2025 IB DP Physics 模擬試題連答案詳解

Let the ground level be \(y = 0\). The position of the first ball thrown upwards is given by \(y_1 = ut - \frac{1}{2}gt^2\). The position of the second ball dropped from rest at height \(H\) is given by \(y_2 = H - \frac{1}{2}gt^2\). When they collide, their positions are equal: \(y_1 = y_2\), which gives \(ut - \frac{1}{2}gt^2 = H - \frac{1}{2}gt^2\). Simplifying this equation gives \(ut = H\), so the time taken is \(t = \frac{H}{u}\).

1 mark for equating the position equations of the two balls and correctly solving for \(t = \frac{H}{u}\).

Let \(v_1\) and \(v_2\) be the final velocities of the blocks of mass \(m\) and \(3m\) respectively. By conservation of momentum: \(mv = mv_1 + 3mv_2 \implies v = v_1 + 3v_2\). For an elastic collision, the relative speed of approach equals the relative speed of separation: \(v = v_2 - v_1\). Adding these two equations yields \(2v = 4v_2 \implies v_2 = \frac{v}{2}\). The kinetic energy of the block of mass \(3m\) after collision is \(E_{k, \text{final}} = \frac{1}{2}(3m)v_2^2 = \frac{1}{2}(3m)\left(\frac{v}{2}\right)^2 = \frac{3}{8}mv^2\). Since the initial kinetic energy is \(E_{k, \text{initial}} = \frac{1}{2}mv^2\), the fraction transferred is \(\frac{E_{k, \text{final}}}{E_{k, \text{initial}}} = \frac{\frac{3}{8}mv^2}{\frac{1}{2}mv^2} = \frac{3}{4}\).

1 mark for using momentum and kinetic energy conservation (or relative velocity relation) to find the final velocity of the \(3m\) block and computing the correct ratio of kinetic energies to get \(\frac{3}{4}\).

In a time interval \(\Delta t\), the volume of water emerging from the pipe is \(\Delta V = A v \Delta t\). The mass of this water is \(\Delta m = \rho \Delta V = \rho A v \Delta t\). The kinetic energy given to this mass is \(\Delta E_k = \frac{1}{2} \Delta m v^2 = \frac{1}{2} (\rho A v \Delta t) v^2 = \frac{1}{2} \rho A v^3 \Delta t\). The minimum power \(P\) delivered is the energy per unit time: \(P = \frac{\Delta E_k}{\Delta t} = \frac{1}{2} \rho A v^3\).

1 mark for correctly setting up the mass flow rate and calculating the kinetic energy per unit time to arrive at \(\frac{1}{2} \rho A v^3\).

For the solid phase heating: the energy delivered is \(Q_1 = P t_1 = m c_s (T_2 - T_1)\), which gives \(c_s = \frac{P t_1}{m (T_2 - T_1)}\). For the melting phase: the energy delivered is \(Q_2 = P \Delta t = m L_f\), which gives \(L_f = \frac{P \Delta t}{m}\). The ratio is \(\frac{L_f}{c_s} = \frac{\frac{P \Delta t}{m}}{\frac{P t_1}{m (T_2 - T_1)}} = \frac{\Delta t (T_2 - T_1)}{t_1}\).

1 mark for writing the expressions for heat absorbed in both stages and correctly calculating the ratio \(\frac{L_f}{c_s}\).

The volume \(V\) of the wire remains constant during stretching, where \(V = A \cdot L\). Since the length increases to \(L' = 1.10 L\), the cross-sectional area must decrease to \(A' = \frac{A}{1.10}\) to keep \(V\) constant. The resistance is given by \(R = \rho \frac{L}{A}\). Therefore, the new resistance is \(R' = \rho \frac{L'}{A'} = \rho \frac{1.10 L}{A / 1.10} = 1.10^2 \rho \frac{L}{A} = 1.21 R\).

1 mark for using volume conservation to determine the new area and using the resistance formula to obtain \(1.21 R\).

The activity is \(A = \lambda N\), where the decay constant \(\lambda = \frac{\ln 2}{T_{1/2}}\). Initially, \(N_X(0) = N_Y(0) = N_0\). After time \(t = 2T\), the number of remaining nuclei of X (which has half-life \(T\)) is \(N_X(2T) = N_0 \left(\frac{1}{2}\right)^{2T/T} = \frac{N_0}{4}\). The number of remaining nuclei of Y (which has half-life \(2T\)) is \(N_Y(2T) = N_0 \left(\frac{1}{2}\right)^{2T/2T} = \frac{N_0}{2}\). The activities at \(t = 2T\) are \(A_X(2T) = \lambda_X N_X = \frac{\ln 2}{T} \frac{N_0}{4} = \frac{\ln 2 \cdot N_0}{4T}\), and \(A_Y(2T) = \lambda_Y N_Y = \frac{\ln 2}{2T} \frac{N_0}{2} = \frac{\ln 2 \cdot N_0}{4T}\). The ratio \(\frac{A_X(2T)}{A_Y(2T)} = 1\).

1 mark for setting up the activity expressions with the correct decay constants and remaining nuclei numbers after \(2T\) to find the ratio is 1.

For a stable circular orbit, the gravitational force provides the centripetal acceleration: \(\frac{G M m}{r^2} = \frac{m v^2}{r} \implies v = \sqrt{\frac{GM}{r}}\). For the second orbit with radius \(4r\), the orbital speed is \(v' = \sqrt{\frac{GM}{4r}} = \frac{1}{2} \sqrt{\frac{GM}{r}} = \frac{v}{2}\).

