2022 AQA AS-Level Physics 7407 模擬試題連答案詳解

(a) During beta-plus decay, a proton decays into a neutron, emitting a positron and an electron neutrino. The equation is: \(^{18}_{9}\text{F} \rightarrow ^{18}_{8}\text{O} + \beta^+ + \nu_e\). (b) Since flavor changes (proton to neutron), this is governed by the weak interaction. The exchange particle is the \(W^+\) boson because positive charge is carried from the proton to the positron. (c) Charge conservation: Initial charge = \(+9e\), Final charge = \(+8e\) (from Oxygen) + \(+1e\) (from positron) + 0 (from neutrino) = \(+9e\). Baryon number conservation: Initial baryon number = 18, Final baryon number = 18 (Oxygen) + 0 (positron) + 0 (neutrino) = 18. Lepton number conservation: Initial lepton number = 0, Final lepton number = 0 (Oxygen) - 1 (positron) + 1 (electron neutrino) = 0.

(a) [3 marks] 1 mark for correct Oxygen symbol with 18 and 8. 1 mark for positron (\(\beta^+\) or \(e^+\)). 1 mark for electron neutrino (\(\nu_e\)). (b) [2 marks] 1 mark for weak interaction. 1 mark for \(W^+\) boson (accept \(W\) boson). (c) [3.75 marks] 1.25 marks for charge conservation equation and calculation. 1.25 marks for baryon number conservation equation and calculation. 1.25 marks for lepton number conservation equation and calculation.

(a) By the conservation of linear momentum: \(m_1 v_1 + m_2 v_2 = (m_1 + m_2) v_f\). Substituting the values: \((0.45 \times 2.8) + (0.15 \times 0) = (0.45 + 0.15) v_f\), which gives \(1.26 = 0.60 v_f\), so \(v_f = 2.1\text{ m s}^{-1}\). (b) Initial kinetic energy: \(E_{k,i} = \frac{1}{2} m_1 v_1^2 = 0.5 \times 0.45 \times (2.8)^2 = 1.764\text{ J}\). Final kinetic energy: \(E_{k,f} = \frac{1}{2} (m_1 + m_2) v_f^2 = 0.5 \times 0.60 \times (2.1)^2 = 1.323\text{ J}\). Kinetic energy lost: \(\Delta E_k = 1.764 - 1.323 = 0.441\text{ J}\) (or \(0.44\text{ J}\) to two significant figures). (c) The lost kinetic energy is transferred to other forms of energy, such as thermal energy in the gliders and surroundings due to the work done during deformation, and sound energy.

(a) [3 marks] 1 mark for stating conservation of momentum. 1 mark for substitution of values. 1 mark for correct final velocity with unit (\(2.1\text{ m s}^{-1}\)). (b) [3 marks] 1 mark for correct initial kinetic energy. 1 mark for correct final kinetic energy. 1 mark for correct kinetic energy lost (\(0.44\text{ J}\)). (c) [2.75 marks] 1.5 marks for stating it is converted into thermal/heat energy. 1.25 marks for stating sound energy (or work done during deformation).

(a) Convert values to SI units: \(d = 0.38 \times 10^{-3}\text{ m}\), \(e = 1.45 \times 10^{-3}\text{ m}\). Calculate Young modulus: \(E = \frac{4 \times 45.0 \times 2.035}{\pi \times (0.38 \times 10^{-3})^2 \times 1.45 \times 10^{-3}} = \frac{366.3}{\pi \times 1.444 \times 10^{-7} \times 1.45 \times 10^{-3}} = \frac{366.3}{6.578 \times 10^{-10}} \approx 1.11 \times 10^{11}\text{ Pa}\). (b) Find percentage uncertainties: \(\%\Delta F = \frac{0.5}{45.0} \times 100\% = 1.11\%\), \(\%\Delta L = \frac{0.002}{2.035} \times 100\% = 0.10\%\), \(\%\Delta d = \frac{0.01}{0.38} \times 100\% = 2.63\%\), \(\%\Delta e = \frac{0.05}{1.45} \times 100\% = 3.45\%\). Total percentage uncertainty: \(\%\Delta E = \%\Delta F + \%\Delta L + 2(\%\Delta d) + \%\Delta e = 1.11\% + 0.10\% + 2(2.63\%) + 3.45\% = 9.92\%\) (accept \(9.8\%\) to \(10.0\%\) depending on rounding). (c) The diameter \(d\) contributes \(2 \times 2.63\% = 5.26\%\) to the overall uncertainty, which is the largest single contribution because its percentage uncertainty is doubled in the formula. This can be reduced by using a digital micrometer with higher precision and taking multiple measurements at different orientations and positions along the wire to calculate a mean diameter.

(a) [2 marks] 1 mark for correct conversion of units. 1 mark for substituting values to show \(1.1 \times 10^{11}\text{ Pa}\). (b) [3 marks] 1 mark for individual percentage uncertainties of all variables. 1 mark for doubling the percentage uncertainty of the diameter. 1 mark for correct total percentage uncertainty of \(9.9\%\) (accept \(9.8\%\) to \(10.0\%\)). (c) [3.75 marks] 1.75 marks for identifying diameter (or diameter squared) as the greatest source of uncertainty and explaining why. 2 marks for suggesting using a higher resolution instrument (e.g. digital micrometer) AND taking multiple measurements at different points/orientations to find a mean.

