2025 IB DP Physics 模擬試題連答案詳解

The motion can be split into vertical and horizontal components:
1. For the vertical component, the initial velocity \(u_y = 0\). Using the kinematic equation:
\(v_y^2 = u_y^2 + 2gy\)
\(v_y^2 = 0 + 2 \times 10 \times 45 = 900\text{ m}^2\text{ s}^{-2}\)
\(v_y = 30\text{ m s}^{-1}\)

2. The horizontal component of the velocity remains constant because there is no horizontal acceleration:
\(v_x = 40\text{ m s}^{-1}\)

3. The final speed \(v\) is the magnitude of the velocity vector:
\(v = \sqrt{v_x^2 + v_y^2} = \sqrt{40^2 + 30^2} = 50\text{ m s}^{-1}\)

Award [1] for the correct answer of 50 m/s.
- Accept calculations using g = 9.81 m/s^2 which yields approximately 49.7 m/s, nearest to 50 m/s.

According to the principle of conservation of linear momentum, the total momentum before the collision must equal the total momentum after the collision.

Taking the initial direction of motion of mass \(m\) as the positive direction:
\(p_{\text{initial}} = m v + 3m (0) = mv\)

After the collision, let \(v'\) be the velocity of the block of mass \(3m\). The block of mass \(m\) rebounds, so its velocity is \(-0.5v\):
\(p_{\text{final}} = m(-0.5v) + 3m(v')\)

Setting the initial and final momentum equal to each other:
\(mv = -0.5mv + 3mv'\)
\(1.5mv = 3mv'\)
\(v' = 0.5v\)

Award [1] for the correct speed of 0.50v.
- Verify that the sign of the rebounding velocity was correctly handled as negative.

Power \(P\) is defined as the rate of doing work, or energy transferred per unit time: \(P = \frac{W}{t}\).
Since the power is constant, the work done on the vehicle in time \(t\) is:
\(W = P t\)

By the work-energy theorem, this work is equal to the change in the vehicle's kinetic energy (since it starts from rest):
\(W = \Delta E_k = \frac{1}{2} m v^2\)

Equating these two expressions:
\(P t = \frac{1}{2} m v^2\)

Rearranging to express speed \(v\) in terms of time \(t\):
\(v^2 = \frac{2P}{m} t \implies v = \sqrt{\frac{2P}{m}} \sqrt{t}\)

Since \(P\) and \(m\) are constant:
\(v \propto \sqrt{t}\)

Award [1] for the correct proportional relation.

From the ideal gas equation \(PV = nRT\), we can relate the initial and final states of the gas:
\(\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}\)

We are given:
- Final volume \(V_2 = 2 V_1\)
- Final absolute temperature \(T_2 = 0.5 T_1\)

Substitute these into the equation:
\(\frac{P_1 V_1}{T_1} = \frac{P_2 (2 V_1)}{0.5 T_1}\)

Simplify by cancelling \(V_1\) and \(T_1\) from both sides:
\(P_1 = \frac{2 P_2}{0.5}\)
\(P_1 = 4 P_2 \implies \frac{P_2}{P_1} = \frac{1}{4} = 0.25\)

Award [1] for the correct ratio of 0.25.

The total resistance in the circuit is:
\(R_{\text{total}} = R + r = 3r + r = 4r\)

The current in the circuit is:
\(I = \frac{\varepsilon}{R_{\text{total}}} = \frac{\varepsilon}{4r\)}

The total electrical power produced by the cell is:
\(P_{\text{total}} = I \varepsilon = \frac{\varepsilon^2}{4r\)}

The power dissipated internally as thermal energy in the internal resistance \(r\) is:
\(P_{\text{internal}} = I^2 r = \left(\frac{\varepsilon}{4r}\right)^2 r = \frac{\varepsilon^2}{16r}\)

The fraction of power dissipated within the cell is:
\(\frac{P_{\text{internal}}}{P_{\text{total}}} = \frac{\frac{\varepsilon^2}{16r}}{\frac{\varepsilon^2}{4r}} = \frac{4}{16} = 0.25\)

This corresponds to \(25\%\).

Award [1] for the correct answer of 25%.