1 mark for using the relationship \(v \propto \frac{1}{\sqrt{r}}\) to find that multiplying the radius by 4 halves the orbital speed.

The energy of the photon emitted in the transition from level 3 to level 2 is \(\Delta E_{32} = E_3 - E_2 = -2.0\text{ eV} - (-6.0\text{ eV}) = 4.0\text{ eV}\). The energy of the photon emitted in the transition from level 2 to level 1 is \(\Delta E_{21} = E_2 - E_1 = -6.0\text{ eV} - (-12.0\text{ eV}) = 6.0\text{ eV}\). Since the wavelength \(\lambda\) is inversely proportional to the energy of the transition (\(\lambda = \frac{hc}{\Delta E}\)), we have \(\lambda_{21} = \lambda \times \frac{\Delta E_{32}}{\Delta E_{21}} = \lambda \times \frac{4.0\text{ eV}}{6.0\text{ eV}} = \frac{2}{3}\lambda\).

1 mark for calculating the energies of the transitions and using the inverse relationship between photon energy and wavelength to find \(\frac{2}{3}\lambda\).

By conservation of linear momentum, \(m v = (m + 3m) v_{\text{final}}\), so the final speed is \(v_{\text{final}} = \frac{v}{4}\).

The initial kinetic energy is:
\(E_{\text{initial}} = \frac{1}{2} m v^2\)

The final kinetic energy of the combined mass is:
\(E_{\text{final}} = \frac{1}{2} (4m) \left(\frac{v}{4}\right)^2 = \frac{1}{2} (4m) \frac{v^2}{16} = \frac{1}{8} m v^2 = \frac{1}{4} E_{\text{initial}}\)

The kinetic energy lost is:
\(E_{\text{loss}} = E_{\text{initial}} - E_{\text{final}} = E_{\text{initial}} - \frac{1}{4} E_{\text{initial}} = \frac{3}{4} E_{\text{initial}}\)

Thus, the fraction of the initial kinetic energy lost is \(\frac{3}{4} = 0.75\).

Award [1] for the correct answer C.

When unpolarized light of intensity \(I_0\) passes through the first polarizer \(A\), its intensity is reduced by half:
\(I_1 = \frac{1}{2} I_0\)

According to Malus's Law, when this linearly polarized light passes through the second polarizer \(B\), the transmitted intensity is:
\(I_2 = I_1 \cos^2(\theta)\)

With \(\theta = 30^\circ\):
\(I_2 = \left(\frac{1}{2} I_0\right) \cos^2(30^\circ) = \frac{1}{2} I_0 \left(\frac{\sqrt{3}}{2}\right)^2 = \frac{1}{2} I_0 \left(\frac{3}{4}\right) = \frac{3}{8} I_0 = 0.375 I_0 \approx 0.38 I_0\)

Award [1] for the correct answer C.

Let \(N_0\) be the initial number of parent nuclei \(X\).
At time \(t\), let \(N_X\) be the remaining number of \(X\) nuclei and \(N_Y\) be the number of decayed daughter nuclei \(Y\).

Since \(N_X + N_Y = N_0\) and the ratio is \(\frac{N_Y}{N_X} = 7\), we can write:
\(N_Y = 7 N_X\)

Substituting this into the total equation:
\(N_X + 7 N_X = N_0 \implies 8 N_X = N_0 \implies N_X = \frac{N_0}{8}\)

The fraction of remaining \(X\) nuclei is:
\(\frac{N_X}{N_0} = \frac{1}{8} = \left(\frac{1}{2}\right)^3\)

Since \(\frac{N_X}{N_0} = \left(\frac{1}{2}\right)^{t/T}\), we have:
\(\frac{t}{T} = 3 \implies t = 3.0 T\)

Award [1] for the correct answer C.

The magnetic force provides the centripetal force:
\(q v B = \frac{m v^2}{r} \implies r = \frac{m v}{q B}\)

Using the relationship between momentum and kinetic energy, \(p = m v = \sqrt{2 m E_k}\), we can write the radius as:
\(r = \frac{\sqrt{2 m E_k}}{q B}\)

Since both particles have the same kinetic energy \(E_k\) and are in the same magnetic field \(B\):
\(r \propto \frac{\sqrt{m}}{q}\)

Let's calculate the ratio \(\frac{r_p}{r_\alpha}\):
\(\frac{r_p}{r_\alpha} = \frac{\sqrt{m_p}/q_p}{\sqrt{m_\alpha}/q_\alpha} = \frac{\sqrt{m}/e}{\sqrt{4m}/(2e)} = \frac{\sqrt{m}/e}{2\sqrt{m}/(2e)} = \frac{\sqrt{m}/e}{\sqrt{m}/e} = 1\)

Therefore, the ratio is \(1:1\).

Award [1] for the correct answer B.

The energy difference between the initial and final energy levels is:
\(\Delta E = E_{\text{initial}} - E_{\text{final}} = -1.5\text{ eV} - (-3.4\text{ eV}) = 1.9\text{ eV}\)

Converting this energy to joules:
\(\Delta E = 1.9 \times 1.60 \times 10^{-19}\text{ J} = 3.04 \times 10^{-19}\text{ J}\)

The relationship between photon energy and wavelength \(\lambda\) is:
\(\Delta E = \frac{hc}{\lambda} \implies \lambda = \frac{hc}{\Delta E}\)

Using Planck's constant \(h = 6.63 \times 10^{-34}\text{ J s}\) and speed of light \(c = 3.00 \times 10^8\text{ m s}^{-1}\):
\(\lambda = \frac{6.63 \times 10^{-34} \times 3.00 \times 10^8}{3.04 \times 10^{-19}} \approx 6.54 \times 10^{-7}\text{ m}\)

Award [1] for the correct answer C.