(a) Total resistance in bright light: \(R_{\text{total}} = 150 + 75 = 225\ \Omega\). Potential difference across LDR: \(V_{\text{LDR}} = \frac{R_{\text{LDR}}}{R_{\text{total}}} \times V_{\text{in}} = \frac{75.0}{225} \times 12.0 = 4.00\text{ V}\). (b) In the dark, \(V_{\text{LDR}} = 9.60\text{ V}\). The potential difference across the fixed resistor is: \(V_R = 12.0 - 9.60 = 2.40\text{ V}\). The current in the circuit is: \(I = \frac{V_R}{R} = \frac{2.40}{150} = 0.0160\text{ A}\). Thus, the resistance of the LDR is: \(R_{\text{LDR}} = \frac{V_{\text{LDR}}}{I} = \frac{9.60}{0.0160} = 600\ \Omega\). (c) When internal resistance is present, total resistance becomes: \(R_{\text{total}} = R + R_{\text{LDR}} + r = 150 + 75.0 + 15.0 = 240\ \Omega\). The current from the battery is: \(I = \frac{\mathcal{E}}{R_{\text{total}}} = \frac{12.0}{240} = 0.050\text{ A}\). The potential difference across the LDR is: \(V_{\text{LDR}} = I \times R_{\text{LDR}} = 0.050 \times 75.0 = 3.75\text{ V}\).

(a) [2 marks] 1 mark for finding total resistance. 1 mark for correct voltage \(4.00\text{ V}\). (b) [3 marks] 1 mark for finding voltage across the fixed resistor (\(2.40\text{ V}\)). 1 mark for calculating current (\(0.016\text{ A}\)). 1 mark for correct resistance of LDR (\(600\ \Omega\)). (c) [3.75 marks] 1 mark for adding internal resistance to find new total resistance (\(240\ \Omega\)). 1 mark for finding new current (\(0.050\text{ A}\)). 1.75 marks for correct potential difference across the LDR (\(3.75\text{ V}\)).

(a) The mass of the wire is \(m = \rho \times V = \rho \times A \times L\). The mass per unit length is \(\mu = \frac{m}{L} = \rho A\). Substituting the values: \(\mu = 7800 \times (3.50 \times 10^{-7}) = 2.73 \times 10^{-3}\text{ kg m}^{-1}\). (b) The speed of transverse waves on a stretched string is given by: \(v = \sqrt{\frac{T}{\mu}}\). Substituting values: \(v = \sqrt{\frac{140}{2.73 \times 10^{-3}}} = \sqrt{51282} \approx 226.46\text{ m s}^{-1}\) (or \(226\text{ m s}^{-1}\) to 3 s.f.). (c) For the first harmonic, the length of the string is half a wavelength, so \(L = \frac{\lambda}{2}\), giving \(\lambda = 2L = 2 \times 1.20 = 2.40\text{ m}\). The frequency is: \(f = \frac{v}{\lambda} = \frac{226.46}{2.40} \approx 94.4\text{ Hz}\) (accept \(94.3\text{ Hz} - 94.4\text{ Hz}\) depending on rounding of wave speed).

(a) [2 marks] 1 mark for realizing that \(\mu = \rho A\). 1 mark for correct calculation to obtain \(2.73 \times 10^{-3}\text{ kg m}^{-1}\). (b) [3 marks] 1 mark for recalling \(v = \sqrt{T/\mu}\). 1 mark for substituting tension and \(\mu\) correctly. 1 mark for correct speed with unit (\(226\text{ m s}^{-1}\)). (c) [3.75 marks] 1 mark for identifying that wavelength is \(2L\) (\(2.40\text{ m}\)). 1 mark for using \(f = v/\lambda\). 1.75 marks for correct frequency value of \(94.4\text{ Hz}\) with unit.

(a) Cross-sectional area: \(A = \pi r^2 = \pi \left(\frac{0.80 \times 10^{-3}}{2}\right)^2 = \pi \times (0.40 \times 10^{-3})^2 \approx 5.027 \times 10^{-7}\text{ m}^2\) (or \(5.03 \times 10^{-7}\text{ m}^2\)). (b) Tensile stress: \(\sigma = \frac{F}{A} = \frac{65}{5.027 \times 10^{-7}} \approx 1.293 \times 10^8\text{ Pa}\) (or \(1.29 \times 10^8\text{ Pa}\)). Tensile strain: \(\epsilon = \frac{\sigma}{E} = \frac{1.293 \times 10^8}{1.20 \times 10^{11}} \approx 1.078 \times 10^{-3}\) (or \(1.08 \times 10^{-3}\)). (c) Extension: \(e = \epsilon \times L = 1.078 \times 10^{-3} \times 2.50 = 2.695 \times 10^{-3}\text{ m}\). Elastic strain energy: \(E_s = \frac{1}{2} F e = 0.5 \times 65 \times 2.695 \times 10^{-3} \approx 0.0876\text{ J}\) (or \(0.088\text{ J}\) to two significant figures).

(a) [2 marks] 1 mark for correct radius in meters. 1 mark for correct area calculation. (b) [4 marks] 1 mark for correct stress formula and substitution. 1 mark for correct stress value with unit. 1 mark for correct strain formula. 1 mark for correct strain value (no unit). (c) [2.75 marks] 1 mark for calculating extension. 1.75 marks for calculating elastic strain energy correctly with unit (accept \(0.087 - 0.089\text{ J}\)).

(a) Slit spacing \(d\) is the reciprocal of the number of lines per meter. \(d = \frac{1}{550 \times 10^3} = 1.818 \times 10^{-6}\text{ m}\) (or \(1.82 \times 10^{-6}\text{ m}\)). (b) Using the diffraction grating formula: \(d \sin \theta = n \lambda\). For the second order (\(n = 2\)): \((1.818 \times 10^{-6}) \times \sin(38.5^\circ) = 2 \lambda\). \(1.818 \times 10^{-6} \times 0.6225 = 2 \lambda\), which gives \(2 \lambda = 1.1317 \times 10^{-6}\text{ m}\), so \(\lambda = 5.659 \times 10^{-7}\text{ m}\) (or \(5.66 \times 10^{-7}\text{ m}\)). (c) For maximum order, \(\theta \le 90^\circ\), which implies \(\sin \theta \le 1\). Therefore, \(n \le \frac{d}{\lambda} = \frac{1.818 \times 10^{-6}}{5.659 \times 10^{-7}} \approx 3.21\). The highest order possible is \(n = 3\). The total number of visible maxima is \(2n + 1 = 2(3) + 1 = 7\) maxima (consisting of the central maximum, 3 orders on each side).

