Physics 5054/22 — October/November 2024

Walkthrough

The question asks for the word equation linking resultant force to momentum. Newton's second law can be stated as $F = \Delta p / t$, which in words is 'resultant force equals the rate of change of momentum' or 'increase in momentum per unit time'.

Key Takeaways

Force is the rate of change of momentum. This is the more general form of Newton's second law, equivalent to $F = ma$ when mass is constant.

Common Mistakes

Writing 'mass times acceleration' is not the word equation for momentum. The scheme specifically accepts 'increase in momentum / unit time' or 'rate of increase in momentum'.

Things to Be Careful About

Ensure the wording matches the mark scheme exactly. 'Change in momentum over time' is acceptable, but 'momentum divided by time' is not precise enough as it ignores the change in momentum.

Walkthrough

The pump increases the speed of $180 \text{ kg}$ of water by $30 \text{ m/s}$ in $2.0 \text{ s}$. The change in momentum is $\Delta p = m \Delta v = 180 \times 30 = 5400 \text{ kg m/s}$. The force is the rate of change of momentum: $F = \Delta p / t = 5400 / 2.0 = 2700 \text{ N}$. Alternatively, calculate the acceleration $a = \Delta v / t = 30 / 2.0 = 15 \text{ m/s}^2$, then use $F = ma = 180 \times 15 = 2700 \text{ N}$.

Key Takeaways

Force can be calculated directly from the mass, change in velocity, and time using $F = m \Delta v / t$. This is equivalent to finding the acceleration first and then using $F = ma$.

Common Mistakes

Forgetting to divide by the time interval, or using the wrong mass (e.g., the mass of the jet ski instead of the water).

Things to Be Careful About

The question asks for the force on the water, so use the mass of the water ($180 \text{ kg}$), not the total mass of the jet ski and student. The unit is newtons (N), which is already provided in the answer blank.

Walkthrough

Newton's third law states that when two bodies interact, the forces they exert on each other are equal in magnitude and opposite in direction. The jet ski's pump pushes the water backwards (the action force). Therefore, the water pushes back on the pump/jet ski with an equal force in the opposite direction (the reaction force), which is forwards.

Key Takeaways

Newton's third law pairs always act on different objects. The action and reaction forces are equal in size and opposite in direction.

Common Mistakes

Saying 'the water pushes back' without specifying that the forces are equal and opposite. Stating that the forces act on the same object (e.g., 'the forces on the water cancel out') is incorrect; they act on different objects.

Things to Be Careful About

The mark scheme specifically looks for the word 'opposite' to describe the direction of the reaction force. 'Forwards' is also acceptable if 'opposite' is used, but 'opposite' is the key scientific term here.

Walkthrough

The forwards force on the jet ski is equal in magnitude to the backwards force on the water, so $F = 2700 \text{ N}$. The total mass being accelerated is the mass of the student plus the mass of the jet ski: $m = 70 + 280 = 350 \text{ kg}$. Using Newton's second law, $a = F / m = 2700 / 350 = 7.714... \text{ m/s}^2$, which rounds to $7.7 \text{ m/s}^2$.

Key Takeaways

When applying $F = ma$, the mass $m$ must be the total mass of the object or system being accelerated by the resultant force $F$.

Common Mistakes

Using only the mass of the jet ski ($280 \text{ kg}$) or only the mass of the student ($70 \text{ kg}$) instead of the total combined mass. Forgetting to add the masses together.

Things to Be Careful About

The force calculated in (a)(ii) is the force on the water. By Newton's third law, the force on the jet ski is equal in magnitude ($2700 \text{ N}$). Always use the total mass of the system when calculating acceleration.

Walkthrough

The question asks how thermal energy is transferred through a solid plastic material. This is conduction.

1. In a solid, particles are closely packed and held in fixed positions by strong forces. They cannot move freely, but they can vibrate about their fixed positions.
2. When thermal energy is added to one end, the particles there gain kinetic energy and vibrate more vigorously.
3. These vibrating particles collide with their neighbouring particles, passing on some of their vibrational energy. This process continues from particle to particle, transferring thermal energy through the material without any overall movement of the particles themselves.

