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Using conservation of energy, both stones start with the same kinetic energy per unit mass \(\frac{1}{2}u^2\) and the same gravitational potential energy per unit mass \(gh\) relative to the sea. Therefore, they must hit the sea with the same kinetic energy per unit mass, which means they have the same final speed \(v = \sqrt{u^2 + 2gh}\).

Alternatively, using the equations of motion:
For stone Y: \(v^2 = u^2 + 2gh\).
For stone X: it rises to a maximum height, stops, and then falls. As it passes the edge of the cliff on its downward path, its speed is again \(u\) downwards. From this point, its motion is identical to that of stone Y, meaning it also hits the sea with speed \(v = \sqrt{u^2 + 2gh}\).

C is the correct response. 1 mark is awarded for identifying that the final speeds of both stones are equal based on either conservation of energy or equations of motion.

Let the final velocities of the mass \(2m\) and mass \(m\) be \(v_1\) and \(v_2\) respectively.

By conservation of linear momentum:
\(2m(v) + m(-v) = 2m(v_1) + m(v_2)\)
\(mv = 2mv_1 + mv_2 \implies v = 2v_1 + v_2 \quad [1]\)

Since the collision is perfectly elastic, the relative speed of approach is equal to the relative speed of separation:
\(u_1 - u_2 = v_2 - v_1\)
\(v - (-v) = v_2 - v_1 \implies 2v = v_2 - v_1 \quad [2]\)

Subtracting equation [1] from equation [2]:
\(2v - v = (v_2 - v_1) - (2v_1 + v_2)\)
\(v = -3v_1 \implies v_1 = -\frac{1}{3}v\)

Substituting \(v_1\) back into equation [2]:
\(v_2 = 2v + v_1 = 2v - \frac{1}{3}v = \frac{5}{3}v\)

A is the correct response. 1 mark is awarded for applying conservation of momentum and the elastic collision relative speed relationship to solve for both final velocities.

For the block to remain at rest, the forces acting in the vertical direction must be in equilibrium.
The only vertical forces acting on the block are:
1. The downward gravitational pull (weight), equal to \(mg\).
2. The upward static frictional force, equal to \(f\).

Since the block remains stationary under both \(F_{\text{min}}\). and \(2F_{\text{min}}\), the acceleration is zero in the vertical direction. Therefore, the upward frictional force must exactly balance the weight of the block:
\(f = mg\)

Increasing the horizontal force increases the maximum possible value of static friction that *could* act before sliding, but the actual frictional force currently exerted remains equal to \(mg\).

B is the correct response. 1 mark is awarded for applying Newton's first law to vertical forces in equilibrium and concluding that frictional force must equal the weight.

We apply the conservation of linear momentum to the system consisting of the rocket and booster:

Initial Momentum = Final Momentum
\(M v = (0.80M)(1.20v) + (0.20M) v_{\text{booster}}\)

Divide each term by the mass \(M\):
\(v = 0.96v + 0.20v_{\text{booster}}\)
\(0.04v = 0.20v_{\text{booster}}\)
\(v_{\text{booster}} = \frac{0.04v}{0.20} = 0.20v\)

Since the sign is positive, the velocity of the ejected booster is \(0.20v\) in the same direction as the rocket's original motion.

A is the correct response. 1 mark is awarded for correctly setting up and solving the conservation of momentum equation.

The relationship between e.m.f., current, load resistance, and internal resistance is given by:
\(E = I(R + r)\)

For the two different settings of the variable resistor, we have:
\(E = I_1(R_1 + r)\) and \(E = I_2(R_2 + r)\)

Equating the two expressions for \(E\):
\(I_1(R_1 + r) = I_2(R_2 + r)\)
\(I_1 R_1 + I_1 r = I_2 R_2 + I_2 r\)

Rearranging to group terms in \(r\) on one side:
\(I_1 r - I_2 r = I_2 R_2 - I_1 R_1\)
\(r(I_1 - I_2) = I_2 R_2 - I_1 R_1\)
\(r = \frac{I_2 R_2 - I_1 R_1}{I_1 - I_2}\)

B is the correct response. 1 mark is awarded for setting up the two simultaneous loop equations and correctly rearranging them to solve for internal resistance \(r\).

1. For a negative temperature coefficient (NTC) thermistor, its resistance increases as the temperature decreases.
2. In a series potential divider, the output voltage across the thermistor is given by:
\(V_{\text{out}} = E \times \frac{R_{\text{thermistor}}}{R_{\text{thermistor}} + R}\)
As the resistance of the thermistor increases, it represents a larger fraction of the total circuit resistance, meaning a larger share of the constant supply e.m.f. is dropped across it. Consequently, \(V_{\text{out}}\) increases.

A is the correct response. 1 mark is awarded for correctly stating that the NTC thermistor resistance increases at lower temperatures, and that this results in an increased output voltage across it.

Applying Kirchhoff's second law around the loop:
\(\sum E = \sum I R\)

Because the two batteries are connected in opposition, the net e.m.f. of the circuit is:
\(E_{\text{net}} = 12.0\text{ V} - 6.0\text{ V} = 6.0\text{ V}\)

The total resistance in the loop is the sum of the external resistance and both internal resistances:
\(R_{\text{total}} = 3.0\,\Omega + 1.0\,\Omega + 2.0\,\Omega = 6.0\,\Omega\)

The current \(I\) in the circuit is:
\(I = \frac{E_{\text{net}}}{R_{\text{total}}} = \frac{6.0\text{ V}}{6.0\,\Omega} = 1.0\text{ A}\)

A is the correct response. 1 mark is awarded for calculating the net opposing e.m.f. and total resistance, and solving for the current.

For vertical motion:
Using \(v_y^2 = u_y^2 + 2g y\), where the initial vertical velocity \(u_y = 0\):
\(v_y^2 = 0 + 2 \times 9.81\text{ m s}^{-2} \times 45\text{ m} = 882.9\text{ m}^2\text{ s}^{-2}\)

For horizontal motion:
The horizontal component of the velocity remains constant since there is no air resistance:
\(v_x = 15\text{ m s}^{-1} \implies v_x^2 = 225\text{ m}^2\text{ s}^{-2}\)

To find the magnitude of the final velocity (speed):
\(v = \sqrt{v_x^2 + v_y^2} = \sqrt{225 + 882.9} = \sqrt{1107.9} \approx 33.3\text{ m s}^{-1}\)

To 2 significant figures, this is \(33\text{ m s}^{-1}\).

C is the correct response. 1 mark is awarded for finding the final vertical velocity, keeping horizontal velocity constant, and vectorially summing them to calculate the final speed.

Taking the upward direction as positive, the initial velocity is \(+u\) and the acceleration of free fall is \(-g\). When the ball hits the ground, its vertical displacement is \(-h\). Using the equation of motion \(s = ut + \frac{1}{2}at^2\), we substitute these values: \(-h = ut + \frac{1}{2}(-g)t^2 = ut - \frac{1}{2}gt^2\). Multiplying both sides by \(-1\) gives \(h = \frac{1}{2}gt^2 - ut\).

1 mark for the correct substitution of directions into the equation of motion \(s = ut + \frac{1}{2}at^2\) to yield \(h = \frac{1}{2}gt^2 - ut\).