(a) [2 marks] 1 mark for converting lines per mm to lines per m. 1 mark for obtaining \(1.82 \times 10^{-6}\text{ m}\). (b) [3 marks] 1 mark for stating \(d \sin \theta = n \lambda\). 1 mark for substituting values. 1 mark for correct wavelength value and unit (\(5.66 \times 10^{-7}\text{ m}\)). (c) [3.75 marks] 1 mark for using condition \(\sin \theta \le 1\). 1 mark for finding the maximum order \(n = 3\). 1.75 marks for multiplying by 2 and adding 1 to get 7 maxima.

(a) Photon energy is given by: \(E = h f\). Using Planck's constant \(h = 6.63 \times 10^{-34}\text{ J s}\): \(E = (6.63 \times 10^{-34}) \times (7.25 \times 10^{14}) = 4.807 \times 10^{-19}\text{ J}\) (or \(4.81 \times 10^{-19}\text{ J}\)). (b) Einstein's photoelectric equation is: \(h f = \phi + E_{k,\text{max}}\), where \(\phi\) is the work function. Rearranging gives: \(\phi = h f - E_{k,\text{max}} = 4.807 \times 10^{-19} - 1.12 \times 10^{-19} = 3.687 \times 10^{-19}\text{ J}\). In electron-volts: \(\phi = \frac{3.687 \times 10^{-19}\text{ J}}{1.60 \times 10^{-19}\text{ J eV}^{-1}} \approx 2.304\text{ eV}\) (or \(2.30\text{ eV}\)). (c) The work function is related to the threshold frequency \(f_0\) by: \(\phi = h f_0\). Thus: \(f_0 = \frac{\phi}{h} = \frac{3.687 \times 10^{-19}}{6.63 \times 10^{-34}} \approx 5.56 \times 10^{14}\text{ Hz}\).

(a) [2 marks] 1 mark for recalling \(E = h f\). 1 mark for substitution and calculating \(4.81 \times 10^{-19}\text{ J}\). (b) [3 marks] 1 mark for recalling photoelectric equation. 1 mark for correct work function in Joules (\(3.69 \times 10^{-19}\text{ J}\)). 1 mark for dividing by \(1.60 \times 10^{-19}\) to get \(2.30\text{ eV}\). (c) [3.75 marks] 1 mark for stating relationship \(\phi = h f_0\). 1 mark for correct rearrangement. 1.75 marks for final frequency value with unit (\(5.56 \times 10^{14}\text{ Hz}\)).

(a) A steel ball bearing has a much greater weight compared to the air resistance acting on it than a table tennis ball of the same size. Therefore, air resistance is negligible for the steel ball, and its motion is a much closer approximation to true free fall under gravity.

(b) 1. Use a set square or a plumb line to ensure the metre ruler is perfectly vertical.
2. View the markings on the ruler at eye level at both the point of release and the trapdoor to eliminate parallax errors.

(c) Mean time, \( t_{\text{mean}} = \frac{0.402 + 0.406 + 0.398 + 0.404}{4} = 0.4025 \text{ s} \approx 0.403 \text{ s} \).
Range in the values of \( t = 0.406 - 0.398 = 0.008 \text{ s} \).
Absolute uncertainty in \( t = \frac{\text{Range}}{2} = \frac{0.008 \text{ s}}{2} = \pm 0.004 \text{ s} \).
So, \( t = 0.403 \pm 0.004 \text{ s} \) (or \( 0.4025 \pm 0.004 \text{ s} \)).

(d) Using \( h = \frac{1}{2}gt^2 \implies g = \frac{2h}{t^2} \).
Using \( t_{\text{mean}} = 0.403 \text{ s} \):
\( g = \frac{2 \times 0.800}{(0.403)^2} = 9.85 \text{ m s}^{-2} \) (or \( 9.88 \text{ m s}^{-2} \) if using the unrounded mean value of \( 0.4025 \text{ s} \)).

To find the percentage uncertainty in \( g \):
\( \% \Delta h = \frac{0.002}{0.800} \times 100\% = 0.25\% \)
\( \% \Delta t = \frac{0.004}{0.403} \times 100\% = 0.99\% \) (or \( 0.994\% \) if using \( 0.4025 \text{ s} \))
\( \% \Delta g = \% \Delta h + 2 \times (\% \Delta t) = 0.25\% + 2 \times (0.99\%) = 2.23\% \approx 2.2\% \) (accept \( 2.2\% \) or \( 2.24\% \) or \( 2\% \)).

(a) [2 marks]
- Steel ball bearing has negligible air resistance / weight is much greater than air resistance (1 mark)
- Motion is much closer to true free fall / acceleration is constant (1 mark)

(b) [2 marks]
- Use of set square or plumb line to ensure the ruler is vertical (1 mark)
- Position eye level with readings to eliminate parallax error (1 mark)

(c) [3 marks]
- Correct calculation of mean time as \( 0.403 \text{ s} \) or \( 0.4025 \text{ s} \) (1 mark)
- Correct identification of range as \( 0.008 \text{ s} \) (1 mark)
- Correct calculation of absolute uncertainty as \( \pm 0.004 \text{ s} \) (1 mark)

(d) [3 marks]
- Correct calculation of \( g = 9.85 \text{ m s}^{-2} \) or \( 9.88 \text{ m s}^{-2} \) (1 mark)
- Correct calculation of percentage uncertainties in \( h \) (\( 0.25\% \)) or \( t \) (\( 0.99\% \)) (1 mark)
- Correct overall percentage uncertainty in \( g = 2.2\% \) or \( 2.24\% \) (accept \( 2\% \)) using the correct absolute summation rule \( \% \Delta g = \% \Delta h + 2 \times \% \Delta t \) (1 mark)

(a) The tension \( T \) in the string is equal to the weight of the hanging mass:
\( T = mg = 0.450 \text{ kg} \times 9.81 \text{ m s}^{-2} = 4.4145 \text{ N} \approx 4.41 \text{ N} \).