The total thermal energy \(Q\) supplied to the substance to melt it is:
\(Q = P \times \Delta t = 120\text{ W} \times 80\text{ s} = 9600\text{ J}\)

During phase change, the heat transferred is related to the specific latent heat of fusion \(L_f\) by:
\(Q = m L_f\)

Therefore:
\(L_f = \frac{Q}{m} = \frac{9600\text{ J}}{0.20\text{ kg}} = 4.8 \times 10^4\text{ J kg}^{-1}\)

Award [1] for the correct answer B.

For a stable circular orbit, the gravitational force acts as the centripetal force:
\(\frac{G M m}{r^2} = \frac{m v^2}{r} \implies v = \sqrt{\frac{G M}{r}}\)

This shows that the orbital speed is inversely proportional to the square root of the orbital radius:
\(v \propto \frac{1}{\sqrt{r}}\)

When the radius is increased by a factor of \(4\) (from \(R\) to \(4R\)):
\(v_{\text{new}} = \sqrt{\frac{G M}{4R}} = \frac{1}{2} \sqrt{\frac{G M}{R}} = \frac{v}{2}\)

Award [1] for the correct answer C.

Since the volume and amount of gas are constant, we can apply Gay-Lussac's (pressure) law:
\(\frac{P_1}{T_1} = \frac{P_2}{T_2}\)

First, convert the temperature from Celsius to kelvins:
\(T_1 = 27 + 273 = 300\text{ K}\)

Now, solve for the final temperature \(T_2\):
\(T_2 = T_1 \times \frac{P_2}{P_1} = 300\text{ K} \times \frac{2P}{P} = 600\text{ K}\)

Convert the temperature back to Celsius:
\(T_{\text{final}} = 600 - 273 = 327^\circ\text{C}\)

Award [1] for the correct answer C.

The initial kinetic energy is given by:
\(E_{k,0} = \frac{1}{2} m u^2\)

At the peak of the trajectory, the vertical component of the velocity is zero, while the horizontal component remains unchanged due to the absence of air resistance:
\(v_{\text{peak}} = u \cos\theta\)

The kinetic energy at the peak is:
\(E_{k,\text{peak}} = \frac{1}{2} m v_{\text{peak}}^2 = \frac{1}{2} m (u \cos\theta)^2 = \frac{1}{2} m u^2 \cos^2\theta\)

The ratio of the peak kinetic energy to the initial kinetic energy is:
\(\frac{E_{k,\text{peak}}}{E_{k,0}} = \frac{\frac{1}{2} m u^2 \cos^2\theta}{\frac{1}{2} m u^2} = \cos^2\theta\)

Award [1] for the correct answer B.
- Correctly identifies that horizontal velocity is constant and equal to \(u \cos\theta\) at the peak.
- Correctly calculates the ratio of kinetic energies.

For an ideal gas, the mean kinetic energy of the molecules is directly proportional to the absolute temperature:
\(\frac{1}{2} m \langle v^2 \rangle = \frac{3}{2} k_B T \implies \langle v^2 \rangle \propto T\)
Therefore, if the absolute temperature triples, the mean square speed \(\langle v^2 \rangle\) also triples (note that the root-mean-square speed \(v_{\text{rms}}\) would increase by a factor of \(\sqrt{3}\), but the question asks for the mean square speed).

From the ideal gas law at constant volume:
\(P \propto T\)
Thus, if the temperature triples, the pressure also triples.

Award [1] for the correct answer A.
- Recognizes that the mean square speed is directly proportional to absolute temperature.
- Recognizes that pressure is directly proportional to absolute temperature under constant volume.

After each half-life, the remaining activity is halved.
After 1 half-life: \(\frac{1}{2} A_0\) remains.
After 2 half-lives: \(\frac{1}{4} A_0\) remains.
After 3 half-lives: \(\frac{1}{8} A_0\) remains.

The fraction of the activity that has decayed is:
\(1 - \frac{1}{8} = \frac{7}{8}\)

Award [1] for the correct answer C.
- Determines the remaining fraction after three half-lives as \(1/8\).
- Subtracts this from 1 to find the fraction decayed: \(7/8\).

The potential \(V\) due to a point charge is given by \(V = \frac{k Q}{r}\). For the net potential to be zero, the sum of the potentials from both charges must be zero:
\(\frac{k q}{r_1} + \frac{k (-4q)}{r_2} = 0 \implies \frac{1}{r_1} = \frac{4}{r_2} \implies r_2 = 4r_1\)

We consider two cases along the line joining the charges:
Case 1: The point lies between the two charges.
\(r_1 + r_2 = d \implies r_1 + 4r_1 = d \implies 5r_1 = d \implies r_1 = \frac{d}{5}\)

Case 2: The point lies outside the charges, on the side of \(+q\) (since the magnitude of the negative charge is larger, the zero potential point must be closer to the smaller charge).
\(r_2 - r_1 = d \implies 4r_1 - r_1 = d \implies 3r_1 = d \implies r_1 = \frac{d}{3}\)

Thus, the potential is zero at distances of \(\frac{d}{5}\) and \(\frac{d}{3}\) from the charge \(+q\).

