2023 Cambridge IAS-Level Physics (9702) 模擬試題連答案詳解

Using the equation of motion \(s = ut + \frac{1}{2}at^2\) with upwards chosen as the positive direction: the initial velocity \(u = +15\text{ m s}^{-1}\), acceleration \(a = -9.81\text{ m s}^{-2}\), and time of flight \(t = 4.0\text{ s}\). Substituting these values gives: \(s = (15)(4.0) + \frac{1}{2}(-9.81)(4.0)^2 = 60 - 78.48 = -18.48\text{ m}\). The negative sign indicates the final position is below the starting point, so the height of the cliff is \(18\text{ m}\) (to two significant figures).

1 mark for the correct calculation of displacement using the kinematic formula, yielding a magnitude of 18 m (option A).

By conservation of momentum, total momentum before collision equals total momentum after: \(3000 \times 4.0 = (3000 + 1000) \times v\), which gives a common velocity \(v = 3.0\text{ m s}^{-1}\). The initial kinetic energy is \(E_{ki} = \frac{1}{2} \times 3000 \times 4.0^2 = 24\,000\text{ J} = 24\text{ kJ}\). The final kinetic energy is \(E_{kf} = \frac{1}{2} \times (3000 + 1000) \times 3.0^2 = 18\,000\text{ J} = 18\text{ kJ}\). The loss in kinetic energy is \(24\text{ kJ} - 18\text{ kJ} = 6.0\text{ kJ}\).

1 mark for calculating the common velocity and using it to find the difference between the initial and final kinetic energies (option A).

By conservation of energy, the loss in gravitational potential energy as the bead moves from height \(h\) to the top of the loop of height \(2R = 2 \times 0.20 = 0.40\text{ m}\) is equal to the gain in kinetic energy: \(mg(h - 2R) = \frac{1}{2}mv^2\). Rearranging gives \(g(h - 2R) = \frac{1}{2}v^2\), so \(9.81(h - 0.40) = \frac{1}{2}(3.0)^2 = 4.5\). This gives \(h - 0.40 = 0.459\text{ m}\), which results in \(h = 0.86\text{ m}\).

1 mark for setting up the conservation of energy equation and correctly solving for the initial height h (option C).

The actual weight of the block is \(W = mg = 3.50 \times 9.81 = 34.335\text{ N}\). The tension reading is \(T = 24.525\text{ N}\). The upthrust is \(U = W - T = 34.335 - 24.525 = 9.81\text{ N}\). Since Upthrust \(U = \rho_{\text{water}} V g\), we have \(9.81 = 1000 \times V \times 9.81\), which gives the volume of the block \(V = 1.00 \times 10^{-3}\text{ m}^3\). The density of the block is \(\rho = \frac{m}{V} = \frac{3.50}{1.00 \times 10^{-3}} = 3.50 \times 10^3\text{ kg m}^{-3}\).

1 mark for finding the upthrust, calculating the volume of the block, and then finding its density (option C).

When unpolarised light of intensity \(I_0\) passes through the first polariser, its intensity becomes \(I_1 = 0.5 I_0\). Using Malus's law for the second polariser, the final intensity is \(I_2 = I_1 \cos^2(\theta) = (0.5 I_0) \cos^2(60.0^\circ)\). Since \(\cos(60.0^\circ) = 0.5\), we have \(\cos^2(60.0^\circ) = 0.25\). Therefore, \(I_2 = 0.5 I_0 \times 0.25 = 0.125 I_0\).

1 mark for recognizing that the intensity of unpolarised light is halved by the first filter, and applying Malus's law correctly to obtain 0.125 of the initial intensity (option A).

The grating spacing is \(d = \frac{1 \times 10^{-3}\text{ m}}{400} = 2.5 \times 10^{-6}\text{ m}\). For maximum order \(n\), we set \(\theta = 90^\circ\) in the grating equation \(d \sin\theta = n\lambda\), giving \(n \le \frac{d}{\lambda} = \frac{2.5 \times 10^{-6}}{5.4 \times 10^{-7}} \approx 4.63\). Thus, the maximum order of diffraction is \(n = 4\). The total number of bright fringes observed is \(2n + 1 = 2(4) + 1 = 9\), corresponding to the zero-order fringe and four fringes on either side.

1 mark for calculating the grating spacing d, finding the maximum order n = 4, and calculating the total number of maxima as 2n + 1 = 9 (option C).

First, find the internal resistance \(r\) using \(E = I(R + r)\): \(6.0 = 2.0(2.0 + r)\), which simplifies to \(3.0 = 2.0 + r\), so \(r = 1.0\ \Omega\). When the variable resistor is changed to \(R = 5.0\ \Omega\), the new current is \(I' = \frac{E}{R + r} = \frac{6.0}{5.0 + 1.0} = 1.0\text{ A}\). The power dissipated in the resistor is \(P = (I')^2 R = (1.0)^2 \times 5.0 = 5.0\text{ W}\).

1 mark for calculating the internal resistance of the battery, finding the new current, and calculating the final power output (option B).

Bismuth-212 has 83 protons and \(212 - 83 = 129\) neutrons. In \(\beta^-\)\ decay, a neutron decays into a proton, an electron, and an electron antineutrino. This increases the proton number by 1 and decreases the neutron number by 1. Therefore, the daughter nucleus (Polonium-212) has 84 protons and 128 neutrons. Since each proton consists of \(2u + 1d\) quarks and each neutron consists of \(1u + 2d\) quarks, the total number of up quarks is \(84 \times 2 + 128 \times 1 = 296\), and the total number of down quarks is \(84 \times 1 + 128 \times 2 = 340\).

1 mark for identifying the daughter nucleus as Polonium-212 (Z=84, N=128) and correctly calculating the total up and down quark counts (option B).