(b) Since \( f \propto \frac{1}{L} \), the graph of \( f \) against \( \frac{1}{L} \) should be a straight line of positive gradient that passes through the origin.

(c) (i) The wave speed on a stretched string is given by \( v = \sqrt{\frac{T}{\mu}} \).
For the first harmonic (fundamental frequency), the length of the string \( L \) is equal to half a wavelength:
\( \lambda = 2L \).
Using the wave equation:
\( v = f\lambda \implies \sqrt{\frac{T}{\mu}} = f(2L) \).
Rearranging this for \( f \) gives:
\( f = \left( \frac{1}{2}\sqrt{\frac{T}{\mu}} \right) \frac{1}{L} \).
Comparing this to the straight-line equation \( y = mx \) (where \( y = f \) and \( x = \frac{1}{L} \)), the gradient \( m \) is indeed \( \frac{1}{2}\sqrt{\frac{T}{\mu}} \).

(ii) Substituting the values:
Gradient \( = \frac{1}{2} \sqrt{\frac{4.4145}{1.25 \times 10^{-3}}} = \frac{1}{2} \sqrt{3531.6} = 0.5 \times 59.427 = 29.71 \approx 29.7 \text{ m s}^{-1} \) (or \( \text{Hz m} \)).

(d) Friction in the pulley is a likely source of systematic error. Friction opposes the downward pull of the hanging mass, meaning the actual tension \( T \) in the vibrating segment of the string is less than the calculated weight \( mg \). Since wave speed \( v \propto \sqrt{T} \), a lower actual tension results in lower wave speed, which leads to a lower fundamental frequency \( f \) for any given length \( L \), thus reducing the gradient of the graph.

(a) [1 mark]
- Correct tension value \( T = 4.41 \text{ N} \) (or \( 4.4 \text{ N} \)) (1 mark)

(b) [2 marks]
- Graph has a straight line with a positive gradient (1 mark)
- Line passes through, or extrapolates directly to, the origin (1 mark)

(c) (i) [2 marks]
- Uses wave speed formula and fundamental wavelength condition \( \lambda = 2L \) to state \( f = \frac{v}{2L} \) (1 mark)
- Correctly substitutes \( v = \sqrt{\frac{T}{\mu}} \) and rearranges to show \( f = \left( \frac{1}{2}\sqrt{\frac{T}{\mu}} \right) \frac{1}{L} \) (1 mark)

(c) (ii) [3 marks]
- Correct substitution of values into the gradient formula (1 mark)
- Correct calculation of numerical value \( 29.7 \) (accept range \( 29.6 - 30.0 \)) (1 mark)
- Correct unit: \( \text{m s}^{-1} \) or \( \text{Hz m} \) (1 mark)

(d) [2 marks]
- Identifies friction in the pulley (1 mark)
- Explains that friction reduces the effective tension in the string compared to \( mg \), which lowers the wave speed and therefore reduces the experimental frequency/gradient (1 mark)

(a) A long, thin wire produces a larger, more measurable extension for a given force, which reduces the percentage uncertainty in the extension measurement. (b) 1. Measure the diameter at several different positions along the wire and at different orientations to calculate an average, accounting for non-uniformity. 2. Ensure the micrometer is closed gently using the ratchet to avoid compressing the wire, and check for a zero error before taking measurements. (c) Area \( A = \frac{\pi d^2}{4} = \frac{\pi (3.8 \times 10^{-4} \text{ m})^2}{4} = 1.134 \times 10^{-7} \text{ m}^2 \). Percentage uncertainty in \( d \) is \( \frac{0.01}{0.38} \times 100\% = 2.63\% \). Since \( A \propto d^2 \), the percentage uncertainty in \( A \) is \( 2 \times 2.63\% = 5.26\% \). Absolute uncertainty in \( A = 5.26\% \times 1.134 \times 10^{-7} \text{ m}^2 = 0.0597 \times 10^{-7} \text{ m}^2 \approx 0.06 \times 10^{-7} \text{ m}^2 \). (d) Young modulus \( E = \frac{F L_0}{A \Delta L} = \frac{35.0 \times 2.45}{1.134 \times 10^{-7} \times 6.3 \times 10^{-3}} = 1.20 \times 10^{11} \text{ Pa} \). Percentage uncertainty in \( L_0 \) is \( \frac{0.01}{2.45} \times 100\% = 0.41\% \). Percentage uncertainty in extension is \( \frac{0.1}{6.3} \times 100\% = 1.59\% \). Total percentage uncertainty in \( E = 5.26\% + 0.41\% + 1.59\% = 7.26\% \approx 7.3\% \).

Part (a): 1 mark for identifying that a larger extension is produced for a given force. 1 mark for explaining that this reduces the percentage/fractional uncertainty in the extension measurement. Part (b): 1 mark for measuring at multiple positions/orientations and averaging. 1 mark for checking for a zero error on the micrometer or using the ratchet to avoid over-tightening. Part (c): 1 mark for correct area calculation (1.13 x 10^-7 m^2). 1 mark for doubling the percentage uncertainty of diameter to get 5.3%. 1 mark for correct absolute uncertainty (0.06 x 10^-7 m^2). Part (d): 1 mark for correct substitution to find E = 1.20 x 10^11 Pa. 1 mark for summing the percentage uncertainties of area, original length, and extension. 1 mark for correct final percentage uncertainty (accept 7.3% or 7%).