Key Takeaways

Conduction in solids occurs through the vibration of particles and the collision of neighbouring particles. It does not involve the bulk movement of the material.

Common Mistakes

• Saying 'particles move through the material'. In conduction, particles only vibrate in place; they do not travel through the solid.
• Forgetting to mention collisions. The transfer of energy relies on particles bumping into each other.

Things to Be Careful About

Use precise language: 'vibrate', 'collide', 'neighbouring particles', and 'transfer thermal energy'. Avoid using the word 'heat' as a noun to describe the energy; use 'thermal energy' instead.

Walkthrough

The question asks how air bubbles in the plastic insulation reduce thermal energy transfer. This involves both stopping convection and minimizing conduction.

1. Trapping the air: The plastic matrix holds the air in small, isolated pockets or bubbles. This prevents the air from flowing freely.
2. Preventing convection: Because the air is trapped and cannot circulate, convection currents cannot form. Convection is a major way heat is lost in unfrozen air spaces, so stopping it significantly reduces energy transfer.
3. Poor conductivity: Air itself is a very poor thermal conductor (a good insulator) because its particles are far apart, making collisions infrequent. By trapping air in small bubbles, the insulation relies on this low conductivity to slow down heat transfer by conduction.

Key Takeaways

Trapped air is an excellent insulator because it stops convection (no bulk fluid movement) and minimizes conduction (air is a poor conductor). This principle is used in many insulating materials, such as fiberglass, foam, and double-glazed windows.

Common Mistakes

• Saying 'air is a good conductor'. Air is a poor conductor; gases are generally good insulators.
• Only mentioning one reason. The scheme awards marks for trapping air, preventing convection, AND air being a bad conductor. All three points are needed for full marks.

Things to Be Careful About

Ensure you clearly separate the two mechanisms: convection is stopped because the air is trapped and cannot move; conduction is reduced because air itself is a poor conductor. Do not just say 'it insulates better'; explain the physics.

Walkthrough

Pressure is defined as force per unit area:

$$p = \frac{F}{A}$$

Rearranging this equation to solve for the force $F$ exerted by the enclosed air on the piston face gives:

$$F = p \times A$$

Substitute the given values for internal air pressure ($p = 1.0 \times 10^{5}\text{ Pa}$) and cross-sectional area of the piston ($A = 3.5 \times 10^{-3}\text{ m}^2$):

$$F = (1.0 \times 10^{5}\text{ Pa}) \times (3.5 \times 10^{-3}\text{ m}^2) = 350\text{ N}$$

Key Takeaways

• Force is calculated from pressure using $F = pA$.
• Always ensure $p$ is in pascals ($\text{Pa}$ or $\text{N/m}^2$) and area $A$ is in square metres ($\text{m}^2$) so that the resulting force is directly in newtons ($\text{N}$).

Common Mistakes

• Dividing pressure by area instead of multiplying ($F \neq p/A$).
• Errors when multiplying powers of ten ($10^5 \times 10^{-3} = 10^2$).

Things to Be Careful About

• The unit $\text{N}$ is already printed on the answer line, so only the numerical value $350$ is needed in the blank.

Walkthrough

For the piston to remain stationary, it must be in mechanical equilibrium according to Newton's first law:

1. The air inside the syringe exerts a pressure of $1.0 \times 10^{5}\text{ Pa}$ on the left face of the piston, pushing it to the right with a force of $350\text{ N}$.
2. The surrounding atmosphere outside the syringe also exerts atmospheric pressure ($1.0 \times 10^{5}\text{ Pa}$) on the right side of the piston, producing an equal force of $350\text{ N}$ directed to the left.
3. Since friction is negligible and these two opposing forces are equal in magnitude and opposite in direction, the resultant (net) force on the piston is zero ($350\text{ N} - 350\text{ N} = 0$). Hence, the piston does not accelerate or move.

Key Takeaways

• An object remains at rest when the resultant force acting on it is zero.
• Atmospheric pressure acts on all exposed surfaces from the outside.