Let \(t\) be the time from launch to collision. The vertical displacement of the projectile is given by \(y_1 = h - \frac{1}{2}gt^2\). The vertical displacement of the target is given by \(y_2 = vt - \frac{1}{2}gt^2\). At the moment of collision, their heights are equal: \(y_1 = y_2\). Therefore, \(h - \frac{1}{2}gt^2 = vt - \frac{1}{2}gt^2\), which simplifies to \(t = \frac{h}{v}\). During this time, the projectile travels horizontally at a constant speed \(u\). The horizontal distance covered is \(d = u t = \frac{uh}{v}\).

1 mark for the correct option B. Awarded for correctly setting up the vertical equations of motion to find the time of flight \(t = \frac{h}{v}\) and multiplying by \(u\) to find the horizontal distance.

The potential difference across the fixed resistor is given by \(V_{\text{out}} = E \times \frac{R}{R + R_T}\), so the ratio is \(\frac{V_{\text{out}}}{E} = \frac{R}{R + R_T}\). Initially, with \(R_T = \frac{1}{3}R\), the ratio is \(\frac{R}{R + \frac{1}{3}R} = \frac{1}{4/3} = 0.75\). Finally, with \(R_T = 3R\), the ratio is \(\frac{R}{R + 3R} = \frac{1}{4} = 0.25\). The decrease in the ratio is \(0.75 - 0.25 = 0.50\).

1 mark for the correct option C. Awarded for calculating the initial ratio (0.75) and final ratio (0.25) and finding the difference.

Let the left-hand end of the beam be at position \(x = 0\text{ m}\). The beam is uniform, so its weight of \(40\text{ N}\) acts at its midpoint, \(x = 0.50\text{ m}\). The second wire is at position \(x = 1.00 - 0.20 = 0.80\text{ m}\), with tension \(T_2\). For vertical equilibrium: \(T_1 + T_2 = W_{\text{beam}} + W_{\text{block}} \implies 20 + T_2 = 40 + 20 \implies T_2 = 40\text{ N}\). Taking moments about \(x = 0\): clockwise moments = anticlockwise moments. \((40 \times 0.50) + (20 \times d) = T_2 \times 0.80 \implies 20 + 20d = 40 \times 0.80 \implies 20 + 20d = 32 \implies 20d = 12 \implies d = 0.60\text{ m}\).

1 mark for the correct option D. Awarded for applying vertical equilibrium to find the second tension (40 N) and the principle of moments about the left end to solve for the block position \(d = 0.60\text{ m}\).

By conservation of energy, the gravitational potential energy of the car at the top of the ramp equals the work done against the constant resistive force on the flat section. For the first experiment: \(m g H = F d \implies d = \frac{mgH}{F}\). For the second experiment, the total mass is \(2m\), so the initial potential energy is \(2mgH\). Since the resistive force \(F\) remains the same, the stopping distance \(d'\) is given by: \(2mgH = F d' \implies d' = \frac{2mgH}{F} = 2d\).

1 mark for the correct option D. Awarded for realizing that the kinetic energy at the bottom doubles when mass is doubled, and since the stopping force is constant, the stopping distance must also double.

Taking the direction to the right as positive: Initial momentum of the system: \(p_{\text{initial}} = (2m)(v) + (3m)(0) = 2mv\). After the collision, glider A moves to the left, so its velocity is \(-0.20v\). Let \(v_B\) be the velocity of glider B. Final momentum of the system: \(p_{\text{final}} = (2m)(-0.20v) + (3m)(v_B) = -0.40mv + 3mv_B\). Using conservation of momentum: \(2mv = -0.40mv + 3mv_B \implies 2.40mv = 3mv_B \implies v_B = 0.80v\).

1 mark for the correct option C. Awarded for using conservation of linear momentum with the correct signs for velocities to obtain \(v_B = 0.80v\).

The original wire has spring constant \(k = \frac{EA}{L}\). For a force \(F\), the extension is \(x = \frac{F}{k}\) and the work done is \(W = \frac{1}{2}Fx\). A half-wire has length \(L/2\), so its spring constant is \(k' = \frac{EA}{L/2} = 2k\). Under the same force \(F\), the new extension is \(x' = \frac{F}{2k} = \frac{x}{2}\). The new work done is \(W' = \frac{1}{2}Fx' = \frac{1}{2}F\left(\frac{x}{2}\right) = \frac{W}{2}\).

1 mark for the correct option A. Awarded for identifying that halving the length doubles the stiffness, which halves the extension for the same force, thus halving the work done.

When unpolarised light of intensity \(I_0\) passes through the first polariser, its intensity is reduced by half, so the transmitted intensity is \(I_1 = \frac{1}{2}I_0\). When this plane-polarised light passes through the second filter (the analyser), we apply Malus's law: \(I_2 = I_1 \cos^2(\theta) = \left(\frac{1}{2}I_0\right) \cos^2(60^\circ)\). Since \(\cos(60^\circ) = 0.5\), we have \(\cos^2(60^\circ) = 0.25\). Therefore, \(I_2 = 0.5 I_0 \times 0.25 = 0.125 I_0\).

1 mark for the correct option A. Awarded for calculating the intensity after the first filter (0.50 \(I_0\)) and applying Malus's law with \(\cos^2(60^\circ) = 0.25\) to get 0.125 \(I_0\).

A proton has a quark composition of \(uud\) (two up quarks, one down quark). A neutron has a quark composition of \(udd\) (one up quark, two down quarks). In a \(\beta^+\)) decay, a proton turns into a neutron: \(uud \rightarrow udd\). This means one up quark changes into a down quark. Consequently, the total number of up quarks in the nucleus decreases by 1, and the total number of down quarks in the nucleus increases by 1.

1 mark for the correct option A. Awarded for identifying the quark compositions of proton (uud) and neutron (udd) and correctly determining that up quarks decrease by 1 while down quarks increase by 1.