(a) Connect the chemical cell in series with an ammeter, a variable resistor (or rheostat), and a switch. Connect a voltmeter in parallel across the terminals of the chemical cell. (b) The terminal potential difference is related to current by \( V = -rI + \varepsilon \). If you plot \( V \) on the vertical y-axis and \( I \) on the horizontal x-axis, the resulting graph is a straight line with a negative gradient. The y-intercept represents the electromotive force (emf) \( \varepsilon \), and the magnitude of the gradient represents the internal resistance \( r \). (c)(i) The gradient of the graph is \( m = \frac{1.12 - 1.48}{0.65 - 0} = \frac{-0.36}{0.65} = -0.554 \text{ V A}^{-1} \). Since \( \text{gradient} = -r \), the internal resistance is \( r = 0.55 \text{ }\Omega \) (or \( 0.554 \text{ }\Omega \)). (c)(ii) Keeping the switch closed causes a continuous current to flow, which heats up the cell. A rise in temperature can change the internal resistance of the cell and run down its chemical energy, leading to systematic errors and inconsistent readings.

Part (a): 1 mark for cell, variable resistor, and ammeter in series (with a switch). 1 mark for voltmeter connected in parallel across the cell. Part (b): 1 mark for stating V is plotted on the y-axis and I on the x-axis. 1 mark for stating that the y-intercept equals the emf. 1 mark for stating that the magnitude of the gradient equals the internal resistance. Part (c)(i): 1 mark for recognizing that gradient = -r. 1 mark for substituting values to find gradient (-0.36 / 0.65). 1 mark for final value of r = 0.55 ohms (accept 0.55 to 0.56 ohms). Part (c)(ii): 1 mark for stating that continuous current causes the cell to heat up. 1 mark for explaining that temperature changes affect the internal resistance or deplete the cell quickly.

The kaon \(K^0\) has strangeness \(S = +1\) and is composed of a down quark and a strange antiquark, \(d\bar{s}\). The anti-neutral kaon, \(\bar{K}^0\), is its antiparticle, meaning its quark structure is the antiparticle equivalent of \(d\bar{s}\), which is \(\bar{d}s\) (strangeness \(S = -1\)). Since it is a meson, its baryon number is \(0\).

1 mark for the correct answer A.

The speed of the ball just before impact is \(v_1 = \sqrt{2gh}\) (directed downwards). The speed of the ball just after rebounding is \(v_2 = \sqrt{2g(0.64h)} = 0.8\sqrt{2gh}\) (directed upwards). Taking the upwards direction as positive, the change in momentum (impulse) is \(I = \Delta p = m v_2 - (- m v_1) = m (0.8\sqrt{2gh} + \sqrt{2gh}) = 1.8 m \sqrt{2gh}\).

1 mark for the correct answer D.

The terminal potential difference is given by \(V = \frac{\varepsilon R}{R+r}\). When \(R=0\), \(V=0\). As \(R\) increases, the fraction \(\frac{R}{R+r}\) increases towards \(1\), so \(V\) increases continuously. The power dissipated in the load resistor is \(P = I^2 R = \frac{\varepsilon^2 R}{(R+r)^2}\). According to the maximum power transfer theorem, \(P\) reaches a maximum when \(R = r\). Therefore, as \(R\) increases from zero, \(P\) first increases to a maximum and then decreases.

1 mark for the correct answer B.

For a string of length \(L\) fixed at both ends vibrating in its third harmonic, there are three half-wavelength loops. Thus, \(L = 3\left(\frac{\lambda}{2}\right)\), which gives \(\lambda = \frac{2L}{3}\). The distance between a node and its adjacent antinode is a quarter of a wavelength: \(d = \frac{\lambda}{4} = \frac{2L}{12} = \frac{L}{6}\).

1 mark for the correct answer B.

The extension of a wire is given by \(\Delta L = \frac{FL}{AE}\). For the second wire, the length is \(2L\) and the area is \(2A\), while the force \(F\) and material (Young modulus \(E\)) are the same. Its extension is \(\Delta L_2 = \frac{F(2L)}{(2A)E} = \frac{FL}{AE} = \Delta L\).

1 mark for the correct answer B.

Using Einstein's photoelectric equation, \(E_k = \frac{hc}{\lambda} - \Phi\), where \(\Phi\) is the work function of the metal. If the wavelength is halved to \(\frac{\lambda}{2}\), the energy of each incident photon doubles to \(\frac{2hc}{\lambda}\). The new maximum kinetic energy is \(E_k' = \frac{2hc}{\lambda} - \Phi = 2\left(E_k + \Phi\right) - \Phi = 2E_k + \Phi\). Since the work function \(\Phi\) must be greater than zero for photoemission to have occurred originally, the new maximum kinetic energy must be greater than \(2E_k\).

1 mark for the correct answer B.

The fringe spacing is given by \(w = \frac{\lambda D}{s}\). Since the distance to the screen \(D\) and the slit separation \(s\) remain constant, the fringe spacing is directly proportional to the wavelength: \(w \propto \lambda\). Therefore, the new fringe spacing is \(w_{\text{red}} = w_{\text{green}} \times \frac{\lambda_{\text{red}}}{\lambda_{\text{green}}} = 1.2\text{ mm} \times \frac{6.3 \times 10^{-7}\text{ m}}{5.4 \times 10^{-7}\text{ m}} = 1.2 \times 1.167 = 1.4\text{ mm}\).

1 mark for the correct answer C.

Rearranging the formula for \(g\) gives \(g = 4\pi^2 \frac{L}{T^2}\). The percentage uncertainty in a calculated quantity is found by summing the fractional (or percentage) uncertainties of the independent variables, multiplied by their powers. Thus, the percentage uncertainty in \(g\) is: \(\frac{\Delta g}{g} = \frac{\Delta L}{L} + 2\frac{\Delta T}{T} = 2\% + 2(3\%) = 8\%\).

1 mark for the correct answer D.

