Physics 5054/21 — October/November 2025

Walkthrough

The goal is to show that $\frac{p^2}{2m}$ is equal to the kinetic energy $E_k$.

1. Recall the definition of momentum: $p = mv$. This allows us to write $p^2 = m^2 v^2$.
2. Substitute $p^2$ into the given expression: $\frac{p^2}{2m} = \frac{m^2 v^2}{2m}$.
3. Simplify the fraction by cancelling one $m$ from the numerator and denominator: $\frac{m^2 v^2}{2m} = \frac{1}{2}mv^2$.
4. Recognise that $\frac{1}{2}mv^2$ is the standard formula for kinetic energy, $E_k$.
5. Therefore, $\frac{p^2}{2m} = E_k$.

Key Takeaways

This derivation is a useful bridge between momentum and kinetic energy. It shows that for a given mass, kinetic energy is proportional to the square of the momentum.

Common Mistakes

• Attempting to start from $E_k = \frac{1}{2}mv^2$ and trying to force $p$ in, which can lead to algebraic errors.
• Failing to show the intermediate step $\frac{m^2 v^2}{2m}$, which is a required mark in the scheme.
• Not explicitly stating that $\frac{1}{2}mv^2$ is equal to $E_k$.

Things to Be Careful About

• Ensure the algebraic cancellation of $m$ is clear: $m^2 / m = m$.
• The mark scheme awards marks for seeing $\frac{m^2 v^2}{2m}$ and for recognising $\frac{1}{2}mv^2 = E_k$.
• Do not introduce $F=ma$ or other unrelated formulas; stick to the definitions of $p$ and $E_k$.

Walkthrough

1. Read the momentum at $t = 6.0\text{ s}$ from Fig. 1.1. The graph is horizontal at this point, indicating constant momentum. Reading the value from the y-axis, it is $39\ 000\text{ kg m/s}$.
2. We are given $E_k = 5.4 \times 10^5\text{ J}$.
3. Use the relationship derived in part (b)(i): $E_k = \frac{p^2}{2m}$.
4. Rearrange the formula to make $m$ the subject: $m = \frac{p^2}{2 E_k}$.
5. Substitute the values: $m = \frac{39000^2}{2 \times 5.4 \times 10^5}$.
6. Calculate the numerator: $39000^2 = 1.521 \times 10^9$.
7. Calculate the denominator: $2 \times 5.4 \times 10^5 = 1.08 \times 10^6$.
8. Divide: $1.521 \times 10^9 / 1.08 \times 10^6 = 1408.333...\text{ kg}$.
9. The mark scheme accepts $1400\text{ kg}$ (2 s.f.), which is appropriate given the precision of the graph reading and the given $E_k$.

Key Takeaways

Graph reading must be combined with algebraic manipulation. When a value is read from a graph, it may not be exact, so the final answer should reflect the appropriate significant figures.

Common Mistakes

• Reading the wrong value from the graph (e.g., $38\ 000$ instead of $39\ 000$). The graph is slightly below $40\ 000$, and $39\ 000$ is the accepted reading.
• Algebraic errors when rearranging $E_k = \frac{p^2}{2m}$ to solve for $m$. A common mistake is $m = \frac{2 E_k}{p^2}$ or $m = \frac{p^2}{E_k}$.
• Calculation errors with large numbers; using scientific notation helps avoid mistakes.
• Forgetting to include units in the final answer.

Things to Be Careful About

• The graph reading at $t = 6.0\text{ s}$ is $39\ 000$. The mark scheme explicitly accepts $39\ 000$. If a candidate reads $38\ 000$, they will get an error carried forward (ecf) mark for the mass calculation, but the final mass must be consistent with their reading.
• Significant figures: $5.4 \times 10^5$ has 2 s.f., so the mass should ideally be given to 2 s.f. ($1400$ or $1.4 \times 10^3$).

Walkthrough

Weight is the force of gravity on an object, calculated by multiplying mass by gravitational field strength ($g$). Using $g = 9.8 \text{ N/kg}$, the weight is $37 \times 9.8 = 362.6 \text{ N}$. Rounding to two significant figures gives $360 \text{ N}$, which matches the mark scheme.

Key Takeaways

Weight is a force and must be calculated using $W = mg$. Always include the unit newtons (N).

Common Mistakes

Forgetting to multiply by $g$ and just writing the mass value. Using the wrong value for $g$ (e.g. not converting if given in different units, though here it is standard).