The potential energy \(E_p\) in SHM at displacement \(x\) is:
\(E_p = \frac{1}{2} k x^2\)

The total energy \(E_T\) is:
\(E_T = \frac{1}{2} k A^2\)

The kinetic energy \(E_k\) is the difference between total energy and potential energy:
\(E_k = E_T - E_p = \frac{1}{2} k (A^2 - x^2)\)

We require \(E_k = 3 E_p\):
\(\frac{1}{2} k (A^2 - x^2) = 3 \left(\frac{1}{2} k x^2\right)\)

Dividing both sides by \(\frac{1}{2} k\):
\(A^2 - x^2 = 3 x^2\)
\(A^2 = 4 x^2 \implies x^2 = \frac{A^2}{4}\)
\(x = \pm \frac{A}{2\)

Award [1] for the correct displacement value.

The gravitational potential \(V\) at radius \(r\) from a planet of mass \(M\) is:
\(V = -\frac{GM}{r}\)

For a circular orbit, the centripetal force is provided by the gravitational attraction:
\(\frac{GMm}{r^2} = \frac{mv^2}{r} \implies mv^2 = \frac{GMm}{r}\)

The kinetic energy \(E_k\) of the satellite is:
\(E_k = \frac{1}{2} mv^2 = \frac{GMm}{2r}\)

Since \(\frac{GM}{r} = -V\), we can substitute this into the kinetic energy equation:
\(E_k = \frac{1}{2} m (-V) = -\frac{1}{2} mV\)

Award [1] for the correct expression.

Determine the number of half-lives \(n\) that have elapsed in \(9.0\text{ hours}\):
\(n = \frac{t}{T_{1/2}} = \frac{9.0}{3.0} = 3\)

The fraction of active nuclei remaining after \(3\) half-lives is:
\(\left(\frac{1}{2}\right)^3 = \frac{1}{8}\)

The question asks for the fraction of nuclei that have *decayed*. This is:
\(1 - \text{fraction remaining} = 1 - \frac{1}{8} = \frac{7}{8}\)

Award [1] for the correct answer of 7/8. Reject 1/8 (which represents the fraction remaining, not decayed).

At the maximum height, the velocity of the projectile is purely horizontal and has a magnitude of \( v = u \cos\theta \). The acceleration of the projectile at this point is solely due to gravity \( g \), directed vertically downwards. Since the velocity is horizontal and acceleration is vertical, the acceleration is perpendicular to the velocity and acts as the centripetal acceleration: \( a_c = g \). The radius of curvature \( R \) is given by the relation \( a_c = \frac{v^2}{R} \). Substituting the values: \( g = \frac{(u \cos\theta)^2}{R} \implies R = \frac{u^2 \cos^2\theta}{g} \).

Award [1] for the correct choice B. Reject all other options.

Let the kinetic coefficient of friction between the block and the belt be \( \mu_k \). The friction force acting on the block is \( f = \mu_k m g \), which causes an acceleration \( a = \mu_k g \). The time taken for the block to accelerate from rest to speed \( v \) is \( t = \frac{v}{a} = \frac{v}{\mu_k g} \). During this time \( t \), the conveyor belt travels a distance \( s_{\text{belt}} = v t = \frac{v^2}{\mu_k g} \). To keep the belt moving at a constant speed \( v \), the belt's motor must exert a force equal and opposite to the friction force, \( F_{\text{motor}} = f = \mu_k m g \). The work done by the motor is therefore: \( W_{\text{motor}} = F_{\text{motor}} \cdot s_{\text{belt}} = (\mu_k m g) \cdot \left(\frac{v^2}{\mu_k g}\right) = m v^2 \).

Award [1] for the correct choice B. Reject all other options.

The total energy given to a mass \( \Delta m \) of water by the pump consists of gravitational potential energy to lift the water to the surface and kinetic energy to eject it: \( \Delta E = \Delta E_{\text{p}} + \Delta E_{\text{k}} = \Delta m g h + \frac{1}{2} \Delta m v^2 \). The power \( P \) is the rate of energy transfer: \( P = \frac{\Delta E}{\Delta t} = \frac{\Delta m}{\Delta t} g h + \frac{1}{2} \frac{\Delta m}{\Delta t} v^2 = R \left( g h + \frac{1}{2} v^2 \right) \).

Award [1] for the correct choice B. Reject all other options.