In decay D, charge is conserved (\(+1 \rightarrow +1 + 0\)), lepton number is conserved (\(0 \rightarrow -1 + 1 = 0\)), and baryon number is conserved (\(0 \rightarrow 0 + 0\)). Strangeness changes from \(+1\) to \(0\), which is permitted in a weak interaction decay. Decay A violates lepton number (\(0 \rightarrow -1 - 1 = -2\)). Decay B violates baryon number (\(0 \rightarrow 1 + 0 = 1\)). Decay C violates charge conservation (\(+1 \rightarrow +1 - 1 + 0 = 0\)).

1 mark for identifying the correct decay equation that satisfies all conservation laws (charge, lepton number, baryon number, and strangeness change of 1 in weak interactions).

The maximum height \(H\) of a projectile is given by \(H = \frac{u^2 \sin^2(30^\circ)}{2g}\). Since \(\sin(30^\circ) = 0.5\), we have \(H = \frac{0.25 u^2}{2g}\). For the new angle \(\theta\), the maximum height is \(2H = \frac{u^2 \sin^2\theta}{2g}\). Substituting the first expression into the second: \(2 \left(\frac{0.25 u^2}{2g}\right) = \frac{u^2 \sin^2\theta}{2g}\), which simplifies to \(\sin^2\theta = 0.5\). Therefore, \(\sin\theta = \sqrt{0.5}\), giving \(\theta = 45^\circ\).

1 mark for the correct calculation showing that the maximum height is proportional to \(\sin^2\theta\), leading to \(\sin\theta = \sqrt{2}\sin(30^\circ)\) and hence \(\theta = 45^\circ\).

Density is given by \(\rho = \frac{m}{V} = \frac{4m}{\pi d^2 L}\). When combining uncertainties for quantities multiplied or divided, we add the percentage uncertainties, multiplying by the power of each term. Therefore, the percentage uncertainty in density is: \(\frac{\Delta \rho}{\rho} \times 100 = \frac{\Delta m}{m} \times 100 + 2 \left(\frac{\Delta d}{d} \times 100\right) + \frac{\Delta L}{L} \times 100 = 1.5\% + 2(2.0\%) + 1.0\% = 6.5\%\).

1 mark for correctly applying the rules for combining uncertainties (adding the percentage uncertainties, with the diameter uncertainty doubled because it is squared), resulting in \(6.5\%\).

Since both wires are copper and have the same mass, they must have the same volume, \(V = A_X L_X = A_Y L_Y\). We are given \(L_Y = 3L_X\), so \(A_X L_X = A_Y (3L_X) \implies A_Y = \frac{1}{3}A_X\). Resistance is given by \(R = \rho \frac{L}{A}\). Thus, the resistance of Y is \(R_Y = \rho \frac{L_Y}{A_Y} = \rho \frac{3L_X}{\frac{1}{3}A_X} = 9 \left(\rho \frac{L_X}{A_X}\right) = 9 R_X\). The ratio \(\frac{R_Y}{R_X}\) is therefore \(9\).

1 mark for identifying that constant volume requires the cross-sectional area of Y to be one-third of X, and using the resistance formula to determine that the resistance is 9 times greater.

For a closed pipe of length \(L\), the fundamental frequency is \(f_{c,1} = \frac{v}{4L} = 220\text{ Hz}\). For an open pipe of the same length \(L\), the fundamental frequency is \(f_{o,1} = \frac{v}{2L} = 2 f_{c,1} = 440\text{ Hz}\). The first overtone of the open pipe is its second harmonic, which has a frequency of \(f_{o,2} = 2 f_{o,1} = 2 \times 440\text{ Hz} = 880\text{ Hz}\).

1 mark for calculating the fundamental frequency of the open pipe as 440 Hz and multiplying by 2 to find the first overtone (second harmonic) as 880 Hz.

The elastic strain energy stored in a wire is \(W = \frac{1}{2} F \Delta L\). Young modulus is defined as \(E = \frac{\text{stress}}{\text{strain}} = \frac{F / A}{\Delta L / L} = \frac{F L}{A \Delta L}\). Rearranging this for the force gives \(F = \frac{E A \Delta L}{L}\). Substituting this expression for \(F\) into the energy equation yields \(W = \frac{1}{2} \left(\frac{E A \Delta L}{L}\right) \Delta L = \frac{E A (\Delta L)^2}{2 L}\).

1 mark for combining the elastic strain energy equation \(W = \frac{1}{2} F \Delta L\) with the expression for force from Young modulus to derive the correct formula.

Using the diffraction grating formula \(d \sin\theta = n\lambda\), we have \(d \sin(60^\circ) = 3\lambda\), which gives \(\frac{d}{\lambda} = \frac{3}{\sin(60^\circ)} = \frac{3}{0.866} \approx 3.46\). The maximum possible order \(n_{\text{max}}\) occurs when \(\sin\theta \le 1\), meaning \(n \le \frac{d}{\lambda} \approx 3.46\). Since the order must be an integer, the maximum observable order is \(n = 3\). The total number of observable maxima is \(2n_{\text{max}} + 1 = 2(3) + 1 = 7\) (the zeroth order, and orders 1, 2, and 3 on both sides).

1 mark for finding the maximum possible order is 3, and then correctly calculating the total number of maxima as \(2 \times 3 + 1 = 7\).

According to Einstein's photoelectric equation, \(h f = \phi + E_k\), so \(h f - \phi = E_k\). When the frequency is doubled, the new maximum kinetic energy \(E_k'\) is given by \(E_k' = h(2f) - \phi = 2(h f) - \phi\). Substituting \(h f = E_k + \phi\) into this equation gives \(E_k' = 2(E_k + \phi) - \phi = 2E_k + 2\phi - \phi = 2E_k + \phi\).

1 mark for using Einstein's equation to relate the old and new frequencies, showing that doubling the frequency results in a new maximum kinetic energy of \(2E_k + \phi\).

