy of the system is \(E_p = -\frac{GMm}{r}\). For a circular orbit, the centripetal force is provided by the gravitational force: \(\frac{mv^2}{r} = \frac{GMm}{r^2}\), which gives the kinetic energy as \(E_k = \frac{1}{2}mv^2 = \frac{GMm}{2r}\). Dividing the kinetic energy by the potential energy yields \(\frac{E_k}{E_p} = \frac{GMm / (2r)}{-GMm / r} = -0.5\).

Award 1 mark for identifying the correct formulas for kinetic and potential energy in a circular orbit and obtaining the ratio of -0.5.

After three half-lives (\(t = 3T_{1/2}\)), the fraction of active nuclei remaining in the sample is \(\left(\frac{1}{2}\right)^3 = \frac{1}{8}\). Therefore, the number of remaining active nuclei is \(\frac{1}{8}N_0\). The number of decayed nuclei is \(N_0 - \frac{1}{8}N_0 = \frac{7}{8}N_0\). Since activity is directly proportional to the number of active nuclei (\(A = \lambda N\)), the activity of the sample is also reduced to \(\frac{1}{8}\) of its initial value, so \(A = \frac{1}{8}A_0\).

Award 1 mark for correctly determining both the decayed fraction as 7/8 and the activity fraction as 1/8.

The magnetic force provides the centripetal acceleration: \(qvB = \frac{mv^2}{r}\), which gives \(r = \frac{mv}{qB}\). The period of the circular motion is \(T = \frac{2\pi r}{v} = \frac{2\pi m}{qB}\). Notice that the period is independent of the velocity \(v\). For the second particle with mass \(2m\) and charge \(2q\), the new period is \(T' = \frac{2\pi (2m)}{(2q)B} = \frac{2\pi m}{qB} = T\).

Award 1 mark for expressing the period of circular motion in a magnetic field and showing that the period remains T.

When unpolarized light passes through the first polarizer, its intensity is halved, so \(I_1 = \frac{I_0}{2}\). The light is now vertically polarized. It then passes through the second polarizer whose axis is at \(60^\circ\) to the vertical. By Malus's Law, the transmitted intensity is \(I = I_1 \cos^2(60^\circ) = \frac{I_0}{2} \times (0.5)^2 = \frac{I_0}{8}\).

Award 1 mark for applying both the factor of 1/2 for the first polarizer and Malus's law for the second polarizer to get the final intensity of I_0 / 8.

When a wave transitions from one medium to another, its frequency remains unchanged, so the frequency in medium 2 is \(f\). The wave speed is given by the wave equation \(v = f\lambda\). Since \(f\) is constant, the wavelength is directly proportional to the wave speed: \(\lambda_2 = \frac{v_2}{f} = \frac{0.80v}{f} = 0.80\lambda\).

Award 1 mark for stating that the frequency remains unchanged and calculating that the wavelength is scaled by the same factor as the speed (0.80).

The energy of the emitted photon is equal to the difference in energy between the two levels: \(\Delta E = E_{\text{initial}} - E_{\text{final}} = -1.51\text{ eV} - (-3.40\text{ eV}) = 1.89\text{ eV}\). Convert this energy into Joules: \(\Delta E = 1.89 \times 1.60 \times 10^{-19}\text{ J} = 3.024 \times 10^{-19}\text{ J}\). Using the photon energy formula \(E = hf\), we find the frequency: \(f = \frac{\Delta E}{h} = \frac{3.024 \times 10^{-19}\text{ J}}{6.63 \times 10^{-34}\text{ J s}} \approx 4.56 \times 10^{14}\text{ Hz}\).

Award 1 mark for calculating the energy difference in Joules and using E = hf to determine the correct frequency of 4.56 * 10^14 Hz.

For a satellite of mass \(m\) in a stable circular orbit of radius \(r\) around a planet of mass \(M\), the orbital speed \(v\) is given by equating the centripetal force to the gravitational force: \(\frac{m v^2}{r} = \frac{G M m}{r^2}\), which gives \(m v^2 = \frac{G M m}{r}\). The kinetic energy of the satellite is therefore \(E_k = \frac{1}{2} m v^2 = \frac{G M m}{2 r}\). The gravitational potential energy of the system is \(E_p = -\frac{G M m}{r} = -2 E_k\). The total mechanical energy of the satellite in its orbit is \(E_{total} = E_k + E_p = E_k - 2 E_k = -E_k\). To completely escape the gravitational field of the planet, the satellite's total energy at infinity must be at least zero. Thus, the minimum additional energy \(\Delta E\) required is \(\Delta E = 0 - E_{total} = 0 - (-E_k) = E_k\).

