2023 Cambridge IAS-Level Physics (9702) Practice Paper with Answers

The density \(\rho\) of the cylinder is given by:

\(\rho = \frac{m}{V} = \frac{4m}{\pi d^2 L}\)

The fractional uncertainty in density is the sum of the fractional uncertainties of the measured quantities:

\(\frac{\Delta \rho}{\rho} = \frac{\Delta m}{m} + 2\frac{\Delta d}{d} + \frac{\Delta L}{L}\)

Now, compute each percentage uncertainty:

- Mass: \(\frac{3}{120} \times 100\% = 2.5\%\)
- Diameter: \(\frac{0.02}{2.00} \times 100\% = 1.0\%\)
- Length: \(\frac{0.05}{5.00} \times 100\% = 1.0\%\)

Substituting these values:

Percentage uncertainty in density = \(2.5\% + 2(1.0\%) + 1.0\% = 5.5\%\).

- 1 mark for the correct answer of 5.5%.

Let the upward direction be positive.

- Initial velocity, \(v_i = +u\)
- Final velocity at the base of the cliff, \(v_f = -3u\)
- Acceleration, \(a = -g\)
- Vertical displacement, \(s = -h\)

Using the equation of motion:

\(v_f^2 = v_i^2 + 2as\)

Substitute the known values:

\((-3u)^2 = u^2 + 2(-g)(-h)\)

\(9u^2 = u^2 + 2gh\)

\(8u^2 = 2gh\)

\(h = \frac{4u^2}{g}\)

- 1 mark for the correct answer of 4u^2 / g.

The total stopping distance consists of two parts: the thinking distance and the braking distance.

1. **Thinking distance (\(d_1\))**:
During the reaction time, the car travels at a constant speed of \(24\text{ m s}^{-1}\).
\(d_1 = v \times t = 24\text{ m s}^{-1} \times 0.50\text{ s} = 12\text{ m}\).

2. **Braking distance (\(d_2\))**:
The car decelerates from \(24\text{ m s}^{-1}\) to \(0\text{ m s}^{-1}\) at a rate of \(-6.0\text{ m s}^{-2}\).
Using \(v^2 = u^2 + 2as\):
\(0^2 = 24^2 + 2(-6.0)d_2\)
\(0 = 576 - 12d_2\)
\(12d_2 = 576 \implies d_2 = 48\text{ m}\).

3. **Total stopping distance**:
\(d_{\text{total}} = d_1 + d_2 = 12\text{ m} + 48\text{ m} = 60\text{ m}\).

- 1 mark for the correct answer of 60 m.

Using Newtons second law, the average force on the ball is equal to the rate of change of momentum of the ball.

Taking the initial direction of motion as positive:
- Initial momentum, \(p_i = mv\)
- Final momentum, \(p_f = -0.6mv\)

Change in momentum of the ball:
\(\Delta p = p_f - p_i = -0.6mv - mv = -1.6mv\)

The average force exerted on the ball by the wall is:
\(F_{\text{on ball}} = \frac{\Delta p}{\Delta t} = -\frac{1.6mv}{\Delta t}\)

By Newtons third law, the force exerted by the ball on the wall is equal in magnitude and opposite in direction:
\(F_{\text{on wall}} = \frac{1.6mv}{\Delta t}\).

- 1 mark for the correct answer of 1.6mv / \Delta t.

The resistance of a wire is given by:
\(R = \rho \frac{L}{A}\)
where \(\rho\) is the resistivity, \(L\) is the length, and \(A\) is the cross-sectional area.

Since the volume \(V = L \times A\) remains constant, we have:
\(L_1 A_1 = L_2 A_2\)

If the length increases by \(10\%\):
\(L_2 = 1.10 L_1\)

Reflecting this, the new cross-sectional area \(A_2\) is:
\(A_2 = \frac{L_1}{L_2} A_1 = \frac{A_1}{1.10}\)

The new resistance \(R_2\) is:
\(R_2 = \rho \frac{L_2}{A_2} = \rho \frac{1.10 L_1}{A_1 / 1.10} = 1.10^2 \times \rho \frac{L_1}{A_1} = 1.21 R_1\).

- 1 mark for the correct answer of 1.21 R.

The resistance \(R\) of a wire is given by:
\(R = \rho \frac{L}{A} = \rho \frac{L}{\pi (d/2)^2} = \frac{4 \rho L}{\pi d^2}\)

Let the resistance of wire X be \(R_X = R\).

For wire Y:
\(R_Y = \frac{4 \rho (2L)}{\pi (2d)^2} = \frac{8 \rho L}{4 \pi d^2} = \frac{2 \rho L}{\pi d^2} = \frac{1}{2} R_X\)

Since they are connected in parallel, they experience the same potential difference \(V\). The current \(I\) in each wire is inversely proportional to its resistance (\(I = \frac{V}{R}\)):

\(\frac{I_X}{I_Y} = \frac{R_Y}{R_X} = \frac{0.5 R_X}{R_X} = 0.5\).

- 1 mark for the correct ratio of 0.50.

The total pressure at the bottom is the sum of atmospheric pressure and the hydrostatic pressures of the oil and water layers:

\(p = p_{\text{atm}} + h_{\text{oil}} \rho_{\text{oil}} g + h_{\text{water}} \rho_{\text{water}} g\)

Calculate the hydrostatic pressure of the oil:
\(p_{\text{oil}} = 0.10\text{ m} \times 800\text{ kg m}^{-3} \times 9.81\text{ m s}^{-2} = 784.8\text{ Pa}\)

Calculate the hydrostatic pressure of the water:
\(p_{\text{water}} = 0.15\text{ m} \times 1000\text{ kg m}^{-3} \times 9.81\text{ m s}^{-2} = 1471.5\text{ Pa}\)

Calculate total pressure:
\(p = 101\,000\text{ Pa} + 784.8\text{ Pa} + 1471.5\text{ Pa} = 103\,256.3\text{ Pa} \approx 1.03 \times 10^5\text{ Pa}\).

- 1 mark for the correct calculation of total pressure of 1.03 x 10^5 Pa.