First, calculate the equivalent resistance \( R_p \) of the parallel combination of the variable resistor \( R = 6.0\ \Omega \) and the fixed resistor \( 12.0\ \Omega \):

\( R_p = \frac{R \times 12.0}{R + 12.0} = \frac{6.0 \times 12.0}{6.0 + 12.0} = \frac{72.0}{18.0} = 4.0\ \Omega \)

Next, calculate the total resistance \( R_{\text{total}} \) of the circuit:

\( R_{\text{total}} = 4.0\ \Omega + R_p = 4.0\ \Omega + 4.0\ \Omega = 8.0\ \Omega \)

Using the potential divider formula, the potential difference \( V_p \) across the parallel combination is:

\( V_p = \text{e.m.f.} \times \frac{R_p}{R_{\text{total}}} = 12\text{ V} \times \frac{4.0\ \Omega}{8.0\ \Omega} = 6.0\text{ V} \)

1 mark for the correct calculation of the parallel branch resistance, total circuit resistance, and potential difference, leading to \( 6.0\text{ V} \) (Option C).

Let the left end \( A \) be at \( x = 0 \). The plank extends from \( x = 0 \) to \( x = 4.0\text{ m} \).
- Rope 1 is at \( x = 0 \) with tension \( T_A = 100\text{ N} \).
- Rope 2 is at \( x = 3.0\text{ m} \) with tension \( T_2 \).
- The uniform plank's weight of \( 150\text{ N} \) acts at its center of gravity, \( x = 2.0\text{ m} \).
- The load of \( 300\text{ N} \) is placed at position \( x \).

For vertical equilibrium:
\( T_A + T_2 = 150\text{ N} + 300\text{ N} = 450\text{ N} \)
\( 100\text{ N} + T_2 = 450\text{ N} \implies T_2 = 350\text{ N} \)

Taking moments about the left end \( A \) (\( x = 0 \)):
\( \text{Clockwise moments} = \text{Counter-clockwise moments} \)
\( (150\text{ N} \times 2.0\text{ m}) + (300\text{ N} \times x) = T_2 \times 3.0\text{ m} \)
\( 300 + 300x = 350 \times 3.0 \)
\( 300 + 300x = 1050 \)
\( 300x = 750 \)
\( x = 2.5\text{ m} \)

1 mark for using rotational and translational equilibrium conditions correctly to solve for the unknown distance, obtaining \( 2.5\text{ m} \) (Option C).

Using the equation of motion for vertical displacement \( s \) with upwards chosen as the positive direction:

\( s = ut + \frac{1}{2}at^2 \)

where:
- \( u = +15.0\text{ m s}^{-1} \)
- \( t = 5.00\text{ s} \)
- \( a = -9.81\text{ m s}^{-2} \)

Substituting these values:

\( s = (15.0 \times 5.00) + \frac{1}{2}(-9.81)(5.00)^2 \)
\( s = 75.0 - 122.625 = -47.625\text{ m} \)

The negative sign indicates that the final position is below the starting point. Thus, the height of the cliff is \( 47.6\text{ m} \).

1 mark for correctly applying the kinematic equation with consistent signs to find the displacement, yielding \( 47.6\text{ m} \) (Option A).

Let rightwards be the positive direction.
Initial states:
- Glider 1: \( m_1 = 0.20\text{ kg} \), \( u_1 = +1.5\text{ m s}^{-1} \)
- Glider 2: \( m_2 = 0.30\text{ kg} \), \( u_2 = -1.0\text{ m s}^{-1} \)

By conservation of momentum:
\( m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2 \)
\( 0.20(1.5) + 0.30(-1.0) = 0.20 v_1 + 0.30 v_2 \)
\( 0 = 0.20 v_1 + 0.30 v_2 \implies v_1 = -1.5 v_2 \)

For a perfectly elastic collision, the relative velocity of approach equals the relative velocity of separation:
\( u_1 - u_2 = v_2 - v_1 \)
\( 1.5 - (-1.0) = v_2 - v_1 \implies 2.5 = v_2 - v_1 \)

Substitute \( v_1 = -1.5 v_2 \) into the relative velocity equation:
\( 2.5 = v_2 - (-1.5 v_2) = 2.5 v_2 \implies v_2 = +1.0\text{ m s}^{-1} \) (to the right).

Then:
\( v_1 = -1.5(1.0) = -1.5\text{ m s}^{-1} \) (to the left).

Thus, the \( 0.20\text{ kg} \) glider moves at \( 1.5\text{ m s}^{-1} \) to the left and the \( 0.30\text{ kg} \) glider moves at \( 1.0\text{ m s}^{-1} \) to the right.

1 mark for using the conservation of momentum and the relative velocity equation for an elastic collision to find the correct final velocities (Option A).

Let \( m \) be the mass of the car.
Initial potential energy at height \( H \) is \( E_p = m g H \).
Since \( 10\% \) of this energy is lost, the remaining mechanical energy is:
\( E_{\text{mech}} = 0.90 m g H \)

At the second hill, the mechanical energy consists of potential energy and kinetic energy:
\( E_{\text{mech}} = m g h + \frac{1}{2} m v^2 \)

Equating the two expressions and canceling \( m \):
\( 0.90 g H = g h + \frac{1}{2} v^2 \)

Substitute the given values:
- \( h = 18\text{ m} \)
- \( v = 12\text{ m s}^{-1} \)
- \( g = 9.81\text{ m s}^{-2} \)

\( 0.90(9.81)H = (9.81 \times 18) + 0.5(12)^2 \)
\( 8.829 H = 176.58 + 72 \)
\( 8.829 H = 248.58 \)
\( H = \frac{248.58}{8.829} \approx 28.2\text{ m} \)

To two significant figures, this is \( 28\text{ m} \).

1 mark for equating the fraction of initial potential energy to the sum of final potential and kinetic energy, and correctly solving for \( H \), yielding \( 28\text{ m} \) (Option C).