Things to Be Careful About

The mark scheme accepts $360 \text{ N}$, which is $362.6 \text{ N}$ rounded to two significant figures. Ensure you use $g = 9.8 \text{ N/kg}$ (or $9.81$) rather than $10$, as $37 \times 10 = 370 \text{ N}$ would not match the expected $360 \text{ N}$ for subsequent marks.

Walkthrough

The useful work done in lifting the boxes is the work done against gravity. This is calculated by multiplying the upward force (the weight of the boxes, $360 \text{ N}$) by the vertical distance moved ($13 \text{ m}$). $360 \times 13 = 4680 \text{ J}$, which rounds to $4700 \text{ J}$ to two significant figures.

Key Takeaways

Work done is force times distance in the direction of the force. For lifting, the force is the weight and the distance is the height.

Common Mistakes

Using the electrical energy from part (b) instead of the useful work done. Forgetting to multiply by the distance.

Things to Be Careful About

Use the rounded value of $360 \text{ N}$ from part (a) if following the mark scheme's path, though using $362.6 \times 13 = 4713.8 \text{ J}$ also rounds to $4700 \text{ J}$. Always round the final answer to the appropriate number of significant figures (two here, based on the given data $13 \text{ m}$ and $37 \text{ kg}$).

Walkthrough

The total electrical energy transferred to the motor is given by the formula $E = VIt$, where $V$ is the potential difference (e.m.f.), $I$ is the current, and $t$ is the time. Substituting the given values: $230 \times 5.6 \times 11 = 14168 \text{ J}$. To two significant figures, this is $1.4 \times 10^4 \text{ J}$.

Key Takeaways

Electrical energy can be calculated using $E = VIt$. Always ensure time is in seconds.

Common Mistakes

Using $P = IV$ and forgetting to multiply by time to get energy. Forgetting to convert minutes to seconds (not applicable here as time is already in seconds).

Things to Be Careful About

The mark scheme expects the answer in standard form to two significant figures: $1.4 \times 10^4 \text{ J}$. Writing $14168 \text{ J}$ or $14000 \text{ J}$ may not be accepted if strict significant figure rules are applied, though $1.4 \times 10^4$ is the safest format.

Walkthrough

Efficiency is the ratio of useful energy output to total energy input, expressed as a percentage. The useful energy output is the work done lifting the boxes ($4680 \text{ J}$), and the total energy input is the electrical energy transferred ($14168 \text{ J}$). Dividing and multiplying by 100 gives $33.03\%$, which rounds to $33\%$.

Key Takeaways

Efficiency is always less than 100% in real processes. It is calculated as (useful output / total input) × 100.

Common Mistakes

Calculating efficiency as (total input / useful output), which gives a number greater than 1. Forgetting to multiply by 100 to get a percentage.

Things to Be Careful About

Use the unrounded values ($4680$ and $14168$) for the calculation to avoid rounding errors, though using $4700 / 14000$ also gives approximately $33.5\%$, which the mark scheme accepts as $33\%$ or $0.33$.

Walkthrough

The water cools from $22.0^\circ\text{C}$ to $6.0^\circ\text{C}$, so the temperature change $\Delta\theta$ is $22.0 - 6.0 = 16.0^\circ\text{C}$. The mass of the water is $0.24\text{ kg}$ and its specific heat capacity is $4200\text{ J / (kg }^\circ\text{C)}$. Using the equation $Q = mc\Delta\theta$, we multiply these three values together to find the total thermal energy lost by the water. $0.24 \times 4200 \times 16 = 16128\text{ J}$. This is correctly expressed in standard form as $1.6 \times 10^4\text{ J}$ to two significant figures.

Key Takeaways

The equation $Q = mc\Delta\theta$ calculates the energy transferred when a substance changes temperature. Always ensure mass is in kg, temperature change is in $^\circ\text{C}$ (or K), and specific heat capacity is in $\text{J / (kg }^\circ\text{C)}$.

Common Mistakes

• Using the wrong temperature change (e.g., using $22.0$ instead of the difference $16.0$).
• Forgetting to convert mass to kg (not an issue here as it is already in kg, but a common trap).
• Failing to express the final answer in standard form or with the correct number of significant figures.