When the block is submerged in the liquid, three forces act on it in equilibrium:
- The tension \(T\) from the spring balance acting upwards (which is the reading of \(15.0\text{ N}\))
- The upthrust \(U\) acting upwards
- The weight \(W\) of the block acting downwards

Thus, \(T + U = W \implies W = T + U\).

Using Archimedes principle, the upthrust is equal to the weight of the displaced liquid:
\(U = \rho_{\text{liquid}} V g = 1200\text{ kg m}^{-3} \times (2.5 \times 10^{-4}\text{ m}^3) \times 9.81\text{ m s}^{-2}\)
\(U = 2.943\text{ N}\)

So, the weight is:
\(W = 15.0\text{ N} + 2.943\text{ N} = 17.943\text{ N} \approx 17.9\text{ N}\).

- 1 mark for the correct weight calculation of 17.9 N.

The formula for resistivity is \(\rho = \frac{R A}{L} = \frac{\pi R d^2}{4 L}\). The fractional uncertainty in \(\rho\) is given by \(\frac{\Delta \rho}{\rho} = \frac{\Delta R}{R} + 2 \frac{\Delta d}{d} + \frac{\Delta L}{L}\). First, calculate individual percentage uncertainties: for resistance, \(\% \Delta R = \frac{0.1}{4.0} \times 100\% = 2.5\%\); for diameter, \(\% \Delta d = \frac{0.01}{0.50} \times 100\% = 2.0\%\); for length, \(\% \Delta L = \frac{0.02}{2.00} \times 100\% = 1.0\%\). Summing these gives \(\% \Delta \rho = 2.5\% + 2(2.0\%) + 1.0\% = 7.5\%\).

1 mark for calculating the correct percentage uncertainty by summing the individual percentage uncertainties, remembering to double the percentage uncertainty of the diameter.

Let the time when they pass be \(t\). The first object is dropped from rest from height \(H\) and falls to a height of \(\frac{H}{3}\), meaning it has fallen a distance of \(\frac{2}{3}H\). Using \(s = \frac{1}{2}gt^2\), we get \(\frac{2}{3}H = \frac{1}{2}gt^2\), which simplifies to \(t^2 = \frac{4H}{3g}\). At the same time \(t\), the second object is at height \(\frac{H}{3}\) with initial speed \(u\), so \(\frac{H}{3} = ut - \frac{1}{2}gt^2\). Substituting \(\frac{1}{2}gt^2 = \frac{2}{3}H\) into this equation yields \(\frac{H}{3} = ut - \frac{2}{3}H\), which gives \(ut = H\). Thus, \(u = \frac{H}{t}\). Substituting \(t = \sqrt{\frac{4H}{3g}}\) into this gives \(u = \sqrt{\frac{3gH}{4}}\).

1 mark for setting up the equation of motion for both objects, equating their collision height, and solving for the initial velocity \(u\).

By conservation of linear momentum, \(m v = m v_1 + 3m v_2\), which simplifies to \(v = v_1 + 3v_2\). For a perfectly elastic head-on collision, the relative velocity of approach equals the relative velocity of separation: \(v - 0 = v_2 - v_1\), so \(v_1 = v_2 - v\). Substituting this into the momentum equation gives \(v = (v_2 - v) + 3v_2\), which simplifies to \(2v = 4v_2\), so \(v_2 = \frac{1}{2}v\). Substituting \(v_2\) back gives \(v_1 = \frac{1}{2}v - v = -\frac{1}{2}v\).

1 mark for applying both the conservation of momentum and the relative velocity relation for elastic collisions, and solving the simultaneous equations correctly.

For the cube to float in equilibrium, the total upward buoyant force (upthrust) must equal its weight. The upthrust is the sum of the buoyant forces from both liquids: \(U = V_2 \rho_2 g + V_1 \rho_1 g\), where \(V\) is the total volume of the cube, \(V_2 = fV\), and \(V_1 = (1-f)V\). The weight of the cube is \(W = V \rho_b g\). Equating upthrust and weight: \(f V \rho_2 g + (1-f) V \rho_1 g = V \rho_b g\). Dividing both sides by \(V g\) gives \(f \rho_2 + (1-f) \rho_1 = \rho_b\). Rearranging this gives \(f(\rho_2 - \rho_1) = \rho_b - \rho_1\), which yields \(f = \frac{\rho_b - \rho_1}{\rho_2 - \rho_1}\).

1 mark for setting up the equilibrium equation where upthrust equals weight, and correctly rearranging to solve for the fraction \(f\).

The resistance of a wire of length \(L\) and diameter \(d\) is given by \(R = \rho \frac{4L}{\pi d^2}\). Thus, resistance is proportional to \(\frac{L}{d^2}\). For wire \(Y\), the length is \(2L_X\) and the diameter is \(0.5d_X\). This means the resistance of wire \(Y\) is \(R_Y = R_X \times \frac{2}{(0.5)^2} = 8 R_X\). Since the wires are connected in parallel across the same e.m.f., they experience the same potential difference \(V\). The power dissipated is \(P = \frac{V^2}{R}\). Therefore, the ratio of power dissipated is \(\frac{P_X}{P_Y} = \frac{V^2 / R_X}{V^2 / R_Y} = \frac{R_Y}{R_X} = 8\).

1 mark for determining that the resistance of wire \(Y\) is 8 times that of wire \(X\), and using the parallel power relationship \(P = V^2 / R\) to find the correct power ratio.

The wire is divided into two parallel paths between points \(P\) and \(Q\). The minor arc has a length proportional to \(\frac{120^\circ}{360^\circ} = \frac{1}{3}\) of the total circumference, so its resistance is \(R_1 = \frac{1}{3} \times 12.0\ \Omega = 4.0\ \Omega\). The major arc has a length proportional to \(\frac{240^\circ}{360^\circ} = \frac{2}{3}\) of the total circumference, so its resistance is \(R_2 = \frac{2}{3} \times 12.0\ \Omega = 8.0\ \Omega\). The effective resistance \(R\) of these two parallel paths is given by \(\frac{1}{R} = \frac{1}{R_1} + \frac{1}{R_2} = \frac{1}{4.0} + \frac{1}{8.0} = \frac{3}{8.0}\), which gives \(R = \frac{8.0}{3} \approx 2.67\ \Omega\).