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

where:
- \( \lambda = 633 \times 10^{-9}\text{ m} \)
- \( n = 3 \)
- \( \theta = 38.0^\circ \)

First, solve for the slit separation \( d \):
\( d = \frac{n \lambda}{\sin\theta} = \frac{3 \times 633 \times 10^{-9}}{\sin(38.0^\circ)} = \frac{1.899 \times 10^{-6}}{0.61566} \approx 3.085 \times 10^{-6}\text{ m} \)

Now, calculate the number of lines per millimetre \( N_{\text{mm}} \):
\( N_{\text{mm}} = \frac{10^{-3}\text{ m}}{d} = \frac{10^{-3}}{3.085 \times 10^{-6}} \approx 324\text{ mm}^{-1} \)

1 mark for using the diffraction grating equation correctly to calculate the slit separation \( d \) and converting it to the number of lines per millimetre, yielding \( 324\text{ mm}^{-1} \) (Option B).

A meson is composed of a quark and an antiquark. Let's analyze the properties of the positive kaon \( \text{K}^+ \):
1. It has a strangeness of \( +1 \). Since a strange quark \( s \) has a strangeness of \( -1 \), the anti-strange quark \( \bar{s} \) must be present because its strangeness is \( +1 \). The charge of \( \bar{s} \) is \( +1/3 e \).
2. The total charge of \( \text{K}^+ \) is \( +1e \). To obtain this, we need another quark \( q \) such that \( Q_q + Q_{\bar{s}} = +1e \implies Q_q + 1/3 e = +1e \implies Q_q = +2/3 e \).
3. The quark with charge \( +2/3 e \) is the up quark \( u \).

Therefore, the quark composition of \( \text{K}^+ \) is \( u\bar{s} \).

1 mark for identifying that a strangeness of \( +1 \) implies an anti-strange quark \( \bar{s} \) and using charge conservation to find that the other quark is an up quark \( u \) (Option B).

When unpolarised light of intensity \( I_0 \) passes through the first polarizing filter, it becomes plane-polarized, and its intensity is halved:
\( I_1 = \frac{1}{2} I_0 \)

When this plane-polarized light of intensity \( I_1 \) passes through the second filter whose transmission axis is at an angle \( \theta = 60^\circ \), the transmitted intensity \( I_2 \) is given by Malus's law:
\( I_2 = I_1 \cos^2(60^\circ) \)

Since \( \cos(60^\circ) = 0.5 \):
\( I_2 = I_1 \times (0.5)^2 = 0.25 I_1 \)

Substitute \( I_1 = \frac{1}{2} I_0 \) into the equation:
\( I_2 = 0.25 \times \left(\frac{1}{2} I_0\right) = \frac{1}{8} I_0 \)

Reflecting this, the ratio \( \frac{I_2}{I_0} \) is \( \frac{1}{8} \).

1 mark for remembering that the first polarizer halves the intensity of unpolarised light and correctly applying Malus's law with \( \theta = 60^\circ \), giving a ratio of \( 1/8 \) (Option A).

The equation can be rearranged to make the thermal conductivity \(k\) the subject:

\(k = \frac{P x}{A \Delta T}\)

Now, we determine the SI base units for each quantity:
- Power \(P\) has the unit of watt (\(\text{W}\)). Since \(\text{W} = \text{J}\ \text{s}^{-1}\) and \(\text{J} = \text{kg}\ \text{m}^2\ \text{s}^{-2}\), the base units of \(P\) are \(\text{kg}\ \text{m}^2\ \text{s}^{-3}\).
- Thickness \(x\) has the unit of metre (\(\text{m}\)).
- Area \(A\) has the unit of square metre (\(\text{m}^2\)).
- Temperature difference \(\Delta T\) has the unit of kelvin (\(\text{K}\)).

Substituting these units into the rearranged equation:

\([k] = \frac{(\text{kg}\ \text{m}^2\ \text{s}^{-3}) \times \text{m}}{\text{m}^2 \times \text{K}} = \text{kg}\ \text{m}\ \text{s}^{-3}\ \text{K}^{-1}\)

Therefore, the correct option is B.

1 mark for correctly expressing the SI base units of power as \(\text{kg}\ \text{m}^2\ \text{s}^{-3}
\) and substituting into the rearranged equation to obtain the correct final SI base units.

Using the equation of motion:

\(s = u t + \frac{1}{2} a t^2\)

Let us define the upward direction as positive.
- The initial velocity is \(u = +15.0\ \text{m}\ \text{s}^{-1}\).
- The acceleration due to gravity is \(a = -9.81\ \text{m}\ \text{s}^{-2}\).
- The total time is \(t = 4.50\ \text{s}\).

Substitute these values into the equation:

\(s = (15.0)(4.50) + \frac{1}{2}(-9.81)(4.50)^2\)

\(s = 67.5 - 4.905 \times 20.25\)

\(s = 67.5 - 99.33 = -31.83\ \text{m}\)

The negative sign indicates that the final position of the stone is below the starting point. Therefore, the height of the cliff is \(31.8\ \text{m}\).

Hence, the correct option is A.

1 mark for using the appropriate equations of motion with correct signs for velocity and acceleration to calculate the cliff's height.

First, apply the law of conservation of momentum:

\(m_X u_X + m_Y u_Y = m_X v_X + m_Y v_Y\)

\((2.0)(4.0) + (3.0)(0) = (2.0)v_X + (3.0)(2.4)\)

\(8.0 = 2.0 v_X + 7.2\)

\(2.0 v_X = 0.80 \implies v_X = 0.40\ \text{m}\ \text{s}^{-1}\)

Since the velocity is positive, trolley X continues to move in its original direction of motion at a speed of \(0.40\ \text{m}\ \text{s}^{-1}\).