The mass of water striking the wall per second is \(\frac{\Delta m}{\Delta t} = \rho A v\). Choosing the direction of the initial jet as positive: initial velocity \(u = +v\) and final velocity \(v_f = -0.4v\). The change in momentum of the water per second is \(\frac{\Delta p}{\Delta t} = \frac{\Delta m}{\Delta t}(v_f - u) = \rho A v (-0.4v - v) = -1.4\rho A v^2\). By Newton's second law, this equals the force on the water. By Newton's third law, the force exerted on the wall has the same magnitude but opposite direction: \(F = 1.4\rho A v^2\).

1 mark for calculating the mass flow rate \(\rho A v\), setting up the correct change in velocity \(\Delta v = 1.4v\), and combining them to get the force \(1.4\rho A v^2\).

Let the total current in the circuit be \(I\). This is the current flowing through the resistor of resistance \(R\), so the power dissipated in it is \(P_1 = I^2 R\). The remaining current splits between the parallel resistors of resistances \(2R\) and \(3R\). The current \(I_2\) through the resistor of resistance \(2R\) is given by the current divider formula: \(I_2 = I \times \frac{3R}{2R + 3R} = 0.6I\). The power dissipated in this resistor is \(P_2 = I_2^2 (2R) = (0.6I)^2 (2R) = 0.72 I^2 R\). The ratio of the powers is: \(\frac{P_1}{P_2} = \frac{I^2 R}{0.72 I^2 R} = \frac{1}{0.72} = \frac{25}{18}\).

1 mark for finding the current through the \(2R\) resistor as \(0.6I\), expressing the powers \(P_1\) and \(P_2\) in terms of \(I\) and \(R\), and calculating the ratio as \(\frac{25}{18}\).

Initially, the upthrust equals the weight of the cylinder: \(F_u = \rho A d g = W_{\text{cyl}}\). When the mass \(m\) is placed on top, the new total downward force is \(W_{\text{cyl}} + mg\). In the new equilibrium, the cylinder is submerged to a depth of \(d+x\), so the new upthrust is \(F_{u,\text{new}} = \rho A (d+x) g\). Since \(F_{u,\text{new}} = W_{\text{cyl}} + mg\), we have \(\rho A (d+x) g = \rho A d g + mg\), which simplifies to \(\rho A x g = mg\). Solving for \(x\) gives \(x = \frac{m}{\rho A}\).

1 mark for setting the additional upthrust equal to the added weight (\(\rho A x g = mg\)) and simplifying to find \(x = \frac{m}{\rho A}\).

At temperature \(T_1\), let the thermistor resistance be \(R_T\). The potential divider equation gives: \(V_{\text{out}} = V_{\text{in}} \times \frac{R_T}{R_T + 10\text{ k}\Omega} \implies 8.0 = 12 \times \frac{R_T}{R_T + 10\text{ k}\Omega}\). Solving this equation: \(8.0(R_T + 10) = 12R_T \implies 8.0R_T + 80 = 12R_T \implies 4.0R_T = 80 \implies R_T = 20\text{ k}\Omega\). At temperature \(T_2\), the thermistor resistance decreases to half, so \(R_{T,\text{new}} = 10\text{ k}\Omega\). The new output voltage is: \(V_{\text{out,new}} = 12 \times \frac{10\text{ k}\Omega}{10\text{ k}\Omega + 10\text{ k}\Omega} = 12 \times 0.5 = 6.0\text{ V}\).

1 mark for determining the initial thermistor resistance (\(20\text{ k}\Omega\)), halving it to find the new resistance (\(10\text{ k}\Omega\)), and calculating the new output voltage as \(6.0\text{ V\).

Let's find the velocity and displacement in each interval. 1) Interval \(t = 0\) to \(2.0\text{ s}\): Initial velocity \(v_0 = 0\). Final velocity \(v_2 = v_0 + a_1 t_1 = 0 + 3.0 \times 2.0 = 6.0\text{ m s}^{-1}\). Displacement \(s_1 = \frac{v_0 + v_2}{2} \times t_1 = \frac{0 + 6.0}{2} \times 2.0 = 6.0\text{ m}\). 2) Interval \(t = 2.0\text{ s}\) to \(6.0\text{ s}\): Initial velocity \(v_2 = 6.0\text{ m s}^{-1}\). Final velocity \(v_6 = v_2 + a_2 t_2 = 6.0 + 1.0 \times 4.0 = 10.0\text{ m s}^{-1}\). Displacement \(s_2 = \frac{v_2 + v_6}{2} \times t_2 = \frac{6.0 + 10.0}{2} \times 4.0 = 32.0\text{ m}\). 3) Interval \(t = 6.0\text{ s}\) to \(8.0\text{ s}\): Initial velocity \(v_6 = 10.0\text{ m s}^{-1}\). Final velocity \(v_8 = v_6 + a_3 t_3 = 10.0 + (-2.0) \times 2.0 = 6.0\text{ m s}^{-1}\). Displacement \(s_3 = \frac{v_6 + v_8}{2} \times t_3 = \frac{10.0 + 6.0}{2} \times 2.0 = 16.0\text{ m}\). Total displacement is \(s = s_1 + s_2 + s_3 = 6.0 + 32.0 + 16.0 = 54.0\text{ m}\).

1 mark for calculating the correct boundary velocities (\(6.0\text{ m s}^{-1}\), \(10.0\text{ m s}^{-1}\), \(6.0\text{ m s}^{-1}\)) and integrating/summing the displacement for each stage to find the total displacement of \(54\text{ m\).

Taking the direction to the right as positive: Initial momentum: \(p_i = m_X u_X + m_Y u_Y = (2.0 \times 4.0) + (3.0 \times (-2.0)) = 8.0 - 6.0 = +2.0\text{ kg m s}^{-1}\). After collision, block X has velocity \(v_X = -1.0\text{ m s}^{-1}\). Let the velocity of block Y be \(v_Y\). By conservation of momentum: \(m_X v_X + m_Y v_Y = p_i \implies (2.0 \times (-1.0)) + 3.0 v_Y = 2.0 \implies 3.0 v_Y = 4.0 \implies v_Y = +1.33\text{ m s}^{-1}\) (to the right). Now we check if kinetic energy is conserved: Initial kinetic energy \(E_{k,i} = \frac{1}{2}(2.0)(4.0)^2 + \frac{1}{2}(3.0)(-2.0)^2 = 16.0 + 6.0 = 22.0\text{ J}\). Final kinetic energy \(E_{k,f} = \frac{1}{2}(2.0)(-1.0)^2 + \frac{1}{2}(3.0)(1.33)^2 \approx 1.0 + 2.67 = 3.67\text{ J}\). Since \(E_{k,f} < E_{k,i}\), kinetic energy is not conserved, meaning the collision is inelastic.

1 mark for using the conservation of momentum with correct signs to find the final velocity of block Y (\(1.3\text{ m s}^{-1}\) to the right), and comparing initial and final kinetic energies to show that the collision is inelastic.

Let \(V_X\) and \(V_Y\) be the volumes of wires X and Y. Since the volume is conserved, we have: \(V_X = V_Y \implies A \cdot L = A_Y \cdot (2L) \implies A_Y = \frac{A}{2}\). The resistance \(R\) of a wire is given by \(R = \frac{\rho \cdot \text{length}}{\text{area}}\). For wire X: \(R = \frac{\rho L}{A}\). For wire Y: \(R_Y = \frac{\rho (2L)}{A_Y} = \frac{\rho (2L)}{\frac{A}{2}} = 4 \left( \frac{\rho L}{A} \right) = 4R\).

