Physics 5054/11 — October/November 2024

Walkthrough

Power tells us how quickly energy is transferred or work is done. The SI unit of power is the watt ($\text{W}$), which is one joule per second. For a large motor such as a crane's motor, the power is usually given in kilowatts ($\text{kW}$).

Look at each option:

• A $\text{kN}$ is a kilonewton, a unit of force. A crane does exert forces, but the question asks for power, not force.
• B $\text{kW}$ is a kilowatt, a unit of power. This is the correct choice.
• C $\text{kN/s}$ is not a unit of power. It mixes force and time, but power is energy per second, not force per second.
• D $\text{kWh}$ is a kilowatt-hour, a unit of energy. It is the energy used by a 1 kW appliance in 1 hour, not the rate of doing work.

So the correct answer is B.

Key Takeaways

• Power is the rate of transferring energy or doing work.
• The unit of power is the watt ($\text{W}$), and $1\ \text{kW} = 1000\ \text{W}$.
• Do not confuse force ($\text{N}$), power ($\text{W}$ or $\text{kW}$) and energy ($\text{J}$ or $\text{kWh}$).

Common Mistakes

• Choosing A because $\text{kN}$ contains "k" and sounds large, but it is a force unit.
• Choosing D because $\text{kWh}$ contains "kW", but it is an energy unit.
• Thinking $\text{kN/s}$ could be a power unit; power is energy per second, not force per second.

Things to Be Careful About

• Read the question carefully: it asks for a unit of power, not a unit of force or energy.
• Remember that $\text{kW}$ is power, while $\text{kWh}$ is energy.
• Always match the unit to the physical quantity being asked about.

Walkthrough

This question tests your understanding of what mass is and its unit.

• What is mass? Mass is a measure of the amount of matter in an object. It is a scalar quantity, meaning it has magnitude (a size) but no direction. For example, a mass of 5 kg is just 5 kg — there is no direction associated with it.
• What is the unit of mass? The SI unit of mass is the kilogram (kg). Other units like grams (g) or tonnes (t) are also used, but the base unit is the kilogram.
• Now look at the options:
  • A says mass is a scalar and its unit is the kilogram — this is correct.
  • B says the unit is the newton. The newton (N) is the unit of force, not mass, so this is wrong.
  • C says mass is a vector. A vector quantity has both magnitude and direction (like velocity or force). Mass has no direction, so this is wrong.
  • D combines the two errors — it is wrong for both reasons.

Key Takeaways

• Mass is a scalar quantity — it has magnitude only.
• The SI unit of mass is the kilogram (kg).
• The newton (N) is the unit of force, not mass.
• Being able to sort quantities into scalars (magnitude only) and vectors (magnitude and direction) is a key skill in physics.

Common Mistakes

• Confusing mass with weight. Weight is a force (measured in newtons) and is a vector; mass is a scalar measured in kilograms.
• Forgetting that the newton is a unit of force. If you see 'newton' in a question about mass, it is almost certainly a distractor.

Things to Be Careful About

• Always associate mass with the kilogram and weight with the newton.
• Remember that scalars have no direction; if a quantity is described as a vector, it must have both magnitude and direction.

Walkthrough

The pendulum bob is held stationary, so it is in equilibrium. This means the resultant force acting on it is zero. There are three forces acting on the bob:

1. The weight $W$ acting vertically downwards.
2. The horizontal pulling force, which is given as equal to $W$.
3. The tension $T$ in the thread, acting along the thread at an angle $\theta$ to the vertical.

For the resultant force to be zero, the horizontal and vertical components of the tension must exactly balance the other two forces.
Horizontally, the tension's component is $T \sin \theta$, and it must balance the horizontal pull $W$:

Vertically, the tension's component is $T \cos \theta$, and it must balance the weight $W$:

To find the angle $\theta$, divide the horizontal equation by the vertical equation:

The angle whose tangent is 1 is $45^\circ$. Therefore, the thread must be at $45^\circ$ to the vertical, which matches diagram C.

Key Takeaways

• An object in equilibrium has a resultant force of zero.
• Forces can be resolved into perpendicular components (horizontal and vertical) to find unknown angles or magnitudes.
• When two perpendicular forces are equal in magnitude, the resultant force acts at $45^\circ$ to each, and the balancing force (tension) must also act at $45^\circ$.

Common Mistakes

• Assuming the angle is $30^\circ$ or $60^\circ$ because of the common $30^\circ$-$60^\circ$-$90^\circ$ triangle, without checking which side is opposite which force.
• Forgetting that both the horizontal pull and the weight have the same magnitude $W$, so their components must be equal.
• Confusing the angle with the horizontal instead of the vertical.

Things to Be Careful About

• The angle $\theta$ is measured from the vertical, so $\sin \theta$ is the horizontal component and $\cos \theta$ is the vertical component.
• Ensure that the forces being resolved are correctly matched to their directions: horizontal pull balances $T \sin \theta$, weight balances $T \cos \theta$.

Walkthrough

Average speed is defined as

It is not the average of the two speeds, because the student spends different times at the two speeds.

First convert each time from minutes to hours:

• 20 minutes = $\frac{20}{60} = \frac{1}{3}$ hour
• 15 minutes = $\frac{15}{60} = \frac{1}{4}$ hour

Now find the distance on each part using distance = speed × time:

• XY: $3.0 \times \frac{1}{3} = 1.0$ km
• YZ: $4.0 \times \frac{1}{4} = 1.0$ km

Total distance = 1.0 + 1.0 = 2.0 km.