Next, check the conservation of kinetic energy:
- Initial total kinetic energy:
\(E_{k,i} = \frac{1}{2} m_X u_X^2 = \frac{1}{2}(2.0)(4.0)^2 = 16.0\ \text{J}\)
- Final total kinetic energy:
\(E_{k,f} = \frac{1}{2} m_X v_X^2 + \frac{1}{2} m_Y v_Y^2 = \frac{1}{2}(2.0)(0.40)^2 + \frac{1}{2}(3.0)(2.4)^2 = 0.16 + 8.64 = 8.80\ \text{J}\)

Since \(E_{k,f} < E_{k,i}\), total kinetic energy is not conserved, meaning the collision is inelastic.

Therefore, the correct option is B.

1 mark for calculating the correct post-collision speed of trolley X and identifying that kinetic energy is not conserved, signifying an inelastic collision.

The useful work done per second by the pump consists of lifting the water (increasing its gravitational potential energy) and accelerating the water (increasing its kinetic energy).

Let the mass rate of water pumped be \(\frac{\Delta m}{\Delta t} = 15\ \text{kg}\ \text{s}^{-1}\).

1. Rate of increase of gravitational potential energy:
\(P_{\text{GPE}} = \frac{\Delta m}{\Delta t} g h = 15 \times 9.81 \times 12 = 1765.8\ \text{W}\)

2. Rate of increase of kinetic energy:
\(P_{\text{KE}} = \frac{1}{2} \frac{\Delta m}{\Delta t} v^2 = \frac{1}{2} \times 15 \times (8.0)^2 = 480\ \text{W}\)

3. Total useful power output from the pump-motor system:
\(P_{\text{out}} = P_{\text{GPE}} + P_{\text{KE}} = 1765.8 + 480 = 2245.8\ \text{W}\)

4. Electrical power input required:
\(\text{Efficiency} = \frac{P_{\text{out}}}{P_{\text{in}}}\)
\(0.75 = \frac{2245.8}{P_{\text{in}}} \implies P_{\text{in}} = \frac{2245.8}{0.75} = 2994.4\ \text{W} \approx 3.0\ \text{kW}\)

Therefore, the correct option is D.

1 mark for calculating both GPE and KE rate changes, summing them to find the total output power, and applying the efficiency formula to find the correct input power.

The strain energy \(E\) stored in a wire stretched within its elastic limit is given by:

\(E = \frac{1}{2} F e\)

where \(F\) is the force and \(e\) is the extension.

From the definition of Young modulus \(E_y\):

\(E_y = \frac{\text{stress}}{\text{strain}} = \frac{F L}{A e}\)

Rearranging this for extension \(e\):

\(e = \frac{F L}{A E_y}\)

Substitute \(e\) back into the strain energy formula:

\(E = \frac{F^2 L}{2 A E_y}\)

Substitute the given values:
- \(F = 180\ \text{N}\)
- \(L = 2.0\ \text{m}\)
- \(A = 1.5 \times 10^{-6}\ \text{m}^2\)
- \(E_y = 1.2 \times 10^{11}\ \text{Pa}\)

\(E = \frac{(180)^2 \times 2.0}{2 \times (1.5 \times 10^{-6}) \times (1.2 \times 10^{11})} = \frac{64800}{3.6 \times 10^5} = 0.18\ \text{J}\)

Therefore, the correct option is C.

1 mark for combining the definition of the Young modulus with the elastic potential energy formula to calculate the correct stored energy.

1. When unpolarised light of intensity \(I_0\) passes through the first polarizing filter, the light becomes plane-polarised, and its intensity is reduced by half:

\(I_1 = \frac{1}{2} I_0\)

2. According to Malus's Law, when this plane-polarised light passes through the analyser, the intensity \(I_2\) of the emerging light is given by:

\(I_2 = I_1 \cos^2(\theta)\)

where \(\theta = 60^\circ\).

3. Calculate the emerging intensity:

\(I_2 = \left(\frac{1}{2} I_0\right) \cos^2(60^\circ)\)

Since \(\cos(60^\circ) = 0.5\), we have \(\cos^2(60^\circ) = 0.25 = \frac{1}{4}\).

\(I_2 = \frac{1}{2} I_0 \times \frac{1}{4} = \frac{1}{8} I_0\)

Therefore, the correct option is A.

1 mark for recognizing the initial reduction in intensity by half for unpolarised light, and correctly applying Malus's law for the analyser.

The diffraction grating equation is:

\(d \sin\theta = n \lambda\)

where:
- \(d\) is the grating spacing (slit separation)
- \(\theta = 28.0^\circ\)
- \(n = 2\) (second-order maximum)
- \(\lambda = 540 \times 10^{-9}\ \text{m}\n\)

Rearranging for \(d\):

\(d = \frac{n \lambda}{\sin\theta} = \frac{2 \times 540 \times 10^{-9}}{\sin(28.0^\circ)}\)

\(d = \frac{1.08 \times 10^{-6}}{0.4695} \approx 2.300 \times 10^{-6}\ \text{m}\)

The number of lines per metre, \(N\), is the reciprocal of \(d\):

\(N = \frac{1}{d} = \frac{1}{2.300 \times 10^{-6}} \approx 4.347 \times 10^5\ \text{lines}\ \text{m}^{-1}\)

Convert this to lines per millimetre (\(N_{\text{mm}}\)):

\(N_{\text{mm}} = \frac{N}{1000} = \frac{4.347 \times 10^5}{1000} \approx 435\ \text{lines}\ \text{mm}^{-1}\)

Therefore, the correct option is B.

1 mark for calculating the slit separation \(d\) correctly and converting it to the correct number of lines per millimetre.

1. **Terminal potential difference (p.d.)**:
The terminal p.d. \(V\) across the battery is given by:

\(V = V_0 \left(\frac{R}{R + r}\right)\)

- When \(R \to 0\), \(V \to 0\).
- As \(R\) increases, the fraction \(\frac{R}{R + r}\) increases continuously towards \(1\). Therefore, terminal p.d. **increases continuously**.