1 mark for recognizing that conserving volume requires the cross-sectional area of wire Y to be half that of wire X, and substituting these dimensions into the resistivity formula to obtain a resistance of \(4R\).

Using the equation of motion \(s = ut + \frac{1}{2}at^2\) with downwards as the positive direction:

- The displacement \(s\) is the final position minus the initial position. Since the stone ends up at the sea below (at a distance \(h\) below the starting point), the displacement is \(s = +h\).
- The initial velocity is upwards, so \(u_{\text{initial}} = -u\).
- The acceleration is downwards, so \(a = +g\).

Substituting these values into the equation of motion gives:
\(h = -uT + \frac{1}{2}gT^2\).

1 mark for the correct substitution of direction-sensitive signs into the equation of motion, identifying that displacement is positive, initial velocity is negative, and acceleration is positive.

Resolve forces parallel to the slope in the direction of acceleration (up the slope):

- The pulling force is \(F\) up the slope.
- The component of the weight acting down the slope is \(mg\sin\theta\).
- The frictional force \(f\) acts down the slope because it opposes the motion.

Applying Newton's second law:
\(\Sigma F = ma\)
\(F - mg\sin\theta - f = ma\)

Solving for \(F\):
\(F = ma + mg\sin\theta + f\)
\(F = m(a + g\sin\theta) + f\).

1 mark for correctly applying Newton's second law parallel to the slope, identifying the correct components of forces acting down the slope (weight component \(mg\sin\theta\) and friction \(f\)), and rearranging to solve for \(F\).

Let the initial direction of motion of X be positive.

Using the principle of conservation of momentum:
Total initial momentum = Total final momentum
\((2m)(3v) + (3m)(-v) = (2m + 3m)V\)
\(6mv - 3mv = 5mV\)
\(3mv = 5mV \implies V = 0.6v\)

Now, calculate the initial kinetic energy:
\(E_{k, \text{initial}} = \frac{1}{2}(2m)(3v)^2 + \frac{1}{2}(3m)(-v)^2 = 9mv^2 + 1.5mv^2 = 10.5mv^2\)

Calculate the final kinetic energy:
\(E_{k, \text{final}} = \frac{1}{2}(5m)(0.6v)^2 = 2.5m(0.36v^2) = 0.9mv^2\)

Therefore, the total loss of kinetic energy is:
\(\Delta E_k = 10.5mv^2 - 0.9mv^2 = 9.6mv^2\).

1 mark for using conservation of momentum to find the correct velocity \(V = 0.6v\), and calculating the difference between the total initial kinetic energy and total final kinetic energy to obtain a loss of \(9.6mv^2\).

For an NTC thermistor, as the temperature increases, its resistance decreases.

The output voltage \(V_{\text{out}}\) across the fixed resistor is given by the potential divider formula:
\(V_{\text{out}} = E \times \frac{R}{R_{\text{thermistor}} + R}\)

As the thermistor resistance \(R_{\text{thermistor}}\) decreases, the total resistance of the circuit decreases, which increases the current. Consequently, the potential difference \(V_{\text{out}}\) across the fixed resistor \(R\) increases.

1 mark for identifying that an NTC thermistor's resistance decreases with temperature, and that this leads to an increase in the output voltage across the fixed resistor.

Let the potential at the lower junction be \(0\text{ V}\) and the potential at the upper junction be \(V\).

According to Kirchhoff's first law at the upper junction, the sum of currents entering equals the sum of currents leaving. Let the currents flowing upwards in the left and right branches be \(I_1\) and \(I_3\) respectively, and the current flowing downwards in the middle branch be \(I_2\).

Applying Ohm's law to each branch:
- Left branch: \(I_1 = \frac{9.0 - V}{3.0}\)
- Middle branch: \(I_2 = \frac{V}{6.0}\)
- Right branch: \(I_3 = \frac{6.0 - V}{2.0}\)

Using Kirchhoff's first law:
\(I_1 + I_3 = I_2\)
\(\frac{9.0 - V}{3.0} + \frac{6.0 - V}{2.0} = \frac{V}{6.0}\)

Multiply the entire equation by \(6.0\) to clear denominators:
\(2(9.0 - V) + 3(6.0 - V) = V\)
\(18.0 - 2V + 18.0 - 3V = V\)
\(36.0 - 5V = V\)
\(6V = 36.0 \implies V = 6.0\text{ V}\)

The current in the middle resistor is:
\(I_2 = \frac{V}{6.0} = \frac{6.0\text{ V}}{6.0\\ \Omega} = 1.0\text{ A}\).

1 mark for setting up the junction potential equation using Kirchhoff's laws and Ohm's law, solving for the potential \(V = 6.0\text{ V}\), and calculating the correct current of \(1.0\text{ A}\).

Since the crate starts from rest and finishes at rest, there is no net change in its kinetic energy. The useful work output is entirely equal to the increase in gravitational potential energy of the crate:

Useful work output \(= \Delta E_p = mgh = 250 \times 9.81 \times 12 = 29430\text{ J}\)

The total electrical energy input to the motor is:
Total energy input \(= \text{Power} \times \text{time} = 2.5 \times 10^3\text{ W} \times 15\text{ s} = 37500\text{ J}\)

The average efficiency \(\eta\) of the motor system is:
\(\eta = \frac{\text{Useful work output}}{\text{Total energy input}} \times 100\\%\)
\(\eta = \frac{29430}{37500} \times 100\\% \approx 78.5\\%\)

This is closest to \(78\\%\).

1 mark for calculating the useful potential energy gain (\(29430\text{ J}\)) and the total input energy (\(37500\text{ J}\)), and dividing the two to find the efficiency of approximately \(78\\%\).

The Young modulus \(E\) of the metal is defined as:
\(E = \frac{\text{Stress}}{\text{Strain}} = \frac{F/A}{x/L} = \frac{FL}{Ax}\)

Therefore, the extension \(x\) is given by:
\(x = \frac{FL}{AE}\)

Since the cross-sectional area of a wire of diameter \(d\) is \(A = \frac{\pi d^2}{4}\), the extension is:
\(x = \frac{4FL}{\pi d^2 E}\)

Because \(F\), \(E\), and \(\pi\) are constant for both wires, the extension is proportional to \(\frac{L}{d^2}\):
\(x \propto \frac{L}{d^2}\)

Let's write the ratio of the extensions:
\(\frac{x_{\text{X}}}{x_{\text{Y}}} = \frac{L_{\text{X}}}{L_{\text{Y}}} \times \left(\frac{d_{\text{Y}}}{d_{\text{X}}}\right)^2\)

Substituting the given values:
\(\frac{x_{\text{X}}}{x_{\text{Y}}} = \frac{L}{2L} \times \left(\frac{2d}{d}\right)^2 = \frac{1}{2} \times 4 = 2\).

1 mark for expressing the extension in terms of length and diameter, substituting the relative dimensions of wires X and Y, and obtaining the correct ratio of 2.

For a tube closed at one end and open at the other, a displacement node must form at the closed end and a displacement antinode must form at the open end.