1 mark for calculating the resistances of the two parallel sections of the circle and finding their combined parallel resistance.

An alpha decay decreases the mass number \(A\) by 4 and the atomic number \(Z\) by 2. This gives a daughter nucleus with \(A = 212 - 4 = 208\) and \(Z = 83 - 2 = 81\). A beta-minus decay does not change the mass number but increases the atomic number by 1. This gives the final nucleus \(Y\) with \(A = 208\) and \(Z = 81 + 1 = 82\). The number of protons is \(Z = 82\), and the number of neutrons is \(A - Z = 208 - 82 = 126\).

1 mark for determining the correct atomic number and mass number of the final nucleus, and identifying the correct proton and neutron counts.

A neutron consists of one up quark and two down quarks (\(\text{udd}\)), while a proton consists of two up quarks and one down quark (\(\text{uud}\)). During beta-minus decay, one of the down quarks in the neutron changes into an up quark to form a proton (\(\text{d} \rightarrow \text{u}\)). This type of decay involves a change in quark flavor, which is uniquely mediated by the weak interaction (weak nuclear force).

1 mark for correctly identifying that a down quark changes to an up quark, and that this process is mediated by the weak interaction.

The percentage uncertainty in \(d\) is: \(\frac{0.02}{0.40} \times 100\% = 5.0\%\). The percentage uncertainty in \(R\) is: \(\frac{0.5}{25.0} \times 100\% = 2.0\%\). The percentage uncertainty in \(L\) is: \(\frac{0.03}{1.50} \times 100\% = 2.0\%\). The relationship for the percentage uncertainty in \(\rho\) is: \(\frac{\Delta \rho}{\rho} \times 100\% = 2 \left(\frac{\Delta d}{d}\right) + \frac{\Delta R}{R} + \frac{\Delta L}{L}\). Therefore, percentage uncertainty in \(\rho = 2(5.0\%) + 2.0\% + 2.0\% = 14\%\).

1 mark for the correct calculation of the individual percentage uncertainties and combining them correctly with the power of 2 for diameter to get 14%.

The resistance of a wire is given by \(R = \rho \frac{4L}{\pi D^2}\). For wire \(X\), \(R_X = R_0\). For wire \(Y\), \(R_Y = \rho \frac{4(2L)}{\pi (2D)^2} = \frac{1}{2} R_0\). Since they are connected in parallel, both wires experience the same potential difference \(V\). The power dissipated is \(P = \frac{V^2}{R}\). Therefore, \(P_X = \frac{V^2}{R_0}\) and \(P_Y = \frac{V^2}{0.5 R_0} = 2 \frac{V^2}{R_0}\). The ratio of the power dissipated is \(\frac{P_X}{P_Y} = 0.5\).

1 mark for correctly determining the resistance of wire Y compared to wire X, and using the parallel circuit condition (constant V) to find the power ratio of 0.5.

Cutting the wire into three equal lengths means each piece has a resistance of \(R_i = \frac{12.0\ \Omega}{3} = 4.0\ \Omega\). When connected in parallel, the total resistance \(R_p\) is calculated using \(\frac{1}{R_p} = \frac{1}{R_i} + \frac{1}{R_i} + \frac{1}{R_i} = \frac{3}{4.0\ \Omega}\), giving \(R_p = \frac{4.0}{3}\ \Omega \approx 1.33\ \Omega\).

1 mark for finding the resistance of each segment (4.0 ohms) and correctly applying the parallel combination formula to get 1.33 ohms.

Using the equation of motion \(s = ut + \frac{1}{2}at^2\) with upwards as positive: \(s = -20.0\text{ m}\), \(u = +15.0\text{ m s}^{-1}\), and \(a = -9.81\text{ m s}^{-2}\). Substituting these values yields: \(-20.0 = 15.0 t - 4.905 t^2\). Rearranging into quadratic form gives: \(4.905 t^2 - 15.0 t - 20.0 = 0\). Using the quadratic formula, the positive root for time is \(t = \frac{15.0 + \sqrt{(-15.0)^2 - 4(4.905)(-20.0)}}{2(4.905)} = \frac{15.0 + \sqrt{225 + 392.4}}{9.81} \approx 4.06\text{ s}\).

1 mark for correctly identifying the displacement as -20.0 m and solving the quadratic kinematics equation to find the total flight time of 4.06 s.

The total pressure \(P\) is the sum of atmospheric pressure and the hydrostatic pressures of the oil and water layers: \(P = P_{\text{atm}} + \rho_{\text{oil}} g h_{\text{oil}} + \rho_{\text{water}} g h_{\text{water}}\). \(P = 1.01 \times 10^5 + (850 \times 9.81 \times 0.40) + (1000 \times 9.81 \times 1.20)\). \(P = 101,000 + 3,335.4 + 11,772 = 116,107.4\text{ Pa} \approx 1.16 \times 10^5\text{ Pa}\).

1 mark for calculating both hydrostatic pressures and adding them to the atmospheric pressure to get 1.16 x 10^5 Pa.

Using conservation of linear momentum (with right as positive): \(m_1 u_1 + m_2 u_2 = (m_1 + m_2) v \Rightarrow (2.0 \times 6.0) + (3.0 \times -2.0) = (2.0 + 3.0) v \Rightarrow 12.0 - 6.0 = 5.0 v \Rightarrow v = 1.2\text{ m s}^{-1}\). The initial kinetic energy is \(E_{ki} = \frac{1}{2} (2.0)(6.0)^2 + \frac{1}{2} (3.0)(-2.0)^2 = 36.0 + 6.0 = 42.0\text{ J}\). The final kinetic energy is \(E_{kf} = \frac{1}{2} (5.0)(1.2)^2 = 3.6\text{ J}\). The kinetic energy lost is \(E_{ki} - E_{kf} = 42.0 - 3.6 = 38.4\text{ J}\).

1 mark for using conservation of momentum to find the common final velocity (1.2 m/s), calculating both initial and final kinetic energies, and subtracting them to obtain the loss of 38.4 J.