2. **Power dissipated in the variable resistor**:
The power \(P\) is:

\(P = I^2 R = \frac{V_0^2 R}{(R+r)^2}\)

- When \(R = 0\), \(P = 0\).
- When \(R \to \infty\), \(I \to 0\) and \(P \to 0\).
- The maximum power transfer theorem states that the power transferred to the variable resistor reaches a maximum when \(R = r\).
Thus, the power **increases to a maximum and then decreases**.

Combining these results, the correct option is B.

1 mark for identifying the correct trends: terminal potential difference increases continuously, and power dissipated reaches a peak and then declines.

Using the equation of motion \(v^2 = u^2 + 2as\). Let downwards be the positive direction. The initial velocity is \(u_i = -u\), the final velocity is \(v = 2u\), the acceleration is \(a = g\), and the displacement is \(s = h\). Substituting these values: \((2u)^2 = (-u)^2 + 2gh\) which simplifies to \(4u^2 = u^2 + 2gh\). This gives \(3u^2 = 2gh\), and thus \(h = \frac{3u^2}{2g}\).

1 mark for the correct option B. 0 marks for other options.

Using conservation of linear momentum with the rightwards direction as positive: \(2m(v) + 3m(-2v) = 2mv_P + 3mv_Q\), which simplifies to \(-4v = 2v_P + 3v_Q\). For a perfectly elastic collision, the relative speed of approach equals the relative speed of separation: \(u_P - u_Q = v_Q - v_P \Rightarrow v - (-2v) = v_Q - v_P\), which yields \(3v = v_Q - v_P\) or \(v_Q = 3v + v_P\). Substituting this into the momentum equation: \(-4v = 2v_P + 3(3v + v_P) \Rightarrow -4v = 5v_P + 9v \Rightarrow -13v = 5v_P \Rightarrow v_P = -2.6v\). The negative sign indicates the direction is to the left.

1 mark for the correct option C. 0 marks for other options.

By Archimedes' principle, the weight of the floating object is equal to the upthrust, which equals the weight of the fluid displaced. When the extra mass \(m\) is placed on top, the increase in downward gravitational force is \(mg\). The equilibrium is maintained by an equal increase in upthrust, which is equal to the weight of the extra fluid displaced: \(\Delta U = \rho (\Delta V) g = \rho (A \Delta h) g\). Equating the two gives \(mg = \rho A \Delta h g\), which simplifies to \(m = \rho A \Delta h\).

1 mark for the correct option C. 0 marks for other options.

Using the conservation of energy, the initial gravitational potential energy at the top is \(mgH\). As the car travels along the track of length \(L\) against a constant frictional force \(F\), the work done against friction is \(W = F \times L\). The remaining energy is converted into kinetic energy at the bottom: \(E_k = mgH - FL\).

1 mark for the correct option B. 0 marks for other options.

The work done in stretching the wire is given by \(W = \frac{1}{2} F x\). The Young modulus is \(E = \frac{\text{Stress}}{\text{Strain}} = \frac{F/A}{x/L} = \frac{FL}{Ax}\). Rearranging this formula for the force gives \(F = \frac{EAx}{L}\). Substituting this expression for \(F\) into the work equation yields \(W = \frac{1}{2} \left(\frac{EAx}{L}\right) x = \frac{EAx^2}{2L}\).

1 mark for the correct option B. 0 marks for other options.

The diffraction grating equation is \(d \sin \theta = n \lambda\). A grating with \(N\) lines per millimetre has a line spacing of \(d = \frac{10^{-3}}{N}\) metres. Here, \(n = 3\) and \(\theta = 30^\circ\). Substituting these values gives \(\frac{10^{-3}}{N} \sin 30^\circ = 3 \lambda\). Since \(\sin 30^\circ = 0.5\), this becomes \(\frac{10^{-3}}{N} \times 0.5 = 3 \lambda\), which simplifies to \(\lambda = \frac{1}{6N} \times 10^{-3}\text{ m}\).

1 mark for the correct option A. 0 marks for other options.

The resistance of a wire is given by \(R = \rho \frac{L}{A}\). Since the volume \(V = A L\) is constant, the cross-sectional area is \(A = \frac{V}{L}\). Substituting this into the resistance equation gives \(R = \rho \frac{L^2}{V}\), so \(R \propto L^2\). When the length is increased by \(10\%\), the new length is \(L' = 1.10 L\). The new resistance is \(R' \propto (1.10 L)^2 = 1.21 L^2\), which means \(R' = 1.21 R\).

1 mark for the correct option C. 0 marks for other options.

A neutron has the quark structure \(udd\) and a proton has the quark structure \(uud\). Therefore, in \(\beta^-\)\ decay, a down quark (\(d\)) is transformed into an up quark (\(u\)). This process is mediated by the weak interaction, which involves the emission of a \(W^-\) boson that subsequently decays into the electron and electron antineutrino.

1 mark for the correct option A. 0 marks for other options.

(a) Velocity is defined as the rate of change of displacement [1]. Acceleration is defined as the rate of change of velocity [1]. (b)(i) During the first stage (constant acceleration): \(u = 0\text{ m s}^{-1}\), \(a = 2.4\text{ m s}^{-2}\), \(t_1 = 3.5\text{ s}\). The final velocity is \(v_1 = u + a t_1 = 2.4 \times 3.5 = 8.4\text{ m s}^{-1}\). The distance is \(s_1 = \frac{1}{2} a t_1^2 = 0.5 \times 2.4 \times 3.5^2 = 14.7\text{ m}\). During the second stage (constant velocity): \(v = 8.4\text{ m s}^{-1}\), \(t_2 = 5.0\text{ s}\). The distance is \(s_2 = v t_2 = 8.4 \times 5.0 = 42.0\text{ m}\). During the third stage (uniform deceleration to rest): \(u = 8.4\text{ m s}^{-1}\), \(v = 0\text{ m s}^{-1}\), \(t_3 = 2.0\text{ s}\). The distance is \(s_3 = \frac{u+v}{2} t_3 = \frac{8.4+0}{2} \times 2.0 = 8.4\text{ m}\). Total distance \(s = s_1 + s_2 + s_3 = 14.7 + 42.0 + 8.4 = 65.1\text{ m}\) [4]. (ii) Average velocity \(v_{\text{avg}} = \frac{\text{Total distance}}{\text{Total time}} = \frac{65.1}{10.5} = 6.2\text{ m s}^{-1}\) [2.5].

(a) 1 mark for defining velocity as rate of change of displacement. 1 mark for defining acceleration as rate of change of velocity. (b)(i) 1 mark for calculating final velocity \(v_1 = 8.4\text{ m s}^{-1}\). 1 mark for calculating \(s_1 = 14.7\text{ m}\) or \(s_3 = 8.4\text{ m}\). 1 mark for calculating \(s_2 = 42.0\text{ m}\). 1 mark for correct sum of all distances to get \(65.1\text{ m}\) (accept \(65\text{ m}\)). (b)(ii) 1 mark for using the formula \(v_{\text{avg}} = \frac{\text{total displacement}}{\text{total time}}\). 1.5 marks for correct final calculation yielding \(6.2\text{ m s}^{-1}\).

(a) The total momentum of a closed system remains constant provided no external force acts on the system [2]. (b)(i) Let the direction to the right be positive. Total initial momentum \(p_i = m_A u_A + m_B u_B = (0.80 \times 3.0) + (1.2 \times (-1.5)) = 2.4 - 1.8 = +0.60\text{ kg m s}^{-1}\). Total final momentum \(p_f = m_A v_A + m_B v_B = (0.80 \times (-0.60)) + (1.2 \times v_B) = -0.48 + 1.2 v_B\). Equating initial and final momentum: \(0.60 = -0.48 + 1.2 v_B \implies 1.08 = 1.2 v_B \implies v_B = +0.90\text{ m s}^{-1}\). Since the value is positive, the direction is to the right [4.5]. (ii) Total initial kinetic energy \(E_{k,i} = \frac{1}{2} m_A u_A^2 + \frac{1}{2} m_B u_B^2 = 0.5 \times 0.80 \times 3.0^2 + 0.5 \times 1.2 \times 1.5^2 = 3.6 + 1.35 = 4.95\text{ J}\). Total final kinetic energy \(E_{k,f} = \frac{1}{2} m_A v_A^2 + \frac{1}{2} m_B v_B^2 = 0.5 \times 0.80 \times (-0.60)^2 + 0.5 \times 1.2 \times 0.90^2 = 0.144 + 0.486 = 0.63\text{ J}\). Since total kinetic energy is not conserved (it decreased from \(4.95\text{ J}\) to \(0.63\text{ J}\)), the collision is inelastic [2].

(a) 1 mark for stating that total momentum is constant. 1 mark for specifying a closed system or no external forces. (b)(i) 1 mark for calculating correct initial momentum (\(+0.60\text{ kg m s}^{-1}\)). 1 mark for correct algebraic expression for final momentum (\(-0.48 + 1.2 v_B\)). 1 mark for equating and solving for \(v_B\). 1.5 marks for specifying the correct magnitude (\(0.90\text{ m s}^{-1}\)) and direction (to the right). (b)(ii) 1 mark for calculating both initial and final kinetic energies correctly. 1 mark for comparing the two values and concluding that the collision is inelastic.

(a) In a fluid, pressure increases with depth according to \(\Delta p = \rho g \Delta h\) [1]. Thus, the pressure at the bottom surface of a submerged object is greater than at its top surface [1]. This pressure difference produces a net upward force on the object, known as upthrust [0.5]. (b)(i) The volume of the cylinder is \(V = A h = 1.5 \times 10^{-3} \times 0.12 = 1.8 \times 10^{-4}\text{ m}^3\). Upthrust \(U = \rho_{\text{fluid}} g V = 850 \times 9.81 \times 1.8 \times 10^{-4} = 1.50\text{ N}\) [3]. (ii) The forces acting on the cylinder are tension \(T\) and upthrust \(U\) acting upwards, and weight \(W\) acting downwards. Since it is in equilibrium, \(T + U = W \implies T = W - U\). The weight is \(W = \rho_{\text{brass}} V g = 8400 \times 1.8 \times 10^{-4} \times 9.81 = 14.83\text{ N}\). Thus, \(T = 14.83 - 1.50 = 13.33\text{ N} \approx 13\text{ N}\) [3].

(a) 1 mark for stating pressure increases with depth. 1 mark for stating that pressure at the bottom is greater than at the top. 0.5 marks for stating this creates a net upward force. (b)(i) 1 mark for calculating volume of cylinder (\(1.8 \times 10^{-4}\text{ m}^3\)). 1 mark for using upthrust formula \(U = \rho g V\). 1 mark for correct final value \(1.5\text{ N}\) (or \(1.50\text{ N}\)). (b)(ii) 1 mark for equilibrium equation \(T = W - U\). 1 mark for calculating weight \(W = 14.8\text{ N}\). 1 mark for correct final value of tension \(13\text{ N}\) (or \(13.3\text{ N}\)).