Total time = 20 + 15 = 35 minutes = $\frac{35}{60} = \frac{7}{12}$ hour.

Therefore

This rounds to 3.4 km/h, so the correct option is B.

The fact that she turns 90° at Y is not needed for average speed, because average speed depends only on total distance along the path, not on the direction of travel.

Key Takeaways

• Average speed is the total distance travelled divided by the total time taken.
• Speeds must be weighted by the time spent at each speed; simply averaging 3.0 and 4.0 gives the wrong answer.
• Always convert minutes to hours before using speeds given in km/h.
• The 90° turn is irrelevant to average speed, but it would matter if the question asked for average velocity or displacement.

Common Mistakes

• Choosing C, 3.5 km/h: this is just the average of 3.0 and 4.0, ignoring the different times spent at each speed.
• Choosing D, 5.0 km/h: this is the resultant speed if the two velocities were combined as perpendicular vectors. It is not the average speed.
• Choosing A, 2.4 km/h: this comes from dividing the straight-line displacement by the total time. Average speed uses the full distance walked, not the displacement.
• Forgetting that 20 minutes is one third of an hour, not 0.2 hours.

Things to Be Careful About

• The units are km/h, so minutes must be written as fractions of an hour.
• Average speed is a scalar, so direction and the right-angle turn do not enter the calculation.
• Rounding: $\frac{24}{7} = 3.428$, which is 3.4 to two significant figures, matching option B.
• If the question had asked for average velocity, you would need displacement and direction, not total distance.

Walkthrough

The car starts from rest, so its initial speed is zero. It has a constant acceleration, which means its speed increases at a steady rate.

• If the graph plotted speed against time, it would be a straight line through the origin with a constant gradient (since $v = at$). Option D is incorrect.
• If the graph plotted distance against time, the distance is given by $s = \frac{1}{2}at^2$. This is a quadratic relationship, so the graph is a curve starting from the origin with an increasing gradient (a parabola). This matches the graph in Fig. 1. Thus, the x-axis is time and the y-axis is distance. Option C is correct.
• If the x-axis were distance and y-axis time (Option A), the gradient (1/speed) would decrease as speed increases, giving a curve with a decreasing gradient. Option A is incorrect.
• Option B (x-axis = speed, y-axis = time) would also give a straight line since $t = v/a$. Option B is incorrect.

Key Takeaways

• For constant acceleration from rest, a speed-time graph is a straight line through the origin.
• A distance-time graph for constant acceleration from rest is a curve with an increasing gradient (parabolic).

Common Mistakes

• Confusing the shapes of distance-time and speed-time graphs for accelerated motion.
• Thinking that any curve starting from the origin must be a speed-time graph.

Things to Be Careful About

• Always check the gradient of the curve. An increasing gradient on a distance-time graph means speed is increasing (acceleration). An increasing gradient on a speed-time graph means acceleration is increasing (jerk), which is not the case here.
• Read the options carefully to match x-axis and y-axis correctly.

Walkthrough

An object falling without air resistance is in free fall. The only force acting on it is its weight, so it experiences a constant acceleration downwards (the acceleration of free fall, $g$).
Because the acceleration is constant and downwards, the object's speed increases continuously as it falls.
The sequence of dots shows the position of the object at equal time intervals. If the speed is constant, the dots would be equally spaced (Option C). If the object were slowing down, the dots would get closer together (Option A). Since the object is speeding up, it covers more distance in each successive equal time interval. Therefore, the gaps between the dots must increase as the object moves downwards.
Looking at the options:

• A shows decreasing gaps (deceleration).
• B shows irregular spacing.
• C shows equal gaps (constant speed / terminal velocity).
• D shows increasing gaps (acceleration).
  Thus, D is the correct sequence.

Key Takeaways

• Free fall is motion under constant acceleration due to gravity.
• For an accelerating object, the distance covered in equal time intervals increases.
• A dot diagram with equal time intervals shows acceleration if the spacing between dots increases in the direction of motion.

Common Mistakes

• Choosing C: assuming the object falls at a constant speed. This would only be true if air resistance balanced the weight (terminal velocity), but the question states 'without air resistance'.
• Choosing A: confusing the direction of acceleration or thinking falling objects slow down.
• Forgetting that the time intervals between dots are equal; the spacing directly reflects the speed at that part of the journey.

Things to Be Careful About

• Read the phrase 'without air resistance' carefully. If air resistance were included, the object would eventually reach terminal velocity, and the bottom dots would be equally spaced.
• Ensure you read the direction of movement (downwards) and match the increasing spacing to the direction of motion.

Walkthrough

When the piston compresses the gas, the same number of gas particles is squeezed into a smaller volume. The kinetic model of gases says the pressure comes from the particles colliding with the container walls. With a smaller volume, the particles collide with the walls more often, so the pressure rises and the frequency of collisions increases. Density is mass divided by volume, so with the same mass in a smaller volume, the density also rises. The only quantity that does not change is the mass itself, because no particles enter or leave the container. So the answer is B.

Key Takeaways

• Compressing a gas at constant temperature increases pressure, collision frequency and density.
• Mass is a measure of the amount of matter and is conserved unless gas enters or leaves the container.
• Density depends on both mass and volume, so it changes when the volume changes.