Award [1] for the correct answer B. Method: Recognize that the potential energy of a satellite in orbit is \(-2 E_k\), meaning the total energy is \(-E_k\). For escape, total energy must be at least zero, requiring an addition of exactly \(E_k\).

(a) At the beginning, heat is still being transferred from the heater to the core of the block, and there is a time lag (thermal inertia) before a steady thermal gradient is established and registered by the thermometer.

(b) From conservation of energy, \(P = mc\frac{\Delta T}{\Delta t}\). The gradient of the graph is \(\frac{\Delta T}{\Delta t} = 0.052\text{ K s}^{-1}\).
Rearranging for \(c\):
\(c = \frac{P}{m \times \text{gradient}} = \frac{50}{1.00 \times 0.052} = 961.54\text{ J kg}^{-1}\text{ K}^{-1}\).
With 2 significant figures, this is \(960\text{ J kg}^{-1}\text{ K}^{-1}\).

(c) (i) Due to heat loss, the actual thermal energy absorbed by the block is less than the electrical energy supplied. This causes the temperature to rise more slowly (smaller gradient). Since \(c\) is inversely proportional to the gradient, a smaller gradient results in a calculated value of \(c\) that is too high (an overestimate).
(ii) Wrap the metal block in lagging/insulating material (such as cotton wool) or polish the outer surface of the block to reduce radiation losses.

(d) The formula is \(c = \frac{P}{m \times \text{gradient}}\).
- Percentage uncertainty in mass: \(\frac{0.01}{1.00} \times 100\% = 1.0\%\)
- Percentage uncertainty in power: \(\frac{2}{50} \times 100\% = 4.0\%\)
- Percentage uncertainty in gradient: \(3.0\%\)
Total percentage uncertainty in \(c\) is:
\(\% \Delta c = \% \Delta P + \% \Delta m + \% \Delta (\text{gradient}) = 4.0\% + 1.0\% + 3.0\% = 8.0\%\).

Absolute uncertainty in \(c\):
\(\Delta c = 8.0\% \times 961.54 = 76.9\text{ J kg}^{-1}\text{ K}^{-1}\).
Rounding appropriately to match the precision of \(c\), \(\Delta c \approx 77\text{ J kg}^{-1}\text{ K}^{-1}\) (or \(80\text{ J kg}^{-1}\text{ K}^{-1}\)).

(a) Award [1] for identifying the thermal lag / time taken for heat to conduct through the block to the thermometer.

(b) Award [1] for using \(P = mc \times \text{gradient}\) or equivalent rearrangement.
Award [1] for correct numerical calculation: \(960\) (or \(961.5\)).
Award [1] for correct unit: \(\text{J kg}^{-1}\text{ K}^{-1}\) or \(\text{J kg}^{-1}\ {^\circ}\text{C}^{-1}\).

(c) (i) Award [1] for stating that the rate of temperature rise (gradient) is reduced because some energy escapes to the environment.
Award [1] for concluding that this leads to an overestimate / higher calculated value of specific heat capacity.
(ii) Award [1] for suggesting insulation (lagging), a lid, or polishing the surface.

(d) Award [1] for calculating individual percentage uncertainties: mass = \(1.0\%\), power = \(4.0\%\).
Award [1] for summing the percentage uncertainties to find total percentage uncertainty = \(8.0\%\).
Award [1] for calculating absolute uncertainty \(\Delta c \approx 77\text{ J kg}^{-1}\text{ K}^{-1}\) or \(80\text{ J kg}^{-1}\text{ K}^{-1}\) (accept \(76.9\)).

(a) The theoretical equation assuming a massless spring is \(T = 2\pi \sqrt{\frac{m}{k}}\). However, the spring itself has mass, and there is also a mass hanger holding the masses. Both of these contribute to the oscillating mass of the system. Therefore, even when the added mass \(m = 0\), the system still oscillates with a non-zero period. This shifted mass translates the graph, giving a positive y-intercept.