We use the potential divider equation or the equation for terminal potential difference:
\(E = V + Ir = V + \left(\frac{V}{R}\right)r = V\left(1 + \frac{r}{R}\right)\)

Setting up the equations for both cases:
1) \(E = V_1\left(1 + \frac{r}{R_1}\right)\)
2) \(E = V_2\left(1 + \frac{r}{R_2}\right)\)

Equating the two expressions for \(E\):
\(V_1\left(1 + \frac{r}{R_1}\right) = V_2\left(1 + \frac{r}{R_2}\right)\)

Expanding both sides:
\(V_1 + \frac{V_1 r}{R_1} = V_2 + \frac{V_2 r}{R_2}\)

Rearranging to group terms with \(r\):
\(r\left(\frac{V_1}{R_1} - \frac{V_2}{R_2}\right) = V_2 - V_1\)

\(r\left(\frac{V_1 R_2 - V_2 R_1}{R_1 R_2}\right) = V_2 - V_1\)

\(r = \frac{(V_2 - V_1)R_1 R_2}{V_1 R_2 - V_2 R_1}\)

Multiplying both the numerator and the denominator by \(-1\) gives:
\(r = \frac{(V_1 - V_2)R_1 R_2}{V_2 R_1 - V_1 R_2}\)

1 mark for the correct option.
- Accept A.
- Reject B, C, D.

Let's check the conservation laws:
1) **Charge (\(Q\)) conservation**:
- Initial: \(Q(\Sigma^+) = +1\)
- Final: \(Q(\pi^+) + Q(\pi^0) + Q_X = +1 + 0 + Q_X = 1 + Q_X\)
- Since \(+1 = 1 + Q_X\), the charge of \(X\) must be \(0\).

2) **Baryon number (\(B\)) conservation**:
- Initial: \(B(\Sigma^+) = +1\) (since it is a baryon)
- Final: \(B(\pi^+) + B(\pi^0) + B_X = 0 + 0 + B_X = B_X\) (since pions are mesons)
- Since \(+1 = B_X\), the baryon number of \(X\) must be \(+1\).

3) **Lepton number (\(L\)) conservation**:
- Initial: \(L(\Sigma^+) = 0\)
- Final: \(L(\pi^+) + L(\pi^0) + L_X = 0 + 0 + L_X = L_X\)
- Since \(0 = L_X\), the lepton number of \(X\) must be \(0\).

Particle \(X\) must be a neutral baryon with zero lepton number. Among the choices:
- A: Neutron (Charge = 0, Baryon number = +1, Lepton number = 0) - Correct.
- B: Proton (Charge = +1) - Incorrect.
- C: Electron neutrino (Baryon number = 0) - Incorrect.
- D: Anti-neutron (Baryon number = -1) - Incorrect.

1 mark for the correct option.
- Accept A.
- Reject B, C, D.

The Young Modulus \(E\) of a material is given by:
\(E = \frac{\text{Stress}}{\text{Strain}} = \frac{F/A}{\Delta L / L} = \frac{F L}{A \Delta L}\)

Rearranging for extension:
\(\Delta L = \frac{F L}{A E}\)

For the first wire: \(\Delta L_1 = \frac{F L}{A E} = \Delta L\).

For the second wire:
- Length is \(2L\).
- The diameter is halved, so the cross-sectional area \(A_2\) is divided by 4: \(A_2 = \frac{A}{4}\).
- The material is the same, so the Young Modulus \(E\) is unchanged.
- The stretching force is \(F\).

Substitute these into the extension equation:
\(\Delta L_2 = \frac{F (2L)}{\left(\frac{A}{4}\right) E} = 8 \left(\frac{F L}{A E}\right) = 8 \Delta L\).

1 mark for the correct option.
- Accept C.
- Reject A, B, D.

The frequency of the \(n\)-th harmonic on a stretched string fixed at both ends is given by:
\(f_n = \frac{n v}{2 L}\)

where the speed of the wave is \(v = \sqrt{\frac{T}{\mu}}\).

This gives:
\(f_n = \frac{n}{2 L} \sqrt{\frac{T}{\mu}}\)

For the third harmonic:
\(f_3 = \frac{3}{2 L} \sqrt{\frac{T}{\mu}}\)

If the tension is doubled to \(2T\), while \(n = 3\), \(L\), and \(\mu\) remain constant, the new frequency is:
\(f_3' = \frac{3}{2 L} \sqrt{\frac{2T}{\mu}} = \sqrt{2} \times \left( \frac{3}{2 L} \sqrt{\frac{T}{\mu}} \right) = \sqrt{2} f_3 \approx 1.41 f_3\).

1 mark for the correct option.
- Accept C.
- Reject A, B, D.

First, calculate the initial kinetic energy (\(E_{ki}\)):
\(E_{ki} = \frac{1}{2} m_1 v_1^2 = \frac{1}{2} (2.0) (4.0)^2 = 16.0\text{ J}\)

Next, use the conservation of linear momentum to find the common final velocity (\(v_f\)):
\(m_1 v_1 = (m_1 + m_2) v_f\)
\((2.0)(4.0) = (2.0 + 3.0) v_f\)
\(8.0 = 5.0 v_f \implies v_f = 1.6\text{ m s}^{-1}\)

Now, calculate the final kinetic energy (\(E_{kf}\)):
\(E_{kf} = \frac{1}{2} (m_1 + m_2) v_f^2 = \frac{1}{2} (5.0) (1.6)^2 = 2.5 \times 2.56 = 6.4\text{ J}\)

The loss in kinetic energy is:
\(\Delta E_k = E_{ki} - E_{kf} = 16.0\text{ J} - 6.4\text{ J} = 9.6\text{ J}\).

1 mark for the correct option.
- Accept B.
- Reject A, C, D.