- In the fundamental mode (first harmonic), the length of the tube \(L\) is equal to a quarter of a wavelength: \(L = \frac{\lambda_1}{4} \implies \lambda_1 = 4L\).
- In the first overtone (the second possible resonant frequency, which is the third harmonic), the next stationary wave pattern has a node-antinode-node-antinode structure. Here, the length \(L\) is equal to three-quarters of a wavelength: \(L = \frac{3\lambda_3}{4} \implies \lambda_3 = \frac{4L}{3}\).

Using the wave equation \(v = f\lambda\), the frequency \(f_3\) of the first overtone is:
\(f_3 = \frac{v}{\lambda_3} = \frac{v}{\frac{4L}{3}} = \frac{3v}{4L}\).

1 mark for identifying that the first overtone in a closed-open tube corresponds to a length of three-quarters of a wavelength, and using the wave speed relation to derive the frequency \(\frac{3v}{4L}\).

Using the equation of motion \(v^2 = u^2 + 2as\):
Taking upwards as the positive direction:
- Initial velocity \(v_i = u\)
- Final velocity \(v_f = -3u\) (downwards)
- Acceleration \(a = -g\)
- Displacement \(s = -h\)

Substituting these values:
\((-3u)^2 = u^2 + 2(-g)(-h)\)
\(9u^2 = u^2 + 2gh\)
\(8u^2 = 2gh\)
\(h = \frac{4u^2}{g}\)

1 mark for identifying the correct kinematic equation and substituting the values correctly to obtain \(h = \frac{4u^2}{g}\).

When the ball is released, it is moving upwards with its maximum speed. Drag force is maximum and acts downwards (opposing the upward motion), in the same direction as gravity. Therefore, the net retarding force is at its maximum, and the magnitude of acceleration is greatest at this instant. At maximum height, velocity is zero so air resistance is zero; the acceleration is equal to \(g\). As the ball falls downwards, speed increases and drag increases upwards, which decreases the net downward force and thus decreases the acceleration magnitude.

1 mark for identifying that maximum speed upwards leads to maximum downward drag, which combines with gravity to give maximum acceleration magnitude at release.

Taking the direction to the right as positive:
Initial momentum \(= m(2v) + 3m(-v) = -mv\).
Let \(v_P\) and \(v_Q\) be the velocities after collision.
Conservation of linear momentum:
\(m v_P + 3m v_Q = -mv \implies v_P + 3v_Q = -v\) (Equation 1)

Since the collision is perfectly elastic, relative speed of approach equals relative speed of separation:
\(u_P - u_Q = v_Q - v_P\)
\(2v - (-v) = v_Q - v_P \implies 3v = v_Q - v_P \implies v_P = v_Q - 3v\) (Equation 2)

Substitute Equation 2 into Equation 1:
\((v_Q - 3v) + 3v_Q = -v \implies 4v_Q = 2v \implies v_Q = +0.5v\) (moving to the right)

Finding \(v_P\):
\(v_P = 0.5v - 3v = -2.5v\) (moving to the left)

1 mark for using both conservation of momentum and the relative speed relationship to find the correct velocities of \(P\) and \(Q\).

Applying Newton's second law to the crate, with the downward direction as positive:
\(mg - R = ma\)
where \(R\) is the upward normal force from the floor on the crate.
\(R = m(g - a)\)
\(R = 80\text{ kg} \times (9.81\text{ m s}^{-2} - 1.5\text{ m s}^{-2})\)
\(R = 80 \times 8.31 = 664.8\text{ N} \approx 660\text{ N}\) (to 2 significant figures).

1 mark for applying \(R = m(g-a)\) and calculating the force correctly to 2 significant figures.

At terminal velocity, the sphere has zero acceleration, so the net vertical force is zero.
Downward force = Upward forces
\(mg = U + kv_t\)
\(kv_t = mg - U\)
\(v_t = \frac{mg - U}{k}\)

1 mark for establishing force equilibrium at terminal velocity and solving for \(v_t\).

Since the resistors are connected in parallel across the battery, the potential difference across each is \(E\).
Power dissipated in resistor \(R\): \(P_R = \frac{E^2}{R}\)
Total power dissipated in the circuit: \(P_{\text{total}} = \frac{E^2}{R} + \frac{E^2}{2R} + \frac{E^2}{3R} = \frac{E^2}{R} \left(1 + \frac{1}{2} + \frac{1}{3}\right) = \frac{E^2}{R} \left(\frac{11}{6}\right)\)
Ratio: \frac{P_R}{P_{\text{total}}} = \frac{E^2/R}{\frac{11E^2}{6R}} = \frac{6}{11}

1 mark for expressing the power of individual components using \(P = V^2/R\) and determining the ratio of the power in \(R\) to the total power as \(\frac{6}{11}\).

At temperature \(T_2\), the potential difference across the fixed resistor \(R_f\) is \(V_f = 8.0\text{ V}\).
By Kirchhoff's second law, the potential difference across the thermistor \(V_{\text{th}}\) is:
\(V_{\text{th}} = E - V_f = 12.0\text{ V} - 8.0\text{ V} = 4.0\text{ V}\).
Since the fixed resistor and thermistor are in series, the ratio of their resistances equals the ratio of their potential differences:
\(\frac{R_{\text{th}}}{R_f} = \frac{V_{\text{th}}}{V_f}\)
\(R_{\text{th}} = 4.0\text{ k}\Omega \times \frac{4.0\text{ V}}{8.0\text{ V}} = 2.0\text{ k}\Omega\).

1 mark for using the potential divider ratio or Kirchhoff's laws to find the thermistor resistance of \(2.0\text{ k}\Omega\).

The terminal potential difference \(V\) is related to the current \(I\) by the equation \(V = E - Ir\). This equation can be rewritten as \(V = -rI + E\), which is the equation of a straight line (\(y = mx + c\)) with a negative gradient equal to \(-r\) and a non-zero vertical axis intercept equal to \(E\).

1 mark for identifying the linear relationship with negative gradient and non-zero intercept from the equation \(V = E - Ir\).

Let the upward direction be positive. The displacement \(s\) from the projection point (top of the cliff) to the ground is \(-h\). Using the kinematic equation of motion \(s = ut + \frac{1}{2}at^2\), where \(a = -g\) and \(t = T\), we obtain:

\(-h = uT - \frac{1}{2}gT^2\)

Multiplying both sides by \(-1\) yields:

\(h = \frac{1}{2}gT^2 - uT\).

1 mark for correctly setting up the SUVAT equation with appropriate signs for displacement and acceleration, and rearranging to solve for h.

The change in velocity is equal to the area under the acceleration-time (\(a-t\)) graph. Since the acceleration decreases linearly from \(6.0\text{ m s}^{-2}\) to zero over a time interval of \(4.0\text{ s}\), the shape under the graph is a triangle.

\(\Delta v = \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 4.0\text{ s} \times 6.0\text{ m s}^{-2} = 12\text{ m s}^{-1}\)

Since the car starts from rest, the initial velocity is zero, so the final velocity at \(t = 4.0\text{ s}\) is \(12\text{ m s}^{-1}\).

1 mark for using the area under the acceleration-time graph to find the change in velocity.

The sphere is in equilibrium, so the upward forces must balance the downward forces.

Upward force: upthrust \(U = \rho V g\).
Downward forces: weight \(W = mg\) and the tension in the string \(T\) pulling downwards.

Equilibrium equation:
\(U = W + T\)
\(\rho V g = mg + T\)

Rearranging for tension \(T\):
\(T = \rho V g - mg\).

