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First, calculate the weight of the rocket: \(W = mg = 0.25\text{ kg} \times 9.8\text{ m/s}^2 = 2.45\text{ N}\).
Next, calculate the resultant upward force on the rocket: \(F_{\text{net}} = \text{Thrust} - W = 6.0\text{ N} - 2.45\text{ N} = 3.55\text{ N}\).
Finally, calculate the acceleration using Newton's second law: \(a = \frac{F_{\text{net}}}{m} = \frac{3.55\text{ N}}{0.25\text{ kg}} = 14.2\text{ m/s}^2\).

1 mark: Calculate the weight of the rocket, \(W = 2.45\text{ N}\).
1 mark: Determine the resultant upward force, \(F = 3.55\text{ N}\).
1 mark: Calculate the acceleration, \(a = 14.2\text{ m/s}^2\) (allow \(14\text{ m/s}^2\) if rounded to two significant figures).

By the law of conservation of momentum, total momentum before collision equals total momentum after collision:
\(m_A u_A + m_B u_B = m_A v_A + m_B v_B\)
Substitute the given values into the equation:
\((2.0 \times 3.5) + (1.5 \times 0) = (2.0 \times 1.1) + (1.5 \times v_B\))
\(7.0 = 2.2 + 1.5 v_B\)
\(1.5 v_B = 4.8\)
\(v_B = \frac{4.8}{1.5} = 3.2\text{ m/s}\) to the right.

1 mark: Recall and apply the conservation of momentum equation.
1 mark: Calculate total initial momentum as \(7.0\text{ kg m/s}\).
1 mark: Correctly rearrange and solve for final velocity of B, \(v_B = 3.2\text{ m/s}\).

First, calculate the useful work output (change in gravitational potential energy):
\(W_{\text{out}} = mgh = 80\text{ kg} \times 9.8\text{ m/s}^2 \times 15\text{ m} = 11760\text{ J}\).
Next, calculate the useful power output:
\(P_{\text{out}} = \frac{W_{\text{out}}}{t} = \frac{11760\text{ J}}{12\text{ s}} = 980\text{ W}\).
Finally, calculate efficiency:
\(\text{Efficiency} = \frac{P_{\text{out}}}{P_{\text{in}}} = \frac{980\text{ W}}{1400\text{ W}} = 0.70\) (or \(70\%\)).

1 mark: Calculate useful work output \(11760\text{ J}\) or useful power output \(980\text{ W}\).
1 mark: Recall and use efficiency formula: \(\text{efficiency} = \frac{\text{useful power output}}{\text{input power}}\).
1 mark: Correctly calculate the efficiency as \(0.70\) or \(70\%\).

First, calculate the thermal energy required to heat the water:
\(Q = mc\Delta\theta = 0.75\text{ kg} \times 4200\text{ J/(kg }^\circ\text{C)} \times (100 - 20)\text{ }^\circ\text{C} = 0.75 \times 4200 \times 80 = 252000\text{ J}\).
Next, write down the heating power in watts:
\(P = 1.8\text{ kW} = 1800\text{ W}\).
Finally, use the power equation to solve for time:
\(t = \frac{Q}{P} = \frac{252000\text{ J}}{1800\text{ W}} = 140\text{ s}\).

1 mark: Calculate the thermal energy required, \(Q = 252000\text{ J}\).
1 mark: Recall and rearrange \(P = \frac{E}{t}\) to \(t = \frac{E}{P}\) with correct conversion of power to \(1800\text{ W}\).
1 mark: Correctly calculate the time as \(140\text{ s}\).

When light passes from one medium to another, its frequency remains unchanged. Therefore, the frequency of the light inside the glass block is also \(4.60 \times 10^{14}\text{ Hz}\).
Using the wave equation \(v = f\lambda\), we can find the wavelength inside the glass block:
\(\lambda = \frac{v}{f} = \frac{1.90 \times 10^8\text{ m/s}}{4.60 \times 10^{14}\text{ Hz}} \approx 4.13 \times 10^{-7}\text{ m}\).

1 mark: State or use that the frequency of light remains constant at \(4.60 \times 10^{14}\text{ Hz}\) in the glass block.
1 mark: Recall and rearrange wave equation \(v = f\lambda\) to \(\lambda = \frac{v}{f}\).
1 mark: Correctly calculate wavelength as \(4.13 \times 10^{-7}\text{ m}\) (accept \(4.1 \times 10^{-7}\text{ m}\)).

First, calculate the current in the circuit using Ohm's law on the resistor:
\(I = \frac{V_R}{R} = \frac{4.5\text{ V}}{6.0\text{ }\Omega} = 0.75\text{ A}\).
In a series circuit, the current is identical at all points, so the current through the thermistor is also \(0.75\text{ A}\).
Next, calculate the potential difference across the thermistor:
\(V_T = V_{\text{supply}} - V_R = 12\text{ V} - 4.5\text{ V} = 7.5\text{ V}\).
Finally, calculate the resistance of the thermistor:
\(R_T = \frac{V_T}{I} = \frac{7.5\text{ V}}{0.75\text{ A}} = 10\text{ }\Omega\).

1 mark: Calculate the current through the series circuit as \(0.75\text{ A}\).
1 mark: Determine the potential difference across the thermistor as \(7.5\text{ V}\) (or construct a correct potential divider ratio equation).
1 mark: Calculate the resistance of the thermistor as \(10\text{ }\Omega\).

First, calculate the change in wavelength:
\(\Delta\lambda = \lambda - \lambda_0 = 494.1\text{ nm} - 486.0\text{ nm} = 8.1\text{ nm}\).
Next, use the redshift equation:
\(\frac{\Delta\lambda}{\lambda_0} = \frac{v}{c}\)
\(\frac{8.1}{486.0} = \frac{v}{3.00 \times 10^8}\)
Solve for the recession speed \(v\):
\(v = 3.00 \times 10^8 \times 0.016667 = 5.0 \times 10^6\text{ m/s}\).

1 mark: Calculate the change in wavelength, \(\Delta\lambda = 8.1\text{ nm}\).
1 mark: Correct substitution into the redshift equation: \(\frac{8.1}{486.0} = \frac{v}{3.00 \times 10^8}\).
1 mark: State the final calculated speed, \(v = 5.0 \times 10^6\text{ m/s}\).

First, calculate the initial and final count rates corrected for background radiation:
\(\text{Corrected initial count rate} = 440 - 24 = 416\text{ counts/minute}\).
\(\text{Corrected final count rate} = 76 - 24 = 52\text{ counts/minute}\).
Determine the fraction of active nuclei remaining:
\(\frac{52}{416} = \frac{1}{8}\).
Since \(\left(\frac{1}{2}\right)^3 = \frac{1}{8}\), exactly \(3\) half-lives have elapsed in the \(15\text{ days}\).
Therefore, the half-life is:
\(T_{1/2} = \frac{15\text{ days}}{3} = 5.0\text{ days}\).

1 mark: Correctly subtract background radiation to find initial and final corrected source activity (\(416\text{ cpm}\) and \(52\text{ cpm}\)).
1 mark: Determine that the activity has reduced by three half-lives (factor of \(8\)).
1 mark: Correctly calculate the half-life as \(5.0\text{ days}\).

1. Calculate the work done in lifting the water: \(W = mgh = 150 \times 9.8 \times 12 = 17640\text{ J}\). 2. Calculate the useful power output of the pump: \(P_{\text{out}} = \frac{W}{t} = \frac{17640}{30} = 588\text{ W}\). 3. Use the efficiency to calculate the electrical power input: \\text{Efficiency} = \\frac{P_{\\text{out}}}{P_{\\text{in}}} \\implies 0.80 = \\frac{588}{P_{\\text{in}}} \\implies P_{\\text{in}} = \\frac{588}{0.80} = 735\\text{ W}.

- Use of \(W = mgh\) to find work done = 17640 J (or \(P_{\text{out}} = \frac{mgh}{t}\) to find output power = 588 W): 1 mark. - Use of \\text{Efficiency} = \\frac{\\text{Power Output}}{\\text{Power Input}}: 1 mark. - Correct final power input of 735 W (or 740 W): 1 mark.

1. Convert the time to seconds: \(t = 2.0\text{ minutes} = 2.0 \times 60 = 120\text{ s}\). 2. Calculate the electric current using the charge and time: \(I = \frac{Q}{t} = \frac{360\text{ C}}{120\text{ s}} = 3.0\text{ A}\). 3. Calculate the resistance of the wire using Ohm's law: \(R = \frac{V}{I} = \frac{12\text{ V}}{3.0\text{ A}} = 4.0\\ \Omega\).

- Conversion of time to seconds (120 s) and use of \(I = \frac{Q}{t}\): 1 mark. - Calculation of current \(I = 3.0\text{ A}\): 1 mark. - Use of \(R = \frac{V}{I}\) to obtain \(4.0\\ \Omega\) (accept 4 \\Omega): 1 mark.

1. Use the orbital speed formula to express the orbital period \(T\) in seconds: \(v = \frac{2 \pi r}{T} \implies T = \frac{2 \pi r}{v}\). 2. Substitute the radius \(r = 7000\text{ km}\) and speed \(v = 7.5\text{ km/s}\): \(T = \frac{2 \pi \times 7000}{7.5} \approx 5864.3\text{ seconds}\). 3. Convert the period into minutes: \(T_{\text{minutes}} = \frac{5864.3}{60} \approx 97.74\text{ minutes}\). Rounding to 2 significant figures gives 98 minutes.

- Recall and use of \(v = \frac{2\pi r}{T}\) (or rearranged form): 1 mark. - Correct calculation of period in seconds (5864 s or \(5.9 \times 10^3\text{ s}\)): 1 mark. - Correct division by 60 and final answer rounded to 2 significant figures (98 minutes, accept 97.7 minutes): 1 mark.

A vacuum is a region that contains no particles (atoms or molecules). Thermal conduction relies on the vibration and collision of adjacent particles, and convection requires a fluid medium to transfer heat through density currents. Because there are no particles in the vacuum gap between the panes, heat transfer via conduction and convection through this gap is prevented. Thermal energy can only transfer across the gap via infrared radiation, which does not require a medium, thereby significantly reducing the overall heat loss.

1 mark: State that a vacuum contains no particles (atoms/molecules). 1 mark: Explain that conduction and convection require a medium/particles to transfer thermal energy. 0.5 mark: Conclude that heat transfer by conduction and convection through the gap is prevented.

At the moment of release, the only force acting on the skydiver is weight downwards, causing maximum initial acceleration. As the skydiver's speed increases, the upward air resistance acting on them increases. This reduces the downward resultant force (Weight - Air Resistance), which in turn decreases the acceleration. Eventually, the upward air resistance becomes equal in magnitude to the downward weight. At this point, the resultant force is zero, the acceleration becomes zero, and the skydiver falls at a constant speed known as terminal velocity.

1 mark: State that air resistance increases as speed increases. 1 mark: Explain that the resultant force decreases, leading to a decrease in acceleration. 0.5 mark: State that terminal velocity is reached when air resistance equals weight, meaning zero resultant force.

In a metal, electrical current is the flow of free (conduction) electrons through a lattice of positive metal ions. When the temperature of the metal wire increases, these positive ions gain thermal energy and vibrate with a larger amplitude and frequency. This increased movement of the ions makes it more difficult for the conduction electrons to pass through, as they collide more frequently with the vibrating lattice ions. These frequent collisions obstruct the flow of charge, which increases the resistance of the metal wire.

1 mark: State that higher temperature causes metal lattice ions to vibrate with greater amplitude/kinetic energy. 1 mark: Explain that this increases the frequency of collisions between conduction electrons and the vibrating ions. 0.5 mark: State that this increased rate of collisions further obstructs the flow of current, thus increasing resistance.

Redshift is the phenomenon where light emitted from distant galaxies is observed to have a longer wavelength (shifted towards the red end of the spectrum) because the galaxies are moving away from the observer. Observations show that light from almost all distant galaxies is redshifted, and the further away a galaxy is, the faster it is moving away (greater redshift). This indicates that the fabric of space itself is expanding in all directions. If the universe is currently expanding, it must have been smaller, hotter, and denser in the past, supporting the Big Bang theory that the universe expanded from a single, infinitely dense point.

1 mark: Define redshift as the increase in observed wavelength of light from moving-away galaxies. 1 mark: Explain that greater redshift in more distant galaxies shows that the universe is expanding. 0.5 mark: Explain that this expansion implies the universe started from a single common point in the past.

When plane (straight) water waves encounter a narrow gap that is of a similar size to their wavelength, they undergo diffraction. This causes the wavefronts to bend and spread out as circular waves on the other side of the gap. If the gap is widened so that it is much larger than the wavelength, the diffraction effect becomes much less pronounced. The waves will pass through the gap largely unaffected, remaining straight in the center and only bending or curving slightly at the edges of the aperture.

1 mark: State that waves diffract and become circular when passing through a narrow gap comparable to the wavelength. 1 mark: State that widening the gap significantly reduces the diffraction effect. 0.5 mark: Explain that the wavefronts remain mostly straight with only slight curving/bending at the edges.

As the bar magnet falls through the copper tube, its magnetic field cuts through the conducting walls of the tube. This change in magnetic flux linkage induces an electromotive force (e.m.f.) and consequently sets up circulating electrical currents (eddy currents) within the copper tube. According to Lenz's law, the direction of any induced current is such that it opposes the change that produced it. Therefore, the induced currents generate a magnetic field of their own that opposes the motion of the falling magnet, creating an upward magnetic force that resists its descent.

1 mark: Identify that the falling magnet's changing magnetic field induces an e.m.f. or current in the copper tube. 1 mark: Name Lenz's law and explain that the induced current/magnetic field opposes the motion of the magnet. 0.5 mark: Conclude that this opposition results in an upward magnetic force acting on the downward-falling magnet.

An alpha particle is a helium nucleus, consisting of two protons and two neutrons, giving it a relatively large mass and a high positive charge of +2. Because of this high charge and larger size, alpha particles interact strongly with the atoms of the medium they pass through, easily stripping electrons from them to cause high ionization. However, each ionization event transfers some of the alpha particle's kinetic energy to the medium. Because they ionize so intensely, they lose all their kinetic energy within a very short distance (usually 3 to 5 cm in air), resulting in a very low penetrating power compared to beta particles, which are much lighter, carry a -1 charge, and ionize less frequently.

1 mark: Explain that alpha particles have a large mass and a high charge (+2), which leads to frequent collisions and high ionizing power. 1 mark: Explain that each ionization event transfers energy away from the alpha particle, causing it to lose kinetic energy rapidly. 0.5 mark: Conclude that this rapid energy loss limits their penetration to a very short distance (few cm in air).

The critical angle is defined as the angle of incidence in an optically denser medium (glass) that results in an angle of refraction of exactly \(90^\circ\) in the less dense medium (air), meaning the refracted ray travels along the boundary. When the angle of incidence inside the glass is increased beyond this critical angle, no refraction can occur. Instead, all of the light is reflected back into the glass. This process is called total internal reflection, where the boundary acts like a perfect mirror, and the angle of reflection equals the angle of incidence.

1 mark: Define the critical angle as the angle of incidence in the denser medium for which the angle of refraction is 90 degrees. 1 mark: State that when the angle of incidence is greater than the critical angle, total internal reflection occurs. 0.5 mark: State that the light is reflected back into the glass obeying the law of reflection (angle of incidence equals angle of reflection).

During evaporation, the molecules in a liquid are in constant random motion and possess a range of kinetic energies. Only the molecules with the highest kinetic energy have enough energy to overcome the attractive forces (intermolecular forces) holding them to neighboring molecules. These high-energy molecules escape from the surface of the liquid into the air. Consequently, the average kinetic energy of the remaining molecules in the liquid decreases. Since temperature is directly related to the average kinetic energy of the molecules, the temperature of the remaining liquid decreases.

1 mark: Identify that only the highest energy molecules have enough energy to overcome intermolecular forces of attraction. 1 mark: Explain that these high-energy molecules escape from the liquid surface. 1 mark: Explain that the average kinetic energy of the remaining molecules decreases, leading to a drop in temperature.

The increase in the observed wavelength of light from distant galaxies is known as redshift. This shift to longer wavelengths occurs because the galaxy is moving away from Earth. Because almost all distant galaxies exhibit redshift, and the further they are, the greater their redshift, it indicates that the galaxies are receding from us and from each other, which provides strong evidence that the Universe is expanding.

1 mark: State that the effect is redshift. 1 mark: Explain that the increase in wavelength indicates the galaxy is moving away from Earth. 1 mark: Explain that this provides evidence that the Universe is expanding.

As the permanent magnet falls through the copper pipe, the magnetic flux linking the pipe changes. This changing magnetic flux induces an electromotive force (e.m.f.) and causes electrical currents (eddy currents) to flow in the conducting copper pipe. According to Lenz's law, the direction of these induced currents must oppose the change that produced them. Therefore, the induced currents create an opposing magnetic field that exerts an upward magnetic force on the falling magnet, acting as a retarding force and slowing its descent.

1 mark: State that the falling magnet creates a changing magnetic flux which induces an electromotive force (e.m.f.) or current in the copper pipe. 1 mark: Explain that the induced current produces its own magnetic field. 1 mark: Explain that this induced magnetic field opposes the falling motion of the magnet, creating a retarding force.

1. The refractive index \(n\) is calculated using Snell's Law:
\(n = \frac{\sin i}{\sin r} = \frac{\sin 40.0^\circ}{\sin 25.0^\circ} = \frac{0.6428}{0.4226} = 1.52\)

2. The critical angle \(c\) is related to the refractive index by the equation:
\(\sin c = \frac{1}{n} = \frac{1}{1.52} = 0.658\)
\(c = \sin^{-1}(0.658) = 41.1^\circ\) (accept range \(41.0^\circ\) to \(41.3^\circ\)).

M1: State or use formula \(n = \frac{\sin i}{\sin r}\) and substitute values to find \(n = 1.52\) (or \(1.5\)) [1 mark]

M2: State or use formula \(\sin c = \frac{1}{n}\) [1 mark]

A1: Correct calculation of critical angle \(c\) in the range \(41.0^\circ\) to \(41.3^\circ\) (or \(41.8^\circ\) if \(n = 1.5\) is used) [1 mark]

1. Deceleration is the magnitude of the gradient of the speed-time graph during the final period:
\(\text{deceleration} = \frac{\text{change in speed}}{\text{time taken}} = \frac{8.0\text{ m/s} - 0\text{ m/s}}{16.0\text{ s} - 12.0\text{ s}} = 2.0\text{ m/s}^2\).

2. The total distance travelled is equal to the area under the speed-time graph. The area of the trapezium is:
\(\text{Distance} = \frac{1}{2} \times (a + b) \times h\)
where \(a = 8.0\text{ s}\) (from \(t = 4.0\text{ s}\) to \(12.0\text{ s}\)), \(b = 16.0\text{ s}\) (the total duration), and \(h = 8.0\text{ m/s}\):
\(\text{Distance} = \frac{1}{2} \times (8.0 + 16.0) \times 8.0 = 12.0 \times 8.0 = 96\text{ m}\).

M1: Correct calculation of deceleration magnitude as \(2.0\text{ m/s}^2\) (accept \(-2.0\text{ m/s}^2\) if specified as acceleration) [1 mark]

M2: Correct method for finding the total distance by summing the areas of the geometric components or using the trapezium formula [1 mark]

A1: Correct total distance of \(96\text{ m}\) with appropriate unit [1 mark]

1. The corrected count rate is found by subtracting the background count rate from the measured count rate:
- At \(t = 0\): \(\text{Corrected count rate} = 680 - 40 = 640\text{ counts/min}\).
- At \(t = 8.0\text{ hours}\): \(\text{Corrected count rate} = 200 - 40 = 160\text{ counts/min}\).

2. Determine the number of half-lives that have passed:
\(\frac{160\text{ counts/min}}{640\text{ counts/min}} = \frac{1}{4} = \left(\frac{1}{2}\right)^2\)
This shows that exactly 2 half-lives have elapsed in \(8.0\text{ hours}\).

Therefore, the half-life \(t_{1/2}\) of the isotope is:
\(t_{1/2} = \frac{8.0\text{ hours}}{2} = 4.0\text{ hours}\).

M1: Correctly calculate both corrected count rates: \(640\text{ counts/min}\) and \(160\text{ counts/min}\) [1 mark]

M2: Identify that the activity has decreased to a quarter of its original value, which corresponds to 2 half-lives (e.g., show process: \(640 \rightarrow 320 \rightarrow 160\)) [1 mark]

A1: Correct half-life value of \(4.0\text{ hours}\) (or \(240\text{ minutes}\)) with appropriate unit [1 mark]