Things to Be Careful About

The question asks for the energy transferred from the water, which is a positive quantity representing the magnitude of energy lost. The temperature change $\Delta\theta$ is always taken as the absolute difference $(\theta_{\text{initial}} - \theta_{\text{final}})$ when calculating energy transferred.

Walkthrough

In solid ice, water molecules are arranged in a fixed, regular, orderly lattice and can only vibrate about fixed positions. When ice melts into liquid water at $6.0^\circ\text{C}$, the molecules gain enough energy to overcome the forces of attraction holding them in the lattice. In the liquid state, the molecules are still close together but have no regular or orderly arrangement. They are free to move around and slide past one another within the volume of the liquid.

Key Takeaways

The transition from solid to liquid involves a change from an ordered, fixed arrangement to a disordered, mobile arrangement. The molecules remain close together in both states, but their freedom of movement increases.

Common Mistakes

• Saying molecules are "further apart" in liquid water (for water, liquid is actually slightly denser than ice, so molecules are closer together on average; even generally, saying they are further apart is not a primary defining feature of melting).
• Saying molecules "move faster" (this is about temperature/kinetic energy, not the structural difference between the phases at a given comparison).
• Confusing the structure of a liquid with that of a gas (molecules in a gas are far apart and move freely in all directions; liquid molecules are still close together).

Things to Be Careful About

The question asks for differences in molecular structure (arrangement and motion), not just temperature or energy. Focus on arrangement (orderly vs. random) and motion (vibrating in place vs. sliding/moving freely).

Walkthrough

The room is at $22.0^\circ\text{C}$ and the water is at $6.0^\circ\text{C}$, so thermal energy is transferred from the room to the water, warming it up. The rate of this transfer depends on the thermal conductivity of the cup's material. Metals are good thermal conductors because they have delocalised electrons that can rapidly transfer kinetic energy through the material. Plastic is a poor thermal conductor (a good thermal insulator) because it lacks these delocalised electrons. Therefore, a metal cup allows thermal energy to enter the water much faster than a plastic cup.

Key Takeaways

The rate of heat transfer by conduction through a material depends on its thermal conductivity. Metals are good conductors; non-metals like plastics and wood are poor conductors (insulators).

Common Mistakes

• Saying "metal conducts heat better" without linking it to the rate of temperature change.
• Saying "metal is colder" or "metal absorbs more heat" (the issue is the rate of transfer, not the total energy or the temperature of the cup itself).
• Using the word "heat" as a substance instead of "thermal energy" or describing the conduction mechanism.

Things to Be Careful About

The question asks to "State why". A single, clear statement about thermal conductivity or insulation is sufficient. Do not overcomplicate with convection or radiation unless specifically asked; the primary difference here is conduction through the cup walls.

Walkthrough

For a thin converging lens, parallel rays of light (paraxial rays) that are parallel to the principal axis will be refracted by the lens and converge at a single point on the other side. This point is called the principal focus or focal point. The mark scheme requires the student to mention both "parallel rays" and "meet after refraction" to earn both marks.

Key Takeaways

• The principal focus is defined specifically for rays parallel to the principal axis.
• It is a real point for a converging lens where light actually meets.

Common Mistakes

• Saying "where all rays meet" — only parallel rays meet at the principal focus; rays from other points meet at other image points.
• Forgetting to say "after refraction" or "through the lens".

Things to Be Careful About

• Ensure the definition includes "parallel to the principal axis". Rays parallel to each other but not to the principal axis will meet at a point on the focal plane, not at the principal focus itself.

Walkthrough

The focal length is a specific distance associated with the principal focus. It is measured from the optical centre (centre) of the thin lens to the principal focus. This is a standard definition.

Key Takeaways

• Focal length ($f$) is a distance, not a point.
• It is measured from the centre of the lens to the principal focus.

Common Mistakes

• Saying "distance from the focus to the object" — that is the object distance ($u$).
• Saying "distance from the lens edge" — for a thin lens, we measure from the centre (optical centre).

Things to Be Careful About

• The unit of focal length is metres (m) or centimetres (cm), but the definition itself does not require units.

Walkthrough

The question gives an object height of 1.8 cm and a position 2.2 cm to the left of the lens centre. Using the grid in Fig. 4.2, where each large square is 1 cm (as indicated by the scale markers), the student must count 2.2 cm to the left of the vertical lens line and draw an arrow 1.8 cm tall starting from the principal axis. The arrow should point upwards.