Common Mistakes

• Stating only that 'atmospheric pressure acts' without specifying that it creates a force to the left or that the forces balance.
• Attributing the lack of motion to friction, even though the question explicitly states that friction is negligible.

Things to Be Careful About

• Both marking points are required: (1) naming the opposing force/pressure from the outside atmosphere acting to the left, and (2) stating that the forces balance / there is no resultant force.

Walkthrough

To explain why gas pressure decreases at constant temperature using the kinetic particle model, address three key logical steps:

1. Origin of pressure: Gas pressure is the force per unit area exerted by gas particles as they constantly collide with and rebound from the inner walls of the container and the piston face.
2. Change in particle density: When the piston is pulled to the right, the volume containing the trapped air increases while the total number of particles remains constant. The particles therefore become more spread out (further apart / less densely packed).
3. Effect on collision rate: Because the particles are more spread out and their average speed is unchanged (since temperature is constant), they take longer to travel between wall collisions. Thus, the frequency of collisions with the walls (number of impacts per unit area per second) decreases, resulting in a lower pressure.

Key Takeaways

• At constant temperature, the average kinetic energy and speed of the gas particles remain constant.
• Increasing the volume reduces particle density and therefore reduces the collision rate with the container walls.

Common Mistakes

• Saying that the particles move slower or have less kinetic energy (incorrect because temperature is constant).
• Stating only 'fewer collisions' without specifying collisions with the walls or collisions per unit time / frequency.
• Omitting the mention that particles are further apart / volume increases.

Things to Be Careful About

• Do not confuse collisions between particles with collisions between particles and the walls. It is the collisions with the walls/piston that generate gas pressure.

Walkthrough

Because the temperature of the trapped air remains constant, Boyle's law applies:

$$p_1 V_1 = p_2 V_2$$

Identify the initial and final conditions from the question:

• Initial pressure, $p_1 = 1.0 \times 10^{5}\text{ Pa}$
• Initial volume, $V_1 = 1.2 \times 10^{-4}\text{ m}^3$
• Final volume, $V_2 = 1.5 \times 10^{-4}\text{ m}^3$
• Final pressure, $p_2 = ?$

Rearrange to make $p_2$ the subject:

$$p_2 = \frac{p_1 V_1}{V_2}$$

Substitute the values:

$$p_2 = \frac{1.0 \times 10^{5} \times 1.2 \times 10^{-4}}{1.5 \times 10^{-4}} = \frac{12}{1.5 \times 10^{-4}} = 8.0 \times 10^{4}\text{ Pa}$$

(Alternatively written as $80\,000\text{ Pa}$ or $80\text{ kPa}$).

Key Takeaways

• For a fixed mass of gas at constant temperature, pressure is inversely proportional to volume ($p_1 V_1 = p_2 V_2$).
• As volume increases, pressure decreases proportionately.

Common Mistakes

• Inverting the volumes, calculating $p_2 = p_1 \times \frac{V_2}{V_1}$, which would incorrectly give an increased pressure.
• Arithmetic mistakes when handling scientific notation and negative exponents.

Things to Be Careful About

• The unit $\text{Pa}$ is provided on the answer line.
• Ensure the answer is written to a sensible number of significant figures (2 s.f., matching the input data: $8.0 \times 10^4\text{ Pa}$ or $80\,000\text{ Pa}$).

Walkthrough

The focal length $f$ is the distance from the optical centre of the lens to the principal focus $F_2$ (or $F_1$). The grid in Fig. 5.1 is full-scale with major squares of 1 cm by 1 cm. Counting the major squares from the centre of the lens to the dot labelled $F_2$, we find it is 1.8 cm away. Thus, the focal length is 1.8 cm.

Key Takeaways

The focal length is a fundamental property of a lens, measured from the optical centre to the principal focus. On a full-scale grid, it can be read directly.

Common Mistakes

Reading the distance to the edge of the lens instead of the centre. Counting the minor grid squares instead of using the major 1 cm squares, leading to an incorrect power of ten.