Award [1] for the correct answer A.
- Sets up the equation for net zero potential.
- Finds both valid positions along the line.

For the block to remain stationary, the forces in both the vertical and horizontal directions must be balanced.
- In the vertical direction, the downward weight \(mg\) must be balanced by the upward static friction force \(f_s\):
\(f_s = m g\)

- In the horizontal direction, the applied push \(F\) must equal the normal force \(N\) exerted by the wall:
\(N = F\)

The maximum static frictional force available is:
\(f_{s,\text{max}} = \mu_s N = \mu_s F\)

To prevent sliding, the required friction force must not exceed the maximum possible static friction force:
\(f_s \le f_{s,\text{max}} \implies m g \le \mu_s F \implies F \ge \frac{m g}{\mu_s}\)

Thus, the minimum force required is \(\frac{m g}{\mu_s}\).

Award [1] for the correct answer B.
- Relates normal force to horizontal force and friction to weight.
- Employs static friction inequality correctly to find minimum force.

For an open-open pipe of length \(L\), the fundamental frequency is:
\(f_{\text{open}} = \frac{v}{2L}\)

For a closed-open pipe of length \(L'\), the possible frequencies are odd harmonics:
\(f_n = \frac{n v}{4L'}\) where \(n = 1, 3, 5, \dots\)
The first overtone is the third harmonic (\(n = 3\)), so:
\(f_{\text{closed, 1st overtone}} = \frac{3v}{4L'}\)

We are given that these two frequencies are equal:
\(\frac{v}{2L} = \frac{3v}{4L'} \implies \frac{1}{2L} = \frac{3}{4L'}\)

Solving for the ratio \(\frac{L'}{L}\):
\(4L' = 6L \implies \frac{L'}{L} = 1.5\)

Award [1] for the correct answer B.
- Uses the correct formulas for fundamental of open-open pipe and first overtone of closed-open pipe.
- Equates and calculates the ratio \(L'/L = 1.5\).

The lifetime \(\tau\) of a star is directly proportional to the total energy it can produce (which depends on its fuel supply, proportional to its mass \(M\)) and inversely proportional to the rate at which it consumes fuel (which is its luminosity \(L\)):
\(\tau \propto \frac{M}{L}\)

Given \(L \propto M^{3.5}\):
\(\tau \propto \frac{M}{M^{3.5}} = M^{1 - 3.5} = M^{-2.5}\)

Award [1] for the correct answer B.
- States the relationship that lifetime is proportional to \(M/L\).
- Substitutes the mass-luminosity relation to find the power of \(-2.5\).

Using the Doppler effect formula for an approaching source:
\(f = f_0 \left(\frac{v}{v - v_s}\right) \implies \frac{f_0}{f} = 1 - \frac{v_s}{v} \implies \frac{v_s}{v} = 1 - \frac{f_0}{f}\)

For the second scenario with double the source speed \(2v_s\):
\(f' = f_0 \left(\frac{v}{v - 2v_s}\right) = f_0 \left(\frac{1}{1 - 2\frac{v_s}{v}}\right)\)

Substitute the expression for \(\frac{v_s}{v}\):
\(f' = f_0 \left(\frac{1}{1 - 2\left(1 - \frac{f_0}{f}\right)}\right) = f_0 \left(\frac{1}{1 - 2 + 2\frac{f_0}{f}}\right) = f_0 \left(\frac{1}{2\frac{f_0}{f} - 1}\right)\)

Simplify the fraction:
\(f' = f_0 \left(\frac{f}{2f_0 - f}\right) = \frac{f f_0}{2f_0 - f}\)

Award [1] for the correct answer A.
- Writes correct equations for both Doppler frequency shifts.
- Eliminates the speed terms \(v_s/v\) to express the final frequency solely in terms of \(f\) and \(f_0\).

First, calculate the useful power output required to lift the mass against gravity at a constant speed: \(P_{\text{out}} = F v = m g v = 20\text{ kg} \times 9.8\text{ m s}^{-2} \times 1.5\text{ m s}^{-1} = 294\text{ W}\). Next, relate the power output to the power input using the efficiency: \(\eta = \frac{P_{\text{out}}}{P_{\text{in}}}\). Therefore, \(P_{\text{in}} = \frac{P_{\text{out}}}{\eta} = \frac{294\text{ W}}{0.60} = 490\text{ W}\).

Award [1] for the correct answer C. Correct method involves calculating the power output \(P = m g v\) and dividing by the efficiency \(0.60\).

(a) Cross-sectional area:
\(A = \frac{\pi d^2}{4} = \frac{\pi (0.38 \times 10^{-3}\text{ m})^2}{4} = 1.134 \times 10^{-7}\text{ m}^2 \approx 1.1 \times 10^{-7}\text{ m}^2\)

Percentage uncertainty in \(d\):
\(\frac{\Delta d}{d} \times 100\% = \frac{0.02}{0.38} \times 100\% = 5.26\%\)

Since \(A\) depends on \(d^2\), the percentage uncertainty in \(A\) is twice that of \(d\):
\(\% \text{ uncertainty in } A = 2 \times 5.26\% = 10.5\% \approx 11\%\)

(b)(i) The resistance of a wire is given by:
\(R = \frac{\rho L}{A}\)
Rearranging for \(R/L\), which is the gradient \(m\):
\(m = \frac{R}{L} = \frac{\rho}{A} \Rightarrow \rho = m A\)