1 mark for setting up the equilibrium equation and correctly identifying the expression for tension.

The impulse delivered by the force is equal to the change in momentum of the body:

\(p = F \Delta t\)

The kinetic energy \(E_k\) of the body can be expressed in terms of momentum \(p\) and mass \(m\) as:

\(E_k = \frac{p^2}{2m}\)

Substituting \(p = F \Delta t\) into this equation gives:

\(E_k = \frac{(F \Delta t)^2}{2m} = \frac{F^2 (\Delta t)^2}{2m}\).

1 mark for linking momentum to impulse and kinetic energy to momentum to derive the correct expression.

Let \(v_X\) and \(v_Y\) be the velocities after the collision.
Using conservation of linear momentum:
\(m(v) + 3m(-v) = m v_X + 3m v_Y\)
\(-2mv = m v_X + 3m v_Y \implies v_X + 3v_Y = -2v\) (Equation 1)

For a perfectly elastic collision, the relative speed of approach equals the relative speed of separation:
\(u_X - u_Y = v_Y - v_X\)
\(v - (-v) = v_Y - v_X \implies 2v = v_Y - v_X \implies v_X = v_Y - 2v\) (Equation 2)

Substituting Equation 2 into Equation 1:
\((v_Y - 2v) + 3v_Y = -2v\)
\(4v_Y = 0 \implies v_Y = 0\)

Using this in Equation 2:
\(v_X = -2v\).

Thus, the final velocity of X is \(-2v\) and the final velocity of Y is \(0\).

1 mark for applying both the conservation of momentum and the elastic collision relative speed relation to solve for the final velocities.

Let \(I\) be the current in the circuit.
The potential difference across the internal resistance of the battery (lost volts) is given by \(E - V\).
The power dissipated in the internal resistance is:
\(P_{\text{int}} = I(E - V)\)

The power delivered to the external resistor is:
\(P_{\text{ext}} = I V\)

The ratio of the two powers is:
\(\frac{P_{\text{int}}}{P_{\text{ext}}} = \frac{I(E - V)}{IV} = \frac{E - V}{V}\).

1 mark for expressing internal power as \(I(E - V)\) and external power as \(IV\), and taking the correct ratio.

First, calculate the equivalent parallel resistance \(R_p\):
\(\frac{1}{R_p} = \frac{1}{4.0} + \frac{1}{6.0} + \frac{1}{12.0} = \frac{3 + 2 + 1}{12.0} = \frac{6.0}{12.0}\)
\(R_p = 2.0\ \Omega\)

The total resistance of the circuit is:
\(R_{\text{total}} = R_p + r = 2.0 + 1.0 = 3.0\ \Omega\)

The total circuit current is:
\(I_{\text{total}} = \frac{E}{R_{\text{total}}} = \frac{12.0\text{ V}}{3.0\ \Omega} = 4.0\text{ A}\)

The potential difference across the parallel network is:
\(V = I_{\text{total}} \times R_p = 4.0\text{ A} \times 2.0\ \Omega = 8.0\text{ V}\)

The current through the \(6.0\ \Omega\) resistor is:
\(I = \frac{V}{6.0\ \Omega} = \frac{8.0\text{ V}}{6.0\ \Omega} \approx 1.33\text{ A} \approx 1.3\text{ A}\).

1 mark for calculating the combined parallel resistance, total current, terminal potential difference, and the individual branch current.

As light intensity increases, the resistance of the LDR, \(R_{\text{LDR}}\), decreases.
The output voltage across the fixed resistor is given by:
\(V_{\text{out}} = \left(\frac{R_{\text{fixed}}}{R_{\text{LDR}} + R_{\text{fixed}}}\right) V_{\text{in}}\)

Since \(R_{\text{LDR}}\) decreases, the total resistance of the circuit decreases, causing the current to increase. This results in a larger potential difference across the fixed resistor. Therefore, \(V_{\text{out}}\) increases.

1 mark for correctly identifying that LDR resistance decreases and that this causes the output voltage across the fixed resistor to increase.

\( \text{(a)} \) The vertical component of the initial velocity \( u_y \) is:
\( u_y = u \sin\theta = 18.0 \sin(35.0^\circ) = 10.32\text{ m s}^{-1} \approx 10.3\text{ m s}^{-1} \).

\( \text{(b)} \) Taking upwards as positive, the vertical displacement \( s_y \) is:
\( s_y = u_y t - \frac{1}{2} g t^2 \)
\( s_y = (10.32 \times 3.60) - (0.5 \times 9.81 \times 3.60^2) \)
\( s_y = 37.15 - 63.57 = -26.42\text{ m} \)
Since \( s_y \) is negative, the ball is \( 26.4\text{ m} \) below its starting height. Thus, the height of the cliff is \( h = 26.4\text{ m} \).

\( \text{(c)} \) The horizontal component of the velocity is constant:
\( v_x = u \cos\theta = 18.0 \cos(35.0^\circ) = 14.74\text{ m s}^{-1} \)
The vertical component of the velocity just before impact is:
\( v_y = u_y - g t = 10.32 - (9.81 \times 3.60) = 10.32 - 35.32 = -25.00\text{ m s}^{-1} \)
The magnitude of the final velocity \( v \) is:
\( v = \sqrt{v_x^2 + v_y^2} = \sqrt{14.74^2 + (-25.00)^2} = \sqrt{217.27 + 625.00} = \sqrt{842.27} = 29.0\text{ m s}^{-1} \)
The angle \( \phi \) with the horizontal is:
\( \phi = \arctan\left(\frac{|v_y|}{v_x}\right) = \arctan\left(\frac{25.00}{14.74}\right) = 59.5^\circ \).

(a) [1 mark]
- Correct calculation of vertical component: 18.0 * sin(35) = 10.3 m s^-1 [1B]

(b) [3 marks]
- Use of s = u*t - 0.5*g*t^2 with correct signs: [1M]
- Correct numerical substitution: (10.32 * 3.60) - (0.5 * 9.81 * 3.60^2) [1M]
- Final height h = 26.4 m (accept 26.3 to 26.5 m): [1A]

(c) [5 marks]
- Calculation of constant horizontal velocity (14.7 m s^-1): [1M]
- Calculation of final vertical velocity (-25.0 m s^-1): [1M]
- Correct use of Pythagoras theorem to find magnitude: [1M]
- Final velocity magnitude v = 29.0 m s^-1: [1A]
- Angle = 59.5 degrees below horizontal: [1A]

\( \text{(a)} \) Linear momentum is defined as the product of an object's mass and its velocity.

\( \text{(b)} \) The impulse delivered by the motor is represented by the total area under the force-time graph from \( t = 0 \) to \( t = 2.00\text{ s} \):
- Area of triangular section (from \( t = 0 \) to \( 1.20\text{ s} \)):
\( \text{Area}_1 = \frac{1}{2} \times \text{base} \times \text{height} = 0.5 \times 1.20 \times 12.0 = 7.20\text{ N s} \)
- Area of rectangular section (from \( t = 1.20\text{ s} \) to \( 2.00\text{ s} \)):
\( \text{Area}_2 = \text{base} \times \text{height} = (2.00 - 1.20) \times 12.0 = 0.80 \times 12.0 = 9.60\text{ N s} \)
- Total impulse = \( 7.20 + 9.60 = 16.8\text{ N s} \).