Things to Be Careful About

Ensure the measurement is taken from the exact centre of the lens (the optical centre) to the centre of the dot representing the focus. The mark scheme accepts 1.8 cm.

Walkthrough

The object is placed 3.0 cm from the centre of the lens. Since the focal length is 1.8 cm, 3.0 cm is between $F_1$ and $2F_1$ on the left side. Measure 3.0 cm to the left of the lens along the principal axis. From that point, draw a vertical arrow upwards with a height of 2.1 cm (just over 2 major grid squares). Label the tip or the arrow 'O'.

Key Takeaways

In ray diagrams, the object is typically drawn as an upright arrow on the principal axis. Its height and distance from the lens are given in the question and must be plotted to scale.

Common Mistakes

Placing the object on the wrong side of the lens (it should be on the side opposite to where the image will form, usually the left). Drawing the arrow with the wrong height (e.g., 2 cm instead of 2.1 cm) or pointing it downwards.

Things to Be Careful About

The arrow must be vertical and perpendicular to the principal axis. The label 'O' should be clearly associated with the arrow.

Walkthrough

To find the image, draw at least two of the three principal rays from the tip of the object arrow O:

1. Ray 1: Draw a line from the tip of O parallel to the principal axis until it hits the lens. After passing through the lens, draw this ray refracting so that it passes through the principal focus $F_2$ on the right side.
2. Ray 2: Draw a line from the tip of O straight through the optical centre of the lens. This ray passes through without deviation.
3. Ray 3 (optional): Draw a line from the tip of O through the principal focus $F_1$ on the left side until it hits the lens. After the lens, draw this ray emerging parallel to the principal axis.

The rays will converge and intersect at a point on the right side of the lens, above the principal axis. This intersection is the tip of the image. Draw a vertical arrow from this intersection point down to the principal axis to represent the image. The image is inverted (pointing downwards) and real.

Key Takeaways

Three principal rays can be used to locate an image formed by a converging lens. Any two of these are sufficient. The intersection of the refracted rays gives the position and size of the image.

Common Mistakes

Drawing the ray through $F_1$ as refracting towards $F_2$ instead of emerging parallel. Drawing the image on the same side as the object. Forgetting to draw the image arrow down to the principal axis, or drawing it upright instead of inverted.

Things to Be Careful About

Use a ruler for all straight lines. The rays must be drawn from the tip of the object arrow, not the base. The image arrow must be drawn from the intersection point down to the principal axis, and it must be inverted.

Walkthrough

Linear magnification is defined as the ratio of the image height to the object height: $m = h_\text{I} / h_\text{o}$. From the ray diagram constructed in part (iii), measure the height of the image arrow. It should be approximately $3.15\ \text{cm}$. Dividing this by the object height of $2.1\ \text{cm}$ gives $3.15 / 2.1 = 1.5$. The mark scheme accepts a range from 1.2 to 1.8 to allow for graphical construction inaccuracies.

Key Takeaways

Linear magnification can be calculated from heights ($h_\text{I} / h_\text{o}$) or from distances ($v / u$). For a real image formed by a converging lens, the magnification is positive when calculated as $v/u$ (using sign convention) or simply as the ratio of heights.

Common Mistakes

Using the wrong formula, such as $h_\text{o} / h_\text{I}$. Reading the image height incorrectly from the grid (e.g., reading 3.0 cm instead of 3.15 cm). Forgetting that magnification is a ratio and has no units.

Things to Be Careful About

The mark scheme allows a range (1.2 to 1.8) because the answer depends on the accuracy of the hand-drawn ray diagram. If a candidate uses the lens formula with the given values ($f=1.8$, $u=3.0$), they will get exactly 1.5, which is well within the accepted range.

Walkthrough

An image is real if it is formed by the actual convergence of light rays and can be projected onto a screen. In the ray diagram, the refracted rays physically intersect on the opposite side of the lens from the object. This intersection point is where the image is formed, making it a real image. Virtual images, by contrast, are formed by the apparent divergence of rays (they appear to come from a point behind the lens) and cannot be projected onto a screen.