(b)(ii) \(\rho = m A = 4.25\text{ }\Omega\text{ m}^{-1} \times 1.134 \times 10^{-7}\text{ m}^2 = 4.82 \times 10^{-7}\text{ }\Omega\text{ m} \approx 4.8 \times 10^{-7}\text{ }\Omega\text{ m}\)

(b)(iii) Percentage uncertainty in \(m\):
\(\frac{0.15}{4.25} \times 100\% = 3.53\%\)

Percentage uncertainty in \(\rho\):
\(\% \text{ uncertainty in } \rho = \% \text{ uncertainty in } m + \% \text{ uncertainty in } A = 3.53\% + 10.53\% = 14.06\% \approx 14\%\)

(b)(iv) If there is a systematic error where all length measurements are shifted by a constant amount \(\Delta L\):
\(L_{\text{measured}} = L + \Delta L \Rightarrow L = L_{\text{measured}} - \Delta L\)
The equation for resistance becomes:
\(R = \frac{\rho}{A}(L_{\text{measured}} - \Delta L) = \frac{\rho}{A}L_{\text{measured}} - \frac{\rho}{A}\Delta L\)
A graph of \(R\) against \(L_{\text{measured}}\) will have a shifted y-intercept, but its gradient remains \(\frac{\rho}{A}\). Therefore, the systematic error does not affect the gradient.

(a)
[1] for calculating the correct cross-sectional area \(A = 1.1 \times 10^{-7}\text{ m}^2\) (accept \(1.13 \times 10^{-7}\text{ m}^2\)).
[1] for finding the percentage uncertainty of diameter \(d\) as \(5.3\%\).
[1] for doubling the percentage uncertainty of \(d\) to find the percentage uncertainty of \(A\) as \(11\%\) (or \(10.5\%\)).

(b)(i)
[1] for \(\rho = m A\).

(b)(ii)
[1] for substituting values: \(4.25 \times 1.13 \times 10^{-7}\).
[1] for correct final answer \(4.8 \times 10^{-7}\text{ }\Omega\text{ m}\) (accept \(4.82 \times 10^{-7}\text{ }\Omega\text{ m}\)). Allow ECF from (a).

(b)(iii)
[1] for calculating percentage uncertainty in \(m\) as \(3.5\%\).
[1] for adding the percentage uncertainties: \(3.5\% + 10.5\% = 14\%\) (accept \(14\% - 15\%\)). Allow ECF from (a).

(b)(iv)
[1] for stating that the gradient remains unaffected.
[1] for explaining that a systematic offset in \(L\) shifts the line parallelly but does not alter the rate of change of \(R\) with \(L\) (the gradient).

(a) Using the kinematic equation \(v^2 = u^2 + 2as\) with \(u = 0\), \(a = g\), and \(s = h\), the final velocity \(v\) just before reaching the light gate is:
\(v^2 = 2gh\)
Assuming that the cylinder is short enough that its velocity \(v\) remains approximately constant during the time interval \(\Delta t\) it takes to pass through the light gate:
\(v \approx \frac{L}{\Delta t}\)
Substituting this into the velocity equation:
\(\left(\frac{L}{\Delta t}\right)^2 = 2gh \Rightarrow \frac{1}{(\Delta t)^2} = \frac{2g}{L^2} h\)

(b)(i) \(\Delta t = 14.6\text{ ms} = 14.6 \times 10^{-3}\text{ s}\)
\(\frac{1}{(\Delta t)^2} = \frac{1}{(14.6 \times 10^{-3}\text{ s})^2} = 4691.3\text{ s}^{-2} \approx 4.69 \times 10^3\text{ s}^{-2}\)

(b)(ii) Percentage uncertainty in \(\Delta t\):
\(\frac{0.1}{14.6} \times 100\% = 0.685\%\)
Since \(y = (\Delta t)^{-2}\), the percentage uncertainty in \(\frac{1}{(\Delta t)^2}\) is twice the percentage uncertainty in \(\Delta t\):
Percentage uncertainty \(= 2 \times 0.685\% = 1.37\% \approx 1.4\%\)

(b)(iii) Absolute uncertainty in \(\frac{1}{(\Delta t)^2}\):
\(\text{Absolute uncertainty} = 4.69 \times 10^3\text{ s}^{-2} \times 0.0137 = 64\text{ s}^{-2}\)

(c)(i) From the relation \(\frac{1}{(\Delta t)^2} = \left(\frac{2g}{L^2}\right) h\), the gradient \(m\) is:
\(m = \frac{2g}{L^2} \Rightarrow g = \frac{m L^2}{2}\)
Using the given values:
\(g = \frac{7.8 \times 10^3 \times (0.050)^2}{2} = 9.75\text{ m s}^{-2} \approx 9.8\text{ m s}^{-2}\)

(c)(ii) Air resistance opposes the motion of the cylinder, reducing its downward acceleration. This results in a smaller velocity \(v\) as the cylinder passes through the light gate. Because \(\Delta t = \frac{L}{v}\), a smaller velocity increases the measured time interval \(\Delta t\). This makes \(\frac{1}{(\Delta t)^2}\) smaller at each height, resulting in a shallower gradient \(m\) and therefore a lower calculated value of \(g\).

(a)
[1] for stating \(v^2 = 2gh\) AND stating the assumption that velocity is approximately constant during the passing of the light gate (or length of cylinder is much smaller than drop height).
[1] for substituting \(v = \frac{L}{\Delta t}\) into \(v^2 = 2gh\) and correctly rearranging to show the given formula.

(b)(i)
[1] for \(4.69 \times 10^3\text{ s}^{-2}\) (accept \(4691\text{ s}^{-2}\)).