(b) Squaring the theoretical equation yields:
\(T^2 = \frac{4\pi^2}{k} m\)
Comparing this to \(y = mx + c\), the gradient of the graph of \(T^2\) against \(m\) is \(\frac{4\pi^2}{k}\).
From the line of best fit, the gradient is \(3.25\text{ s}^2\text{ kg}^{-1}\).
Therefore:
\(3.25 = \frac{4\pi^2}{k}\)
\(k = \frac{4\pi^2}{3.25} = 12.15\text{ N m}^{-1}\).
To 3 significant figures, \(k = 12.1\text{ N m}^{-1}\) (or \(\text{kg s}^{-2}\)).

(c) (i) The total time \(t = 23.2\text{ s}\) has an absolute uncertainty of \(\Delta t = 0.4\text{ s}\).
The percentage uncertainty in the total time is:
\(\% \Delta t = \frac{0.4}{23.2} \times 100\% \approx 1.72\%\).
Since \(T = \frac{t}{20}\), dividing the time by a constant factor of 20 does not change its percentage uncertainty. Thus, the percentage uncertainty in \(T\) is also \(1.7\%\) (or \(1.72\%\)).
(ii) Since \(T^2 = T \times T\), the percentage uncertainties are added:
\(\% \Delta (T^2) = 2 \times \% \Delta T = 2 \times 1.72\% = 3.44\% \approx 3.4\%\).

(d) Human reaction time error (typically \(\approx \pm 0.1\text{ s}\) to \(\pm 0.2\text{ s}\)) occurs at the start and end of the timing process and remains relatively constant regardless of the total duration measured. By timing 20 oscillations instead of 1, this fixed absolute reaction error is divided by 20 to find the error in one period (e.g., \(\frac{0.2\text{ s}}{20} = 0.01\text{ s}\)). This dramatically reduces the percentage (relative) uncertainty in the measured period.

(a) Award [1] for identifying that the spring itself and/or the hanger have mass.
Award [1] for explaining that this additional mass oscillates even when the added mass \(m = 0\), resulting in a non-zero period and thus a positive y-intercept.

(b) Award [1] for equating the gradient of the line to \(\frac{4\pi^2}{k}\).
Award [1] for substituting the gradient value \(3.25\).
Award [1] for the final correct value and unit: \(12.1\text{ N m}^{-1}\) (accept range \(12\) to \(12.2\); accept \(\text{kg s}^{-2}\)).

(c) (i) Award [1] for using the formula \(\% \text{ uncertainty} = \frac{\text{absolute uncertainty}}{\text{measured value}} \times 100\%\).
Award [1] for \(1.7\%\) (accept \(1.72\%\)).
(ii) Award [1] for doubling the percentage uncertainty of \(T\) to get \(3.4\%\) (accept \(3.44\%\)).

(d) Award [1] for stating that the absolute uncertainty due to human reaction time is roughly constant/independent of the total time measured.
Award [1] for explaining that timing more oscillations spreads this fixed uncertainty over a larger total time, thus reducing the relative (percentage) uncertainty in the calculated period.

(a) The kinetic energy of the block initially is:
\( E_{k1} = \frac{1}{2} m v_0^2 = \frac{1}{2} (0.50) (4.0)^2 = 4.0 \text{ J} \)

The force of dynamic friction acting on the block is:
\( F_f = \mu_d m g = 0.25 \times 0.50 \times 9.81 = 1.23 \text{ N} \)

The work done by friction over the distance \( d = 1.2 \text{ m} \) is:
\( W_f = F_f \times d = 1.23 \times 1.2 = 1.47 \text{ J} \)

According to the work-energy theorem, the kinetic energy of the block right before collision is:
\( E_{k2} = E_{k1} - W_f = 4.0 - 1.47 = 2.53 \text{ J} \)

The speed \( v_2 \) of the block is:
\( v_2 = \sqrt{\frac{2 E_{k2}}{m}} = \sqrt{\frac{2 \times 2.53}{0.50}} = \sqrt{10.12} = 3.18 \text{ m s}^{-1} \approx 3.2 \text{ m s}^{-1} \).

(b) As the spring is compressed by a distance \( x \), work is done against both the spring force and the friction force over this distance.
Using the conservation of energy:
\( E_{k2} = \frac{1}{2} k x^2 + F_f x \)
\( 2.53 = \frac{1}{2} (120) x^2 + 1.23 x \)
\( 60 x^2 + 1.23 x - 2.53 = 0 \)

Solving this quadratic equation for \( x \):
\( x = \frac{-1.23 \pm \sqrt{1.23^2 - 4(60)(-2.53)}}{120} \)
\( x = \frac{-1.23 \pm \sqrt{1.51 + 607.2}}{120} = \frac{-1.23 \pm 24.67}{120} \)
Taking the positive root:
\( x = 0.195 \text{ m} \approx 0.20 \text{ m} \).

(c) Hooke's Law states that the force exerted by the spring is proportional to the compression, \( F = -kx \).
As the block compresses the spring from \( x = 0 \) to \( x = x_{\max} \), the magnitude of the force increases linearly from \( 0 \) to a maximum value of \( k x_{\max} = 120 \times 0.195 \approx 23.4 \text{ N} \).
The direction of this force is opposite to the direction of motion of the block.

(a) [3 marks]
- Correct calculation of initial kinetic energy (4.0 J) [1M]
- Correct calculation of friction force (1.23 N) or work done by friction (1.47 J) [1M]
- Equating remaining KE to calculate speed showing at least 2 significant figures (3.18 or 3.2 m s^-1) [1M]

(b) [3 marks]
- Setting up correct energy conservation equation involving both elastic potential energy and friction work [1M]
- Substituting correct values into quadratic equation [1M]
- Finding correct positive solution (0.20 m or 0.195 m) [1M]

(c) [2.33 marks]
- Stating that force increases linearly with compression/displacement (or mentioning Hooke's law) [1M]
- Stating that force reaches a maximum at maximum compression [1M]
- Identifying that the force opposes the block's motion [0.33M]

(a) The cross-sectional area of the wire is:
\( A = \frac{\pi d^2}{4} = \frac{\pi (0.38 \times 10^{-3})^2}{4} = 1.134 \times 10^{-7} \text{ m}^2 \)

Using the resistivity formula \( \rho = \frac{R A}{L} \):
\( \rho = \frac{4.50 \times 1.134 \times 10^{-7}}{1.250} = 4.08 \times 10^{-7} \ \Omega \text{ m} \).

(b) The formula for resistivity is \( \rho = \frac{\pi R d^2}{4 L} \).
Therefore, the fractional uncertainty in \( \rho \) is:
\( \frac{\Delta \rho}{\rho} = \frac{\Delta R}{R} + 2 \frac{\Delta d}{d} + \frac{\Delta L}{L} \)

Calculating individual fractional uncertainties:
\( \frac{\Delta R}{R} = \frac{0.05}{4.50} \approx 0.0111 \) (or \( 1.11\% \))
\( 2 \frac{\Delta d}{d} = 2 \times \frac{0.02}{0.38} \approx 0.1053 \) (or \( 10.53\% \))
\( \frac{\Delta L}{L} = \frac{0.002}{1.250} = 0.0016 \) (or \( 0.16\% \))

Total fractional uncertainty:
\( \frac{\Delta \rho}{\rho} = 0.0111 + 0.1053 + 0.0016 = 0.1180 \) (or \( 11.8\% \))

Absolute uncertainty:
\( \Delta \rho = 0.1180 \times 4.08 \times 10^{-7} = 4.82 \times 10^{-8} \ \Omega \text{ m} \approx 0.5 \times 10^{-7} \ \Omega \text{ m} \).

(c) The diameter measurement contributes the most to the uncertainty.
This is because its fractional uncertainty is the largest (approx. \( 5.3\% \)) and its contribution is doubled (approx. \( 10.5\% \)) since diameter is squared in the area calculation.

(a) [3 marks]
- Correct calculation of cross-sectional area (1.13 x 10^-7 m^2) [1M]
- Correct substitution into resistivity formula [1M]
- Correct value of resistivity (4.08 x 10^-7 \Omega m or 4.1 x 10^-7 \Omega m) with correct units [1M]

(b) [3.33 marks]
- Recognizing that the fractional uncertainty of diameter must be doubled [1M]
- Summing all fractional uncertainties correctly (0.118 or 11.8%) [1M]
- Multiplying fractional uncertainty by calculated resistivity [1M]
- Expressing final absolute uncertainty to 1 significant figure (0.5 x 10^-7 \Omega m or 5 x 10^-8 \Omega m) [0.33M]

(c) [2 marks]
- Stating that diameter contributes the most [1M]
- Providing justification based on its large fractional uncertainty AND that the effect is doubled because diameter is squared [1M]

(a) Energy to raise temperature of ice to \( 0^\circ\text{C} \):
\( Q_1 = m c_{\text{ice}} \Delta T_1 = 0.40 \times 2100 \times 10 = 8400 \text{ J} \).