(a) Gravitational potential energy is the energy stored due to an object's position in a gravitational field [1], while kinetic energy is the energy of an object due to its motion [1]. (b)(i) Using conservation of mechanical energy: \(m g h_A + \frac{1}{2} m v_A^2 = m g h_B + \frac{1}{2} m v_B^2\). Dividing by \(m\): \(9.81 \times 22 + 0.5 \times 6.0^2 = 9.81 \times 4.0 + 0.5 v_B^2 \implies 215.82 + 18.0 = 39.24 + 0.5 v_B^2 \implies 194.58 = 0.5 v_B^2 \implies v_B^2 = 389.16 \implies v_B = 19.7\text{ m s}^{-1}\) [3.5]. (ii) Total initial energy \(E_A = m g h_A + \frac{1}{2} m v_A^2 = 450 \times 9.81 \times 22 + 0.5 \times 450 \times 6.0^2 = 105219\text{ J}\). Work done against resistive force \(W_f = F d = 280 \times 48 = 13440\text{ J}\). Final energy \(E_B = E_A - W_f = 105219 - 13440 = 91779\text{ J}\). Thus, \(m g h_B + \frac{1}{2} m v_B^2 = 91779 \implies 450 \times 9.81 \times 4.0 + 225 v_B^2 = 91779 \implies 17658 + 225 v_B^2 = 91779 \implies 225 v_B^2 = 74121 \implies v_B = 18.1\text{ m s}^{-1}\) [3].

(a) 1 mark for correct definition of gravitational potential energy. 1 mark for correct definition of kinetic energy. (b)(i) 1 mark for stating conservation of energy equation. 1.5 marks for correct substitution of numbers. 1 mark for the correct final speed of \(19.7\text{ m s}^{-1}\) (accept \(20\text{ m s}^{-1}\)). (b)(ii) 1 mark for calculating work done against resistive force (\(13440\text{ J}\)). 1 mark for correct energy conservation relation \(E_B = E_A - W_f\). 1 mark for calculating final speed as \(18.1\text{ m s}^{-1}\) (accept \(18\text{ m s}^{-1}\)).

(a) A plane-polarised wave is a transverse wave in which oscillations are restricted to a single plane [1] perpendicular to the direction of wave propagation [1]. (b) Electromagnetic waves are transverse waves and can oscillate in any direction perpendicular to the propagation direction, so their oscillations can be restricted to a single plane [1]. Sound waves are longitudinal waves with oscillations only parallel to the direction of propagation, which cannot be restricted further, so they cannot be polarised [0.5]. (c)(i) The intensity after the first filter is \(I_1 = 0.5 I_0\) [1]. (ii) Using Malus's Law: \(I_2 = I_1 \cos^2\theta\). Given \(I_2 = 0.125 I_0\): \(0.125 I_0 = 0.5 I_0 \cos^2\theta \implies \cos^2\theta = 0.25 \implies \cos\theta = 0.5 \implies \theta = 60^\circ\) [4].

(a) 1 mark for stating oscillations are in a single plane. 1 mark for stating this plane is perpendicular to the direction of wave travel. (b) 1 mark for linking polarisation to transverse waves (e.g., electromagnetic waves). 0.5 marks for stating longitudinal waves (sound) cannot be polarised. (c)(i) 1 mark for \(0.5 I_0\). (c)(ii) 1 mark for quoting Malus's Law. 1 mark for equating to \(0.125 I_0\). 1 mark for finding \(\cos^2\theta = 0.25\) or \(\cos\theta = 0.5\). 1 mark for final angle \(60^\circ\) (or \(\frac{\pi}{3}\text{ rad}\)).

(a) Electromotive force is the energy converted from other forms to electrical energy per unit charge [1.5]. Potential difference is the electrical energy converted to other forms of energy per unit charge [1]. (b)(i) Total resistance of circuit \(R_{\text{total}} = R + r\). Current \(I = \frac{E}{R_{\text{total}}} = \frac{E}{R + r}\). Power in external resistor \(P = I^2 R\). Substituting \(I\): \(P = \left(\frac{E}{R+r}\right)^2 R = \frac{E^2 R}{(R + r)^2}\) [2]. (ii) Using \(E = I(R + r)\): For \(R = 4.0\\ \Omega\): \(E = 1.0(4.0 + r)\). For \(R = 10.0\\ \Omega\): \(E = 0.50(10.0 + r)\). Equating both: \(4.0 + r = 5.0 + 0.50r \implies 0.50r = 1.0 \implies r = 2.0\\ \Omega\). Substituting \(r\) back: \(E = 1.0(4.0 + 2.0) = 6.0\text{ V}\) [4].

(a) 1.5 marks for e.m.f. definition (energy transfer per unit charge from chemical/other to electrical). 1 mark for p.d. definition (energy transfer per unit charge from electrical to other forms). (b)(i) 1 mark for writing correct expression for current \(I\). 1 mark for substituting into \(P = I^2 R\) to obtain the target equation. (b)(ii) 1 mark for setting up first equation. 1 mark for setting up second equation. 1 mark for solving for internal resistance \(r = 2.0\\ \Omega\). 1 mark for solving for e.m.f. \(E = 6.0\text{ V}\).

(a) A fundamental particle is a particle with no internal structure that cannot be split into smaller components [1.5]. The two main classes are quarks and leptons [1]. (b)(i) \({}_0^1\text{n} \rightarrow {}_1^1\text{p} + {}_{-1}^0\text{e} + \bar{\nu}_e\) [2]. (ii) Weak interaction (or weak nuclear force) [1]. (iii) A neutron has quark composition \(udd\) and a proton has quark composition \(uud\) [1]. During the decay, a down (\(d\)) quark changes into an up (\(u\)) quark [1], emitting an electron and an electron antineutrino [1].

(a) 1.5 marks for defining fundamental particle. 1 mark for naming quarks and leptons. (b)(i) 1 mark for correct proton and nucleon numbers for neutron and proton. 1 mark for correct representation of electron (\({}_{-1}^0\text{e}\)) and antineutrino. (b)(ii) 1 mark for naming weak interaction. (b)(iii) 1 mark for stating compositions of neutron (\(udd\)) and proton (\(uud\)). 1 mark for stating a down quark changes to an up quark. 1 mark for stating emission of electron and electron antineutrino.