(b)(ii)
[1] for percentage uncertainty in \(\Delta t\) as \(0.68\%\) or \(0.69\%\).
[1] for doubling this value to get \(1.4\%\) (accept \(1.37\%\)).

(b)(iii)
[1] for absolute uncertainty as \(64\text{ s}^{-2}\) (accept \(60\text{ s}^{-2}\) to \(65\text{ s}^{-2}\)). Allow ECF from (b)(i) and (b)(ii).

(c)(i)
[1] for expressing \(g = \frac{m L^2}{2}\).
[1] for \(9.8\text{ m s}^{-2}\) (accept \(9.75\text{ m s}^{-2}\)).

(c)(ii)
[1] for stating that air resistance decreases the velocity, thereby increasing \(\Delta t\) (or decreasing \(\frac{1}{(\Delta t)^2}\)).
[1] for concluding that the gradient is reduced, leading to a smaller calculated value of \(g\).

(a) Sound waves from the tuning fork travel down the tube and reflect at the water surface (which acts as a fixed boundary). The incident and reflected waves, which have the same frequency and amplitude, superpose/interfere. At some frequencies, this superposition forms a standing wave with nodes (zero displacement) and antinodes (maximum displacement).

(b) (i) For a tube closed at one end, the first resonance occurs at a quarter-wavelength condition (including end correction \(e\)): \(L_1 + e = \frac{\lambda}{4}\). The second resonance occurs at the next resonance position: \(L_2 + e = \frac{3\lambda}{4}\). Subtracting these equations: \((L_2 + e) - (L_1 + e) = \frac{3\lambda}{4} - \frac{\lambda}{4} \implies L_2 - L_1 = \frac{\lambda}{2}\). Thus, the end correction cancels out, and the difference is exactly half a wavelength.

(ii) \(L_2 - L_1 = 56.5\text{ cm} - 18.5\text{ cm} = 38.0\text{ cm} = 0.380\text{ m}\).
\(\frac{\lambda}{2} = 0.380\text{ m} \implies \lambda = 0.760\text{ m}\).
Using \(v = f\lambda\):
\(v = 440\text{ Hz} \times 0.760\text{ m} = 334.4\text{ m s}^{-1} \approx 334\text{ m s}^{-1}\).

(iii) From \(L_1 + e = \frac{\lambda}{4}\):
\(0.185\text{ m} + e = \frac{0.760\text{ m}}{4} = 0.190\text{ m}\).
\(e = 0.190\text{ m} - 0.185\text{ m} = 0.005\text{ m} = 0.5\text{ cm}\).

(c) Since \(v = f\lambda\), and the frequency \(f\) of the tuning fork remains constant, a higher speed of sound \(v\) in helium requires a larger wavelength \(\lambda\). Since the first resonance length \(L_1\) is approximately proportional to \(\lambda\) (\(L_1 \approx \frac{\lambda}{4}\)), the first resonance length \(L_1\) will increase.

(a)
- Incident and reflected waves travel in opposite directions and interfere/superpose [1]
- Standing wave is formed with nodes (at water surface) and antinodes (near open end) [1]

(b)(i)
- For first resonance: \(L_1 + e = \frac{\lambda}{4}\) AND for second resonance: \(L_2 + e = \frac{3\lambda}{4}\) [1]
- Subtracting gives \(L_2 - L_1 = \frac{\lambda}{2}\) showing that end correction \(e\) cancels out [1]

(b)(ii)
- \(\lambda = 2 \times (0.565 - 0.185) = 0.760\text{ m}\) [1]
- \(v = f\lambda = 440 \times 0.760 = 334\text{ m s}^{-1}\) (accept \(334.4\text{ m s}^{-1}\)) [1]

(b)(iii)
- \(\frac{\lambda}{4} = 0.190\text{ m}\) [1]
- \(e = 0.190 - 0.185 = 0.005\text{ m}\) or \(0.5\text{ cm}\) [1]

(c)
- Since \(f\) is constant, a higher speed \(v\) means a longer wavelength \(\lambda\) [1]
- Since \(L_1 \approx \frac{\lambda}{4}\), the length \(L_1\) must increase [1]

(a) (i) Proton number = 53, Neutron number = 131 - 53 = 78.

(ii) The nuclear equation is:
\(^{131}_{\phantom{0}53}\text{I} \rightarrow {}^{131}_{\phantom{0}54}\text{Xe} + {}^{0}_{-1}\beta^- + \bar{\nu}_e\) (or \(^{131}_{\phantom{0}53}\text{I} \rightarrow {}^{131}_{\phantom{0}54}\text{Xe} + e^- + \bar{\nu}_e\)).

(b) (i) Convert half-life to seconds:
\(T_{1/2} = 8.0\text{ days} = 8.0 \times 24 \times 3600 = 6.912 \times 10^5\text{ s}\).
\(\lambda = \frac{\ln 2}{T_{1/2}} = \frac{0.6931}{6.912 \times 10^5\text{ s}} = 1.003 \times 10^{-6}\text{ s}^{-1} \approx 1.0 \times 10^{-6}\text{ s}^{-1}\).

(ii) Using \(A_0 = \lambda N_0\):
\(N_0 = \frac{A_0}{\lambda} = \frac{8.0 \times 10^7\text{ Bq}}{1.003 \times 10^{-6}\text{ s}^{-1}} = 7.97 \times 10^{13} \approx 8.0 \times 10^{13}\) nuclei.