\( \text{(c)} \) The weight of the rocket also exerts a downward impulse over the same time interval of \( 2.00\text{ s} \):
\( \text{Downward impulse} = W \times \Delta t = m g \Delta t = 0.350 \times 9.81 \times 2.00 = 6.87\text{ N s} \)
Using the impulse-momentum theorem, the net upward impulse is:
\( \Delta p_{\text{net}} = \text{Motor Impulse} - \text{Downward Impulse} \)
\( \Delta p_{\text{net}} = 16.8 - 6.87 = 9.93\text{ N s} \)
Since the rocket starts from rest, its final momentum is \( p = 9.93\text{ kg m s}^{-1} \).
Its final speed is:
\( v = \frac{p}{m} = \frac{9.93}{0.350} = 28.4\text{ m s}^{-1} \).

(a) [1 mark]
- Product of mass and velocity: [1B]

(b) [3 marks]
- Recognises impulse as area under F-t graph: [1M]
- Calculates triangular or rectangular areas correctly: [1M]
- Obtains total impulse of 16.8 N s: [1A]

(c) [4 marks]
- Calculates weight W = mg = 3.43 N: [1M]
- Calculates downward gravitational impulse = 6.87 N s: [1M]
- Calculates net impulse = 9.93 N s: [1M]
- Obtains speed = 28.4 m s^-1 (accept range 28.3 to 28.5): [1A]

\( \text{(a)} \) The principle of conservation of momentum states that the total momentum of a closed system of interacting objects remains constant. This only applies provided there is no external force acting on the system.

\( \text{(b)} \) Define the direction to the right as positive.
Before the collision:
\( u_A = +2.50\text{ m s}^{-1} \), \( u_B = -1.20\text{ m s}^{-1} \)
After the collision:
\( v_A = -0.50\text{ m s}^{-1} \)
By conservation of momentum:
\( m_A u_A + m_B u_B = m_A v_A + m_B v_B \)
\( (0.400 \times 2.50) + (0.600 \times -1.20) = (0.400 \times -0.50) + (0.600 \times v_B) \)
\( 1.00 - 0.72 = -0.20 + 0.600 v_B \)
\( 0.28 = -0.20 + 0.600 v_B \)
\( 0.48 = 0.600 v_B \)
\( v_B = +0.80\text{ m s}^{-1} \)
Since the result is positive, the velocity of Glider B is \( 0.80\text{ m s}^{-1} \) to the right.

\( \text{(c)} \) Check total kinetic energy before and after the collision:
Total initial kinetic energy:
\( E_{ki} = \frac{1}{2} m_A u_A^2 + \frac{1}{2} m_B u_B^2 = 0.5(0.400)(2.50)^2 + 0.5(0.600)(-1.20)^2 = 1.25 + 0.432 = 1.682\text{ J} \)
Total final kinetic energy:
\( E_{kf} = \frac{1}{2} m_A v_A^2 + \frac{1}{2} m_B v_B^2 = 0.5(0.400)(-0.50)^2 + 0.5(0.600)(0.80)^2 = 0.05 + 0.192 = 0.242\text{ J} \)
Since the final kinetic energy (\( 0.242\text{ J} \)) is not equal to the initial kinetic energy (\( 1.682\text{ J} \)), kinetic energy is not conserved, meaning the collision is inelastic.

(a) [2 marks]
- Total momentum of a system remains constant: [1B]
- Provided no external force acts on the system: [1B]

(b) [3 marks]
- Substitution into momentum conservation equation with correct signs: [1M]
- Algebraic simplification to find v_B: [1M]
- Final velocity 0.80 m s^-1 to the right: [1A]

(c) [3 marks]
- Calculates initial kinetic energy (1.68 J): [1M]
- Calculates final kinetic energy (0.242 J): [1M]
- States that because kinetic energy is not conserved, the collision is inelastic: [1A]

\( \text{(a)} \) Initially, at the moment of release, \( v = 0 \), so the viscous drag force is zero. The downward weight of the sphere is greater than the upward upthrust, producing a net downward force that causes the sphere to accelerate downwards.
As the speed of the sphere increases, the upward drag force increases. This reduces the net downward force acting on the sphere, and hence its acceleration decreases.
Eventually, the sum of the upward forces (upthrust + drag) equals the downward weight. The net force becomes zero, meaning acceleration is zero, and the sphere continues downwards at a constant terminal velocity.

\( \text{(b)} \) At terminal velocity, the net force is zero:
\( W = U + F_D \implies m g = U + k v_{\text{term}} \)
\( W = 0.0450\text{ kg} \times 9.81\text{ m s}^{-2} = 0.44145\text{ N} \)
\( 0.44145 = 0.054 + k(1.40) \)
\( 1.40 k = 0.38745 \)
\( k = 0.27675\text{ N s m}^{-1} \approx 0.277\text{ kg s}^{-1} \)
To find the SI base units of \( k \):
\( [k] = \frac{[F_D]}{[v]} = \frac{\text{kg m s}^{-2}}{\text{m s}^{-1}} = \text{kg s}^{-1} \).

\( \text{(c)} \) At release, the drag force \( F_D = 0 \).
The net downward force is:
\( F_{\text{net}} = W - U = 0.44145 - 0.054 = 0.38745\text{ N} \)
Using Newton's second law:
\( a = \frac{F_{\text{net}}}{m} = \frac{0.38745}{0.0450} = 8.61\text{ m s}^{-2} \).

(a) [3 marks]
- Explains that initially weight exceeds upthrust, causing downward acceleration: [1B]
- Explains that drag increases with speed, reducing net force and acceleration: [1B]
- States that terminal velocity is reached when upward forces equal downward weight (F_net = 0, a = 0): [1B]

(b) [4 marks]
- Equates forces: mg = U + k*v_term: [1M]
- Calculates weight W = 0.441 N: [1M]
- Calculates k = 0.277 (accept 0.28): [1A]
- States correct SI base units of kg s^-1: [1B]

(c) [2 marks]
- Calculates net force at release (F_net = W - U = 0.387 N): [1M]
- Obtains acceleration = 8.61 m s^-2: [1A]

\( \text{(a)} \) Kirchhoff's second law states that the sum of the e.m.f.s around any closed loop is equal to the sum of the potential differences around that same loop (\( \sum E = \sum I R \)). It is based on the principle of the conservation of energy.

\( \text{(b)} \) Let \( I_1 \) be the current flowing upwards from battery \( E_1 \) to junction A, let \( I_3 \) be the current flowing upwards from battery \( E_2 \) to junction A, and let \( I_2 \) be the downward current flowing through the central resistor \( R_2 \).
Applying Kirchhoff's First Law at junction A:
\( I_1 + I_3 = I_2 \)
Applying Kirchhoff's Second Law to the left loop (containing \( E_1 \), \( R_1 \), and \( R_2 \)):
\( E_1 - I_1 R_1 - I_2 R_2 = 0 \implies 9.0 = 15 I_1 + 10 I_2 \)
Applying Kirchhoff's Second Law to the right loop (containing \( E_2 \), \( R_3 \), and \( R_2 \)):
\( E_2 - I_3 R_3 - I_2 R_2 = 0 \implies 6.0 = 22 I_3 + 10 I_2 \)

\( \text{(c)} \) Let the electrical potential at junction B be \( 0\text{ V} \), and the potential at junction A be \( V_A \).
We can write the currents in terms of the potential \( V_A \):
\( I_1 = \frac{9.0 - V_A}{15} \)
\( I_3 = \frac{6.0 - V_A}{22} \)
\( I_2 = \frac{V_A}{10} \)
Substitute these into Kirchhoff's First Law:
\( \frac{9.0 - V_A}{15} + \frac{6.0 - V_A}{22} = \frac{V_A}{10} \)
Multiply the equation by 330 to clear denominators:
\( 22(9.0 - V_A) + 15(6.0 - V_A) = 33 V_A \)
\( 198 - 22 V_A + 90 - 15 V_A = 33 V_A \)
\( 288 = 70 V_A \)
\( V_A = 4.11\text{ V} \)
The potential difference across \( R_2 \) is therefore \( 4.11\text{ V} \approx 4.1\text{ V} \).