From the ideal gas equation, we have \( p V = n R T \). Since the process is described by \( p = \alpha V^2 \), we can substitute \( p \) into the ideal gas equation: \( (\alpha V^2) V = n R T \implies \alpha V^3 = n R T \). This shows that the absolute temperature \( T \) is proportional to \( V^3 \). When the volume is doubled from \( V_0 \) to \( 2V_0 \), the new temperature \( T' \) is related to the initial temperature \( T \) by: \( T' \propto (2V_0)^3 = 8 V_0^3 \implies T' = 8 T \). Thus, the temperature increases by a factor of 8.

Award [1] for the correct choice C. Reject all other options.

The power dissipated in the external resistor \( R \) is given by \( P = I^2 R = \left(\frac{\varepsilon}{R+r}\right)^2 R = \frac{\varepsilon^2 R}{(R+r)^2} \). According to the maximum power transfer theorem, the power dissipated is maximized when \( R = r \). The maximum power is: \( P_{\text{max}} = \frac{\varepsilon^2 r}{(r+r)^2} = \frac{\varepsilon^2}{4r} \). When \( R = 3r \), the power is: \( P = \frac{\varepsilon^2 (3r)}{(3r+r)^2} = \frac{3 \varepsilon^2 r}{16 r^2} = \frac{3 \varepsilon^2}{16 r} \). The ratio of the power to the maximum power is: \( \frac{P}{P_{\text{max}}} = \frac{\frac{3 \varepsilon^2}{16 r}}{\frac{\varepsilon^2}{4 r}} = \frac{3}{16} \times 4 = \frac{3}{4} = 0.75 \).

Award [1] for the correct choice C. Reject all other options.

The total mechanical energy \( E \) of a satellite of mass \( m \) in a stable circular orbit of radius \( r \) is given by the sum of its kinetic and gravitational potential energies: \( E = E_{\text{k}} + E_{\text{p}} = \frac{GMm}{2r} - \frac{GMm}{r} = -\frac{GMm}{2r} \). For the initial orbit of radius \( r \), the total energy is \( E_{\text{initial}} = -\frac{GMm}{2r} \). For the final orbit of radius \( 2r \), the total energy is \( E_{\text{final}} = -\frac{GMm}{2(2r)} = -\frac{GMm}{4r} \). The change in the total mechanical energy is: \( \Delta E = E_{\text{final}} - E_{\text{initial}} = -\frac{GMm}{4r} - \left(-\frac{GMm}{2r}\right) = +\frac{GMm}{4r} \).

Award [1] for the correct choice B. Reject all other options.

Let \( A_0 \) be the initial activity of isotope Y, so the initial activity of isotope X is \( 8A_0 \). The activity of a sample after time \( t \) is given by \( A(t) = A_0 (2)^{-\frac{t}{T_{1/2}}} \). For isotope X, \( A_{\text{X}}(t) = 8 A_0 (2)^{-\frac{t}{T}} \). For isotope Y, \( A_{\text{Y}}(t) = A_0 (2)^{-\frac{t}{2T}} \). Setting the two activities equal: \( 8 A_0 (2)^{-\frac{t}{T}} = A_0 (2)^{-\frac{t}{2T}} \implies 2^3 \cdot 2^{-\frac{t}{T}} = 2^{-\frac{t}{2T}} \). Equating the exponents: \( 3 - \frac{t}{T} = -\frac{t}{2T} \implies 3 = \frac{t}{T} - \frac{t}{2T} = \frac{t}{2T} \implies t = 6T \).

Award [1] for the correct choice C. Reject all other options.

Using the mass-luminosity relation \( L \propto M^4 \), we can write the ratio of the luminosities of the two stars as: \( \frac{L_{\text{X}}}{L_{\text{Y}}} = \left(\frac{M_{\text{X}}}{M_{\text{Y}}}\right)^4 \). Given that \( \frac{L_{\text{X}}}{L_{\text{Y}}} = 256 \), we have: \( \left(\frac{M_{\text{X}}}{M_{\text{Y}}}\right)^4 = 256 \). Taking the fourth root of both sides gives: \( \frac{M_{\text{X}}}{M_{\text{Y}}} = (256)^{1/4} = 4 \).

Award [1] for the correct choice B. Reject all other options.