(b) Energy to melt the ice at \( 0^\circ\text{C} \):
\( Q_2 = m L_f = 0.40 \times 3.3 \times 10^5 = 1.32 \times 10^5 \text{ J} \)

Energy to raise the temperature of the water to \( 20^\circ\text{C} \):
\( Q_3 = m c_{\text{water}} \Delta T_2 = 0.40 \times 4200 \times 20 = 33600 \text{ J} \)

Total energy required:
\( Q_{\text{total}} = Q_1 + Q_2 + Q_3 = 8400 + 132000 + 33600 = 1.74 \times 10^5 \text{ J} \)

Time taken \( t \):
\( t = \frac{Q_{\text{total}}}{P} = \frac{1.74 \times 10^5}{150} = 1160 \text{ s} \) (or approximately \( 19.3 \text{ minutes} \)).

(c) In a real experiment, thermal energy is lost to the surroundings (such as the air and container).
Thus, the heater must supply more total energy than calculated, resulting in a longer heating time.

(a) [2 marks]
- Correct substitution of values into Q = mc\Delta T [1M]
- Correct calculation of energy (8400 J) [1M]

(b) [4.33 marks]
- Correct calculation of energy for phase change (1.32 x 10^5 J) [1M]
- Correct calculation of energy for water heating (33600 J) [1M]
- Adding all three energies correctly to get 1.74 x 10^5 J [1M]
- Using t = Q/P to find 1160 s (or 19 minutes / 19.3 minutes) [1.33M]

(c) [2 marks]
- Identifying that energy is lost to the surroundings [1M]
- Explaining that more energy must be supplied to compensate, hence taking longer [1M]

(a) The gravitational force provides the centripetal force for the circular orbit:
\( \frac{G M m}{r^2} = m \omega^2 r = m \left(\frac{2\pi}{T}\right)^2 r \)

Rearranging for \( r^3 \):
\( r^3 = \frac{G M T^2}{4 \pi^2} \)

Substitute the given values:
\( r^3 = \frac{6.67 \times 10^{-11} \times 6.4 \times 10^{23} \times (7200)^2}{4 \pi^2} \)
\( r^3 = \frac{4.269 \times 10^{13} \times 5.184 \times 10^7}{39.48} \approx 5.605 \times 10^{19} \text{ m}^3 \)
\( r = (5.605 \times 10^{19})^{1/3} \approx 3.83 \times 10^6 \text{ m} \approx 3.8 \times 10^6 \text{ m} \).

(b) Gravitational potential energy \( E_p \) is given by:
\( E_p = -\frac{G M m}{r} \)
\( E_p = -\frac{6.67 \times 10^{-11} \times 6.4 \times 10^{23} \times 850}{3.83 \times 10^6} \)
\( E_p = -9.47 \times 10^9 \text{ J} \approx -9.5 \times 10^9 \text{ J} \).

(c) The kinetic energy will decrease.
For a stable circular orbit, \( \frac{G M m}{r^2} = \frac{m v^2}{r} \), which yields \( v^2 = \frac{G M}{r} \).
Kinetic energy is \( E_k = \frac{1}{2} m v^2 = \frac{G M m}{2r} \).
Since \( E_k \) is inversely proportional to \( r \), an increase in orbital radius \( r \) leads to a decrease in kinetic energy.

(a) [3 marks]
- Equating gravitational force to centripetal force [1M]
- Expressing orbital radius r in terms of T, M, and G [1M]
- Substituting values correctly to show r \approx 3.8 x 10^6 m [1M]

(b) [2.33 marks]
- Using correct equation for gravitational potential energy (including the negative sign) [1M]
- Correct substitution of values using the calculated radius [1M]
- Finding correct answer with unit (-9.5 x 10^9 J or -9.47 x 10^9 J) [0.33M] (accept positive value only if explicitly defined as magnitude)

(c) [3 marks]
- Stating that kinetic energy decreases [1M]
- Equating forces to show that v^2 = GM/r (or E_k = GMm / 2r) [1M]
- Explaining that as r increases, velocity and thus kinetic energy must decrease to maintain orbit [1M]

(a) The fraction of remaining nuclei is:
\( \frac{N}{N_0} = \frac{5.0 \times 10^{17}}{4.0 \times 10^{18}} = 0.125 = \frac{1}{8} \)

Since \( \frac{1}{8} = \left(\frac{1}{2}\right)^3 \), the sample has undergone exactly \( 3 \) half-lives.
Therefore:
\( 3 \times T_{1/2} = 15.0 \text{ hours} \implies T_{1/2} = 5.0 \text{ hours} \).