(a) [2 marks]
- Sum of e.m.f.s = sum of p.d.s in a closed loop: [1B]
- Conservation of energy: [1B]

(b) [4 marks]
- Correct First Law junction equation (e.g., I_1 + I_3 = I_2): [1M]
- Correct loop 1 equation: 9.0 = 15 * I_1 + 10 * I_2: [1M]
- Correct loop 2 equation: 6.0 = 22 * I_3 + 10 * I_2: [1M]
- Consistent system of equations: [1M]

(c) [2 marks]
- Suitable method to solve simultaneous equations: [1M]
- Correct potential difference of 4.1 V (accept 4.11 V): [1A]

\( \text{(a)} \) The fixed resistor and LDR are connected in series across the terminals of the power supply. The output voltage is tapped across the LDR.
In bright sunlight, the LDR resistance is \( 450\,\Omega \) and \( R = 1800\,\Omega \):
\( V_{\text{out}} = V_{\text{in}} \times \frac{R_{\text{LDR}}}{R + R_{\text{LDR}}} = 12.0 \times \frac{450}{1800 + 450} = 2.40\text{ V} \).

\( \text{(b)} \) In darkness, the LDR resistance is \( 75.0\text{ k}\Omega = 75000\,\Omega \):
\( V_{\text{out}} = 12.0 \times \frac{75000}{1800 + 75000} = 12.0 \times \frac{75000}{76800} = 11.72\text{ V} \approx 11.7\text{ V} \).

\( \text{(c)} \) When the warning light is connected in parallel with the LDR, the combined resistance of the LDR and warning light is:
\( \frac{1}{R_{\text{p}}} = \frac{1}{75.0\text{ k}\Omega} + \frac{1}{10.0\text{ k}\Omega} \)
\( R_{\text{p}} = \frac{75.0 \times 10.0}{75.0 + 10.0} = 8.824\text{ k}\Omega = 8824\,\Omega \)
Using the potential divider equation with this new parallel resistance:
\( V_{\text{out}}' = 12.0 \times \frac{R_{\text{p}}}{R + R_{\text{p}}} = 12.0 \times \frac{8824}{1800 + 8824} = 9.97\text{ V} \).

(a) [3 marks]
- Describes series connection of resistors across supply with output across LDR: [1B]
- Correct potential divider equation: [1M]
- V_out = 2.40 V: [1A]

(b) [2 marks]
- Calculation using 75.0 k\Omega: [1M]
- V_out = 11.7 V: [1A]

(c) [3 marks]
- Calculates equivalent resistance of parallel combination (8.82 k\Omega): [1M]
- Uses potential divider formula with parallel resistance: [1M]
- Calculates output voltage = 9.97 V (accept 10.0 V): [1A]

\( \text{(a)} \) The equation is:
\( V = E - I r \) (or \( E = V + I r \)).

\( \text{(b)} \) Substituting the given values into the equation gives two simultaneous equations:
- Equation 1 (for \( I = 0.50\text{ A} \)):
\( 1.35 = E - 0.50 r \)
- Equation 2 (for \( I = 1.20\text{ A} \)):
\( 1.14 = E - 1.20 r \)
Subtracting Equation 2 from Equation 1 to eliminate \( E \):
\( 1.35 - 1.14 = -0.50 r - (-1.20 r) \)
\( 0.21 = 0.70 r \)
\( r = 0.30\,\Omega \)
Substituting \( r = 0.30\,\Omega \) back into Equation 1:
\( E = 1.35 + 0.50(0.30) = 1.50\text{ V} \).

\( \text{(c)} \) The electrical power \( P \) dissipated in the external resistor when \( I = 1.20\text{ A} \) is:
\( P = V I = 1.14 \times 1.20 = 1.368\text{ W} \approx 1.37\text{ W} \)
The total power produced by the cell is:
\( P_{\text{total}} = E I = 1.50 \times 1.20 = 1.80\text{ W} \)
The efficiency \( \eta \) of power transfer is:
\( \eta = \frac{\text{Useful power output}}{\text{Total power input}} = \frac{V I}{E I} = \frac{V}{E} = \frac{1.14}{1.50} = 0.76 \) (or \( 76\% \)).

(a) [1 mark]
- States V = E - Ir or equivalent: [1B]

(b) [4 marks]
- Sets up two correct simultaneous equations: [1M]
- Method of subtraction to find r: [1M]
- Calculates internal resistance r = 0.30 \Omega: [1A]
- Calculates e.m.f. E = 1.50 V: [1A]

(c) [3 marks]
- Calculates power output = 1.37 W (accept 1.368 W): [1M]
- Identifies efficiency as V / E or P_out / P_in: [1M]
- Obtains efficiency of 76% (or 0.76): [1A]

A sample set of experimental results is shown below:
* \(h = 5.0\text{ cm} \implies 1/h = 0.200\text{ cm}^{-1}\), \(t_1 = 3.65\text{ s}, t_2 = 3.61\text{ s} \implies t_{\text{mean}} = 3.63\text{ s} \implies t^2 = 13.18\text{ s}^2\)
* \(h = 8.0\text{ cm} \implies 1/h = 0.125\text{ cm}^{-1}\), \(t_1 = 2.87\text{ s}, t_2 = 2.89\text{ s} \implies t_{\text{mean}} = 2.88\text{ s} \implies t^2 = 8.29\text{ s}^2\)
* \(h = 12.0\text{ cm} \implies 1/h = 0.0833\text{ cm}^{-1}\), \(t_1 = 2.34\text{ s}, t_2 = 2.36\text{ s} \implies t_{\text{mean}} = 2.35\text{ s} \implies t^2 = 5.52\text{ s}^2\)
* \(h = 16.0\text{ cm} \implies 1/h = 0.0625\text{ cm}^{-1}\), \(t_1 = 2.02\text{ s}, t_2 = 2.04\text{ s} \implies t_{\text{mean}} = 2.03\text{ s} \implies t^2 = 4.12\text{ s}^2\)
* \(h = 20.0\text{ cm} \implies 1/h = 0.0500\text{ cm}^{-1}\), \(t_1 = 1.82\text{ s}, t_2 = 1.80\text{ s} \implies t_{\text{mean}} = 1.81\text{ s} \implies t^2 = 3.28\text{ s}^2\)
* \(h = 25.0\text{ cm} \implies 1/h = 0.0400\text{ cm}^{-1}\), \(t_1 = 1.61\text{ s}, t_2 = 1.63\text{ s} \implies t_{\text{mean}} = 1.62\text{ s} \implies t^2 = 2.62\text{ s}^2\)