Key Takeaways

Real images are formed by converging rays and are always inverted (for a single lens). Virtual images are formed by diverging rays (extended backwards) and are always upright.

Common Mistakes

Saying 'real because it is on the right side' without mentioning the convergence of rays. Saying 'virtual' because the object is inside the focal length (which is not the case here; $u = 3.0\ \text{cm} > f = 1.8\ \text{cm}$).

Things to Be Careful About

The explanation must link the classification ('real') to the physical reason ('formed by converging rays' or 'actual intersection of rays'). Simply stating 'real' may not earn the mark if an explanation is implicitly required by the word 'Explain'.

Walkthrough

The heater operates at a constant power of $P = 40 \text{ W}$. The time it is switched on is $t = 2.0 \text{ minutes}$. Energy transferred is calculated using $E = Pt$. First, convert the time into seconds because the watt is joules per second: $2.0 \text{ minutes} \times 60 = 120 \text{ s}$. Then multiply: $40 \text{ W} \times 120 \text{ s} = 4800 \text{ J}$.

Key Takeaways

Power is the rate of energy transfer. To find total energy, multiply power by time, ensuring time is in seconds when power is in watts.

Common Mistakes

Forgetting to convert minutes to seconds, which would give an answer of 80 J instead of 4800 J.
Using the wrong formula, such as $E = VIt$ without calculating $I$ first, though that is also valid if done correctly.

Things to Be Careful About

Always check the units. The mark scheme explicitly expects the answer in joules, so time must be in seconds. Give the answer to 2 or 3 significant figures; 4800 is exact here.

Walkthrough

The heater is connected to a 12 V supply and dissipates 40 W. To find the resistance, we can use the power equation $P = IV$ to find the current, and then Ohm's law $R = V/I$.
First, calculate the current: $I = P / V = 40 / 12 = 3.333... \text{ A}$.
Next, calculate the resistance: $R = V / I = 12 / 3.333... = 3.6 \text{ } \Omega$.
Alternatively, you could use $P = V^2 / R$ directly to find $R = V^2 / P = 144 / 40 = 3.6 \text{ } \Omega$, but the mark scheme outlines the two-step method via current.

Key Takeaways

Power, voltage, and current are linked by $P = IV$. Resistance is linked to voltage and current by $R = V/I$. These can be combined to find resistance directly from power and voltage using $R = V^2 / P$.

Common Mistakes

Rounding the intermediate current value too early (e.g., using $I = 3.3 \text{ A}$ gives $R = 12 / 3.3 = 3.636... \text{ } \Omega$, which might be accepted but carries a rounding error). It is best to keep the fraction $40/12$ in the calculator.
Using $P = I^2R$ without first finding $I$ correctly.

Things to Be Careful About

The question asks for resistance in ohms ($\Omega$). Ensure the final answer has the correct unit if writing it out, though the blank provides the unit. Keep at least 3 significant figures for intermediate values to avoid rounding errors in the final answer.

Walkthrough

A transformer requires a changing magnetic field in the core to induce a voltage in the secondary coil. Because the primary coil is connected to an a.c. supply, the current and therefore the magnetic field constantly change direction and magnitude. Iron is a soft magnetic material, meaning it is easily magnetised and demagnetised. This allows the magnetic flux in the core to follow the alternating current without retaining permanent magnetism, which is essential for the transformer to work.

Key Takeaways

Transformer cores must be made of a soft magnetic material (like iron) that can be easily magnetised and demagnetised to match the alternating current.

Common Mistakes

• Saying iron is a 'good conductor of electricity' (this is irrelevant and does not explain why it is used for the magnetic core).
• Saying 'iron is magnetic' without specifying that it is a temporary magnetic material.

Things to Be Careful About

• Use the exact phrase 'temporary magnetic material' or explain that it is easily magnetised and demagnetised. Just saying 'magnetic' is not sufficient.

Walkthrough

Electromagnetic induction is the process of inducing a voltage across a conductor when it is exposed to a changing magnetic field. In a transformer:

1. The a.c. mains supply causes an alternating current to flow through the primary coil.
2. This alternating current produces a magnetic field that is constantly changing in magnitude and direction within the iron core.
3. The secondary coil is wrapped around the same core, so it is exposed to this changing magnetic field. According to the principle of electromagnetic induction, a changing magnetic flux through a coil induces an e.m.f. (voltage) across it.