During alpha decay, the nucleus emits a helium nucleus (\(\text{}^{4}_{2}\text{He}\)), reducing the nucleon number by 4 and the proton number by 2: \(\text{}^{212}_{83}\text{Bi} \rightarrow \text{}^{208}_{81}\text{X} + \text{}^{4}_{2}\alpha\). During beta-minus decay, a neutron decays into a proton and an electron, leaving the nucleon number unchanged and increasing the proton number by 1: \(\text{}^{208}_{81}\text{X} \rightarrow \text{}^{208}_{82}\text{Y} + \text{}^0_{-1}\beta^-\). Thus, the nucleon number is 208 and the proton number is 82.

1 mark for tracing both decay stages correctly to obtain a nucleon number of 208 and a proton number of 82.

A neutron has a quark composition of \(\text{udd}\). A proton has a quark composition of \(\text{uu}d\). Therefore, in the beta-minus decay of a neutron, one of the down (\(\text{d}\)) quarks is transformed into an up (\(\text{u}\)) quark.

1 mark for identifying the correct quark configurations of the neutron and proton and identifying that a down quark changes to an up quark.

The density $\rho$ of the cylinder is given by:

$$\rho = \frac{m}{V} = \frac{4m}{\pi d^2 L}$$

The fractional uncertainty in density is the sum of the fractional uncertainties of the independent measurements:

$$\frac{\Delta \rho}{\rho} = \frac{\Delta m}{m} + 2\frac{\Delta d}{d} + \frac{\Delta L}{L}$$

Calculating each individual percentage uncertainty:

$$\frac{\Delta m}{m} \times 100\% = \frac{0.3}{24.0} \times 100\% = 1.25\%$$
$$2\frac{\Delta d}{d} \times 100\% = 2 \times \frac{0.02}{1.20} \times 100\% \approx 3.33\%$$
$$\frac{\Delta L}{L} \times 100\% = \frac{0.05}{5.00} \times 100\% = 1.00\%$$

Adding these together:

$$\frac{\Delta \rho}{\rho} \times 100\% = 1.25\% + 3.33\% + 1.00\% = 5.58\% \approx 5.6\%$$

1 mark for the correct calculation of the percentage uncertainty, accounting for the factor of 2 in the diameter term and summing the percentage uncertainties correctly.

The equation of motion for the falling object is:

$$mg - kv^2 = ma \implies a = g - \frac{k}{m}v^2$$

This shows that $a$ is a quadratic function of $v$. At $v = 0$, the acceleration is $a = g$.

The gradient of the graph of $a$ against $v$ is given by the derivative:

$$\frac{da}{dv} = -\frac{2kv}{m}$$

At $v = 0$, the gradient $\frac{da}{dv} = 0$, which means the graph starts with a horizontal tangent at the vertical intercept $(0, g)$. As $v$ increases, the gradient becomes increasingly negative (steeper and bending downwards) until the object reaches terminal velocity where $a = 0$.

1 mark for identifying that the equation of motion yields a parabolic relation starting with a horizontal gradient at the y-intercept.

Let the horizontal boundary interface between the oil and water be the reference level.

In the arm containing the oil, the height of the oil above this interface is $L = 15.0\text{ cm}$.

Since the top of the oil column is $2.0\text{ cm}$ higher than the top of the water column in the other arm, the height of the water column above the same interface level is:

$$h = 15.0\text{ cm} - 2.0\text{ cm} = 13.0\text{ cm}$$

At the interface level, the pressure exerted by the oil column equals the pressure exerted by the water column:

$$\rho g L = \rho_{\text{water}} g h \implies \rho L = \rho_{\text{water}} h$$

$$\rho \times 15.0\text{ cm} = (1.0 \times 10^3\text{ kg m}^{-3}) \times 13.0\text{ cm}$$

$$\rho = \frac{13.0}{15.0} \times 1.0 \times 10^3\text{ kg m}^{-3} \approx 8.7 \times 10^2\text{ kg m}^{-3}$$

1 mark for setting up the hydrostatic pressure balance at the interface level with the correct height of the water column (13.0 cm).

Using Newton's second law, the average force is the rate of change of momentum:

$$F = \frac{\Delta p}{\Delta t} = \frac{m(v - u)}{\Delta t}$$

Taking the direction to the right as positive:
- Initial velocity $u = -24\text{ m s}^{-1}$
- Final velocity $v = +36\text{ m s}^{-1}$

$$\Delta p = 0.15\text{ kg} \times (36 - (-24))\text{ m s}^{-1} = 0.15 \times 60 = 9.0\text{ kg m s}^{-1}$$

Now calculate the average force:

$$F = \frac{9.0\text{ kg m s}^{-1}}{4.5 \times 10^{-3}\text{ s}} = 2000\text{ N} = 2.0\text{ kN}$$

(If the sign change of the velocity was neglected, the calculated change in momentum would be $1.8\text{ kg m s}^{-1}$ yielding $0.40\text{ kN}$, which is incorrect).

1 mark for using the correct signs for opposite velocities to calculate the change in momentum and dividing by the collision time.

Let the original length be $L$ and the original cross-sectional area be $A$. The original resistance is:

$$R = \rho\frac{L}{A}$$

The new length is $L' = 1.25 L$.

Since the volume $V = L \times A$ is constant, the new cross-sectional area $A'$ must satisfy:

$$L A = L' A' \implies A' = \frac{L}{L'} A = \frac{A}{1.25}$$

The new resistance $R'$ is:

$$R' = \rho\frac{L'}{A'} = \rho\frac{1.25 L}{A / 1.25} = 1.25^2 \left(\rho\frac{L}{A}\right) = 1.5625 R \approx 1.56 R$$

1 mark for showing that resistance is proportional to the square of the length when volume is constant, and calculating the factor $1.25^2 = 1.56$.