Common Mistakes

• Choosing C (frequency of particle collision) — this does increase, because the particles have less distance to travel between the walls.
• Choosing D (density) — density is mass over volume; since the volume decreases, the density increases.
• Confusing the constancy of mass with the constancy of density.

Things to Be Careful About

• The question asks which property does not change. Read the negative phrasing carefully.
• Remember that temperature is constant here, so the average kinetic energy of the particles stays the same — this is why only the collision frequency, not the particle speed, changes.
• Do not confuse the number of particles (constant) with the density (which changes because the volume changes).

Walkthrough

The question gives the car's weight as $8000\ \text{N}$ and asks for its mass. Weight and mass are different quantities: weight is the force of gravity on an object, measured in newtons, while mass is the amount of matter in the object, measured in kilograms. They are linked by the equation

where $W$ is the weight in newtons, $m$ is the mass in kilograms, and $g$ is the gravitational field strength, which is $9.8\ \text{N/kg}$ on Earth's surface.

To find the mass, rearrange the equation to make $m$ the subject:

This rounds to $820\ \text{kg}$, which is option B.

Looking at the other options: A ($810\ \text{kg}$) is the value you would get if you divided by $9.9$ or made a rounding error. C ($8000\ \text{kg}$) is just the weight with the wrong unit — confusing newtons with kilograms. D ($78\,000\ \text{kg}$) comes from multiplying $8000$ by $9.8$ instead of dividing.

Key Takeaways

• Weight is a force ($W = mg$), measured in newtons; mass is a measure of the amount of matter, measured in kilograms.
• To find mass from weight, divide by the gravitational field strength: $m = W/g$.
• The gravitational field strength on Earth is approximately $9.8\ \text{N/kg}$.

Common Mistakes

• Confusing mass and weight — the most common error here is picking C, which gives the weight a mass unit.
• Multiplying instead of dividing: $8000 \times 9.8 = 78\,400$, which is option D.
• Using the wrong value of $g$ — the question does not give $g$, so you must recall it is $9.8\ \text{N/kg}$ on Earth.

Things to Be Careful About

• Always check the units: weight in newtons, mass in kilograms, gravitational field strength in N/kg.
• Rounding: $816\ \text{kg}$ rounds to $820\ \text{kg}$, matching option B exactly.

Walkthrough

The lorry experiences two horizontal forces: the driving force of $80\ \text{kN}$ forwards and the total resistive force of $20\ \text{kN}$ backwards. Since these act along the same straight line in opposite directions, the resultant (net) force is found by subtracting the smaller from the larger:

This resultant force points in the direction of the larger force, which is forwards, i.e. in the same direction as the lorry is moving. A resultant force in the direction of motion causes the lorry to speed up, so the motion is an acceleration.

Thus the correct row is A.

Key Takeaways

• When forces act along the same line, find the resultant by vector addition, treating one direction as positive.
• The direction of the resultant force determines whether the object accelerates or decelerates: if the resultant is in the direction of motion, the object accelerates; if opposite, it decelerates.
• Units must be consistent; here both forces are already in $\text{kN}$, so no conversion is needed.

Common Mistakes

• Adding the forces to get $100\ \text{kN}$ instead of subtracting: the two forces act in opposite directions, so they must be subtracted.
• Confusing the direction of the resultant with the direction of motion: the resultant is forward, so the lorry accelerates, not decelerates.
• Thinking that a large resistive force always means deceleration; it is the direction of the resultant force that matters.

Things to Be Careful About

• Always state the direction of a vector quantity. Here the resultant is $60\ \text{kN}$ forwards.
• If the driving force were smaller than the resistive force, the resultant would be backward and the lorry would decelerate.
• The question asks for both the motion and the resultant force; make sure both parts of the answer are correct.

Walkthrough

This question asks for the cause of the energy transfer when a cyclist brakes. The cyclist is moving, so the bicycle has energy in its kinetic store. To slow down, the brakes must remove this kinetic energy. The brake blocks press on the wheel rims and friction acts between them. Friction opposes the motion and does work, so kinetic energy is transferred to the thermal store — the brakes and wheel become warm.

Now look at the options:

• A friction — correct, because friction is the force that does work against the motion and transfers the energy.
• B mass — mass is a property of the cyclist and bicycle; it does not cause an energy transfer by itself.
• C thrust — thrust is a forward driving force; it would increase the kinetic store, not transfer energy to the thermal store when braking.
• D weight — weight is the gravitational force pulling downwards; it does not cause the braking energy transfer.

So the answer is A.

Key Takeaways

• Friction is a contact force that opposes relative motion between two surfaces.
• When friction acts, it does work and transfers energy from the kinetic store to the thermal store.
• Energy is not destroyed when braking: it is transferred from one store to another.

Common Mistakes

• Choosing weight because it is a force: weight acts vertically downwards and is not the force causing the braking energy transfer.
• Choosing thrust because the cyclist is moving: thrust is a driving force, not the braking force.
• Confusing the property mass with a force or cause of energy transfer.

Things to Be Careful About

• The question asks for the cause of the transfer, not the store that energy is transferred to.
• Use the correct energy-store language: kinetic store and thermal store, not "heat energy" as a store.
• Friction always transfers energy to the thermal store when it does work against motion.