Key Takeaways

A transformer works on the principle of electromagnetic induction. An alternating current in the primary coil creates a changing magnetic field, which induces a voltage in the secondary coil.

Common Mistakes

• Stating that the magnetic field is 'constant' or 'static'. It must be changing or alternating.
• Saying 'current flows from the primary to the secondary coil'. The coils are electrically isolated; only the magnetic field links them.
• Forgetting to use the word 'induced' for the voltage in the secondary coil.

Things to Be Careful About

• Each of the three points (alternating current, changing magnetic field, induced voltage) is a separate mark. Ensure all three links in the causal chain are stated clearly.

Walkthrough

The transformer turns-ratio equation relates the voltages across the primary and secondary coils to the number of turns on each coil:

$$\frac{V_S}{V_P} = \frac{N_S}{N_P}$$

where $V_S$ is the secondary voltage, $V_P$ is the primary voltage, $N_S$ is the number of turns on the secondary coil, and $N_P$ is the number of turns on the primary coil. This equation applies to maximum (peak) voltages as well as r.m.s. voltages, as long as the units are consistent. We are given:

• $V_P = 300$ V (maximum)
• $N_P = 750$ turns
• $N_S = 60$ turns

Rearranging for $V_S$:

$$V_S = 300 \times \frac{60}{750} = 24 \text{ V}$$

Key Takeaways

The transformer equation $V_S / V_P = N_S / N_P$ can be used with either maximum (peak) voltages or r.m.s. voltages. A step-down transformer has fewer turns on the secondary coil than on the primary coil, resulting in a lower output voltage.

Common Mistakes

• Inverting the turns ratio (e.g., calculating $300 \times 750 / 60$).
• Using r.m.s. values when the question asks for maximum values, or vice versa (though here only maximum values are given, so no conversion is needed).
• Forgetting to include the unit 'V' in the final answer.

Things to Be Careful About

• Ensure the equation is rearranged correctly before substituting the values.
• The answer must be given with the correct unit (V).

Walkthrough

An alpha particle ($\alpha$-particle) is emitted from an unstable nucleus during alpha decay. It is identical to a helium nucleus, meaning it is made up of 2 protons and 2 neutrons bound together. It contains no electrons or any other particles.

Key Takeaways

Alpha particles are helium nuclei consisting of exactly 2 protons and 2 neutrons. They have a nucleon number of 4 and a proton number of 2.

Common Mistakes

Writing 'helium atoms' instead of 'helium nuclei' or 'alpha particles'. An alpha particle has no electrons, so it is not a neutral atom. Writing '2 protons and 2 electrons' is incorrect; electrons are not present in the nucleus.

Things to Be Careful About

The mark scheme requires all three elements: 2 neutrons, 2 protons, and the absence of other particles (like electrons). Omitting 'no other particles' can cost a mark.

Walkthrough

Because an alpha particle contains 2 protons and 2 neutrons and no electrons, it is exactly the nucleus of a helium-4 atom. Therefore, the particle identical in composition to an alpha particle is a helium nucleus.

Key Takeaways

Alpha particles and helium nuclei are the same thing. Remember to specify 'nucleus' rather than 'atom' because there are no electrons.

Common Mistakes

Answering 'helium atom'. A helium atom has 2 electrons orbiting the nucleus; an alpha particle does not.

Things to Be Careful About

The word 'nucleus' is essential. 'Helium' on its own is not precise enough to score the mark.

Walkthrough

When thorium-230 ($^{230}_{90}\text{Th}$) undergoes alpha decay, it emits an alpha particle ($^{4}_{2}\text{He}$). The nucleon number (top number) decreases by 4, and the proton number (bottom number) decreases by 2. This gives a new isotope with nucleon number $230 - 4 = 226$ and proton number $90 - 2 = 88$. The number of neutrons is found by subtracting the proton number from the nucleon number: $226 - 88 = 138$.