The resistance $R$ of a wire of resistivity $\rho$, length $L$ and diameter $d$ is given by:

$$R = \rho \frac{L}{A} = \rho \frac{L}{\pi \left(\frac{d}{2}\right)^2} = \frac{4\rho L}{\pi d^2}$$

Thus, resistance is proportional to $\frac{L}{d^2}$. We can express the ratio as:

$$\frac{R_X}{R_Y} = \left(\frac{L_X}{L_Y}\right) \times \left(\frac{d_Y}{d_X}\right)^2$$

Given that $L_X = 0.5 L_Y$ and $d_X = 2 d_Y$:

$$\frac{R_X}{R_Y} = (0.5) \times \left(\frac{1}{2}\right)^2 = 0.5 \times \frac{1}{4} = 0.125 = \frac{1}{8}$$

1 mark for correctly applying the relationship $R \propto L/d^2$ and substituting the given proportions to find the ratio.

For the first decay:

$$^{212}_{83}\text{Bi} \to ^{212}_{84}\text{Po} + X$$

The nucleon number remains unchanged (212) and the proton number increases by 1 (from 83 to 84). To conserve charge, the emitted particle must have a proton number (charge) of $-1$ and nucleon number of $0$. This is a $\beta^-$ particle (electron).

For the second decay:

$$^{212}_{83}\text{Bi} \to ^{208}_{81}\text{Tl} + Y$$

The nucleon number decreases by 4 (from 212 to 208) and the proton number decreases by 2 (from 83 to 81). This is characteristic of the emission of an $\alpha$ particle ($^4_2\text{He}$).

Thus, $X$ is a $\beta^-$ particle and $Y$ is an $\alpha$ particle.

1 mark for identifying the first decay as beta-minus emission and the second decay as alpha emission by applying conservation of proton and nucleon numbers.

A neutron has the quark composition $udd$.
A proton has the quark composition $uud$.

During the beta-minus decay, one of the down ($d$) quarks in the neutron is transformed into an up ($u$) quark.

The charge of a down quark is $-\frac{1}{3}e$ and the charge of an up quark is $+\frac{2}{3}e$.

The change in total quark charge is:

$$\Delta Q = \left(+\frac{2}{3}e\right) - \left(-\frac{1}{3}e\right) = +e$$

Thus, a down quark changes to an up quark, and the total charge of the quarks increases by $e$ (from $0$ to $+e$).

1 mark for identifying that a down quark changes to an up quark and stating that the total quark charge increases by $e$.

The formula for resistivity is \(\rho = \frac{\pi d^2 R}{4L}\). The fractional uncertainty in resistivity is given by \(\frac{\Delta \rho}{\rho} = 2 \frac{\Delta d}{d} + \frac{\Delta R}{R} + \frac{\Delta L}{L}\). Substituting the percentage uncertainties: \(2(4\%) + 2.5\% + 0.2\% = 8\% + 2.5\% + 0.2\% = 10.7\%\). Therefore, the percentage uncertainty is 10.7%.

1 mark for the correct calculation: 2 * 4% + 2.5% + 0.2% = 10.7%. Award 0 marks for neglecting the factor of 2 for diameter (which gives 6.7%).

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

Now, use the resistivity formula:
\(\rho = \frac{R A}{L} = \frac{3.6 \times 1.134 \times 10^{-7}}{2.4} = 1.70 \times 10^{-7}\ \Omega\text{ m}\) (to 2 s.f.).

(b) Let the initial length be \(L\) and the initial cross-sectional area be \(A\).
Since mass and density are constant, the volume \(V = A L\) must remain constant.
If the length is doubled to \(2L\), the cross-sectional area must halve to \(0.5A\).

Using the formula for resistance:
\(R' = \rho \frac{2L}{0.5A} = 4 \left( \rho \frac{L}{A} \right) = 4 R\)
\(R' = 4 \times 3.6 = 14.4\ \Omega\).

(a) [Total: 4 marks]
- Formula for cross-sectional area or correct substitution: \(A = 1.13 \times 10^{-7}\text{ m}^2\) [1 mark]
- Resistivity formula: \(\rho = R A / L\) [1 mark]
- Correct substitution of values: \(\rho = 3.6 \times 1.134 \times 10^{-7} / 2.4\) [1 mark]
- Correct final answer with units: \(1.7 \times 10^{-7}\ \Omega\text{ m}\) [1 mark]

(b) [Total: 4.57 marks]
- Recognition that volume is constant, so area halving when length doubles [1.57 marks]
- Expressing new resistance formula in terms of new length and area: \(R' = \rho (2L) / (0.5A)\) [1 mark]
- Finding factor of 4 increase in resistance [1 mark]
- Correct final calculation: \(14.4\ \Omega\) (allow \(14\ \Omega\)) [1 mark]

(a) Express the resistance of each wire in terms of \(L\) and \(A\):
\(R_P = \rho_P \frac{L}{A} = 1.5 \times 10^{-8} \frac{L}{A}\)
\(R_Q = \rho_Q \frac{2.0L}{1.5A} = 4.5 \times 10^{-8} \times \frac{2.0}{1.5} \frac{L}{A} = 6.0 \times 10^{-8} \frac{L}{A}\)

Take the ratio of the two resistances:
\(\frac{R_P}{R_Q} = \frac{1.5 \times 10^{-8}}{6.0 \times 10^{-8}} = 0.25\) (which is shown).

(b) Since P and Q are in series, they carry the same current \(I\).
The total resistance is:
\(R_{\text{total}} = R_P + R_Q = R_P + 4.0 R_P = 5.0 R_P\)

The potential difference across P is given by the potential divider formula:
\(V_P = V_{\text{total}} \times \frac{R_P}{R_{\text{total}}} = 6.0 \times \frac{R_P}{5.0 R_P} = 1.2\text{ V}\).

(a) [Total: 4.57 marks]
- Expression for resistance of P: \(R_P = 1.5 \times 10^{-8} (L/A)\) [1 mark]
- Expression for resistance of Q: \(R_Q = 4.5 \times 10^{-8} (2.0L/1.5A)\) [1.57 marks]
- Calculation of coefficient for Q: \(6.0 \times 10^{-8}\) [1 mark]
- Division of the two expressions to get exactly 0.25 [1 mark]

(b) [Total: 4 marks]
- Relation between current and series connection / Potential Divider principle [1 mark]
- Equation for total resistance in terms of \(R_P\) or relative units: \(R_{\text{total}} = 5.0 R_P\) [1 mark]
- Correct substitution of values: \(V_P = 6.0 / 5.0\) [1 mark]
- Final answer: \(1.2\text{ V}\) [1 mark]

(a) At maximum height, the final vertical velocity is \(v = 0\).
Using the equation of motion:
\(v^2 = u^2 + 2as\)
\(0 = (12.5)^2 + 2(-9.81)s\)
\(19.62 s = 156.25\)
\(s = 7.96\text{ m} \approx 8.0\text{ m}\) (which is shown).