Walkthrough

1. Recall Hooke's Law: Hooke's Law states that the extension of an elastic object is directly proportional to the force (load) applied to it, as long as the limit of proportionality is not exceeded. Mathematically, $F = kx$ or $x = \frac{1}{k} F$.
2. Interpret the graph axes: The vertical axis is extension and the horizontal axis is load. A proportional relationship ($y \propto x$) appears as a straight line through the origin on a graph.
3. Locate the linear region: In Fig. 1, the line is straight from the origin (0,0) through point A and continues straight up to point B.
4. Identify the limit of proportionality: The limit of proportionality is the point beyond which the graph is no longer a straight line. At point B, the line starts to curve upwards (towards C and D). This means that for loads greater than the load at B, the extension is no longer proportional to the load.
5. Conclusion: Point B marks the end of the straight-line section, so B is the limit of proportionality.

Note: Point C is often close to the elastic limit (the point beyond which the spring does not return to its original length when the load is removed), but strictly speaking, the limit of proportionality is where the linearity ends, which is B.

Key Takeaways

• Hooke's Law: Extension $\propto$ Load up to the limit of proportionality.
• Graph shape: A load-extension graph is a straight line through the origin up to the limit of proportionality. Beyond that, it curves.
• Key points: Origin (0,0), Limit of proportionality (end of straight line), Elastic limit (end of elastic deformation, often very close to limit of proportionality), Yield point (where large extension occurs with little extra load), Fracture point.

Common Mistakes

• Confusing limit of proportionality with elastic limit: Students often pick the point where the curve gets steep (like C) or where it breaks. The limit of proportionality is specifically where the straight line ends.
• Reading the axes wrong: If axes were swapped (load on y, extension on x), the straight line would still indicate proportionality, but the gradient would be the spring constant $k$. Here, gradient is $1/k$.
• Choosing A: Point A is just a point on the linear section, not the limit.

Things to Be Careful About

• Definition: Ensure you know the definition: limit of proportionality is the point up to which Hooke's Law is obeyed (straight line graph).
• Graph reading: Look carefully at where the straight line stops. In this diagram, the straight line clearly ends at B.
• Units: Not applicable here as it is a conceptual identification question.

Walkthrough

A satellite moving in a circular orbit is constantly changing direction. A force is needed to change direction, and for circular motion this force acts towards the centre of the circle. Here, the centre of the circle is the centre of the Earth, so the force on the satellite is towards the centre of the Earth. This force is the gravitational pull of the Earth on the satellite.

Key Takeaways

• An object moving in a circle is accelerating because its direction is changing.
• The force causing circular motion always points towards the centre of the circle.
• For a satellite, this force is the Earth's gravitational attraction.

Common Mistakes

• Choosing B: thinking the force is in the direction of motion. A force in the direction of motion would speed the satellite up, not make it turn.
• Choosing A: thinking the force is away from the centre. This is the fictitious 'centrifugal' feeling, not a real force acting on the satellite.
• Choosing C: thinking the force opposes motion. That would slow the satellite down, not keep it in orbit.

Things to Be Careful About

• The satellite is not moving away from the Earth even though it is moving sideways: the force towards the centre continuously changes its direction.
• 'Centripetal' means towards the centre. There is no outward force acting on the satellite.

Walkthrough

1. Identify the pivot and the condition for balance: The beam is pivoted at its centre. Since it is uniform, its own weight acts downwards through the pivot point. The perpendicular distance from the pivot to the line of action of the beam's weight is zero, so the beam's weight creates no moment. We only need to consider the three hanging weights.
2. Apply the principle of moments: For an object to be in rotational equilibrium (balanced), the total clockwise moment about the pivot must equal the total anticlockwise moment about the pivot. Moment = Force × perpendicular distance from pivot.
3. Calculate the clockwise moment: On the right side, there is a 2.0 N weight at a distance of 30 cm from the pivot. $$\text{Clockwise moment} = 2.0 \text{ N} \times 30 \text{ cm} = 60 \text{ N cm}$$
4. Calculate the anticlockwise moments: On the left side, there are two weights.
   • A 3.0 N weight at 5.0 cm from the pivot. Moment = $3.0 \times 5.0 = 15 \text{ N cm}$.
   • A 2.0 N weight. The diagram shows it is at a distance $x$ from the 3.0 N weight. Since the 3.0 N weight is 5.0 cm from the pivot, the total distance of the 2.0 N weight from the pivot is $(5.0 + x)$ cm. Moment = $2.0 \times (5.0 + x) \text{ N cm}$.
5. Set up the equation and solve: $$60 = 15 + 2.0(5.0 + x)$$ Expand the bracket: $$60 = 15 + 10 + 2x$$ $$60 = 25 + 2x$$ Subtract 25 from both sides: $$35 = 2x$$ Divide by 2: $$x = 17.5 \text{ cm}$$

Key Takeaways

• Principle of Moments: $\text{Clockwise moments} = \text{Anticlockwise moments}$ for a balanced object.
• Distance from pivot: Moments are calculated using the perpendicular distance from the pivot, not from another object. If a distance $x$ is given between two objects, you must add the distance of the first object from the pivot to find the total distance.
• Uniform beam at centre pivot: The weight of a uniform beam pivoted at its centre of gravity produces no turning effect (moment = 0).

Common Mistakes

• Using $x$ as the distance from the pivot: A common error is to calculate the moment of the leftmost 2.0 N weight as $2.0 \times x$. The distance $x$ is between the two left weights, so the distance from the pivot is $(5.0 + x)$.
• Including the beam's weight: Students sometimes try to include the weight of the beam in the calculation. Since the beam is uniform and pivoted at its centre, its weight acts through the pivot and creates no moment.
• Unit errors: While it is good practice to convert to metres and Newton-metres (Nm), using cm and N (N cm) is perfectly valid here as long as units are consistent on both sides of the equation. Converting unnecessarily can introduce arithmetic errors.