The time of flight \(t\) for a horizontally projected object is determined by its vertical fall: \(h = \frac{1}{2}gt^2 \implies t = \sqrt{\frac{2h}{g}}\).

The horizontal distance is \(d = v \cdot t = v \sqrt{\frac{2h}{g}}\).

For the second ball:
- The height is \(4h\), so the time of flight is \(t' = \sqrt{\frac{2(4h)}{g}} = 2\sqrt{\frac{2h}{g}} = 2t\).
- The horizontal speed is \(2v\).
- The new horizontal distance is \(d' = (2v)(t') = (2v)(2t) = 4vt = 4d\).

[1 mark] for identifying that the time of flight doubles when the height is quadrupled. [1 mark] for multiplying the doubled speed by the doubled time of flight to find that the distance is quadrupled, leading to option B.

By Newton's second law, the average force is equal to the rate of change of momentum: \(F_{\text{avg}} = \frac{\Delta p}{\Delta t}\).

Taking the initial direction of motion as positive:
- Initial momentum: \(p_{\text{initial}} = mv\)
- Final momentum: \(p_{\text{final}} = -m\left(\frac{v}{2}\right)\)
- Change in momentum of the block: \(\Delta p = p_{\text{final}} - p_{\text{initial}} = -\frac{1}{2}mv - mv = -\frac{3}{2}mv\)

The magnitude of the change in momentum is \(\frac{3}{2}mv\).
By Newton's third law, the magnitude of the average force exerted on the wall is equal to the magnitude of the average force on the block: \(F_{\text{avg}} = \frac{3mv}{2\Delta t}\).

[1 mark] for correctly calculating the change in momentum as having a magnitude of \(1.5mv\) and using Newton's second/third laws to select option C.

From the ideal gas equation, \(P = \frac{N k_B T}{V}\), where \(N\) is the number of molecules and \(V\) is the volume.

Since the container is rigid, the volume \(V\) remains constant.

Initially, \(P_1 = \frac{N_1 k_B T_1}{V}\).

After half of the gas escapes, \(N_2 = \frac{1}{2}N_1\).
When the temperature is doubled, \(T_2 = 2T_1\).

The new pressure is: \(P_2 = \frac{N_2 k_B T_2}{V} = \frac{\left(\frac{1}{2}N_1\right) k_B (2T_1)}{V} = \frac{N_1 k_B T_1}{V} = P_1\).

[1 mark] for using the ideal gas law to show that halving the number of molecules and doubling the temperature leaves the pressure unchanged, corresponding to option B.

The total resistance of the circuit is \(R_{\text{total}} = R + r\).

Given \(R = r\), the total resistance is \(2r\).

The current \(I\) in the circuit is: \(I = \frac{V}{R_{\text{total}}} = \frac{V}{2r}\).

The terminal potential difference is the potential difference across the external resistor \(R\):
\(V_{\text{terminal}} = I \cdot R = \left(\frac{V}{2r}\right) \cdot r = \frac{V}{2}\).

[1 mark] for using Ohm's law and potential divider relationship to find that half the total emf is dropped across the external load when its resistance equals the internal resistance, leading to option B.

The total energy in simple harmonic motion is given by \(E_T = \frac{1}{2}kx_0^2\).

The potential energy at displacement \(x\) is \(E_p = \frac{1}{2}kx^2\).

The kinetic energy is \(E_k = E_T - E_p = \frac{1}{2}k(x_0^2 - x^2)\).

We are given that \(E_k = 3E_p\):
\(\frac{1}{2}k(x_0^2 - x^2) = 3 \left(\frac{1}{2}kx^2\right)\)

Simplifying this equation:
\(x_0^2 - x^2 = 3x^2\)
\(x_0^2 = 4x^2\)
\(x = \pm \frac{x_0}{2}\).

[1 mark] for correctly relating kinetic energy, potential energy, and amplitude, setting up the equation, and solving for displacement to obtain option B.

Let \(x\) be the distance from the charge \(+q\) where the electric field is zero. Since the charges are of the same sign, this point must lie between them, so \(0 < x < d\).