(b) Taking upwards as the positive direction, the vertical displacement \(s\) at the cliff base is \(-35.0\text{ m}\).
Using the equation of motion:
\(s = ut + \frac{1}{2}at^2\)
\(-35.0 = 12.5 t + \frac{1}{2}(-9.81)t^2\)
\(4.905 t^2 - 12.5 t - 35.0 = 0\)

Apply the quadratic formula:
\(t = \frac{-(-12.5) \pm \sqrt{(-12.5)^2 - 4(4.905)(-35.0)}}{2(4.905)}\)
\(t = \frac{12.5 \pm \sqrt{156.25 + 686.7}}{9.81}\)
\(t = \frac{12.5 \pm \sqrt{842.95}}{9.81} = \frac{12.5 \pm 29.03}{9.81}\)

Since time must be positive:
\(t = \frac{12.5 + 29.03}{9.81} = 4.23\text{ s} \approx 4.2\text{ s}\) (to 2 s.f.).

(a) [Total: 3.57 marks]
- Uses \(v^2 = u^2 + 2as\) with \(v=0\) and \(a=-9.81\text{ m s}^{-2}\) [1.57 marks]
- Correct substitution: \(0 = 12.5^2 - 2 \times 9.81 \times s\) [1 mark]
- Final value shown as \(7.96\text{ m}\) which rounds to \(8.0\text{ m}\) [1 mark]

(b) [Total: 5 marks]
- Uses \(s = ut + 0.5 a t^2\) with displacement \(-35.0\text{ m}\) (or calculates time in two stages) [1 mark]
- Correct substitution of values with consistent signs: \(-35 = 12.5t - 4.905t^2\) [1 mark]
- Formulates standard quadratic equation or correctly calculates peak-to-base stage displacement/time [1 mark]
- Solves quadratic equation or adds two-stage times correctly (e.g., \(1.27\text{ s} + 2.96\text{ s}\)) [1 mark]
- Correct final answer: \(4.2\text{ s}\) (accept \(4.23\text{ s}\)) [1 mark]

(a) The horizontal component \(u_x\) of the initial velocity is:
\(u_x = u \cos \theta = 24.0 \cos(35.0^\circ) = 19.66 \approx 19.7\text{ m s}^{-1}\)

The vertical component \(u_y\) of the initial velocity is:
\(u_y = u \sin \theta = 24.0 \sin(35.0^\circ) = 13.77 \approx 13.8\text{ m s}^{-1}\)

(b) Find the time of flight \(T\). Since the ground is horizontal, the vertical displacement over the complete flight is zero:
\(s_y = u_y T - \frac{1}{2} g T^2 = 0\)
\(T = \frac{2 u_y}{g} = \frac{2 \times 13.77}{9.81} = 2.807\text{ s}\)

The horizontal distance traveled (range) is:
\(R_x = u_x T = 19.66 \times 2.807 = 55.2\text{ m}\) (to 3 s.f.).

(a) [Total: 4 marks]
- Formula for horizontal component: \(u_x = u \cos \theta\) [1 mark]
- Value for \(u_x = 19.7\text{ m s}^{-1}\) (or \(19.66\text{ m s}^{-1}\)) [1 mark]
- Formula for vertical component: \(u_y = u \sin \theta\) [1 mark]
- Value for \(u_y = 13.8\text{ m s}^{-1}\) (or \(13.77\text{ m s}^{-1}\)) [1 mark]

(b) [Total: 4.57 marks]
- Use of vertical kinematic equation to determine time of flight: \(0 = u_y T - 0.5 g T^2\) or time to max height \(t = u_y/g\) [1.57 marks]
- Correct calculation of total time \(T = 2.81\text{ s}\) [1 mark]
- Use of horizontal distance equation: \(s_x = u_x T\) [1 mark]
- Correct final answer: \(55.2\text{ m}\) (accept \(55.0\text{ m}\) to \(55.3\text{ m}\) depending on rounding) [1 mark]

(a) When the buoy is floating in equilibrium, the upthrust \(U\) equals the weight \(W\) of the buoy:
\(U = W\)
\(\rho_{\text{water}} V_{\text{submerged}} g = m g\)
\(\rho_{\text{water}} A h = m\)
\(1025 \times 0.65 \times h = 140\)
\(666.25 h = 140\)
\(h = 0.210\text{ m} \approx 0.21\text{ m}\) (which is shown).

(b) When the vertical force of \(450\text{ N}\) is applied, the new equilibrium state requires the new upthrust \(U'\) to equal the sum of the buoy's weight and the applied force:
\(U' = m g + F\)
\(\rho_{\text{water}} A h' g = m g + F\)
\(1025 \times 0.65 \times 9.81 \times h' = (140 \times 9.81) + 450\)
\(6535.9 h' = 1373.4 + 450\)
\(6535.9 h' = 1823.4\)
\(h' = 0.279\text{ m} \approx 0.28\text{ m}\).

(a) [Total: 4.57 marks]
- Principle of flotation: Upthrust = Weight of buoy [1 mark]
- Formula for upthrust: \(U = \rho V g\) [1 mark]
- Equating and simplifying: \(\rho A h = m\) [1.57 marks]
- Correct calculation leading to \(0.21\text{ m}\) [1 mark]

(b) [Total: 4 marks]
- New equilibrium equation: \(U' = W + F\) [1 mark]
- Calculates total downward force: \(1373\text{ N} + 450\text{ N} = 1823\text{ N}\) (or uses change in upthrust \(\Delta U = 450\text{ N}\) to find \(\Delta h = 0.069\text{ m}\)) [1 mark]
- Correct substitution of values into the formula [1 mark]
- Final answer: \(0.28\text{ m}\) (accept \(0.279\text{ m}\)) [1 mark]

(a) Use the equation of motion for a body falling from rest:
\(h = \frac{1}{2} g t^2 \implies g = \frac{2h}{t^2}\)
\(g = \frac{2 \times 1.80}{0.61^2} = \frac{3.60}{0.3721} = 9.675\text{ m s}^{-2} \approx 9.7\text{ m s}^{-2}\).