Key Takeaways

In alpha decay, the parent nucleus loses 4 nucleons and 2 protons. The daughter nucleus always has a nucleon number 4 less and a proton number 2 less than the parent.

Common Mistakes

Forgetting to subtract the alpha particle's nucleon number from the total before calculating neutrons. Using the proton number of thorium (90) instead of radium (88) when calculating neutrons.

Things to Be Careful About

Ensure you are calculating the neutrons for the daughter nucleus (radium), not the parent (thorium). The question asks for the number of neutrons in the isotope of radium.

Walkthrough

The proton number of the radium isotope was found to be 88. In a neutral atom, the number of negatively charged electrons exactly balances the number of positively charged protons. Therefore, a neutral atom of this radium isotope has 88 electrons.

Key Takeaways

For any neutral atom, number of electrons = proton number (atomic number).

Common Mistakes

Assuming the number of electrons changes during radioactive decay. Only the nucleus changes; the electron cloud adjusts later if the atom becomes an ion, but the question specifies a 'neutral atom', so electrons = protons.

Things to Be Careful About

The question specifies 'neutral atom'. If it asked for the number of electrons immediately after decay (before any electron is captured or lost), it might be different, but 'neutral atom' always means electrons = protons.

Walkthrough

Gamma radiation ($\gamma$-radiation) is high-frequency electromagnetic radiation. It carries no electric charge. Electric fields only exert forces on charged particles (like alpha or beta particles). Since gamma rays are uncharged, they experience no force in the electric field and travel in a straight line, completely undeflected.

Key Takeaways

Gamma rays are uncharged and pass straight through electric and magnetic fields without being deflected. Alpha particles deflect towards the negative plate, and beta particles deflect towards the positive plate.

Common Mistakes

Drawing a slight curve for gamma radiation. Drawing deflection in the same direction as alpha or beta particles. Forgetting to extend the line through the entire region between the plates.

Things to Be Careful About

The line must be drawn straight through the field region. Any curvature, even slight, will not score the mark. The line should continue horizontally past the field region as well.

Walkthrough

Magnetic fields only exert forces on moving charged particles (and on magnetic materials). Since gamma radiation is uncharged electromagnetic radiation, it is completely unaffected by the magnetic field. It will travel in a straight line, passing through the field region without any deviation.

Key Takeaways

Neither electric nor magnetic fields can deflect gamma radiation because it has no charge. This is a key way to distinguish gamma rays from alpha and beta particles in experiments.

Common Mistakes

Drawing a curved path for gamma radiation. Applying Fleming's left-hand rule incorrectly to gamma radiation. Drawing deflection upwards or downwards.

Things to Be Careful About

Just like in the electric field, the path must be a perfectly straight horizontal line through the entire magnetic field region. The direction of the magnetic field (into the page) is irrelevant for gamma rays.

Walkthrough

The question asks for the speed of an object at the equator. Speed is defined as distance divided by time ($v = s/t$). The object moves in a circle around the Earth's axis. At the equator, the radius of this circular path is the radius of the Earth itself, $r = 6.4 \times 10^6\ \text{m}$.

First, calculate the distance travelled in one full rotation. This is the circumference of the circle at the equator:

$$s = 2\pi r = 2 \times \pi \times 6.4 \times 10^6\ \text{m} \approx 4.02 \times 10^7\ \text{m}$$

Next, identify the time taken for one rotation. The problem states this is 24 hours. Since the required speed is in m/s, convert hours to seconds:

$$t = 24\ \text{hours} \times 60\ \text{min/hour} \times 60\ \text{s/min} = 86\,400\ \text{s}$$

Finally, substitute these values into the speed equation:

$$v = \frac{4.02 \times 10^7}{86\,400} \approx 465.4\ \text{m/s}$$

The radius $6.4 \times 10^6$ is given to 2 significant figures, so the answer should be rounded to 2 significant figures: $470\ \text{m/s}$.