(iii) \(t = 24\text{ days}\) is exactly 3 half-lives (\(24 / 8.0 = 3\)).
\(A = \frac{A_0}{2^3} = \frac{8.0 \times 10^7\text{ Bq}}{8} = 1.0 \times 10^7\text{ Bq}\).

(c) The number of half-value thicknesses is \(n = \frac{3.6\text{ cm}}{1.2\text{ cm}} = 3\).
Fraction of gamma radiation transmitted = \(\left(\frac{1}{2}\right)^3 = \frac{1}{8} = 0.125\) or \(12.5\%\).
Percentage absorbed = \(100\% - 12.5\% = 87.5\%\) (or \(88\%\) using two significant figures).

(a)(i)
- Both proton number (53) and neutron number (78) correct [1]

(a)(ii)
- Xenon and beta-minus particle shown correctly: \(^{131}_{\phantom{0}54}\text{Xe} + {}^{0}_{-1}\beta^-\) [1]
- Electron antineutrino \(\bar{\nu}_e\) included and conservation of numbers verified [1]

(b)(i)
- Correct conversion of half-life to seconds (\(6.9 \times 10^5\text{ s}\)) [1]
- \(\lambda = 1.0 \times 10^{-6}\text{ s}^{-1}\) [1]

(b)(ii)
- Recall of \(A = \lambda N\) [1]
- Correct substitution leading to \(8.0 \times 10^{13}\) nuclei (accept \(7.97 \times 10^{13}\)) [1]

(b)(iii)
- \(1.0 \times 10^7\text{ Bq}\) [1]

(c)
- Identification of 3 half-value thicknesses leading to \(12.5\%\) transmission (or \(0.125\)) [1]
- Subtraction from \(100\%\) to get \(87.5\%\) (or \(88\%\)) absorption [1]

(a) By the principle of conservation of linear momentum:
\(m_A v_{A,i} + m_B v_{B,i} = m_A v_{A,f} + m_B v_{B,f}\)
Substitute the values:
\((0.50 \times 2.0) + (0.30 \times 0) = (0.50 \times v_{A,f}) + (0.30 \times 1.5)\)
\(1.0 = 0.50 v_{A,f} + 0.45\)
\(0.50 v_{A,f} = 0.55 \implies v_{A,f} = 1.1\text{ m s}^{-1}\).
Since this is positive, glider A continues in its original direction with speed \(1.1\text{ m s}^{-1}\).

(b) (i) Impulse is defined as the product of the average force acting on an object and the time duration of this force, which is equal to the change in momentum of the object (\(I = F \Delta t = \Delta p\)).

(ii) Change in momentum of glider B:
\(\Delta p_B = m_B (v_{B,f} - v_{B,i}) = 0.30 \times (1.5 - 0) = 0.45\text{ N s}\).
Using \(F_{\text{avg}} = \frac{\Delta p}{\Delta t}\):
\(F_{\text{avg}} = \frac{0.45\text{ N s}}{0.080\text{ s}} = 5.625\text{ N} \approx 5.6\text{ N}\).

(c) Calculate total initial kinetic energy:
\(E_{k,i} = \frac{1}{2} m_A v_{A,i}^2 = \frac{1}{2} (0.50\text{ kg}) (2.0\text{ m s}^{-1})^2 = 1.0\text{ J}\).
Calculate total final kinetic energy:
\(E_{k,f} = \frac{1}{2} m_A v_{A,f}^2 + \frac{1}{2} m_B v_{B,f}^2 = \frac{1}{2} (0.50) (1.1)^2 + \frac{1}{2} (0.30) (1.5)^2\)
\(E_{k,f} = 0.3025 + 0.3375 = 0.64\text{ J}\).
Since \(E_{k,f} < E_{k,i}\) (\(0.64\text{ J} < 1.0\text{ J}\)), kinetic energy is not conserved, which shows that the collision is inelastic.

(d) Newton's third law states that when glider A exerts a force on glider B, glider B exerts a force on glider A that is equal in magnitude and opposite in direction. This is true at every instant during the contact period.

(a)
- Statement or formula for conservation of momentum [1]
- Correct calculation showing \(v_{A,f} = 1.1\text{ m s}^{-1}\) [1]

(b)(i)
- Definition of impulse as product of force and time OR change in momentum [1]

(b)(ii)
- Calculation of momentum change of glider B (\(0.45\text{ N s}\)) [1]
- \(F = \frac{0.45}{0.080} = 5.6\text{ N}\) (accept \(5.63\text{ N}\)) [1]

(c)
- Correct calculation of initial \(E_k = 1.0\text{ J}\) [1]
- Correct calculation of final \(E_k = 0.64\text{ J}\) [1]
- Appropriate conclusion stating that kinetic energy is lost/not conserved, hence inelastic [1]

(d)
- Forces are equal in magnitude and opposite in direction [1]
- Force acts on different bodies (A on B and B on A) simultaneously [1]

(a) The required conditions are extremely high temperature (around \(1.5 \times 10^7\text{ K}\)) and extremely high density/pressure. High temperature is necessary because protons must have sufficient kinetic energy to overcome the electrostatic (Coulomb) repulsive force between them. High density/pressure is necessary to ensure a sufficiently high collision rate so that fusion can occur at a sustainable rate. When the nuclei get extremely close (approx. \(10^{-15}\text{ m}\)), the strong nuclear force overcomes the electrostatic repulsion and binds them together.

(b) (i) The two opposing forces are:
1. Gravitational force: pulls the material of the star inward, tending to compress the star.
2. Thermal/radiation pressure: created by the energy released in nuclear fusion in the core, pushing the material outward.

(ii) If the fusion rate increases, more energy is released, which increases the core temperature and pressure. This causes the core to expand against gravity. As the core expands, it cools and its density decreases, which subsequently reduces the rate of fusion back to its equilibrium value.

(c) (i) Since \(L \propto M^{3.5}\):
\(\frac{L}{L_{\odot}} = \left(\frac{3.0 \, M_{\odot}}{M_{\odot}}\right)^{3.5} = 3.0^{3.5} \approx 46.77 \approx 47\).
So, the luminosity is \(47 \, L_{\odot}\).

(ii) Using \(\tau \propto \frac{M}{L}\):
\(\frac{\tau}{\tau_{\odot}} = \frac{M/M_{\odot}}{L/L_{\odot}} = \frac{3.0}{46.77} \approx 0.0641 \approx 0.064\).
Therefore, the lifetime is \(0.064\) (or \(6.4\%\)) of the Sun's lifetime.

(a)
- High temperature required for nuclei to have high kinetic energy to overcome Coulomb/electrostatic repulsion [1]
- High density/pressure required to ensure high collision rate [1]
- Mention of strong nuclear force taking over at short range (approx \(10^{-15}\text{ m}\)) [1]

(b)(i)
- Inward force: gravitational force / pull [1]
- Outward force: thermal gas pressure / radiation pressure [1]

(b)(ii)
- Increased fusion increases temperature/pressure, causing expansion [1]
- Expansion leads to cooling and lower density, which reduces the fusion rate [1]

(c)(i)
- Correct calculation: \(3.0^{3.5} = 47 \, L_{\odot}\) (accept range 46.8 to 47) [1]

(c)(ii)
- Clear formula \(\frac{\tau}{\tau_{\odot}} = \frac{M/M_{\odot}}{L/L_{\odot}}\) [1]
- Evaluation to \(0.064\) (or \(6.4\%\)) [1]

(a) (i) During this interval, the temperature increases, meaning the average molecular kinetic energy increases. The molecular potential energy remains nearly constant. Therefore, the internal energy (the sum of molecular kinetic and potential energies) increases due to the gain in kinetic energy.

(ii) During this interval, the temperature remains constant, so the average molecular kinetic energy does not change. The heat supplied is used to overcome intermolecular forces / break bonds to change state from solid to liquid, which increases the molecular potential energy. Thus, the internal energy increases due to the gain in potential energy.

(b) (i) In the interval \(t = 0\) to \(t = 60\text{ s}\):
Energy supplied \(Q = P \Delta t = 120\text{ W} \times 60\text{ s} = 7200\text{ J}\).
Temperature change \(\Delta T = 0 - (-15) = 15^\circ\text{C}\) (or \(15\text{ K}\)).
Using \(Q = m c_{\text{solid}} \Delta T\):
\(c_{\text{solid}} = \frac{Q}{m \Delta T} = \frac{7200}{0.24 \times 15} = \frac{7200}{3.6} = 2000\text{ J kg}^{-1}\text{ K}^{-1}\).

(ii) In the interval \(t = 60\text{ s}\) to \(t = 240\text{ s}\):
Time interval \(\Delta t = 240 - 60 = 180\text{ s}\).
Energy supplied \(Q = P \Delta t = 120\text{ W} \times 180\text{ s} = 21600\text{ J}\).
Using \(Q = m L_f\):
\(L_f = \frac{Q}{m} = \frac{21600}{0.24} = 90000\text{ J kg}^{-1} = 9.0 \times 10^4\text{ J kg}^{-1}\).

(iii) In the interval \(t = 240\text{ s}\) to \(t = 360\text{ s}\):
Time interval \(\Delta t = 360 - 240 = 120\text{ s}\).
Energy supplied \(Q = P \Delta t = 120\text{ W} \times 120\text{ s} = 14400\text{ J}\).
Temperature change \(\Delta T = 40 - 0 = 40^\circ\text{C}\) (or \(40\text{ K}\)).
Using \(Q = m c_{\text{liquid}} \Delta T\):
\(c_{\text{liquid}} = \frac{Q}{m \Delta T} = \frac{14400}{0.24 \times 40} = \frac{14400}{9.6} = 1500\text{ J kg}^{-1}\text{ K}^{-1}\).

(a)(i)
- Temperature increases, so molecular kinetic energy increases [1]
- Potential energy is constant, so overall internal energy increases [1]

(a)(ii)
- Temperature is constant, so molecular kinetic energy is constant [1]
- Intermolecular bonds are broken / potential energy increases [1]

(b)(i)
- Calculation of energy supplied \(Q = 7200\text{ J}\) [1]
- Correct calculation leading to \(2000\text{ J kg}^{-1}\text{ K}^{-1}\) (accept \(2.0 \times 10^3\text{ J kg}^{-1}\text{ K}^{-1}\)) [1]

(b)(ii)
- Calculation of energy supplied during melting \(Q = 21600\text{ J}\) [1]
- Correct calculation of \(L_f = 9.0 \times 10^4\text{ J kg}^{-1}\) [1]

(b)(iii)
- Calculation of energy supplied \(Q = 14400\text{ J}\) [1]
- Correct calculation of \(c_{\text{liquid}} = 1500\text{ J kg}^{-1}\text{ K}^{-1}\) (accept \(1.5 \times 10^3\text{ J kg}^{-1}\text{ K}^{-1}\)) [1]