Using Einstein's photoelectric equation:
\(hf = \Phi + E_{k, \text{max}}\)

Since \(f = \frac{c}{\lambda}\), we can write:
1) For wavelength \(\lambda\):
\(\frac{hc}{\lambda} = \Phi + E_k\)

2) For wavelength \(\frac{\lambda}{2}\):
\(\frac{hc}{\lambda / 2} = \Phi + 3 E_k \implies 2 \left(\frac{hc}{\lambda}\right) = \Phi + 3 E_k\)

Substitute the expression for \(\frac{hc}{\lambda}\) from the first equation into the second equation:
\(2 (\Phi + E_k) = \Phi + 3 E_k\)
\(2 \Phi + 2 E_k = \Phi + 3 E_k\)
\(\Phi = E_k\) (or \(1.0 E_k\)).

1 mark for the correct option.
- Accept B.
- Reject A, C, D.

The formula for the fringe width \(w\) in a double-slit experiment is:
\(w = \frac{\lambda D}{d}\)

In the second setup:
- New wavelength \(\lambda' = 2\lambda\)
- New screen distance \(D' = 2D\)
- New slit separation \(d' = \frac{d}{2}\)

Calculating the new fringe width \(w'\):
\(w' = \frac{\lambda' D'}{d'} = \frac{(2\lambda)(2D)}{\frac{d}{2}} = 8 \left( \frac{\lambda D}{d} \right) = 8 w\).

1 mark for the correct option.
- Accept C.
- Reject A, B, D.

First, find the percentage uncertainties for each measured quantity:

1) Percentage uncertainty in \(R\):
\(\%\Delta R = \frac{0.3}{15.0} \times 100\% = 2.0\%\)

2) Percentage uncertainty in \(d\):
\(\%\Delta d = \frac{0.01}{0.50} \times 100\% = 2.0\%\)

3) Percentage uncertainty in \(L\):
\(\%\Delta L = \frac{0.02}{2.00} \times 100\% = 1.0\%\)

Using the rules for combining uncertainties (where quantities are multiplied/divided, and powers multiply the percentage uncertainty):
\(\%\Delta \rho = \%\Delta R + 2(\%\Delta d) + \%\Delta L\)
\(\%\Delta \rho = 2.0\% + 2(2.0\%) + 1.0\% = 2.0\% + 4.0\% + 1.0\% = 7.0\%\).

1 mark for the correct option.
- Accept C.
- Reject A, B, D.

Strangeness is conserved in strong and electromagnetic interactions, but can change by \(\pm 1\) in weak interactions. Let us examine the options: In A: \(\pi^- + \text{p} \rightarrow \text{K}^+ + \Sigma^-\) has initial strangeness \(0 + 0 = 0\) and final strangeness \((+1) + (-1) = 0\) (conserved). In B: \(\text{K}^- \rightarrow \pi^- + \pi^0\) has initial strangeness \(-1\) and final strangeness \(0 + 0 = 0\) (not conserved, this is a weak decay). In C: \(\gamma + \text{p} \rightarrow \pi^0 + \text{p}\) has initial strangeness \(0\) and final strangeness \(0\) (conserved). In D: \(\text{p} + \overline{\text{p}} \rightarrow \pi^+ + \pi^-\) has initial strangeness \(0\) and final strangeness \(0\) (conserved).

1 mark for selecting the correct option B.

Resolving forces parallel to the slope up the incline: \(\Sigma F = F - mg\sin\theta - f_{\text{k}} = ma\). The normal reaction force perpendicular to the slope is \(R = mg\cos\theta\). Thus, the frictional force is \(f_{\text{k}} = \mu R = \mu mg\cos\theta\). Substituting this into the equation of motion gives: \(F - mg\sin\theta - \mu mg\cos\theta = ma\). Dividing both sides by \(m\) gives: \(a = \frac{F}{m} - g(\sin\theta + \mu\cos\theta)\).

1 mark for selecting the correct option A.

The volume of the wire is \(V = AL\). Since volume is constant, if the length increases by \(10\%\) to \(L' = 1.1L\), the new cross-sectional area \(A'\) must satisfy \(A'L' = AL\), meaning \(A' = \frac{AL}{1.1L} = \frac{A}{1.1}\). The original resistance is \(R = \rho \frac{L}{A}\). The new resistance is \(R' = \rho \frac{L'}{A'} = \rho \frac{1.1L}{A/1.1} = 1.21 \left(\rho \frac{L}{A}\right) = 1.21R\).

1 mark for selecting the correct option C.

In the third harmonic, the stationary wave has three loops, which correspond to three half-wavelengths. Therefore, \(L = \frac{3\lambda}{2}\), which gives \(\lambda = \frac{2L}{3}\). Using the wave speed equation \(v = f\lambda\), we get \(v = f \left(\frac{2L}{3}\right) = \frac{2fL}{3}\).

1 mark for selecting the correct option B.

The elastic strain energy stored is given by \(U = \frac{1}{2} F \Delta L\), where \(\Delta L\) is the extension of the wire. From the definition of Young modulus, \(E = \frac{\text{Stress}}{\text{Strain}} = \frac{F/A}{\Delta L/L} = \frac{FL}{A\Delta L}\). Rearranging for \(\Delta L\) gives \(\Delta L = \frac{FL}{AE}\). Substituting this back into the energy equation gives \(U = \frac{1}{2} F \left(\frac{FL}{AE}\right) = \frac{F^2 L}{2AE}\).

1 mark for selecting the correct option B.

According to Einstein's photoelectric equation, the initial maximum kinetic energy is \(E_{\text{k,max1}} = hf - \phi\). When the frequency is doubled, the new maximum kinetic energy is \(E_{\text{k,max2}} = h(2f) - \phi = 2hf - \phi\). Expressing this in terms of the initial kinetic energy: \(E_{\text{k,max2}} = 2(hf - \phi) + \phi = 2E_{\text{k,max1}} + \phi\). Since the work function \(\phi\) is positive, \(E_{\text{k,max2}} > 2E_{\text{k,max1}}\). Therefore, the maximum kinetic energy increases to more than double its original value.

1 mark for selecting the correct option B.