Key Takeaways

• Speed in circular motion is the circumference divided by the period of rotation.
• Always ensure units are consistent; convert hours to seconds when the required unit is m/s.
• Respect significant figures in the final answer based on the data provided.

Common Mistakes

• Forgetting to calculate the circumference and using the diameter or radius as the distance.
• Forgetting to convert 24 hours into seconds (using 24 directly would give a wildly incorrect answer).
• Using $v = \omega r$ without calculating $\omega$ first (though valid, $v=s/t$ is more direct for O Level).

Things to Be Careful About

• The mark scheme accepts 470 m/s (2 s.f.) or 465 m/s (3 s.f.). Since the input data ($6.4$) has 2 s.f., 470 is the strictly correct rounding.
• Do not include the unit in the final answer blank if the unit is already printed next to the blank line, but ensure it is clear in the working.

Walkthrough

The question asks why speed decreases as an object moves away from the equator (towards the poles).

Recall the definition of speed for uniform circular motion: $v = s/t$, where $s$ is the distance travelled in one cycle (circumference) and $t$ is the time for one cycle (period).

1. Distance ($s$): At the equator, the path is the largest circle (the equator itself). As you move north or south, the circular path around the axis gets smaller. At the poles, the path is a point (radius 0). So, the circumference of the path decreases as the distance from the equator increases.
2. Time ($t$): The Earth is a rigid body rotating on its axis. Every point on the Earth completes one full rotation in the same amount of time (24 hours). The period is constant.

Since $v = s/t$, if $s$ decreases and $t$ is constant, then $v$ must decrease.

Key Takeaways

• All points on a rotating rigid body have the same angular speed and period.
• Linear speed depends on the radius of the circular path ($v = 2\pi r / T$).

Common Mistakes

• Saying "the radius of the Earth decreases" (the Earth's radius is constant; the radius of the path decreases).
• Forgetting to mention that the time period is constant; just saying the distance decreases is not enough for full marks.

Things to Be Careful About

• Use precise language: "circumference of the path" or "radius of the circular path", not just "distance".
• The mark scheme awards one mark for the decreasing path length and one mark for the constant time period.

Walkthrough

The Earth moves in a curved path (orbit) around the Sun. According to Newton's first law, an object will move in a straight line at constant speed unless acted upon by a resultant force. Since the Earth's direction is constantly changing, there must be a resultant force acting on it.

This force is directed towards the Sun and is the gravitational force of attraction between the Sun and the Earth. The Sun's gravity pulls the Earth, keeping it in orbit.

Key Takeaways

• Gravity is the force that keeps planets in orbit around the Sun and moons around planets.
• It is an attractive force acting at a distance.

Common Mistakes

• Saying "centrifugal force" (this is a fictitious force and not the provider of the orbital force in this context).
• Saying "gravity" without specifying it is due to the Sun (though often accepted, "gravitational attraction due to the Sun" is precise).
• Saying "solar wind" or "radiation pressure".

Things to Be Careful About

• The question asks to "State what provides this force". Two marks are allocated, likely for "gravitational attraction/field" and "due to Sun".
• Do not just write "gravity"; be specific about the source (the Sun).

Walkthrough

The orbit is an ellipse, meaning the distance between the Earth and the Sun changes throughout the year.

Gravitational force depends on distance (though the inverse square law $F = GmM/r^2$ is A Level, the qualitative relationship is expected at O Level: force increases as distance decreases).

• When the Earth is moving away from the Sun (towards aphelion), the distance increases, so the gravitational force decreases.
• When the Earth is moving towards the Sun (towards perihelion), the distance decreases, so the gravitational force increases.

Therefore, the magnitude of the force is not constant; it fluctuates as the distance changes.

Key Takeaways

• In an elliptical orbit, the distance between the two bodies changes.
• Gravitational force is stronger at closer distances and weaker at larger distances.

Common Mistakes

• Saying the force is constant (this is only true for a circular orbit).
• Saying the force increases then decreases without explaining the link to distance.

Things to Be Careful About

• The mark scheme asks for a description of what happens to the magnitude. "Fluctuates" or "varies" is not enough on its own; you must explain how it varies with position (closer/farther).