(b) First, convert the half-life to seconds:
\( T_{1/2} = 5.0 \text{ hours} = 5.0 \times 3600 \text{ s} = 1.8 \times 10^4 \text{ s} \)

The decay constant \( \lambda \) is:
\( \lambda = \frac{\ln(2)}{T_{1/2}} = \frac{0.6931}{1.8 \times 10^4} = 3.85 \times 10^{-5} \text{ s}^{-1} \approx 3.9 \times 10^{-5} \text{ s}^{-1} \).

(c) The initial activity \( A_0 \) is:
\( A_0 = \lambda N_0 \)
\( A_0 = 3.85 \times 10^{-5} \text{ s}^{-1} \times 4.0 \times 10^{18} = 1.54 \times 10^{14} \text{ Bq} \approx 1.5 \times 10^{14} \text{ Bq} \).

(a) [3 marks]
- Calculating the remaining fraction (0.125 or 1/8) [1M]
- Relating this fraction to 3 half-lives [1M]
- Calculating the half-life as 5.0 hours [1M]

(b) [2.33 marks]
- Converting half-life to seconds (1.8 x 10^4 s) [1M]
- Correctly using the formula \lambda = ln(2)/T_1/2 [1M]
- Calculating decay constant as 3.9 x 10^-5 s^-1 (or 3.85 x 10^-5 s^-1) [0.33M]

(c) [3 marks]
- Recalling the activity formula A_0 = \lambda N_0 [1M]
- Substituting decay constant and initial number of nuclei [1M]
- Calculating correct activity (1.5 x 10^14 Bq or 1.54 x 10^14 Bq) with units [1M]

(a) For a single slit diffraction, the angle to the first minimum is:
\( \theta \approx \frac{y}{D} \)
where \( y = 1.8 \text{ cm} = 0.018 \text{ m} \) and \( D = 2.4 \text{ m} \).

\( \theta = \frac{0.018}{2.4} = 7.5 \times 10^{-3} \text{ rad} \)

According to the single slit diffraction formula:
\( \theta = \frac{\lambda}{b} \implies b = \frac{\lambda}{\theta} \)

\( b = \frac{632.8 \times 10^{-9}}{7.5 \times 10^{-3}} = 8.44 \times 10^{-5} \text{ m} \approx 8.4 \times 10^{-5} \text{ m} \) (or \( 84 \ \mu\text{m} \)).

(b) For a circular aperture, the first minimum occurs at an angle:
\( \theta = \frac{1.22 \lambda}{d} \)
where the diameter \( d = b = 8.44 \times 10^{-5} \text{ m} \).

\( \theta = \frac{1.22 \times 632.8 \times 10^{-9}}{8.44 \times 10^{-5}} = 9.15 \times 10^{-3} \text{ rad} \approx 9.2 \times 10^{-3} \text{ rad} \).

(c) The Rayleigh criterion states that two sources are just resolved when the central maximum of one diffraction pattern falls on the first minimum of the other.
The minimum resolvable angular separation is given by \( \theta_{\min} = \frac{1.22 \lambda}{d} \).
If the diameter \( d \) of the aperture is doubled, the minimum resolvable angular separation is halved.
Therefore, the resolution improves because the system can distinguish sources that are closer together in angular separation.

(a) [3 marks]
- Calculating the diffraction angle \theta (7.5 x 10^-3 rad) [1M]
- Recalling the single slit diffraction formula \theta = \lambda/b [1M]
- Finding the correct slit width (8.4 x 10^-5 m or 8.44 x 10^-5 m) [1M]

(b) [2.33 marks]
- Recalling the circular aperture formula \theta = 1.22 \lambda/d [1M]
- Correct substitution of values using previously calculated b [1M]
- Finding correct angle (9.2 x 10^-3 rad or 9.15 x 10^-3 rad) [0.33M]

(c) [3 marks]
- Stating the Rayleigh criterion (central maximum of one image lies on the first minimum of the other) [1M]
- Explaining that doubling diameter halves the minimum angular separation (\theta_{\min} is halved) [1M]
- Concluding that the resolution is improved [1M]