**Graphing:**
* Plotting \(t^2\) on the y-axis against \(1/h\) on the x-axis yields a straight line.
* Scale choice: \(x\)-axis from \(0\) to \(0.22\text{ cm}^{-1}\), \(y\)-axis from \(0\) to \(14.0\text{ s}^2\).
* Gradient calculation:
\[\text{gradient} = \frac{13.18 - 2.62}{0.200 - 0.040} = \frac{10.56}{0.160} = 66.0\text{ s}^2\text{ cm}\]
* y-intercept calculation:
\[y = m x + c \implies 2.62 = (66.0)(0.040) + B \implies B = 2.62 - 2.64 = -0.02\text{ s}^2\]

**Constants:**
* \(A = 66.0\text{ s}^2\text{ cm}\) (or \(0.66\text{ s}^2\text{ m}\))
* \(B = -0.02\text{ s}^2\)

**Table of Results [6 marks]**
* **Range & Number (1 mark):** At least 6 sets of readings with \(h\) ranging over at least \(15.0\text{ cm}\).
* **Column Headings (1 mark):** Correct headings with units for each column: \(h/\text{cm}\), \(t/\text{s}\), \(t^2/\text{s}^2\), \(1/h\ (\text{cm}^{-1})\).
* **Consistency of Raw Readings (1 mark):** \(h\) recorded to nearest mm (e.g., \(10.0\text{ cm}\)) and \(t\) to \(0.01\text{ s}\).
* **Significant Figures (1 mark):** Significant figures for \(1/h\) must match or be one more than the raw significant figures of \(h\).
* **Calculations (1 mark):** Correct calculation of \(1/h\) and \(t^2\).
* **Repeated Readings (1 mark):** Clear evidence of duplicate timing measurements for each height.

**Graph [5 marks]**
* **Axes (1 mark):** Sensible scales where points occupy more than half the grid, with axes labeled clearly with quantities and units.
* **Plotting (1 mark):** Correctly plotted points to within half a small square.
* **Line of Best Fit (1 mark):** Single, sharp line of best fit, balanced about the points.
* **Quality of Data (2 marks):** Points lie very close to the trend line (2 marks for minimal scatter, 1 mark for acceptable scatter).

**Gradient and Intercept [3 marks]**
* **Gradient (2 marks):** Derived using a large triangle where hypotenuse \(> 50\%\) of the drawn line length.
* **y-intercept (1 mark):** Determined by substitution of a point on the line into \(y=mx+c\) or read directly from the y-axis if \(x=0\).

**Constants [6 marks]**
* **Value of \(A\) (1 mark):** Equated to the gradient.
* **Value of \(B\) (1 mark):** Equated to the y-intercept.
* **Units (2 marks):** Correct unit for \(A\) (e.g., \(s^2 cm\) or \(s^2 m\)) and \(B\) (\(s^2\)).
* **Working Consistency (2 marks):** Mathematical work is clear, correct, and matching the graphical analysis.

A sample set of experimental results is shown below (assuming \(E = 1.50\text{ V}\), \(R_0 = 150\ \Omega\), \(R_X = 100\ \Omega\)):
* \(n = 1 \implies 1/n = 1.00\), \(I = 6.0\text{ mA} = 0.0060\text{ A} \implies 1/I = 167\text{ A}^{-1}\)
* \(n = 2 \implies 1/n = 0.50\), \(I = 8.6\text{ mA} = 0.0086\text{ A} \implies 1/I = 116\text{ A}^{-1}\)
* \(n = 3 \implies 1/n = 0.33\), \(I = 10.0\text{ mA} = 0.0100\text{ A} \implies 1/I = 100\text{ A}^{-1}\)
* \(n = 4 \implies 1/n = 0.25\), \(I = 10.9\text{ mA} = 0.0109\text{ A} \implies 1/I = 91.7\text{ A}^{-1}\)
* \(n = 5 \implies 1/n = 0.20\), \(I = 11.5\text{ mA} = 0.0115\text{ A} \implies 1/I = 87.0\text{ A}^{-1}\)
* \(n = 6 \implies 1/n = 0.17\), \(I = 12.0\text{ mA} = 0.0120\text{ A} \implies 1/I = 83.3\text{ A}^{-1}\)

**Graphing:**
* Plot \(1/I\) (on y-axis in \(A^{-1}\)) against \(1/n\) (on x-axis, dimensionless).
* Scale: \(x\)-axis from \(0\) to \(1.1\), \(y\)-axis from \(0\) to \(200\text{ A}^{-1}\).
* Gradient calculation:
\[\text{gradient} = \frac{167 - 83.3}{1.00 - 0.17} = \frac{83.7}{0.83} = 101\text{ A}^{-1}\]
* y-intercept calculation:
\[1/I = \text{gradient} \times (1/n) + \text{intercept} \implies 167 = 101 \times 1.00 + \text{intercept} \implies \text{intercept} = 66\text{ A}^{-1}\]

**Constants:**
* Given \(R_0 = 150\ \Omega\):
* \(E = R_0 / \text{gradient} = 150 / 101 = 1.49\text{ V}\)
* \(R_X = E \times \text{intercept} = 1.49 \times 66 = 98.3\ \Omega\)

**Table of Results [6 marks]**
* **Number of Readings (1 mark):** Readings recorded for all six values of \(n\) (1 to 6).
* **Column Headings (1 mark):** Headings with correct units: \(n\) (dimensionless), \(1/n\) (dimensionless), \(I\) with unit (e.g., \(I/\text{mA}\) or \(I/\text{A}\)), \(1/I\) with unit (e.g., \(1/I\ (\text{A}^{-1})\) or \(1/I\ (\text{mA}^{-1})\)).
* **Consistency (1 mark):** Current \(I\) values recorded to consistent decimal places (matching the digital display precision, e.g., \(0.1\text{ mA}\)).
* **Significant Figures (1 mark):** \(1/I\) must have the same or one more significant figure than raw \(I\).
* **Calculations (1 mark):** Correct calculation of \(1/n\) and \(1/I\).
* **Quality & Control (1 mark):** Demonstrated care, such as recording a repeat reading of \(n=1\) at the end of the experiment to check for cell decay, and getting a consistent value.

**Graph [5 marks]**
* **Axes (1 mark):** Linear, sensible scales ensuring points cover at least \(50\%\) of the grid, labeled with variables and units.
* **Plotting (1 mark):** Points plotted to within half a small square accuracy.
* **Line of Best Fit (1 mark):** Straight line of best fit, single clean line showing balanced points on both sides.
* **Quality of Data (2 marks):** All six points lie very close to the trend line (2 marks for minimal scatter, 1 mark for slight scatter).

**Gradient and Intercept [3 marks]**
* **Gradient (2 marks):** Triangle hypotenuse \(> 50\%\) of the drawn line, reading of coordinates correct.
* **y-intercept (1 mark):** Correctly read from y-axis or calculated algebraically.

**Constants [6 marks]**
* **Value of \(E\) (1 mark):** Correctly calculated as \(E = R_0 / \text{gradient}\).
* **Value of \(R_X\) (1 mark):** Correctly calculated as \(R_X = E \times \text{intercept}\).
* **Units (2 marks):** \(E\) in \(V\) and \(R_X\) in \(\Omega\).
* **Working and Consistency (2 marks):** Clear steps shown, with values consistent with the gradient and intercept derived.