Things to Be Careful About

• Read the diagram carefully: Ensure you identify which distance is from the pivot and which is between weights. In this case, 30 cm and 5.0 cm are from the pivot, but $x$ is between the 3.0 N and 2.0 N weights on the left.
• Signs and directions: Clockwise is right-side-down (or left-side-up), anticlockwise is left-side-down (or right-side-up). Here, right-side weights cause clockwise rotation, left-side weights cause anticlockwise rotation.
• Significant figures: The data is given to 2 significant figures (2.0, 30, 5.0, 3.0). The answer 17.5 is exact from the calculation, but 18 cm (2 s.f.) would also be acceptable in some contexts, though 17.5 matches the option exactly.

Walkthrough

1. Identify the key principle: When a body is suspended freely and is at rest, the only two forces acting on it are its weight (acting downwards through the centre of gravity) and the reaction force at the pivot (acting upwards through the pivot). For the body to be in equilibrium, these forces must be equal, opposite, and collinear. Therefore, the centre of gravity must lie on the vertical line passing directly below the point of suspension.
2. Analyze the given diagram: The vertical dashed line is the plumb line. All four crosses (A, B, C, D) lie on or near this line. This confirms they are all candidates based on the suspension principle alone.
3. Distinguish between the candidates: Since the card is uniform in thickness and density, its centre of gravity is the centroid of its area. The card is irregular, being narrower at the top and wider at the bottom. This means more mass is distributed in the lower half. Consequently, the centre of gravity must be located in the lower half of the card, below the geometric midpoint.
4. Evaluate the points:
   • A is near the pin, too high.
   • B is in the upper half, above the bulk of the mass.
   • D is at the bottom edge; the centre of gravity must be inside the body, balancing the mass above and below it.
   • C is in the lower middle, correctly positioned to balance the greater mass at the bottom.
5. Conclude that C is the centre of gravity.

Key Takeaways

• A freely suspended object at rest has its centre of gravity directly below the point of suspension, on the vertical plumb line.
• For a uniform object, the centre of gravity is the centroid, which shifts towards regions of greater mass or width.

Common Mistakes

• Choosing A or B by assuming the centre of gravity is near the top or at the geometric centre of the bounding shape, ignoring the actual mass distribution.
• Choosing D by placing the centre of gravity at the lowest edge of the object; the centre of gravity must lie within the body where the mass is balanced above and below.

Things to Be Careful About

• Remember that a single plumb line only tells you the centre of gravity lies somewhere on that line. To find it exactly for an irregular object, you would need to suspend it from a second point and draw another plumb line; the intersection of the two lines is the exact centre of gravity. Here, the correct point must be estimated from the mass distribution.

Walkthrough

The two trolleys collide and stick together, which is a perfectly inelastic collision. We must first find the velocity of the combined mass after the collision using the principle of conservation of momentum.

Step 1: Find the final velocity.
Define the right direction as positive. Then $u_X = +2.0\ \text{m/s}$ and $u_Y = -1.0\ \text{m/s}$. The total initial momentum is:

After the collision, the trolleys move together with a combined mass of $m_X + m_Y = 3.0\ \text{kg}$. Let $v$ be their common final velocity:

Equating initial and final momentum gives $3.0 v = 3.0$, so $v = 1.0\ \text{m/s}$ to the right.

Step 2: Find the change in momentum of trolley X.
Trolley X starts at $2.0\ \text{m/s}$ and ends at $1.0\ \text{m/s}$. Its change in momentum is:

The magnitude of the change in momentum is $2.0\ \text{kg m/s}$.

Step 3: Calculate the average force.
Force is the rate of change of momentum. Over the collision time of $0.20\ \text{s}$, the average force on trolley X is:

Key Takeaways

• In a collision where objects stick together, use conservation of momentum with the combined mass to find the final velocity.
• Always assign a positive direction and use negative signs for velocities in the opposite direction.
• The average force on an object during a collision is its change in momentum divided by the collision time ($F = \Delta p / t$).

Common Mistakes

• Ignoring the direction of velocity: Treating both velocities as positive gives a total initial momentum of $5.0\ \text{kg m/s}$, leading to a final velocity of $1.67\ \text{m/s}$ and a wrong force of $3.3\ \text{N}$ (Option A).
• Using the total change in momentum instead of the change for one trolley: Taking the change in momentum of the whole system (which is zero) or incorrectly summing the momentum changes gives incorrect forces like $15\ \text{N}$ or $25\ \text{N}$.
• Forgetting to add the masses: Using the initial mass of X alone for the final momentum calculation instead of the combined mass $m_X + m_Y$.

Things to Be Careful About

• Sign conventions: Momentum is a vector. You must define a positive direction (e.g., right is positive) and apply it consistently to all velocities. Trolley Y moving left has a negative velocity.
• Sticking together: The phrase "stick together" means the final mass is the sum of both masses ($2.0 + 1.0 = 3.0\ \text{kg}$), and they share a single final velocity.
• Force on X vs. Force on Y: By Newton's third law, the force on Y is equal and opposite to the force on X. Both have a magnitude of $10\ \text{N}$, but the force on X is directed to the left (negative), slowing it down from $2.0\ \text{m/s}$ to $1.0\ \text{m/s}$.