(b) The formula for \(g\) is \(g = \frac{2h}{t^2}\).
The fractional uncertainty equation is:
\(\frac{\Delta g}{g} = \frac{\Delta h}{h} + 2 \frac{\Delta t}{t}\)

Calculate each fractional uncertainty term:
\(\frac{\Delta h}{h} = \frac{0.02}{1.80} = 0.0111\) (or \(1.11\%\))
\(\frac{\Delta t}{t} = \frac{0.02}{0.61} = 0.0328\) (or \(3.28\%\))

Now combine them:
\(\frac{\Delta g}{g} = 0.0111 + 2(0.0328) = 0.0111 + 0.0656 = 0.0767\) (or \(7.67\%\) which is \(7.7\%\) or \(8\%\)).

(c) First, find the absolute uncertainty \(\Delta g\):
\(\Delta g = g \times 0.0767 = 9.675 \times 0.0767 = 0.742\text{ m s}^{-2}\).

An absolute uncertainty is usually quoted to 1 significant figure, so:
\(\Delta g \approx 0.7\text{ m s}^{-2}\).

This means \(g\) must be rounded to match the precision (one decimal place):
\(g = 9.7 \pm 0.7\text{ m s}^{-2}\).

(a) [Total: 2.57 marks]
- Recalls and rearranges formula: \(g = 2h/t^2\) [1.57 marks]
- Correct calculation: \(9.7\text{ m s}^{-2}\) (accept \(9.67\text{ m s}^{-2}\)) [1 mark]

(b) [Total: 3 marks]
- Formula for combining uncertainties: \(\frac{\Delta g}{g} = \frac{\Delta h}{h} + 2\frac{\Delta t}{t}\) [1 mark]
- Correct substitution of values: \(\frac{0.02}{1.80} + 2 \left(\frac{0.02}{0.61}\right)\) [1 mark]
- Correct final percentage: \(7.7\%\) (or \(8\%\)) [1 mark]

(c) [Total: 3 marks]
- Calculates absolute uncertainty: \(0.74\text{ m s}^{-2}\) [1 mark]
- Rounds absolute uncertainty to 1 sig fig: \(\pm 0.7\text{ m s}^{-2}\) [1 mark]
- States final value of \(g\) with matching decimal precision: \(9.7 \pm 0.7\text{ m s}^{-2}\) [1 mark]

(a) A neutron consists of one up (u) quark and two down (d) quarks, so its composition is \(udd\).
Its charge is:
\(+\frac{2}{3}e - \frac{1}{3}e - \frac{1}{3}e = 0\).

A proton consists of two up (u) quarks and one down (d) quark, so its composition is \(uud\).
Its charge is:
\(+\frac{2}{3}e + \frac{2}{3}e - \frac{1}{3}e = +1e\).

(b) During beta-minus decay, a neutron (\(udd\)) is converted into a proton (\(uud\)).
Therefore, one down (\(d\)) quark changes its flavor to an up (\(u\)) quark.

(c) Beta-minus decay is a weak interaction process, so the fundamental interaction responsible is the weak interaction (or weak nuclear force).

(a) [Total: 4.57 marks]
- Quark composition of neutron: \(udd\) [1 mark]
- Quark charge of neutron: \(0\) [1 mark]
- Quark composition of proton: \(uud\) [1.57 marks]
- Quark charge of proton: \(+1e\) (or \(+e\)) [1 mark]

(b) [Total: 2 marks]
- Identifies down quark change to up quark [1 mark]
- Writes this explicitly as \(d \rightarrow u\) [1 mark]

(c) [Total: 2 marks]
- Identifies weak interaction / weak nuclear force [2 marks] (Do not accept 'strong force' or 'gravity')

### Typical Experimental Readings & Calculation

**Part (a)**
* Wire diameter measured at different points:
\(d_1 = 0.315\text{ mm}\), \(d_2 = 0.312\text{ mm}\), \(d_3 = 0.318\text{ mm}\).
Average \(d = 0.315\text{ mm}\).

**Part (d)**
Typical table of results:

| \(x\ /\text{ m}\) | \(V\ /\text{ V}\) | \((1/x)\ /\text{ m}^{-1}\) | \((1/V)\ /\text{ V}^{-1}\) |
|---|---|---|---|
| 0.200 | 0.29 | 5.00 | 3.45 |
| 0.300 | 0.39 | 3.33 | 2.56 |
| 0.400 | 0.47 | 2.50 | 2.13 |
| 0.500 | 0.53 | 2.00 | 1.89 |
| 0.600 | 0.58 | 1.67 | 1.72 |
| 0.800 | 0.66 | 1.25 | 1.52 |

**Part (e) & (f)**
* Plot \(1/V\) against \(1/x\).
* Gradient calculation:
Using two points on the line of best fit, e.g., \((5.00, 3.45)\) and \((1.25, 1.52)\):
$$\text{Gradient} = \frac{3.45 - 1.52}{5.00 - 1.25} = \frac{1.93}{3.75} \approx 0.515\text{ m V}^{-1}$$
* y-intercept calculation:
Using \(y = mx + C\):
$$1.52 = 0.515(1.25) + B \implies B = 1.52 - 0.64 = 0.88\text{ V}^{-1}$$

**Part (g)**
* Comparing the equation \(\frac{1}{V} = \frac{A}{x} + B\) to the straight-line equation \(y = mx + c\):
* \(A = \text{gradient} = 0.515\text{ m V}^{-1}\)
* \(B = \text{y-intercept} = 0.88\text{ V}^{-1}\)

**1. Quality of measurements and collection of data (8 marks)**
* **Diameter measurement (2 marks):**
* 1 mark: Raw value of \(d\) recorded to the nearest 0.01 mm or 0.001 mm with unit.
* 1 mark: Repeated measurements in different positions/directions and average shown.
* **Table of results (6 marks):**
* 1 mark: Six sets of readings of \(x\) and \(V\) recorded with no empty cells.
* 1 mark: Range of \(x\) spans at least \(0.500\text{ m}\).
* 1 mark: Column headings with correct units: \(x/\text{m}\), \(V/\text{V}\), \((1/x)/\text{m}^{-1}\), and \((1/V)/\text{V}^{-1}\).
* 1 mark: Raw values of \(x\) recorded to the nearest millimetre (0.001 m) and \(V\) to the nearest 0.01 V.
* 1 mark: Calculated values of \(1/x\) and \(1/V\) calculated to the same number of significant figures as (or one more than) raw data.
* 1 mark: Quality of data: points on the graph lie close to the best-fit straight line.