Key Takeaways

• Ray diagrams must be drawn to scale if a grid is provided.
• Object distance is measured from the centre of the lens.

Common Mistakes

• Placing the object at the wrong distance (e.g., measuring from the edge of the lens instead of the centre).
• Drawing the arrow downwards instead of upwards.
• Not starting the arrow on the principal axis.

Things to Be Careful About

• Read the scale carefully. The diagram shows 1 cm markers. 2.2 cm is 2 small squares and 1/5 of a small square if there are 5 subdivisions, or just estimate carefully. Ensure the height is exactly 1.8 cm.

Walkthrough

Since the object is placed inside the focal length (2.2 cm < focal length, which appears to be around 4-5 cm from the diagram), the image formed is virtual, upright, and magnified. To locate it:

• Ray 1: From the tip of O, draw a ray parallel to the principal axis to the lens. After refraction, it passes through F₂ on the right. Extend this refracted ray backwards (dashed line) to the left.
• Ray 2: From the tip of O, draw a ray straight through the optical centre of the lens. It continues in a straight line without deviation. Extend this backwards (dashed line) to the left.
• Ray 3: From the tip of O, draw a ray towards F₁ on the left. After refraction, it emerges parallel to the principal axis on the right. Extend this backwards (dashed line) to the left.

The three dashed lines will meet at a point on the left side, further away from the lens than the object. This is the tip of the virtual image I. Draw an arrow from the principal axis to this point and label it I.

Key Takeaways

• When the object is inside the focal point of a converging lens, the image is virtual and upright.
• Three standard rays are used to locate the image: parallel-to-axis, through-centre, and through-focus.
• Virtual images are located by extending refracted rays backwards with dashed lines.

Common Mistakes

• Drawing the refracted rays as solid lines continuing forward without extending them backwards.
• Forgetting to label the image I.
• Drawing the image arrow pointing downwards (inverted) — it must be upright.
• Not drawing the rays to scale; the intersection point must be accurate.

Things to Be Careful About

• Use a sharp pencil and ruler for accurate ray construction.
• Dashed lines must be used for the backward extensions; solid lines for actual light paths.
• The image will be larger than the object and on the same side as the object.

Walkthrough

A real image is formed when light rays actually converge at a point. A virtual image is formed when light rays diverge after passing through the lens, and they only appear to come from a point behind the lens (or on the same side as the object for a single lens). In this case, the refracted rays diverge, so we extend them backwards to find the image. Since the rays do not actually meet at the image location, it is a virtual image. Additionally, the image is upright and the object is inside the focal length, both of which are characteristics of a virtual image formed by a converging lens.

Key Takeaways

• Real images: rays actually meet, can be projected on a screen, usually inverted.
• Virtual images: rays only appear to meet (extended backwards), cannot be projected on a screen, usually upright.
• For a converging lens, a virtual image is formed when the object is inside the focal length.

Common Mistakes

• Saying "real" because the rays meet — they only meet if extended backwards with dashed lines.
• Giving no reason, or giving an incorrect reason like "it is on the same side as the object" without explaining why that makes it virtual.

Things to Be Careful About

• The mark scheme accepts multiple valid reasons: rays do not pass through the image, object distance < focal length, or image is upright. Any one of these paired with "virtual" scores the mark.

Walkthrough

An induction heater works by passing a high-frequency alternating current (a.c.) through a coil of wire. This a.c. produces a magnetic field that is constantly changing in magnitude and direction. When this changing magnetic field passes through the conductive iron base of the beaker, it induces an electromotive force (e.m.f.) in the base according to Faraday's law of electromagnetic induction. Because the iron base is a conductor, this induced e.m.f. drives a current (eddy currents) through it, which heats the base due to electrical resistance.

Key Takeaways

• A.c. in a coil produces a changing magnetic field.
• A changing magnetic field through a conductor induces an e.m.f.
• The induced e.m.f. drives a current in the conductor.

Common Mistakes

• Forgetting to mention that the magnetic field must be changing or alternating. A steady magnetic field does not induce an e.m.f.
• Saying 'induces a current' without first stating that an e.m.f. is induced. The e.m.f. is the cause, the current is the effect.

Things to Be Careful About

• The mark scheme awards marks for: (1) magnetic field of coil, (2) magnetic field is changing/alternating, (3) e.m.f. in container. Ensure all three are stated clearly in order.