At this point, the magnitudes of the electric fields due to both charges are equal:
\(\frac{k q}{x^2} = \frac{k (4q)}{(d-x)^2}\)

Simplifying the equation:
\(\frac{1}{x^2} = \frac{4}{(d-x)^2}\)

Taking the square root of both sides (since \(x > 0\) and \(d - x > 0\)):
\(\frac{1}{x} = \frac{2}{d-x}\)

Solving for \(x\):
\(d - x = 2x\)
\(3x = d \implies x = \frac{d}{3}\).

[1 mark] for equating the electric fields, simplifying the resulting expression, and solving for \(x\) to obtain option C.

Let the initial number of nuclei of each isotope be \(N_0\).

For isotope X:
- The half-life is \(2\text{ hours}\).
- After \(6\text{ hours}\), the number of elapsed half-lives is \(n_{\text{X}} = \frac{6}{2} = 3\).
- The remaining number of nuclei is \(N_{\text{X}} = N_0 \left(\frac{1}{2}\right)^3 = \frac{N_0}{8}\).

For isotope Y:
- The half-life is \(3\text{ hours}\).
- After \(6\text{ hours}\), the number of elapsed half-lives is \(n_{\text{Y}} = \frac{6}{3} = 2\).
- The remaining number of nuclei is \(N_{\text{Y}} = N_0 \left(\frac{1}{2}\right)^2 = \frac{N_0}{4}\).

The ratio is:
\(\frac{N_{\text{X}}}{N_{\text{Y}}} = \frac{N_0/8}{N_0/4} = \frac{4}{8} = \frac{1}{2}\).

[1 mark] for calculating the remaining fraction of both isotopes after 6 hours and dividing them to find the ratio \(1/2\), which is option B.

According to the Stefan-Boltzmann law, the luminosity \(L\) of a star behaving as a black body is given by:
\(L = 4\pi R^2 \sigma T^4\)

Therefore, luminosity is proportional to \(R^2 T^4\).

The ratio of the luminosities is:
\(\frac{L_{\text{A}}}{L_{\text{B}}} = \left(\frac{R_{\text{A}}}{R_{\text{B}}}\right)^2 \left(\frac{T_{\text{A}}}{T_{\text{B}}}\right)^4\)

Given that \(R_{\text{A}} = 2R_{\text{B}}\) and \(T_{\text{A}} = 2T_{\text{B}}\):
\(\frac{L_{\text{A}}}{L_{\text{B}}} = (2)^2 \cdot (2)^4 = 4 \cdot 16 = 64\).

[1 mark] for correctly applying the Stefan-Boltzmann proportionalities \(L \propto R^2 T^4\) to find a ratio of 64, corresponding to option D.

To find the minimum power required, we consider the total energy given to the liquid per unit time. In a time interval \(\Delta t\), the volume of liquid pumped out is \(\Delta V = A v \Delta t\). The mass of this volume of liquid is \(\Delta m = \rho \Delta V = \rho A v \Delta t\). The energy given to this mass consists of two parts: 1) the gravitational potential energy needed to lift it by height \(h\): \(\Delta E_p = \Delta m g h = \rho A v g h \Delta t\), and 2) the kinetic energy given to the ejected liquid: \(\Delta E_k = \frac{1}{2} \Delta m v^2 = \frac{1}{2} \rho A v^3 \Delta t\). The total work done by the pump in this time interval is \(W = \Delta E_p + \Delta E_k = \rho A v \Delta t \left(gh + \frac{1}{2}v^2\right)\). The minimum power is the rate of doing work: \(P = \frac{W}{\Delta t} = \rho A v \left(gh + \frac{1}{2}v^2\right)\). Therefore, option B is correct.

[1 mark] Award 1 mark for the correct choice B. Award 0 marks for incorrect choices.

(a) Emf is the total work done per unit charge by the cell. When a current flows, some energy is dissipated as thermal energy due to the internal resistance of the cell (often called 'lost volts'). Therefore, the terminal potential difference, which is the energy delivered to the external circuit per unit charge, is reduced such that \(V = E - Ir\). (b) Using the relationship \(V = E - Ir\), we set up two simultaneous equations: (1) \(1.35 = E - 0.50r\) and (2) \(1.00 = E - 1.20r\). Subtracting equation (2) from (1) gives \(0.35 = 0.70r\), which yields \(r = 0.50\ \Omega\). Substituting this back into equation (1) gives \(1.35 = E - 0.50 \times 0.50 \implies 1.35 = E - 0.25 \implies E = 1.60\text{ V}\).