Walkthrough

The object's mass and kinetic energy are given, and its speed is required. Kinetic energy is the energy an object has because it is moving. For an object of mass $m$ moving at speed $v$,

Here $E_k = 10\ \text{J}$ and $m = 5.0\ \text{kg}$, so we need to rearrange the equation to make $v$ the subject.

Multiply both sides by 2:

Divide both sides by $m$:

Then take the square root:

Substitute the values:

This gives option C.

The other options are too small or too large: $4.0\ \text{m/s}$ would give $E_k = \frac{1}{2} \times 5.0 \times 4.0^2 = 40\ \text{J}$, not $10\ \text{J}$. Only $2.0\ \text{m/s}$ gives kinetic energy $10\ \text{J}$.

Key Takeaways

• The kinetic energy of a moving object is $E_k = \frac{1}{2} m v^2$.
• To find speed from kinetic energy and mass, use $v = \sqrt{\frac{2E_k}{m}}$.
• Always work in SI units: energy in joules, mass in kilograms, speed in m/s.
• Speed is a scalar quantity, so it has no direction.

Common Mistakes

• Forgetting the factor $\frac{1}{2}$ in the kinetic energy equation.
• Forgetting to take the square root after finding $v^2$.
• Mixing up mass and weight: here the mass is $5.0\ \text{kg}$, not a force.
• Quoting the answer without a unit; the unit m/s is required.

Things to Be Careful About

• Make sure the energy is in joules and the mass is in kilograms before substituting.
• The answer should be given to an appropriate number of significant figures; $2.0\ \text{m/s}$ is suitable.
• In a multiple-choice question, check that your final value appears as an option before selecting the letter. Here the correct option is C.

Walkthrough

A wind-up radio requires a person to do work to wind the handle. This transfers energy from the person's chemical energy store (from food) into the elastic potential energy store of the wound spring. This storage step is intentional and useful, making option C a useful transfer in the overall process.

When the radio is turned on, the spring unwinds. The elastic potential energy is transferred into kinetic energy as the spring drives the coil to rotate. This is the mechanical input to the generator, so option A is useful.

The rotating coil moves through a magnetic field, which induces an electrical current via electromagnetic induction. The kinetic energy is thus transferred into electrical energy. This electrical energy is then sent to the speaker. The speaker converts the electrical energy into sound energy (the useful output) and some thermal energy (waste). The speaker's diaphragm vibrates (kinetic energy) to produce the sound waves, so the final stage involves a transfer to sound energy, making option D a useful part of the output chain.

Throughout the device, moving parts experience friction and the electrical circuit has resistance. These factors cause some of the kinetic and electrical energy to be transferred into thermal energy stores in the surroundings. This heating effect is wasted energy that does not contribute to producing sound or storing energy, so option B is not useful.

Key Takeaways

• Every energy transfer chain in a device has a useful path (input $\rightarrow$ storage $\rightarrow$ conversion $\rightarrow$ useful output) and a wasteful path (usually to thermal energy stores via friction or resistance).
• Identifying which transfer is 'not useful' means spotting the waste energy pathway, which is almost always the generation of thermal energy.
• Chemical energy from food is the ultimate source of energy for a wind-up radio, transferred first into elastic potential energy.

Common Mistakes

• Confusing the input energy source: some candidates think the radio has no useful input because it doesn't use mains electricity, forgetting that the person's chemical energy is the intended input.
• Misidentifying the speaker's action: thinking that kinetic energy is not involved in sound production, whereas the speaker cone's vibration (kinetic) is what directly creates the sound waves.
• Assuming all thermal energy is useful: while some devices (like electric heaters) use thermal energy as their useful output, in a radio it is purely waste.

Things to Be Careful About

• Read the question carefully: it asks for the transfer that is not useful, not the one that is useful.
• Remember that 'useful' means contributing to the primary function of the device (producing sound from a stored charge). Any energy that ends up as waste heat is not useful for this purpose.
• In energy transfer questions, thermal energy is almost always the waste output unless the device is specifically designed to heat something (e.g., a toaster, an electric heater).

Walkthrough

Step 1: Identify the total power input. The lamp is rated at $60\ \text{W}$. This is the total power it draws from the supply, so it is the denominator of the efficiency ratio.

Step 2: Identify the wasted power. The lamp wastes $46\ \text{W}$. This is the power that does not produce light.

Step 3: Find the useful power output. The useful power is the part of the input that is not wasted:

Step 4: Apply the efficiency formula. Efficiency is always useful output divided by total input:

This matches option B.

Key Takeaways

• Efficiency is a ratio of useful output to total input, never the other way round.
• In this question the total power is given directly (60 W), so the useful power has to be found by subtracting the wasted power from it.
• The units (watts) cancel in the ratio, so efficiency is a dimensionless number, often expressed as a fraction, decimal, or percentage.

Common Mistakes

• Using the wasted power as the useful output — option D uses $46/60$, which is the fraction of power wasted, not the efficiency.
• Adding the powers — option A uses $60 + 46 = 74$ in the denominator, which has no physical meaning here.
• Reversing the ratio — option C uses $14/46$, which is the ratio of useful to wasted power, not the efficiency.

Things to Be Careful About

• Always make sure the numerator is the useful output and the denominator is the total input.
• The efficiency is less than 1 because some energy is always wasted as heat.
• No unit conversion is needed here since both powers are already in watts.