Walkthrough

When the liquid at point P is heated, thermal energy is transferred to the particles, increasing their kinetic energy. As they move faster, they collide more forcefully and push each other further apart. This increased separation between particles means the overall volume of the liquid increases. Density is defined as mass divided by volume ($\rho = m/V$). Since the mass of the liquid does not change but its volume increases, the density must decrease.

Key Takeaways

• Heating increases particle kinetic energy and speed.
• Faster particles push each other apart, increasing volume (thermal expansion).
• Constant mass with increased volume results in decreased density.

Common Mistakes

• Saying 'particles expand'. Particles themselves do not expand; the space between them increases.
• Forgetting to mention that mass is constant when explaining why density decreases.

Things to Be Careful About

• The question asks for an explanation 'in terms of particles'. You must mention particle speed/motion and the space between them. The mark scheme awards marks for: (1) speed of particles increases, (2) particles push each other apart, (3) volume of liquid increases.

Walkthrough

Because the liquid at point P (near the hot base) has decreased in density, it becomes lighter than the cooler liquid above it. Buoyancy causes the less dense, heated liquid to rise towards the surface. As it rises, the cooler, denser liquid at the surface moves down to take its place. This continuous cycle of rising warm fluid and sinking cool fluid is called a convection current, and it is the primary mechanism for transferring thermal energy through liquids and gases.

Key Takeaways

• Convection occurs due to density differences in fluids.
• Less dense (heated) fluid rises; denser (cooler) fluid sinks.
• This creates a convection current that transfers thermal energy.

Common Mistakes

• Saying 'heat rises'. Heat does not rise; the less dense fluid rises, carrying thermal energy with it.
• Forgetting to mention that the cooler/denser liquid sinks to replace the rising liquid.

Things to Be Careful About

• The mark scheme awards marks for: (1) less dense/heated liquid rises, (2) denser/cooler liquid sinks or convection current established. Ensure both the rising and sinking aspects are covered.

Walkthrough

The circuit consists of resistor P ($8.0\ \Omega$) and the $4.0\ \Omega$ resistor connected in series. The total resistance is the sum of the two: $R_{\text{total}} = 8.0 + 4.0 = 12.0\ \Omega$. Using Ohm's law ($I = V/R$), the current flowing from the $9.0\text{ V}$ battery is $9.0 / 12.0 = 0.75\text{ A}$. The ammeter reads this current.

Key Takeaways

• Resistors in series add together: $R_{\text{total}} = R_1 + R_2$.
• The current is the same everywhere in a series circuit and is given by $I = V/R_{\text{total}}$.

Common Mistakes

• Forgetting to add the two resistances and using only $8.0\ \Omega$ or $4.0\ \Omega$ in the denominator.
• Using the wrong voltage value (e.g., using the p.d. across one resistor instead of the battery e.m.f.).

Things to Be Careful About

• Ensure units are consistent: resistance in $\Omega$ and voltage in V gives current in A.
• The ammeter is in series, so it reads the total circuit current.

Walkthrough

The power transferred to resistor P can be calculated using $P = IV$. We already know the current $I = 0.75\text{ A}$. The p.d. across resistor P is $V_P = I \times R_P = 0.75 \times 8.0 = 6.0\text{ V}$. Multiplying these gives $P = 0.75 \times 6.0 = 4.5\text{ W}$. Alternatively, $P = I^2 R = (0.75)^2 \times 8.0 = 4.5\text{ W}$.

Key Takeaways

• Power can be calculated as $P = IV$, $P = I^2R$, or $P = V^2/R$.
• The voltage $V$ in $P = IV$ must be the p.d. across the specific component (resistor P), not the total battery voltage.

Common Mistakes

• Using the battery voltage ($9.0\text{ V}$) instead of the p.d. across resistor P ($6.0\text{ V}$) in the $P = IV$ formula.
• Forgetting to square the current if using $P = I^2R$.

Things to Be Careful About

• The question asks for power transferred to resistor P specifically, so use the values relevant to P.

Walkthrough

The resistance of a wire is given by $R = \rho l / A$, where $\rho$ is the resistivity, $l$ is the length, and $A$ is the cross-sectional area. The volume of the cylinder is $V = l \times A$. Since the volume is constant and the new length is $l/2$, the new area must be $2A$ to keep $V$ the same. Substituting these into the resistance formula: $R_{\text{new}} = \rho (l/2) / (2A) = (1/4) \times (\rho l / A) = R_{\text{old}} / 4$. Thus, $R_{\text{new}} = 8.0 / 4 = 2.0\ \Omega$.

Key Takeaways

• Resistance is directly proportional to length and inversely proportional to cross-sectional area.
• If volume is constant, halving the length doubles the area, reducing resistance by a factor of 4.

Common Mistakes

• Forgetting that area also changes when length changes for a constant volume.
• Assuming resistance is halved just because length is halved.

Things to Be Careful About

• The material is the same, so resistivity $\rho$ is unchanged. Only geometry changes.

Walkthrough

The voltmeter is connected in parallel across the $4.0\ \Omega$ resistor (resistor Q). This is a potential divider circuit. The voltage across resistor Q is given by $V_Q = E \times R_Q / (R_{\text{new}} + R_Q)$. Substituting the values: $V_Q = 9.0 \times 4.0 / (2.0 + 4.0) = 36.0 / 6.0 = 6.0\text{ V}$. Alternatively, find the new current: $I = 9.0 / (2.0 + 4.0) = 1.5\text{ A}$, then $V_Q = I \times R_Q = 1.5 \times 4.0 = 6.0\text{ V}$.

Key Takeaways

• A series circuit acts as a potential divider: the voltage across a resistor is proportional to its resistance.
• The potential divider formula is $V_{\text{out}} = V_{\text{in}} \times R_2 / (R_1 + R_2)$.

Common Mistakes

• Using the old resistance of resistor P ($8.0\ \Omega$) instead of the new resistance ($2.0\ \Omega$).
• Calculating the voltage across resistor P instead of the voltmeter reading (which is across the $4.0\ \Omega$ resistor).

Things to Be Careful About

• The voltmeter is across the $4.0\ \Omega$ resistor, not resistor P. Read the diagram carefully.

Walkthrough

The fundamental difference between a conductor and an insulator at the particle level is the mobility of the charge carriers. In conducting materials (like metals), the outer electrons of the atoms are not bound to any single atom and can move freely throughout the material. In insulators, all electrons are tightly bound to their atoms and cannot move from place to place.

Key Takeaways

Conductors have free electrons that can move; insulators do not. This mobility is what allows charge to redistribute in conductors.

Common Mistakes

Saying 'current can flow' or 'resistance is low' does not answer the question 'in terms of particles'. The answer must specifically mention free electrons.

Things to Be Careful About

The question specifically asks for the explanation 'in terms of particles', so mentioning free electrons is essential.

Walkthrough

The ball has a conducting surface, meaning it contains both positive charges (protons in the nuclei) and negative charges (free electrons). When the negatively charged rod is brought near the left side of the ball, the negative charges in the ball experience a repulsive force because like charges repel. Since the ball is a conductor, these free electrons move away from the rod to the right-hand side of the ball. The positive charges (protons) are bound in the atomic nuclei and cannot move. As the electrons leave the left side, that region is left with an excess of positive charge.

Key Takeaways

Electrostatic induction in a conductor causes a separation of charge: free electrons move away from a like external charge, leaving the near side oppositely charged.

Common Mistakes

Saying 'positive charges move to the left' is incorrect; protons do not move in solids. Saying 'the ball becomes positively charged' is incorrect; the ball as a whole remains neutral, the charge is just redistributed.

Things to Be Careful About

Be precise about which charges move and which do not. The question states the ball contains both, so you must address the behaviour of both. Mentioning 'like charges repel' provides the physical reason for the electron movement.

Walkthrough

After the charge separation in part (ii), the left side of the ball is positive and the right side is negative. The negatively charged rod attracts the positive charges and repels the negative charges. However, the positive charges are physically closer to the rod than the negative charges. The electrostatic force between two charges decreases with increasing distance. Because the distance to the positive charges is smaller, the attractive force is stronger than the repulsive force acting on the more distant negative charges. The resultant (net) force is therefore attractive, pulling the ball towards the rod.

Key Takeaways

Even a neutral object is attracted to a charged object due to electrostatic induction and the distance-dependence of the electrostatic force. The closer opposite charge always experiences a stronger force than the further like charge.

Common Mistakes

Saying 'the ball is neutral so it is attracted' is not a full explanation. Saying 'the rod pulls the positive charges' without mentioning the repulsion of the negative charges and the distance argument misses the mark. Simply stating 'unlike charges attract' is only one of the two required marks.

Things to Be Careful About

The question asks why the resultant force is towards the rod. You must mention both the attractive force on the closer positive charges and the repulsive force on the further negative charges, and explain why the attraction wins (distance).

Walkthrough

In the Rutherford alpha-particle scattering experiment:

• Position K is directly in line behind the thin gold foil. The vast majority of alpha particles pass straight through the gold sheet without any deflection, resulting in a very high count rate at detector position K.
• Position J is at a large angle (backwards/side deflection). Only a very tiny fraction of alpha particles encounter a close electrostatic repulsion with a gold nucleus and are deflected through large angles (greater than $90^\circ$), so detector J registers a very low (but non-zero) count rate.

Key Takeaways

• Most alpha particles pass straight through with little or no deflection.
• A very small proportion are deflected through large angles.

Common Mistakes

• Stating that zero particles reach position J (it must be noted that some particles are detected, i.e. count rate is not zero).
• Confusing the count rates at J and K.

Things to Be Careful About

• Clearly distinguish between position J (large-angle deflection / side) and position K (straight through).

Walkthrough

The observations from the scattering experiment lead to fundamental conclusions about atomic structure:

1. Most alpha particles pass straight through undeflected (Position K): This proves that the atom consists mostly of empty space.
2. A small number of alpha particles are deflected through large angles (Position J): Since alpha particles are positively charged and relatively massive, large deflections can only be caused by a strong repulsive electrostatic force. This shows that the positive charge and most of the atom's mass are concentrated in a very small, dense central region called the nucleus.

Key Takeaways

• Straight-through transmission $\rightarrow$ atom is mostly empty space.
• Large-angle deflection / backscattering $\rightarrow$ mass and positive charge are concentrated in a tiny, dense nucleus.

Common Mistakes

• Claiming the nucleus contains electrons.
• Stating that the atom is completely solid or dense throughout (the plum pudding model).

Things to Be Careful About

• Awarding two distinct marks: one for the atom being mostly empty space, and one for the concentrated mass/charge (nucleus).

Walkthrough

Velocity is a vector quantity, having both magnitude (speed) and direction. In a circular orbit, even if the speed is constant, the direction of motion is continuously changing. A change in velocity means there is an acceleration. According to Newton's first law, a resultant force is required to produce this acceleration. Thus, a force must act to continuously change the direction of Venus's motion, preventing it from moving off in a straight line tangent to the orbit.

Key Takeaways

Circular motion requires a force because the direction of velocity is constantly changing.

Common Mistakes

Saying 'to keep it moving' or 'to overcome inertia'. Inertia is not a force. The force is needed to change the direction, not to keep it moving.

Things to Be Careful About

Be precise: it's a change in direction of motion (or velocity), not necessarily speed.

Walkthrough

The force that keeps planets in orbit is the gravitational attraction between the planet and the Sun. This is a gravitational field force exerted by the Sun. Because gravity is an attractive force, it pulls the planet towards the source of the field, which is the centre of the Sun (or the centre of the circular orbit).

Key Takeaways

Planetary orbits are maintained by the Sun's gravitational pull directed towards the Sun.

Common Mistakes

Saying 'gravity' without specifying it is the gravitational field of the Sun. Saying 'centrifugal force' (not a real force in this context). Giving the direction as 'outwards' or 'away from the Sun'.

Things to Be Careful About

The mark scheme awards marks for 'gravitational field', 'gravitational attraction', or 'gravity', provided it is specified as being of the Sun. The direction must be 'towards the centre' or 'towards the Sun'.

Walkthrough

The orbit is approximately circular, so the distance travelled in one complete orbit is the circumference of the circle, $2\pi r$. The time taken is the orbital period, $T$. Speed is distance divided by time. First, convert the time from hours to seconds: $5400 \text{ hours} \times 3600 \text{ s/hour} = 1.944 \times 10^7 \text{ s}$. Then calculate the circumference: $2 \times \pi \times 1.1 \times 10^{11} \text{ m} \approx 6.91 \times 10^{11} \text{ m}$. Finally, divide the distance by the time: $6.91 \times 10^{11} / 1.944 \times 10^7 \approx 3.555 \times 10^4 \text{ m/s}$. Rounding to 2 significant figures gives $3.6 \times 10^4 \text{ m/s}$.