**2. Graph plotting and analysis (4 marks)**
* 1 mark: Linear axes. Chosen scales must allow plotted points to occupy more than half of the grid in both x and y directions. No awkward scales.
* 1 mark: Correct plotting of all points to within half a small square.
* 1 mark: Best-fit straight line drawn, with a balanced distribution of points about the line.

**3. Gradient and Intercept (4 marks)**
* 1 mark: Gradient triangle hypotenuse is at least half the length of the drawn line.
* 1 mark: Correct coordinate readings and gradient calculation.
* 1 mark: y-intercept read directly from the y-axis (if \(x=0\) is included) or calculated using a point on the line in \(y = mx + c\).
* 1 mark: Intercept calculation mathematically correct.

**4. Constants and Units (4 marks)**
* 1 mark: Constant \(A\) equated to the gradient.
* 1 mark: Constant \(B\) equated to the y-intercept.
* 1 mark: Correct units given: \(\text{m V}^{-1}\) (or \(\text{cm V}^{-1}\)) for \(A\) and \(\text{V}^{-1}\) for \(B\).
* 1 mark: Values of \(A\) and \(B\) written to 2 or 3 significant figures.

### Typical Experimental Readings & Calculation

**Part (a)**
* Thickness \(t = 6.35\text{ mm} = 6.35 \times 10^{-3}\text{ m}\)
* Width \(w = 25.1\text{ mm} = 2.51 \times 10^{-2}\text{ m}\)
* Percentage uncertainty in \(t\):
Using absolute uncertainty \(\Delta t = 0.1\text{ mm}\) (since wooden rules have some surface unevenness):
$$\%\text{ uncertainty} = \frac{0.1}{6.35} \times 100\% \approx 1.6\%$$

**Part (b) & (c)**
* For \(L_1 = 0.750\text{ m}\):
\(h_0 = 42.5\text{ cm}\), \(h_1 = 34.2\text{ cm}\)
\(y_1 = 42.5 - 34.2 = 8.3\text{ cm} = 0.083\text{ m}\)
* For \(L_2 = 0.550\text{ m}\):
\(h_0 = 42.5\text{ cm}\), \(h_1 = 39.4\text{ cm}\)
\(y_2 = 42.5 - 39.4 = 3.1\text{ cm} = 0.031\text{ m}\)

**Part (d)**
* Calculate \(C_1\):
$$C_1 = \frac{y_1}{L_1^3} = \frac{0.083}{0.750^3} = 0.197\text{ m}^{-2}$$
* Calculate \(C_2\):
$$C_2 = \frac{y_2}{L_2^3} = \frac{0.031}{0.550^3} = 0.186\text{ m}^{-2}$$
* Percentage difference between \(C_1\) and \(C_2\):
$$\%\text{ difference} = \frac{|0.197 - 0.186|}{0.197} \times 100\% = 5.6\%$$
* Conclusion:
The percentage difference of 5.6% is less than the standard criterion of 10%. Therefore, the suggested relationship is supported by the experimental results.

**1. Measurements and Uncertainty (5 marks)**
* 1 mark: Width \(w\) and thickness \(t\) recorded to the correct decimal places (nearest 0.1 mm for caliper or 0.01 mm for micrometer) with units.
* 1 mark: Percentage uncertainty in \(t\) calculated correctly, using an absolute uncertainty in the range of \(0.01\text{ mm}\) to \(0.1\text{ mm}\).
* 1 mark: Raw heights \(h_0\) and \(h_1\) recorded with units (mm or cm) and to the nearest millimetre.
* 1 mark: First deflection value \(y_1\) correctly calculated with unit.
* 1 mark: Second deflection value \(y_2\) correctly calculated for \(L = 0.550\text{ m}\), showing \(y_2 < y_1\).

**2. Data Analysis and Conclusion (4 marks)**
* 1 mark: Two values of the constant \(C\) calculated correctly with appropriate significant figures (typically 2 or 3 SF).
* 1 mark: Correctly calculated percentage difference between the two values of \(C\).
* 1 mark: Clear, explicit conclusion stating whether the results support the relationship.
* 1 mark: Justification of the conclusion based on a specified numerical percentage criterion (e.g. comparing the percentage difference against 10% or 20%).

**3. Limitations and Sources of Uncertainty (4 marks)**
* 1 mark per identified limitation, up to a maximum of 4:
1. Two sets of readings are not enough to draw a valid conclusion.
2. The wooden rule twists or curves slightly sideways when loaded.
3. Difficult to align the vertical scale accurately perpendicular to the table, causing cosine/parallax errors.
4. Parallax error when reading the horizontal rule position against the vertical rule scale.
5. The clamp compresses the wooden rule or shifts slightly under load, altering the zero level.

**4. Suggested Improvements (7 marks)**
* 1 mark per suggested improvement, up to a maximum of 7 (must match identified limitations):
1. Take more readings for several different values of \(L\) and plot a graph of \(y\) against \(L^3\).
2. Use a stiffer material for the cantilever to avoid twisting (or guide the load vertically).
3. Use a set square or plumb line to ensure the vertical scale rule is perfectly vertical.
4. Attach a pointer (e.g., a needle or fine pin) to the end of the cantilever to read off the vertical scale to eliminate parallax error.
5. Use a rigid metal clamp with protective pads to distribute pressure and prevent slipping at the support.
6. Use a traveling microscope to measure vertical deflection instead of a vertical metre rule.