Walkthrough

The question asks for the pressure exerted by a liquid on the bottom of a cylindrical container. Pressure is defined as the force applied perpendicular to a surface per unit area, given by the equation $P = F/A$.

First, identify the force $F$ acting on the base. For a liquid in a cylindrical container with vertical walls, the force on the base is simply the total weight of the liquid. Since the mass of the liquid is $m$, the weight is $F = mg$.

Next, determine the area $A$ of the base. The base is a circle with diameter $d$. The radius $r$ is half the diameter, so $r = d/2$. The area of a circle is $A = \pi r^2$. Substituting $r = d/2$ gives:

Now substitute $F$ and $A$ into the pressure formula:

Dividing by a fraction is the same as multiplying by its reciprocal, so:

This matches option C.

Alternatively, you can use the liquid pressure formula $P = \rho g h$. First find the density $\rho$:

Then substitute into $P = \rho g h$:

Both methods yield the same result.

Key Takeaways

• Pressure is force per unit area: $P = F/A$.
• For a liquid in a cylindrical container, the force on the base is the weight of the liquid ($mg$).
• The area of a circle can be expressed in terms of its diameter as $A = \pi d^2 / 4$.
• When given mass instead of density, you can either use $P = F/A$ directly or find the density first and use $P = \rho g h$.

Common Mistakes

• Forgetting the factor of 4: Using $A = \pi d^2$ instead of $A = \pi d^2 / 4$ leads to option A. This happens when a candidate confuses diameter with radius in the area formula.
• Including $h$ unnecessarily: Option D includes $h$ in the numerator. A candidate might mistakenly combine $P = F/A$ and $P = \rho g h$ without cancelling $h$, or think that depth always appears in the pressure formula regardless of whether mass is given.
• Confusing mass and density: Option B uses $mgh$ in the numerator, which is not a standard pressure expression. It likely comes from misremembering $P = \rho g h$ and substituting $m$ for $\rho$.

Things to Be Careful About

• Always check whether the given dimension is the radius or the diameter. The formula $A = \pi r^2$ requires the radius, so if the diameter $d$ is given, you must substitute $r = d/2$ and square the entire fraction, yielding $\pi d^2 / 4$.
• When dividing by a fraction, remember to multiply by the reciprocal. $\frac{mg}{\pi d^2 / 4}$ becomes $\frac{4mg}{\pi d^2}$, not $\frac{mg}{4\pi d^2}$.
• The depth $h$ is not needed in the final expression when the total mass $m$ is already known, because the weight $mg$ already accounts for the total mass acting on the base.

Walkthrough

1. Identify the principle: A mercury barometer measures atmospheric pressure by balancing it against the pressure exerted by a column of mercury. The pressure at the level of the mercury surface in the reservoir (inside the tube) is equal to the atmospheric pressure acting on the surface of the mercury in the reservoir (outside the tube).

2. Select the correct height ($h$): The pressure depends only on the vertical height of the liquid column above the free surface of the liquid in the reservoir. Looking at the diagram:

   • The height from the reservoir surface to the top of the mercury column is explicitly labelled as 760 mm.
   • Convert this to metres: $h = 760 \text{ mm} = 0.76 \text{ m}$.
   • The other dimensions are distractors:
     • 20 mm is the height of the vacuum space above the column.
     • 150 mm is the depth of the mercury in the reservoir.
     • 850 mm is the total height from the bottom of the reservoir to the top of the tube.

3. Apply the formula: The pressure exerted by a liquid column is given by $p = \rho g h$.

   • Density $\rho = 1.4 \times 10^4 \text{ kg/m}^3$.
   • Gravitational field strength $g = 9.8 \text{ N/kg}$ (using 9.8 gives the exact match to the options; using 10 gives ~106 kPa, which is close but 9.8 is standard for precise barometer calculations).
   • Height $h = 0.76 \text{ m}$.

4. Calculate:

   $$p = (1.4 \times 10^4) \times 9.8 \times 0.76$$ $$p = 14000 \times 9.8 \times 0.76 = 104272 \text{ Pa}$$

   Convert to kPa: $104272 \text{ Pa} \approx 104 \text{ kPa}$.

5. Check distractors (optional but good for verification):

   • Option A (2.7 kPa): Uses $h = 20 \text{ mm} = 0.02 \text{ m}$. $p = 14000 \times 9.8 \times 0.02 \approx 2744 \text{ Pa} = 2.7 \text{ kPa}$. Incorrect height.
   • Option B (21 kPa): Uses $h = 150 \text{ mm} = 0.15 \text{ m}$. $p = 14000 \times 9.8 \times 0.15 \approx 20580 \text{ Pa} = 21 \text{ kPa}$. Incorrect height.
   • Option D (117 kPa): Uses $h = 850 \text{ mm} = 0.85 \text{ m}$. $p = 14000 \times 9.8 \times 0.85 \approx 116620 \text{ Pa} = 117 \text{ kPa}$. Incorrect height.

Key Takeaways

• In a barometer, the atmospheric pressure is supported by the column of liquid above the reservoir level, not the total height of the tube or the depth of the reservoir.
• The formula $p = \rho g h$ applies to the pressure difference between two points in a fluid; here, the difference is between the top of the column (zero pressure, vacuum) and the reservoir surface (atmospheric pressure).
• Always convert units to SI (mm to m) before calculating pressure in Pascals.

