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Redshift shows that the wavelength of light from distant galaxies increases, meaning they are receding from us. The relationship between distance and speed of recession indicates that the entire universe is expanding. Working backwards in time, this suggests that the universe originated from a single, extremely hot and dense point (the Big Bang).

1.0 mark: State that redshift shows galaxies are moving away (receding) / the universe is expanding. 0.238 marks: Explain that more distant galaxies recede faster, suggesting a common origin in the past.

Force is equal to the rate of change of momentum: \(F = \frac{\Delta p}{\Delta t}\). The mass of water striking the wall per unit time is given by \(\frac{\Delta m}{\Delta t} = \rho A v\). Since the water comes to a complete stop, the change in velocity is \(v\). Therefore, the force is \(F = \left(\frac{\Delta m}{\Delta t}\right) \times v = (\rho A v) \times v = \rho A v^2\).

0.5 marks: Correctly express the mass of water striking the wall per second as \(\rho A v\). 0.5 marks: State that force equals rate of change of momentum (or \(F = \frac{\Delta m}{\Delta t} \times v\)). 0.238 marks: State the final formula \(F = \rho A v^2\).

Infrared radiation is a primary mechanism of heat transfer. Shiny, light-colored surfaces are poor emitters and excellent reflectors of infrared radiation, minimizing thermal transfer. Conversely, black, matte surfaces are excellent emitters and absorbers of infrared radiation, which would quickly cool down hot liquids inside the flask.

0.5 marks: State that shiny silver surfaces are poor emitters and/or good reflectors of infrared radiation. 0.5 marks: State that black matte surfaces are excellent emitters and/or absorbers of infrared radiation. 0.238 marks: Connect this to minimizing energy transfer to keep the flask contents at a constant temperature.

When the magnet approaches the copper ring, the magnetic flux through the ring changes, inducing an e.m.f. and thus a current. By Lenz's law, the direction of the induced current opposes the change in magnetic flux that produces it. This creates a magnetic field with a matching magnetic pole facing the approaching magnet, resulting in an upward repulsive force that opposes its downward motion.

0.5 marks: State that the changing magnetic flux induces an e.m.f. and current in the ring. 0.5 marks: Explain that the induced current's magnetic field opposes the downward movement of the magnet (Lenz's law). 0.238 marks: Identify that this opposition results in an upward magnetic force.

According to the principles of gravity, the gravitational force between two bodies is directly proportional to the product of their masses (the mass of the planet and the mass of the Sun) and inversely proportional to the square of the distance between their centers.

0.5 marks: State the masses of the Sun and the planet. 0.5 marks: State the distance between their centers. 0.238 marks: Express the relationship clearly (force increases with mass, decreases with distance).

To detect an internal defect, the wavelength of the sound wave must be comparable to or smaller than the size of the flaw. Since ultrasound has a very high frequency, its wavelength is extremely short. This allows the wave to reflect off the small flaws rather than diffracting around them, which audible sound (with its longer wavelengths) would do.

0.5 marks: State that ultrasound has a shorter wavelength (or higher frequency) than audible sound. 0.5 marks: Explain that shorter wavelengths undergo less diffraction around small features. 0.238 marks: Relate this to the ability to reflect off and clearly resolve small flaws.

Alpha particles are heavy, highly charged (\(+2e\)) helium nuclei that ionize air molecules very efficiently, losing energy rapidly over a short distance (a few centimeters). Beta particles are lightweight, high-speed electrons with a charge of only \(-1e\). They are less ionizing, meaning they interact less frequently with matter and travel much further (several meters in air) before coming to rest.

0.5 marks: Contrast the physical properties (beta has lower charge/mass than alpha). 0.5 marks: Connect these properties to beta being less strongly ionizing. 0.238 marks: Conclude that less ionization results in a greater penetration range before energy is fully dissipated.

An electric current is a flow of free (delocalized) electrons. As these electrons move through the metallic lattice under an applied potential difference, they repeatedly collide with the fixed positive metal ions. During these collisions, kinetic energy is transferred from the moving electrons to the lattice ions. This increases the internal kinetic energy (vibrational energy) of the lattice, resulting in an increase in temperature.

0.5 marks: Identify that moving electrons collide with the fixed positive metal ions in the lattice. 0.5 marks: State that collisions transfer kinetic energy to the lattice ions. 0.238 marks: State that the increased vibration of the lattice ions corresponds to an increase in temperature.

Use the formula \(I = \frac{Q}{t}\). First, convert the time from milliseconds to seconds: \(15\text{ ms} = 15 \times 10^{-3}\text{ s}\). Substitute the values: \(I = \frac{1.8 \times 10^{-6}\text{ C}}{15 \times 10^{-3}\text{ s}} = 1.2 \times 10^{-4}\text{ A}\). Convert this current into microamperes: \(1.2 \times 10^{-4}\text{ A} = 120\text{ }\mu\text{A}\).

1.000 mark for using \(I = Q/t\) with correct unit conversion of time. 0.238 marks for the correct final calculation of current in \(\mu\text{A}\).

For uniform deceleration, the average speed is given by \(\frac{u + v}{2}\). The distance travelled is \(s = \frac{u + v}{2} \times t\). Rearranging for time gives \(t = \frac{2s}{u + v}\). Substituting the given values: \(t = \frac{2 \times 150}{80 + 20} = \frac{300}{100} = 3.0\text{ s}\).

1.000 mark for stating or applying the formula relating distance, average speed, and time. 0.238 marks for the correct calculation of time (3.0 s).

The primary mode of thermal energy transfer addressed by the silvered coating is radiation (specifically infrared radiation). Highly reflective, shiny silver surfaces are poor emitters of radiation (reducing radiation leaving the inner flask) and excellent reflectors or poor absorbers of radiation (reflecting radiation back and preventing it from entering).

1.000 mark for correctly identifying radiation as the mode of transfer. 0.238 marks for explaining that shiny or silvered surfaces are poor emitters or good reflectors of radiation.

The orbital radius is the sum of the Earth's radius and the altitude: \(r = 6400\text{ km} + 600\text{ km} = 7000\text{ km}\). The orbital period in seconds is \(T = 96 \times 60 = 5760\text{ s}\). The orbital speed is \(v = \frac{2\pi r}{T}\). Substituting the values: \(v = \frac{2 \times \pi \times 7000}{5760} \approx 7.64\text{ km/s}\).

1.000 mark for calculating the correct orbital radius (7000 km) and using the formula \(v = 2\pi r / T\). 0.238 marks for the correct final answer between 7.6 and 7.64 km/s.

Impulse is equal to the change in momentum: \(\Delta p = m(v - u)\). Taking the initial direction of motion as positive (\(u = 22\text{ m/s}\)), the rebound velocity is negative (\(v = -18\text{ m/s}\)). Therefore, \(\Delta p = 0.060 \times (-18 - 22) = 0.060 \times (-40) = -2.4\text{ N s}\). The magnitude of the impulse is \(2.4\text{ N s}\).

1.000 mark for using the change in momentum formula and accounting for the opposite direction of velocity. 0.238 marks for the correct final value of 2.4 N s (or kg m/s).

For an echo, the sound travels a total distance of \(2d\) (to the cliff and back). The relationship is \(2d = v \times t\), which gives \(d = \frac{v \times t}{2}\). Substituting the given values: \(d = \frac{1500 \times 0.80}{2} = \frac{1200}{2} = 600\text{ m}\).

1.000 mark for using the echo equation \(2d = v \times t\) or equivalent. 0.238 marks for the correct calculation of 600 m.

The relationship between the refractive index \(n\) of the denser medium and the critical angle \(c\) is given by \(n = \frac{1}{\sin(c)}\). Substituting the given critical angle: \(n = \frac{1}{\sin(42.0^\circ)} = \frac{1}{0.6691} \approx 1.49\).

1.000 mark for stating or applying the formula \(n = 1/\sin(c)\). 0.238 marks for the correct refractive index value of 1.49.

First, determine the total elapsed time in hours: \(2.5\text{ days} = 2.5 \times 24\text{ hours} = 60\text{ hours}\). Next, find the number of half-lives that have elapsed: \(n = \frac{60\text{ hours}}{12\text{ hours}} = 5\). The remaining active nuclei is given by \(N = N_0 \times (1/2)^n = 3.2 \times 10^{15} \times (1/2)^5 = \frac{3.2 \times 10^{15}}{32} = 1.0 \times 10^{14}\).

1.000 mark for calculating the number of half-lives (5 half-lives). 0.238 marks for the correct final number of remaining nuclei (\(1.0 \times 10^{14}\)).

The motion consists of three stages:
1. Acceleration stage:
\(d_1 = \text{average velocity} \times t_1 = \frac{0 + 6.0}{2} \times 4.0 = 12.0\text{ m}\)

2. Constant speed stage:
\(d_2 = v \times t_2 = 6.0 \times 12.0 = 72.0\text{ m}\)

3. Deceleration stage:
\(d_3 = \text{average velocity} \times t_3 = \frac{6.0 + 0}{2} \times 3.0 = 9.0\text{ m}\)

Total distance = \(d_1 + d_2 + d_3 = 12.0 + 72.0 + 9.0 = 93\text{ m}\)

1 mark: Correct calculation of distance for each of the three stages (e.g. 12 m, 72 m, 9 m) or a correct area-under-graph method.
0.238 marks: Correct final sum of 93 m with unit.

Identify the forces and their distances from the pivot at one end (distance = \(0\text{ m}\)):
- The weight of the uniform beam acts at its center of mass, which is at the midpoint: \(d_w = 1.0\text{ m}\).
- Clockwise moment = \(15\text{ N} \times 1.0\text{ m} = 15\text{ N m}\).
- The force \(F\) is applied at \(0.50\text{ m}\) from the free end, which is at \(2.0 - 0.50 = 1.5\text{ m}\) from the pivot.
- Counter-clockwise moment = \(F \times 1.5\text{ m}\).

For equilibrium, sum of clockwise moments = sum of counter-clockwise moments:
\(F \times 1.5 = 15\)
\(F = 10\text{ N}\)

1 mark: Equating clockwise and counter-clockwise moments correctly with the weight acting at 1.0 m.
0.238 marks: Correct value of 10 N with unit.

Let the direction to the right be positive.
Initial momentum of the system:
\(p_i = m_1 u_1 + m_2 u_2 = (0.50 \times 2.0) + (0.30 \times 0) = 1.0\text{ kg m/s}\)

Final momentum of the system:
\(p_f = m_1 v_1 + m_2 v_2 = (0.50 \times [-0.40]) + (0.30 \times v_2) = -0.20 + 0.30 v_2\)

Using conservation of momentum (\(p_i = p_f\)):
\(1.0 = -0.20 + 0.30 v_2\)
\(1.2 = 0.30 v_2\)
\(v_2 = 4.0\text{ m/s}\)

Since the velocity is positive, the direction is to the right.

1 mark: Correct conservation of momentum equation taking direction/signs into account.
0.238 marks: Correct magnitude (4.0 m/s) and direction (to the right).

In copper (a metal), thermal conduction occurs via two processes:
1. Free electrons near the hot stove gain kinetic energy and move rapidly through the structure of the metal. They collide with ions and other electrons in cooler parts of the metal, transferring thermal energy quickly.
2. Copper ions near the heat source vibrate more rapidly. They pass these vibrations along to neighboring ions through lattice vibrations.

1 mark: Explanation of free (delocalized) electrons gaining kinetic energy and colliding with ions to transfer thermal energy.
0.238 marks: Mention of lattice vibrations being passed from atom to atom / ion to ion.

Using Snell's Law:
\(n = \frac{\sin(i)}{\sin(r)}\)

Substitute the values given:
\(1.5 = \frac{\sin(40^\circ)}{\sin(r)}\)

\(\sin(r) = \frac{\sin(40^\circ)}{1.5}\)
\(\sin(40^\circ) \approx 0.6428\)
\(\sin(r) = \frac{0.6428}{1.5} \approx 0.4285\)

\(r = \sin^{-1}(0.4285) \approx 25.37^\circ\)

Rounding to 2 significant figures gives \(25^\circ\).

1 mark: Correct substitution of values into Snell's Law equation.
0.238 marks: Correct calculated angle (25° or 25.4°).

First, calculate the electric power \(P\) dissipated by the resistor:
\(P = I^2 R\)
\(P = (0.50)^2 \times 12 = 0.25 \times 12 = 3.0\text{ W}\)

Convert time to seconds:
\(t = 3.0\text{ minutes} = 3.0 \times 60 = 180\text{ s}\)

Calculate energy \(E\):
\(E = P \times t = 3.0 \times 180 = 540\text{ J}\)

1 mark: Correct calculation of power (3.0 W) or correct formula for energy \(E = I^2 R t\) with correct time conversion (180 s).
0.238 marks: Correct final answer of 540 J.

During beta-minus (\(\beta^-\)) decay, a neutron inside the parent nucleus decays into a proton and an electron (the beta particle).
- The proton number increases by 1: \(6 + 1 = 7\).
- The nucleon number remains unchanged because the total number of protons and neutrons remains 14.
Therefore, the nitrogen nucleus is \(^{14}_{7}\text{N}\).

1 mark: State the correct nucleon number (14).
0.238 marks: State the correct proton number (7).

Redshift occurs when light from distant galaxies has its wavelength stretched (shifted towards the red end of the electromagnetic spectrum) because the galaxies are moving away from the observer. This Doppler-like effect provides evidence that galaxies are receding from us and thus the Universe is expanding.

1 mark: Define redshift as an increase in wavelength (or shift to red/lower frequency end of the spectrum).
0.238 marks: State that it indicates the galaxies are moving away from us / receding.

Using the equation for distance with constant acceleration: \(s = \frac{1}{2}(u + v)t\). Given that the cyclist starts from rest, \(u = 0\), and reaches a final speed \(v = 6.0\text{ m/s}\) in \(t = 4.0\text{ s}\). Substituting the values: \(s = \frac{1}{2}(0 + 6.0) \times 4.0 = 3.0 \times 4.0 = 12\text{ m}\).

1 mark for correct formula or calculation of average speed (\(3.0\text{ m/s}\)); 0.238 marks for the correct final answer of \(12\text{ m}\) with appropriate unit.

Using the principle of conservation of momentum: total initial momentum = total final momentum. \(m_1 u_1 + m_2 u_2 = (m_1 + m_2)v\). Substituting the known values: \((1200 \times 4.0) + (800 \times 0) = (1200 + 800) \times v\). \(4800 = 2000 \times v\). Solve for \(v\): \(v = \frac{4800}{2000} = 2.4\text{ m/s}\).

1 mark for conservation of momentum equation or calculation of initial momentum (\(4800\text{ kg m/s}\)); 0.238 marks for the correct final velocity of \(2.4\text{ m/s}\).

Copper is a metal with a lattice of positive ions and a sea of delocalised (free) electrons. When heated, the free electrons gain kinetic energy and move faster. They diffuse through the metal lattice, colliding with cooler ions and transferring energy quickly. At the same time, ions in the hot region vibrate more vigorously and pass this vibrational energy to neighboring ions.

1 mark for explaining the role of free electrons moving through the lattice and colliding with other particles; 0.238 marks for mentioning the transmission of lattice vibrations between neighboring ions.

The sound wave travels to the seabed and back, so the total distance travelled by the pulse is \(d = v \times t = 1500 \times 0.80 = 1200\text{ m}\). The depth is half of this total distance: \(\text{depth} = \frac{1200}{2} = 600\text{ m}\).

1 mark for calculating the total distance (\(1200\text{ m}\)) or using the correct formula taking half the time; 0.238 marks for the correct depth of \(600\text{ m}\).

Using Snell's Law: \(n = \frac{\sin i}{\sin r}\). Substituting the values: \(1.50 = \frac{\sin 45^\circ}{\sin r}\). \(\sin r = \frac{\sin 45^\circ}{1.50} = \frac{0.7071}{1.50} \approx 0.4714\). \(r = \sin^{-1}(0.4714) \approx 28.1^\circ\).

1 mark for rearranging Snell's law correctly (\(\sin r = \frac{\sin i}{n}\)); 0.238 marks for the correct angle of refraction of \(28.1^\circ\) (accept \(28^\circ\)).

First, convert the time from minutes to seconds: \(t = 1.5 \times 60 = 90\text{ s}\). Then use the formula for current: \(I = \frac{Q}{t} = \frac{18}{90} = 0.20\text{ A}\).

1 mark for converting time to seconds (\(90\text{ s}\)) and stating the current formula; 0.238 marks for the correct answer of \(0.20\text{ A}\).

The force that provides the centripetal acceleration to keep planets in orbit around the Sun is the gravitational force (gravity). According to gravitational principles, the strength of this force decreases as the distance between the two bodies increases.

1 mark for identifying the gravitational force; 0.238 marks for stating that the force decreases with increasing distance.

Determine the number of half-lives that have elapsed: \(800 \xrightarrow{1 \text{ half-life}} 400 \xrightarrow{2 \text{ half-lives}} 200 \xrightarrow{3 \text{ half-lives}} 100\text{ Bq}\). Since \(3\) half-lives equal \(15\text{ hours}\), the length of one half-life is: \(T_{1/2} = \frac{15}{3} = 5.0\text{ hours}\).

1 mark for identifying that three half-lives have elapsed; 0.238 marks for calculating the correct half-life of \(5.0\text{ hours}\).

When the skydiver first jumps, the only significant force is weight acting downwards, causing maximum acceleration. As the speed increases, air resistance (drag) acting upwards increases. The resultant force downwards is the difference between weight and air resistance: \(F_{\text{resultant}} = W - F_{\text{drag}}\). Because air resistance increases, the resultant force decreases. According to Newton's second law, \(F = ma\), a smaller resultant force acting on a constant mass results in a smaller acceleration.

[0.5 marks] For stating that air resistance increases as speed increases. [0.5 marks] For stating that the resultant downward force decreases. [0.238 marks] For linking the smaller resultant force to a lower acceleration using \(F = ma\).

In copper (a metal), thermal conduction occurs via two mechanisms: the vibration of atoms and the movement of free (delocalised) electrons. The free electrons near the heat source gain kinetic energy, move rapidly through the metal lattice, and transfer energy to cooler regions by colliding with ions. In contrast, glass (an insulator) has no free electrons, so thermal energy can only be transferred slowly through the physical vibrations of adjacent atoms (lattice vibrations).

[0.5 marks] For stating that copper has free (delocalised) electrons which can move through the lattice, whereas glass does not. [0.5 marks] For explaining that these free electrons gain kinetic energy and rapidly transfer it by colliding with ions/atoms. [0.238 marks] For noting that glass relies only on slower atom-to-atom lattice vibrations.

An electric field is defined as a region of space where an electric charge experiences a force. The direction of the electric field at any point is defined as the direction of the force that would act on a small positive test charge placed at that point. Since like charges repel, a positive test charge would be pushed away from an isolated positive point charge. Thus, the electric field lines must point radially outwards.

[0.5 marks] For defining an electric field as a region where a charge experiences a force. [0.5 marks] For stating that the field lines point radially outwards. [0.238 marks] For explaining that the direction is defined by the repulsive force acting on a positive test charge.

Initially, both skaters are stationary, so the total initial momentum is zero. According to the law of conservation of momentum, in the absence of external forces, the total final momentum must also be zero. Therefore, \(m_1 v_1 + m_2 v_2 = 0\), which means \(m_1 v_1 = -m_2 v_2\). The negative sign shows they move in opposite directions. Since the classmate has a smaller mass (40 kg compared to 60 kg), they must move with a higher velocity to have a momentum of equal magnitude to the heavier skater.

[0.5 marks] For stating that the initial total momentum is zero, so final total momentum must be zero due to conservation of momentum. [0.5 marks] For stating that they move in opposite directions. [0.238 marks] For explaining that the 40 kg classmate moves faster because their mass is smaller, keeping the magnitudes of their momenta equal.

A planet moving in a circular orbit requires a centripetal force directed towards the center of the orbit (the Sun). This centripetal force is provided by the gravitational force of attraction between the mass of the Sun and the mass of the planet. As the distance from the Sun increases, the gravitational field strength becomes weaker, requiring a lower orbital speed for a stable orbit.

[0.5 marks] For identifying gravitational force of attraction as the provider of the centripetal force. [0.5 marks] For stating that orbital speed decreases as distance from the Sun increases. [0.238 marks] For explaining that a larger distance results in a weaker gravitational pull, requiring less speed to maintain orbit.

When light travels from an optically denser medium (glass) towards a less dense medium (air), it normally refracts away from the normal. However, if the angle of incidence exceeds the critical angle, refraction is no longer possible. Instead, total internal reflection occurs, meaning all light is reflected back into the glass block, obeying the law of reflection where the angle of reflection equals the angle of incidence.

[0.5 marks] For identifying that total internal reflection occurs. [0.5 marks] For stating that no light is refracted/transmitted into the air. [0.238 marks] For stating that the angle of reflection equals the angle of incidence.

Alpha particles are positively charged and experience a magnetic force, deflecting in one direction with a relatively large radius (small deflection) due to their large mass. Beta-minus particles are negatively charged, so they experience a force in the opposite direction and, having a very small mass, deflect much more sharply. Gamma rays are uncharged electromagnetic waves, so they do not experience any magnetic force and pass straight through without deflection.

[0.5 marks] For stating that gamma radiation is not deflected because it is uncharged. [0.5 marks] For stating that alpha and beta-minus particles deflect in opposite directions due to opposite charges. [0.238 marks] For explaining that beta-minus particles deflect more than alpha particles because they have a much smaller mass.

Light from distant galaxies is observed to have longer wavelengths than expected, shifted towards the red end of the spectrum (redshift). This Doppler-like shift indicates that the galaxies are moving away from Earth. Furthermore, observations show that the further away a galaxy is, the faster it is receding (greater redshift). This proportional relationship (Hubble's Law) indicates that the universe is expanding uniformly in all directions.

[0.5 marks] For explaining that redshift means light's wavelength is stretched / shifted to the red end, showing galaxies are moving away. [0.5 marks] For stating that more distant galaxies show a greater redshift / are moving away faster. [0.238 marks] For concluding that this speed-distance relationship shows space itself is expanding.

1. Identify the formula for orbital speed: \(v = \frac{2\pi r}{T}\). 2. Find the total orbital radius \(r\) by adding the radius of the Earth and the altitude: \(r = 6400\text{ km} + 600\text{ km} = 7000\text{ km} = 7.0 \times 10^6\text{ m}\). 3. Convert the orbital period \(T\) from minutes to seconds: \(T = 96 \times 60 = 5760\text{ s}\). 4. Calculate the orbital speed: \(v = \frac{2 \times \pi \times 7.0 \times 10^6}{5760} \approx 7636\text{ m/s}\). State the answer to 2 significant figures (\(7600\text{ m/s}\)) or 3 significant figures (\(7640\text{ m/s}\)).

Award 1 mark for correct calculation of total radius (\(7.0 \times 10^6\text{ m}\)) and time conversion (\(5760\text{ s}\)). Award 0.238 marks for correct application of orbital speed formula resulting in a final answer in the range \(7600\text{ m/s}\) to \(7640\text{ m/s}\).

1. Apply the principle of conservation of momentum: total momentum before collision equals total momentum after collision, \(m_1 v_1 + m_2 v_2 = (m_1 + m_2) v\). 2. Calculate the total initial momentum: \(p_i = (1.2\text{ kg} \times 2.5\text{ m/s}) + (0.80\text{ kg} \times 0\text{ m/s}) = 3.0\text{ kg m/s}\). 3. Determine the total mass after the collision: \(M = 1.2\text{ kg} + 0.80\text{ kg} = 2.0\text{ kg}\). 4. Solve for the final common velocity \(v\): \(3.0\text{ kg m/s} = 2.0\text{ kg} \times v \implies v = 1.5\text{ m/s}\).

Award 1 mark for finding the initial momentum of \(3.0\text{ kg m/s}\) (or writing the correct equation for conservation of momentum with substituted values). Award 0.238 marks for the correct final velocity of \(1.5\text{ m/s}\).

First, determine the deceleration \(a\) of the car using Newton's second law: \(F = ma \implies a = \frac{F}{m} = \frac{1.2\text{ N}}{0.80\text{ kg}} = 1.5\text{ m/s}^2\). Next, use the equation of motion \(v^2 = u^2 - 2ad\) to find the stopping distance \(d\), where \(v = 0\text{ m/s}\) and \(u = 6.0\text{ m/s}\): \(0 = 6.0^2 - 2 \times 1.5 \times d \implies 3.0 \times d = 36 \implies d = 12\text{ m}\). Alternatively, using energy conservation: initial kinetic energy \(E_k = \frac{1}{2} m v^2 = \frac{1}{2} \times 0.80 \times 6.0^2 = 14.4\text{ J}\). Work done by the resistive force \(W = F \times d = 1.2 \times d = 14.4\text{ J} \implies d = 12\text{ m}\).

- Correct formula used for deceleration or energy: \(a = F/m\) or \(E_k = \frac{1}{2}mv^2\) [1 mark]
- Correct calculation of deceleration (\(1.5\text{ m/s}^2\)) or initial kinetic energy (\(14.4\text{ J}\)) [1 mark]
- Correct rearrangement of equations to solve for distance: \(d = 12\) [1 mark]
- Correct unit of metres (\(\text{m}\)) stated [0.5 marks]

The relationship between refractive index \(n\) and critical angle \(c\) is given by: \(\sin c = \frac{1}{n} = \frac{1}{1.45} \approx 0.6897\). Therefore, critical angle \(c = \arcsin(0.6897) \approx 43.6^\circ\). Since the angle of incidence (\(45^\circ\)) is greater than the critical angle (\(43.6^\circ\)), and the ray is travelling from a more optically dense medium (plastic) to a less optically dense medium (air), the ray will undergo total internal reflection.

- Correct formula used: \(\sin c = 1/n\) [1 mark]
- Calculation of critical angle \(c = 43.6^\circ\) (accept \(44^\circ\)) [1/mark]
- Comparison of incidence angle to critical angle (\(45^\circ > 43.6^\circ\)) [1 mark]
- Correct conclusion that total internal reflection will occur [0.5 marks]

By conservation of momentum, the total momentum before collision equals the total momentum after collision: \(m_1 u_1 + m_2 u_2 = (m_1 + m_2) v \implies 0.20 \times 120 + 1.8 \times 0 = (0.20 + 1.8) \times v \implies 24 = 2.0 \times v \implies v = 12\text{ m/s}\). Next, calculate the kinetic energies. Initial kinetic energy: \(E_{ki} = \frac{1}{2} m_1 u_1^2 = \frac{1}{2} \times 0.20 \times 120^2 = 1440\text{ J}\). Final kinetic energy: \(E_{kf} = \frac{1}{2} (m_1 + m_2) v^2 = \frac{1}{2} \times 2.0 \times 12^2 = 144\text{ J}\). Kinetic energy lost: \(\Delta E_k = E_{ki} - E_{kf} = 1440 - 144 = 1296\text{ J}\) (or \(1.3 \times 10^3\text{ J}\)).

- Applying conservation of momentum equation to find \(v = 12\text{ m/s}\) [1.5 marks]
- Correct calculation of initial kinetic energy (\(1440\text{ J}\)) and final kinetic energy (\(144\text{ J}\)) [1 mark]
- Correct subtraction to obtain kinetic energy lost of \(1296\text{ J}\) or \(1300\text{ J}\) [1 mark]

First, calculate the total power incident on the panel: \(P_{\text{incident}} = 800\text{ W/m}^2 \times 1.5\text{ m}^2 = 1200\text{ W}\). Next, calculate the power absorbed by the water: \(P_{\text{absorbed}} = 1200\text{ W} \times 0.70 = 840\text{ W}\). The time duration is \(10\text{ minutes} = 10 \times 60 = 600\text{ s}\). The total thermal energy absorbed is: \(E = P_{\text{absorbed}} \times t = 840\text{ W} \times 600\text{ s} = 504000\text{ J}\) (or \(504\text{ kJ}\)).

- Calculate total power incident: \(800 \times 1.5 = 1200\text{ W}\) [1 mark]
- Calculate absorbed power using efficiency: \(1200 \times 0.7 = 840\text{ W}\) [1 mark]
- Convert minutes to seconds (\(10\text{ min} = 600\text{ s}\)) and multiply power by time [1 mark]
- Correct final energy value \(504000\text{ J}\) (or \(504\text{ kJ}\) or \(5.0 \times 10^5\text{ J}\)) with unit [0.5 marks]

The formula for orbital speed is \(v = \frac{2\pi r}{T}\). The orbital period \(T = 1.8\text{ hours} = 1.8 \times 3600\text{ s} = 6480\text{ s}\). The orbital distance \(2\pi r = 2 \times \pi \times 7.2 \times 10^6\text{ m} \approx 4.5239 \times 10^7\text{ m}\). Calculate orbital speed: \(v = \frac{4.5239 \times 10^7\text{ m}}{6480\text{ s}} \approx 6981\text{ m/s}\). This rounds to \(6980\text{ m/s}\) (or \(7.0 \times 10^3\text{ m/s}\)).

- Statement of formula \(v = \frac{2\pi r}{T}\) [1 mark]
- Conversion of period to seconds: \(1.8 \times 3600 = 6480\text{ s}\) [1 mark]
- Correct calculation of orbital distance \(2\pi r = 4.52 \times 10^7\text{ m}\) [1 mark]
- Correct calculation of final speed in range \(6970\text{ m/s}\) to \(7000\text{ m/s}\) with correct unit [0.5 marks]

First, convert current and time to standard SI units: \(I = 15\text{ mA} = 0.015\text{ A}\), and \(t = 4.0\text{ minutes} = 240\text{ s}\). To find the charge: \(Q = I \times t = 0.015\text{ A} \times 240\text{ s} = 3.6\text{ C}\). To find the energy transferred: \(E = V \times Q\) or \(E = V \times I \times t = 5.0\text{ V} \times 3.6\text{ C} = 18\text{ J}\).

- Use of \(Q = I \times t\) with correct units for current and time [1 mark]
- Calculation of charge \(Q = 3.6\text{ C}\) with unit [1 mark]
- Use of \(E = V I t\) or \(E = Q V\) [1 mark]
- Calculation of energy \(E = 18\text{ J}\) with unit [0.5 marks]

First, find the equivalent resistance \(R_p\) of the parallel combination: \(\frac{1}{R_p} = \frac{1}{6.0} + \frac{1}{12.0} = \frac{2}{12.0} + \frac{1}{12.0} = \frac{3}{12.0} \implies R_p = 4.0\\ \Omega\). Alternatively, \(R_p = \frac{6.0 \times 12.0}{6.0 + 12.0} = \frac{72.0}{18.0} = 4.0\\ \Omega\). Now, add the series resistor to get the total resistance: \(R_{\text{total}} = R_p + 4.0\\ \Omega = 4.0 + 4.0 = 8.0\\ \Omega\). Finally, find the total current supplied: \(I = \frac{V}{R_{\text{total}}} = \frac{12\text{ V}}{8.0\\ \Omega} = 1.5\text{ A}\).

- Correct use of parallel resistor formula to get \(R_p = 4.0\\ \Omega\) [1 mark]
- Calculation of total series resistance \(R_{\text{total}} = 8.0\\ \Omega\) [1 mark]
- Use of Ohm's law \(I = V/R\) [1/mark]
- Correct current \(1.5\text{ A}\) with correct unit [0.5 marks]

First, subtract the background radiation to find the initial count rate of the source alone: \(424 - 24 = 400\text{ counts/minute}\). Next, determine the number of half-lives that elapse in \(12\text{ hours}\): \(\text{Number of half-lives} = \frac{12\text{ hours}}{4.0\text{ hours}} = 3\text{ half-lives}\). After \(3\text{ half-lives}\), the activity of the source alone is reduced by a factor of \(2^3 = 8\): \(\text{Activity of source alone} = \frac{400}{8} = 50\text{ counts/minute}\). Finally, add back the constant background radiation to get the expected total measured count rate: \(50 + 24 = 74\text{ counts/minute}\).

- Subtract background from initial rate: \(424 - 24 = 400\text{ counts/minute}\) [1 mark]
- Determine number of half-lives (\(3\)) and divide source activity by \(8\) to get \(50\text{ counts/minute}\) [1.5 marks]
- Add background radiation back to source activity: \(50 + 24 = 74\text{ counts/minute}\) [1 mark]