[Part a] 1 mark for stating that work is done/energy is lost overcoming the internal resistance of the cell. 1 mark for stating that this reduces the energy available per unit charge to the external circuit. [Part b] 1 mark for setting up the simultaneous equations or recognizing the gradient of the V-I graph represents \(-r\). 1 mark for correct calculation of internal resistance \(r = 0.50\ \Omega\). 1 mark for substituting \(r\) to find emf. 1 mark for correct calculation of emf \(E = 1.60\text{ V}\).

(a) Comparing the given equation to the general SHM displacement equation \(x = A \cos(\omega t)\), we find the amplitude \(A = 0.080\text{ m}\) and angular frequency \(\omega = 15\text{ rad s}^{-1}\). The maximum velocity is given by \(v_{\max} = \omega A = 15 \times 0.080 = 1.2\text{ m s}^{-1}\). (b) The total mechanical energy is equal to the maximum kinetic energy: \(E_T = \frac{1}{2} m v_{\max}^2 = \frac{1}{2} \times 0.25 \times (1.2)^2 = 0.18\text{ J}\). (c) The magnitude of acceleration in SHM is given by \(a = \omega^2 x\). At \(x = 0.040\text{ m}\), \(a = (15)^2 \times 0.040 = 225 \times 0.040 = 9.0\text{ m s}^{-2}\).

[Part a] 1 mark for identifying \(\omega = 15\text{ rad s}^{-1}\) and \(A = 0.080\text{ m}\). 1 mark for calculating \(v_{\max} = 1.2\text{ m s}^{-1}\). [Part b] 1 mark for using the correct total energy formula (either \(\frac{1}{2} m v_{\max}^2\) or \(\frac{1}{2} m \omega^2 A^2\)). 1 mark for correct calculation showing \(0.18\text{ J}\). [Part c] 1 mark for using \(a = \omega^2 x\). 1 mark for correct calculation showing \(9.0\text{ m s}^{-2}\).

(a) Using the mass-luminosity relation, \(L = (4.0)^{3.5} L_{\odot} = (2^2)^{3.5} L_{\odot} = 2^7 L_{\odot} = 128 L_{\odot}\). (b) The lifetime \(\tau\) is proportional to \(M/L\). Therefore, \(\tau_{\text{star}} = \tau_{\odot} \times \left(\frac{M_{\text{star}}}{M_{\odot}}\right) \times \left(\frac{L_{\odot}}{L_{\text{star}}}\right) = 1.0 \times 10^{10} \times \left(\frac{4.0}{128}\right) = 1.0 \times 10^{10} \times 0.03125 = 3.125 \times 10^8\text{ years}\), which rounds to \(3.1 \times 10^8\text{ years}\). (c) Since the initial mass of the star (\(4.0 M_{\odot}\)) is less than the limit for a supernova progenitor (typically around \(8 M_{\odot}\)), its core remnant will remain below the Chandrasekhar limit. Therefore, after shedding its outer layers as a planetary nebula, the remaining core will end its life as a white dwarf.

[Part a] 1 mark for recalling/applying the power law. 1 mark for final value of \(128 L_{\odot}\). [Part b] 1 mark for setting up the ratio equation \(\tau_{\text{star}} = \tau_{\odot} \times \frac{M}{L}\). 1 mark for substituting the values correctly. 1 mark for final calculation yielding \(3.1 \times 10^8\text{ years}\) or \(3.125 \times 10^8\text{ years}\). [Part c] 1 mark for stating white dwarf (accept white dwarf surrounded by planetary nebula).

(a) The total linear momentum of a closed system remains constant, provided no external force acts on it.

(b) Using conservation of momentum: \( m_1 v = (m_1 + m_2) V \). Substituting the values: \( 1.5 \times 4.0 = (1.5 + 2.5) \times V \), which gives \( 6.0 = 4.0 \times V \), hence \( V = 1.5\text{ m s}^{-1} \).

(c) The friction force acting on the combined blocks is \( F_f = \mu (m_1 + m_2) g \). The deceleration is \( a = \frac{F_f}{m_1 + m_2} = \mu g = 0.35 \times 9.81 = 3.43\text{ m s}^{-2} \). Using the kinematic equation \( v^2 = u^2 + 2ad \) where \( v = 0 \) and \( u = 1.5\text{ m s}^{-1} \): \( 0 = 1.5^2 - 2 \times 3.43 \times d \). Solving for \( d \) gives \( d = \frac{2.25}{6.86} \approx 0.33\text{ m} \).

[Part a]: 1 mark for stating total momentum is constant/conserved, 1 mark for specifying a closed system / no external forces.

[Part b]: 1 mark for correct conservation of momentum equation, 1 mark for substituting values, 1 mark for final answer 1.5 m s^-1.

[Part c]: 1 mark for stating friction force formula, 1 mark for calculating deceleration (3.43 m s^-2), 1 mark for using appropriate kinematic equation, 1 mark for substituting values, 1 mark for correct final answer of 0.33 m (accept 0.328 m).

(a) The electromotive force (emf) is the total energy per unit charge supplied by the cell, or the work done per unit charge in moving a charge completely around a circuit.

(b) The power dissipated in \( R \) is \( P = I^2 R \). Since the current in the circuit is given by \( I = \frac{E}{R+r} \), substituting this into the power equation yields: \( P = \left(\frac{E}{R+r}\right)^2 R = \frac{E^2 R}{(R+r)^2} \).

(c) We can set up two equations from \( E = I(R + r) \):
1) \( E = 1.2(4.0 + r) \)
2) \( E = 0.60(9.0 + r) \)
Equating the two expressions for \( E \):
\( 1.2(4.0 + r) = 0.60(9.0 + r) \)
Dividing both sides by \( 0.60 \) gives:
\( 2(4.0 + r) = 9.0 + r \implies 8.0 + 2r = 9.0 + r \implies r = 1.0\ \Omega \).
Substituting \( r = 1.0\ \Omega \) back into the first equation:
\( E = 1.2(4.0 + 1.0) = 6.0\text{ V} \).

[Part a]: 1 mark for 'work done per unit charge' or 'energy supplied per unit charge', 1 mark for 'moving a charge around a complete circuit' or 'by the cell'.

[Part b]: 1 mark for using P = I^2 * R, 1 mark for stating Ohm's law / current equation I = E / (R + r), 1 mark for correct algebraic steps leading to the final expression.

[Part c]: 1 mark for setting up the equation E = I(R + r) for both cases, 1 mark for equating the two expressions, 1 mark for correct algebraic manipulation, 1 mark for finding r = 1.0 ̘, 1 mark for finding E = 6.0 V.

(a) Simple harmonic motion occurs when the acceleration of an object is directly proportional to its displacement from its equilibrium position and is always directed towards that equilibrium position.

(b)(i) The angular frequency is given by \( \omega = \sqrt{\frac{k}{m}} \). Substituting the values: \( \omega = \sqrt{\frac{40}{0.25}} = \sqrt{160} \approx 12.65\text{ rad s}^{-1} \approx 13\text{ rad s}^{-1} \).

(b)(ii) The maximum velocity is given by \( v_{\text{max}} = \omega x_0 \). Substituting the values: \( v_{\text{max}} = 12.65 \times 0.080 \approx 1.01\text{ m s}^{-1} \) (or \( 1.0\text{ m s}^{-1} \)).

(b)(iii) The total mechanical energy is \( E_{\text{total}} = \frac{1}{2} k x_0^2 = 0.5 \times 40 \times 0.080^2 = 0.128\text{ J} \). The potential energy at displacement \( x = 0.040\text{ m} \) is \( E_p = \frac{1}{2} k x^2 = 0.5 \times 40 \times 0.040^2 = 0.032\text{ J} \). The kinetic energy is \( E_k = E_{\text{total}} - E_p = 0.128 - 0.032 = 0.096\text{ J} \).

[Part a]: 1 mark for stating acceleration is proportional to displacement, 1 mark for stating direction is opposite to displacement / towards equilibrium.

[Part b(i)]: 1 mark for substituting values into \( \omega = \sqrt{k/m} \), 1 mark for showing value of ~12.6 or 13 rad s^-1.

[Part b(ii)]: 1 mark for using \( v_{\text{max}} = \omega x_0 \), 1 mark for correct value 1.0 m s^-1 (accept 1.01 m s^-1).

[Part b(iii)]: 1 mark for calculating total energy (0.128 J), 1 mark for calculating potential energy (0.032 J), 1 mark for stating relationship E_k = E_total - E_p, 1 mark for final kinetic energy of 0.096 J.

(a) Electric field strength is the electric force per unit positive charge acting on a small test charge placed at that point.

(b)(i) The electric field strength between parallel plates is \( E = \frac{V}{d} \). Substituting the values: \( E = \frac{50}{0.020} = 2500\text{ V m}^{-1} \).

(b)(ii) The force is given by \( F = qE \). Substituting the values: \( F = 1.6 \times 10^{-19} \times 2500 = 4.0 \times 10^{-16}\text{ N} \). Since the electron is negatively charged, the force is directed upward, in the opposite direction to the downward electric field.

(b)(iii) The horizontal velocity is \( v_x = 2.5 \times 10^7\text{ m s}^{-1} \). The time spent in the electric field is \( t = \frac{L}{v_x} = \frac{0.15}{2.5 \times 10^7} = 6.0 \times 10^{-9}\text{ s} \). The vertical acceleration is \( a_y = \frac{F}{m} = \frac{4.0 \times 10^{-16}}{9.11 \times 10^{-31}} \approx 4.39 \times 10^{14}\text{ m s}^{-2} \). The vertical deflection is \( y = \frac{1}{2} a_y t^2 = 0.5 \times (4.39 \times 10^{14}) \times (6.0 \times 10^{-9})^2 = 7.9 \times 10^{-3}\text{ m} = 7.9\text{ mm} \).

[Part a]: 1 mark for 'force per unit charge', 1 mark for 'acting on a positive test charge'.

[Part b(i)]: 1 mark for formula E = V/d, 1 mark for correct substitution and final answer 2500 V m^-1 (or N C^-1).

[Part b(ii)]: 1 mark for formula F = qE, 1 mark for correct calculation of magnitude 4.0 × 10^-16 N, 1 mark for correctly identifying the direction as upward.

[Part b(iii)]: 1 mark for finding the travel time (6.0 × 10^-9 s), 1 mark for calculating the vertical acceleration (4.39 × 10^14 m s^-2), 1 mark for correct final vertical deflection of 7.9 mm (accept 7.9 × 10^-3 m).

(a) Random decay means that it is impossible to predict when a particular nucleus will decay, although there is a constant probability of decay per unit time. Spontaneous decay means that the decay process is unaffected by any external physical factors (such as temperature, pressure, or chemical state).

(b)(i) The half-life \( T_{1/2} \) in seconds is \( 15.0 \times 3600 = 54000\text{ s} \). The decay constant is \( \lambda = \frac{\ln(2)}{T_{1/2}} = \frac{0.69315}{54000} \approx 1.28 \times 10^{-5}\text{ s}^{-1} \).

(b)(ii) The initial activity is \( A_0 = \lambda N_0 \). Substituting the values: \( A_0 = 1.28 \times 10^{-5} \times 3.5 \times 10^{18} = 4.48 \times 10^{13}\text{ Bq} \approx 4.5 \times 10^{13}\text{ Bq} \).

(b)(iii) The number of half-lives in \( 45.0\text{ hours} \) is \( n = \frac{45.0}{15.0} = 3.0 \). The remaining fraction is \( (1/2)^3 = 1/8 = 0.125 \) (or \( 12.5\% \)).

[Part a]: 1 mark for defining random (cannot predict when a specific nucleus decays), 1 mark for defining spontaneous (unaffected by external conditions).

[Part b(i)]: 1 mark for converting half-life to seconds (54000 s), 1 mark for stating the relationship \( \lambda = \frac{\ln(2)}{T_{1/2}} \), 1 mark for correct value 1.28 × 10^-5 s^-1.

[Part b(ii)]: 1 mark for using A = \lambda * N, 1 mark for correct value 4.5 × 10^13 Bq (accept 4.48 × 10^13 Bq).

[Part b(iii)]: 1 mark for identifying that 3 half-lives have elapsed, 1 mark for using the decay factor (1/2)^n, 1 mark for correct fraction 0.125 (or 1/8 or 12.5%).