Common Mistakes

• Using the wrong height: Candidates often pick a number from the diagram without understanding what it represents. Using the total tube height (850 mm) gives 117 kPa (Option D). Using the vacuum space (20 mm) gives 2.7 kPa (Option A). Using the reservoir depth (150 mm) gives 21 kPa (Option B).
• Forgetting to convert mm to m: If $h = 760$ is used directly, the result is in the thousands, leading to a wrong order of magnitude.
• Confusing mass and weight: Not applicable here directly, but ensuring $\rho$ is in kg/m³ is crucial.

Things to Be Careful About

• Diagram interpretation: The diagram is marked 'NOT TO SCALE'. Do not estimate lengths from the drawing; use the labelled values. The key is to identify which labelled dimension represents the height of the liquid column above the free surface.
• Value of $g$: The mark scheme implies $g = 9.8 \text{ N/kg}$ to get exactly 104 kPa. If you use $g = 10 \text{ N/kg}$, you get $106.4 \text{ kPa}$, which is closest to C but slightly off. Always check if the options require a specific precision or value of $g$.
• Units: The density is given as $1.4 \times 10^4 \text{ kg/m}^3$ (standard form). Ensure calculation handles the power of 10 correctly. $1.4 \times 10^4 = 14000$.
• Final unit: The options are in kPa. Remember that $1 \text{ kPa} = 1000 \text{ Pa}$.

Walkthrough

The question describes a gas being compressed by adding mass to a piston. We evaluate each statement in turn.

Statement C: The piston is in equilibrium. The downward forces are atmospheric pressure on top of the piston plus the weight of the mass. The upward force is the pressure of the trapped air. Since there is an extra downward force from the $1.0\ \text{kg}$ mass, the trapped air pressure must be greater than atmospheric pressure to balance it. Thus, C is false.

Statement D: The container is insulated, meaning no heat can escape. When the mass is placed on the piston, the gas is compressed. Work is done on the gas by the piston. This work increases the internal energy of the gas, which means its temperature increases, not decreases. Thus, D is false.

Statements A and B: These require looking at the pressure-volume relationship. Assuming the gas eventually returns to room temperature (or using Boyle's law $P_1V_1 = P_2V_2$ as a guide for the relationship), we can find the pressure increment $\Delta P$ caused by the $1.0\ \text{kg}$ mass.
Let $P_1 = P_{\text{atm}}$ and $V_1 = 100\ \text{cm}^3$. After adding the mass, $P_2 = P_{\text{atm}} + \Delta P$ and $V_2 = 90\ \text{cm}^3$.

Adding another $1.0\ \text{kg}$ adds another $\Delta P$ to the pressure, so $P_3 = P_{\text{atm}} + 2\Delta P = \frac{11}{9} P_{\text{atm}}$.
Using Boyle's law again to find the new volume $V_3$:

The volume decreases from $90\ \text{cm}^3$ to $81.8\ \text{cm}^3$, a change of $8.2\ \text{cm}^3$. This is less than $10\ \text{cm}^3$, so A is correct.

For B, adding $9.0\ \text{kg}$ more mass (total $10.0\ \text{kg}$) would increase the pressure by $10\Delta P = \frac{10}{9} P_{\text{atm}}$, giving a total pressure of $\frac{19}{9} P_{\text{atm}}$. The volume would be $\frac{900}{19} \approx 47.4\ \text{cm}^3$, not zero. Gases cannot be compressed to zero volume because the particles themselves occupy space and repel each other at short distances. Thus, B is false.

Key Takeaways

• Gas pressure inside a container with a movable piston must balance the external atmospheric pressure plus any additional pressure from masses on the piston.
• Compressing a gas in an insulated container does work on the gas, increasing its temperature.
• For a gas at constant temperature, pressure and volume are inversely proportional ($P \propto 1/V$). Equal increments in pressure produce progressively smaller decrements in volume.
• Gases cannot be compressed to zero volume.

Common Mistakes

• Assuming the pressure of the trapped air is less than atmospheric pressure because the volume decreased (confusing cause and effect; the added mass increases the pressure, which causes the volume to decrease).
• Thinking the temperature decreases during compression (forgetting that work done on the gas increases its internal energy, especially in an insulated system).
• Assuming a linear relationship between mass added and volume decreased (e.g., thinking $1\ \text{kg}$ gives $10\ \text{cm}^3$ decrease, so $10\ \text{kg}$ gives $100\ \text{cm}^3$ decrease to zero volume). Gas laws are non-linear ($P_1V_1 = P_2V_2$).

Things to Be Careful About

• The container is insulated, which is a key detail for evaluating statement D. If it were not insulated, the temperature might return to room temperature, but it would never decrease below room temperature during compression.
• When using $P_1V_1 = P_2V_2$, remember that $P_1$ is the total absolute pressure (atmospheric + pressure from the mass), not just the pressure from the mass. Atmospheric pressure is approximately $10^5\ \text{Pa}$, which is much larger than the pressure from a $1\ \text{kg}$ mass on a small piston, making the fractional pressure change small and the volume change non-linear.
• Always check the physical limits: volume cannot be negative or zero for a real gas.

Walkthrough

The melting point is the temperature at which a solid changes into a liquid. The boiling point is the temperature at which a liquid changes into a gas.

Here the melting point is $-17\,^\circ\text{C}$ and the boiling point is $117\,^\circ\text{C}$. This tells us: