Physics 9702/14 — May/June 2025

Background Concept

A “reasonable estimate” question is a Fermi calculation: you approximate a complicated quantity using simple, sensible orders of magnitude. Here the idea is:

$$N \approx \frac{\text{volume of handset}}{\text{volume per atom}}.$$

If an atom has a typical diameter of order $10^{-10}\ \text{m}$, then an order-of-magnitude estimate for the volume occupied by one atom is the cube of this length scale:

$$V_{\text{atom}} \sim (10^{-10}\ \text{m})^3 = 10^{-30}\ \text{m}^3.$$

This is not exact (atoms are not cubes and solids have empty space), but it is suitable for an order-of-magnitude choice among widely separated options.

Understanding the Question

You are told to estimate the number of atoms in a mobile phone handset by dividing:

• an approximate handset volume (from typical dimensions), by
• an approximate atomic volume (from a typical atomic size).

The multiple-choice options are powers of ten, so you only need the correct order of magnitude.

Approach

1. Estimate typical handset dimensions and hence its volume.
2. Convert that volume to $\text{m}^3$.
3. Use atomic diameter $\sim 10^{-10}\ \text{m}$ to estimate $V_{\text{atom}} \sim 10^{-30}\ \text{m}^3$.
4. Divide to get $N$ and match to the nearest option.

Step-by-Step Reasoning

1. A typical handset might be about $10\ \text{cm}$ by $5\ \text{cm}$ by $1\ \text{cm}$.

So the volume is

$$V_{\text{handset}} \sim 10 \times 5 \times 1\ \text{cm}^3 = 50\ \text{cm}^3.$$

1. Convert to $\text{m}^3$:

$$1\ \text{cm}^3 = (10^{-2}\ \text{m})^3 = 10^{-6}\ \text{m}^3$$

so

$$50\ \text{cm}^3 = 50 \times 10^{-6}\ \text{m}^3 = 5 \times 10^{-5}\ \text{m}^3.$$

1. Atomic volume estimate using a typical atomic diameter $d \sim 10^{-10}\ \text{m}$:

$$V_{\text{atom}} \sim d^3 \sim (10^{-10})^3 = 10^{-30}\ \text{m}^3.$$

1. Number of atoms:

$$N \sim \frac{V_{\text{handset}}}{V_{\text{atom}}} \sim \frac{5 \times 10^{-5}}{10^{-30}} = 5 \times 10^{25} \approx 10^{26}.$$

So the closest option is $10^{26}$.

Key Takeaways

• For order-of-magnitude estimates, use typical dimensions and typical microscopic length scales.
• Always convert units consistently before dividing.
• Cubing a length scale changes powers of ten dramatically: $(10^{-10})^3 = 10^{-30}$.

Common Mistakes

• Forgetting that $1\ \text{cm}^3 = 10^{-6}\ \text{m}^3$ (not $10^{-2}$ or $10^{-3}$).
• Using atomic radius instead of diameter without consistency (either is fine, but it changes the estimate by only a factor of about $8$, not by many orders of magnitude).
• Using $10^{-9}\ \text{m}$ for atomic size (that is closer to a small molecule or some lattice spacings; typical atomic diameters are around $10^{-10}\ \text{m}$).

Things to Be Careful About

• The options differ by many orders of magnitude, so you should not overthink factors like shape or packing; powers of ten dominate.
• Keep track of indices carefully when dividing powers of ten:

$$10^{-5} \div 10^{-30} = 10^{25}.$$

• If you estimate the handset volume slightly differently (e.g. $100\ \text{cm}^3$ instead of $50\ \text{cm}^3$), you still land near $10^{26}$, which is the point of the question.

Background Concept

Energy is measured in joules (J). Many different-looking units are equivalent to the joule because they are built from other definitions:

• Work done (energy transferred):

$$W = Fs$$

So $1\ \text{J} = 1\ \text{N m}$.

• Potential difference is energy per unit charge:

$$V = \frac{E}{Q}$$

So $1\ \text{V} = 1\ \text{J C}^{-1}$.

• Power is energy per unit time:

$$P = \frac{E}{t}$$

So $1\ \text{W} = 1\ \text{J s}^{-1}$.

A common trap is confusing force units (newtons) with energy units (joules). A newton is:

$$1\ \text{N} = 1\ \text{kg m s}^{-2}$$

Understanding the Question

You are given four possible units and asked which one is not equivalent to an energy unit (joule). So we check whether each option can be rewritten as $\text{J}$.

Approach

Take each option and use standard physics definitions to rewrite it in joules:

• Use $W = Fs$ for $\text{N m}$.
• Use $V = E/Q$ for $\text{V C}$.
• Use $P = E/t$ for $\text{W s}$.
• Recognise $\text{kg m s}^{-2}$ as a newton (force), which is not energy unless multiplied by distance.

Step-by-Step Reasoning

1. Option A: $\text{N m}$

From $W = Fs$, work (energy) = force $\times$ distance. So the unit is newton metre:

$$\text{N m} = \text{J}$$

So A is an energy unit.

1. Option B: $\text{V C}$

From $V = E/Q$, rearrange to $E = VQ$. Multiply units:

$$\text{V C} = \left(\frac{\text{J}}{\text{C}}\right)\text{C} = \text{J}$$

So B is an energy unit.

1. Option C: $\text{W s}$

From $P = E/t$, rearrange to $E = Pt$. Multiply units:

$$\text{W s} = \left(\frac{\text{J}}{\text{s}}\right)\text{s} = \text{J}$$

So C is an energy unit.

1. Option D: $\text{kg m s}^{-2}$

This is the base-unit form of the newton:

$$\text{kg m s}^{-2} = \text{N}$$

A newton is a unit of force, not energy. To become energy you would need to multiply by distance (metres) to get $\text{N m}$.

Therefore, D is the unit that is not equivalent to energy.

Key Takeaways

• Check unit equivalence using defining equations (e.g. $W=Fs$, $V=E/Q$, $P=E/t$).
• $\text{J}$ can appear as $\text{N m}$, $\text{V C}$, or $\text{W s}$.
• $\text{kg m s}^{-2}$ is $\text{N}$ (force), not energy.

Common Mistakes

• Thinking any derived SI unit must be energy: force ($\text{N}$) is not energy.
• Forgetting that $\text{V} = \text{J C}^{-1}$, so $\text{V C}$ is energy.
• Mixing up $\text{W}$ (power) with energy; you must multiply by time to get energy.

Things to Be Careful About

• Always reduce to base definitions: $\text{J} = \text{kg m}^2\text{ s}^{-2}$, while $\text{N} = \text{kg m s}^{-2}$ (missing one factor of $\text{m}$).
• Check whether a unit represents “rate of change” (like $\text{W} = \text{J s}^{-1}$) versus the quantity itself.

Background Concept

The wall thickness of a hollow cylindrical tube is the radial difference between the outer and inner radii.

If the external diameter is $D_\text{ext}$ and the internal diameter is $D_\text{int}$, then the external radius is $D_\text{ext}/2$ and the internal radius is $D_\text{int}/2$. So the wall thickness $t$ is

$$t = \frac{D_\text{ext}}{2} - \frac{D_\text{int}}{2} = \frac{D_\text{ext}-D_\text{int}}{2}.$$

For uncertainty propagation at AS level (worst-case method):

• when adding or subtracting quantities, add absolute uncertainties;
• when multiplying/dividing by a constant, the absolute uncertainty is multiplied/divided by the same constant.

Understanding the Question

You are given two caliper measurements:

• internal diameter: $(10.0 \pm 0.1)\ \text{mm}$
• external diameter: $(12.0 \pm 0.1)\ \text{mm}$

You must find the thickness of the glass wall (one side only), not the difference in diameters.

The options differ both in the thickness value (1.0 mm or 2.0 mm) and in the stated uncertainty (0.1 mm or 0.2 mm), so you must compute both correctly.

Approach

1. Use geometry: thickness is half the difference between external and internal diameters.
2. Find the best estimate of $t$ from the best estimates of the diameters.
3. Propagate uncertainty:
   • first find uncertainty in $(D_\text{ext}-D_\text{int})$ by adding absolute uncertainties,
   • then divide that uncertainty by 2 (because thickness is half the difference).
4. Match to the closest option.

Step-by-Step Reasoning

1. Compute thickness:

$$t = \frac{D_\text{ext}-D_\text{int}}{2} = \frac{12.0 - 10.0}{2} = \frac{2.0}{2} = 1.0\ \text{mm}.$$

1. Uncertainty in the diameter difference:

Because $D_\text{ext}$ and $D_\text{int}$ are subtracted, the worst-case absolute uncertainty is

$$\Delta(D_\text{ext}-D_\text{int}) = \Delta D_\text{ext} + \Delta D_\text{int} = 0.1 + 0.1 = 0.2\ \text{mm}.$$

1. Uncertainty in thickness:

Since $t$ is the difference divided by 2,

$$\Delta t = \frac{0.2}{2} = 0.1\ \text{mm}.$$

So the thickness is

$$t = (1.0 \pm 0.1)\ \text{mm},$$

which corresponds to Option A.

Key Takeaways

• For a tube, wall thickness is half the difference between external and internal diameters: $t=(D_\text{ext}-D_\text{int})/2$.
• For subtraction, add absolute uncertainties.
• Scaling a quantity by a constant scales its absolute uncertainty by the same constant.

Common Mistakes

• Using $D_\text{ext}-D_\text{int}$ as the thickness (giving $2.0\ \text{mm}$ instead of $1.0\ \text{mm}$).
• Quoting uncertainty of $\pm 0.2\ \text{mm}$ for the thickness (forgetting to divide the uncertainty by 2 as well).
• Mixing percentage uncertainties here (not needed; absolute is simplest for sums/differences).

Things to Be Careful About

• The thickness is on one side of the tube, so you must divide the diameter difference by 2.
• Keep units consistent (all in mm here, so no conversion needed).
• In MCQs, the uncertainty often distinguishes the correct option: propagate it through every operation, including the final division by 2.

Background Concept

Tension forces act along the line of the elastic. When two equal forces act symmetrically about a direction, their components perpendicular to that direction cancel and their components along that direction add.

To find the force in a particular direction (here, direction $P$), resolve each force into components using trigonometry:

• component along the chosen direction $= F\cos\theta$ (if $\theta$ is the angle between the force and that direction)
• component perpendicular $= F\sin\theta$

Understanding the Question

Two fixed pins $Y$ and $Z$ are vertically separated by $80\ \text{mm}$, and an elastic band runs from $Y$ to a trolley pin $X$ and then to $Z$. The arrangement is symmetric, with $X$ at the centre of the elastic.

Each sloping section $XY$ and $XZ$ has length $50\ \text{mm}$. The horizontal distance from $X$ to the vertical line through $Y$ and $Z$ is $30\ \text{mm}$. The tension in the elastic is $4.0\ \text{N}$ in each sloping section.

We want the resultant force on the trolley in direction $P$ (horizontal, at right angles to $YZ$) when released.

Approach

1. Draw the forces acting on pin $X$: two equal tension forces along $XY$ and $XZ$.
2. Use symmetry: vertical components cancel (one up, one down), while horizontal components (towards the right) add.
3. Find the horizontal component of one tension using the given triangle dimensions, then double it.

Step-by-Step Reasoning

1. There are two tension forces on $X$, each of magnitude $T = 4.0\ \text{N}$, one directed along $XY$ and one along $XZ$.

2. Consider one side, say $XY$. The geometry forms a right triangle:

• hypotenuse $XY = 50\ \text{mm}$
• horizontal adjacent side $= 30\ \text{mm}$
• vertical opposite side $= 40\ \text{mm}$ (half of $80\ \text{mm}$), confirming consistency since $30^2 + 40^2 = 50^2$.

Let $\theta$ be the angle between the tension and the horizontal direction $P$. Then

$$\cos\theta = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{30}{50} = 0.60$$

1. Horizontal component of one tension is

$$T\cos\theta = 4.0 \times 0.60 = 2.4\ \text{N}$$

1. The second tension has the same horizontal component (also to the right). So the total force in direction $P$ is

$$F = 2 \times 2.4 = 4.8\ \text{N}$$

This matches option C.

Key Takeaways

• Tension acts along the elastic, not horizontally/vertically by default.
• In symmetric situations, components perpendicular to the symmetry direction cancel.
• Use triangle ratios from the diagram: here $\cos\theta = 30/50$.
• Resultant in the required direction is the sum of components in that direction.

Common Mistakes

• Using $\sin\theta$ instead of $\cos\theta$ for the horizontal component (depends on which angle you define).
• Adding the full tensions $4.0 + 4.0 = 8.0\ \text{N}$ without resolving into components.
• Forgetting there are two elastic segments, so only taking $2.4\ \text{N}$ instead of doubling.
• Using $80\ \text{mm}$ instead of $40\ \text{mm}$ as the vertical side for one segment.

Things to Be Careful About

• Identify clearly which side of the triangle is the hypotenuse (it is the elastic length $50\ \text{mm}$).
• The required force is specifically in direction $P$, so only horizontal components matter.
• Check symmetry: one tension pulls slightly upward, the other slightly downward, so vertical components cancel exactly.

Background Concept

Distance is a scalar quantity representing the total path length covered by an object. Unlike displacement, which is a vector and can decrease if an object returns towards its starting point, distance travelled never decreases over time. On a distance-time graph, the gradient represents the speed of the object. A steeper gradient indicates a higher speed, and a horizontal line indicates the object is stationary.

Understanding the Question

The student's journey has three stages:

1. Cycling uphill from home to a shop: takes 10 minutes. Distance increases from 0 to some value (let's say 1 km based on the graphs). Speed is $v$.
2. Spending time in the shop: takes 5 minutes. The student is stationary, so distance travelled remains constant. Time goes from 10 to 15 minutes.
3. Cycling home downhill: speed is $2v$. Since distance = speed \times time, and the speed is doubled, the time taken to cover the same 1 km distance is halved (from 10 min to 5 min). The total time is $15 + 5 = 20$ minutes. The total distance travelled is $1 \text{ km} + 1 \text{ km} = 2 \text{ km}$. Distance continues to increase, it does not decrease.

Approach

First, eliminate graphs where distance decreases (A and B), as distance travelled is cumulative and cannot decrease. Second, check the time intervals and gradients for the remaining graphs (C and D) to see which matches the 10-minute uphill ride, 5-minute stop, and 5-minute downhill ride at twice the speed.

Step-by-Step Reasoning

• Stage 1 (0 to 10 min): The student cycles uphill for 10 minutes. The distance increases linearly. Looking at the graphs, both C and D show a linear increase, but C reaches 1 km at 5 minutes, while D reaches 1 km at 10 minutes. Since the ride takes 10 minutes, D is correct and C is incorrect.
• Stage 2 (10 to 15 min): The student spends 5 minutes in the shop. The distance travelled does not change, so the graph must be horizontal. Graph D is horizontal from $t=10$ to $t=15$ min at a distance of 1 km.
• Stage 3 (15 to 20 min): The student cycles home at twice the initial speed. Since distance = speed \times time, and the speed is doubled, the time taken to cover the same 1 km distance is halved (from 10 min to 5 min). The total time is $15 + 5 = 20$ minutes. The total distance travelled is $1 \text{ km} + 1 \text{ km} = 2 \text{ km}$. The gradient of this section is twice as steep as the first section. Graph D shows a steeper linear increase from 1 km at $t=15$ min to 2 km at $t=20$ min.
• Graphs A and B show the distance decreasing back to 0, which would be correct for a displacement-time graph, but the question specifically asks for distance travelled.

Key Takeaways

• Distance is a scalar and always increases or stays constant; it never decreases. Displacement can decrease.
• The gradient of a distance-time graph gives the speed. A steeper gradient means a higher speed.
• Always read the question carefully to ensure you are plotting the correct quantity (distance vs. displacement).

Common Mistakes

• Choosing graph A or B: confusing distance travelled with displacement. If the question asked for displacement from home, A or B might be considered (though A has the wrong times). Distance travelled is cumulative.
• Choosing graph C: misreading the time axis or not noticing that the first leg takes 10 minutes, not 5 minutes.

Things to Be Careful About

• The exact wording "distance travelled" vs "displacement". This is a classic trap in kinematics questions.
• Reading the axes correctly: the time axis is in minutes, and the distance axis is in km. Ensure the slopes and time intervals match the problem description.

Background Concept

When air resistance is negligible, an object in free fall near the Earth’s surface experiences a constant downward acceleration equal to the gravitational field strength $g \approx 9.81\ \text{m s}^{-2}$. This is an example of uniformly accelerated motion, so we can use the SUVAT equations.

One useful equation that connects displacement $s$, acceleration $a$, and time $t$ (without needing final velocity) is:

$$s = ut + \frac{1}{2}at^2$$

where:

• $s$ is displacement in $\text{m}$,
• $u$ is initial velocity in $\text{m s}^{-1}$,
• $a$ is acceleration in $\text{m s}^{-2}$,
• $t$ is time in $\text{s}$.

Understanding the Question

A ball is released (so initial speed is zero) from a height of $12\ \text{m}$ above the ground. The question asks for the time taken to reach the ground. Since the ball starts from rest and accelerates uniformly under gravity, we model the vertical motion with constant acceleration.

Known:

• $s = 12\ \text{m}$ downward
• $u = 0\ \text{m s}^{-1}$
• $a = g = 9.81\ \text{m s}^{-2}$
  Unknown:
• $t$

Approach

Choose a SUVAT equation that includes $s$, $u$, $a$, and $t$. Since we are not asked about final velocity $v$, use:

$$s = ut + \frac{1}{2}at^2$$

Substitute $u=0$ and $a=g$, then rearrange to solve for $t$, and compare the calculated value with the options.

Step-by-Step Reasoning

Start with:

$$s = ut + \frac{1}{2}at^2$$

Here $u=0$, $a=g$, so:

$$s = \frac{1}{2}gt^2$$

Rearrange for $t$:

$$t^2 = \frac{2s}{g}$$ $$t = \sqrt{\frac{2s}{g}}$$

Substitute $s=12\ \text{m}$ and $g=9.81\ \text{m s}^{-2}$:

$$t = \sqrt{\frac{2 \times 12}{9.81}} = \sqrt{\frac{24}{9.81}} = \sqrt{2.447} = 1.56\ \text{s}$$

Rounded to match the options, $t \approx 1.6\ \text{s}$, which corresponds to option C.

Key Takeaways

• Negligible air resistance means constant acceleration $g$ (uniformly accelerated motion).
• “Released from rest” implies $u=0$.
• For free-fall time from a known height, a quick route is $t = \sqrt{2s/g}$.

Common Mistakes

• Using $s = gt^2$ instead of $s = \tfrac{1}{2}gt^2$ (missing the factor $\tfrac{1}{2}$).
• Taking $g = 10\ \text{m s}^{-2}$ and then not matching the closest option carefully.
• Mixing up distance and displacement signs; if you choose upward as positive, then $s=-12\ \text{m}$ and $a=-g$ (the negatives cancel if done consistently).

Things to Be Careful About

• Keep units consistent: $s$ in $\text{m}$ and $g$ in $\text{m s}^{-2}$ gives $t$ in seconds.
• Rounding: the computed value $1.56\ \text{s}$ rounds to $1.6\ \text{s}$ (2 s.f.), matching the provided choices.

Background Concept

Force is connected to momentum by Newton’s second law in its most general form:

$$F = \frac{\Delta p}{\Delta t}$$

where $p = mv$ is momentum. This means force depends not only on how much momentum changes, but also on how quickly it changes.

Other useful related definitions:

• Moment (torque) about a pivot:

$$\text{moment} = F \times (\text{perpendicular distance})$$

• Work done by a force along its line of action:

$$W = F \times s$$

(where $s$ is displacement in the direction of the force).

Understanding the Question

You are given four “word equations” for force. Three of them match standard definitions/formulae. One is incorrect. The task is to identify the incorrect one.

Approach

Check each option against the standard physics relationship:

1. For momentum, ensure “rate of change” (includes time) is present.
2. For $F=ma$, moment, and work, check rearrangements are consistent.

Step-by-Step Reasoning

• Option A: “force = change of momentum”

Correct physics is:

$$F = \frac{\Delta p}{\Delta t}$$

Option A is missing division by time, so it is incorrect.

• Option B: “force = mass $\times$ acceleration”

This is Newton’s second law for constant mass:

$$F = ma$$

So B is correct.

• Option C: “force = moment / perpendicular distance from the pivot”

From

$$\text{moment} = F \times d$$

rearrange to

$$F = \frac{\text{moment}}{d}$$

So C is correct.

• Option D: “force = work done / displacement in the direction of the force”

From

$$W = Fs$$

rearrange to

$$F = \frac{W}{s}$$

So D is correct.

Therefore the only incorrect word equation is A.

Key Takeaways

• Force equals the rate of change of momentum, not just the change.
• Many MCQs test whether you remember the correct wording (“rate of change”) and the implied time factor.
• Rearranging standard definitions (moment, work) can quickly verify options.

Common Mistakes

• Forgetting the time term and writing $F=\Delta p$ instead of $F=\Delta p/\Delta t$.
• Thinking “change of momentum” automatically implies “per second” (it does not unless explicitly stated).
• Confusing moment (torque) with force without including the perpendicular distance.

Things to Be Careful About

• The exact phrase matters: “rate of change of momentum” is the correct definition.
• For momentum questions, always check whether time (or an equivalent rate) appears.
• Ensure the distance in moments is perpendicular to the line of action of the force, as stated in option C.

Background Concept

Terminal velocity occurs when an object falls through a fluid (air) and the resistive (drag) force has increased to equal the weight.

• Weight: $W = mg$ acts downward and is (approximately) constant.
• Drag force: $D$ acts upward (opposes motion) and increases with speed (often roughly proportional to $v$ or $v^2$).
• Newton's second law: resultant force $F_{\text{res}} = ma$. So if $F_{\text{res}} = 0$, then $a = 0$ and the object moves at constant velocity.

When conditions change (e.g. opening a parachute), the drag changes, so the resultant force changes and the object accelerates (which may mean it slows down if the acceleration is opposite to the direction of motion).

Understanding the Question

The skydiver is initially falling at terminal velocity, so they are moving downward at constant speed.

The parachute then opens fully. For the first few seconds after this, we must decide:

• whether the speed increases or decreases,
• whether the acceleration is uniform (constant) or non-uniform (changing).

Approach

1. Use terminal velocity to state the initial force balance.
2. Immediately after the parachute opens, the drag force becomes much larger than before, so compare drag with weight to find the direction of the resultant force and hence the direction of acceleration.
3. Decide whether the speed is increasing or decreasing by comparing the direction of acceleration with the direction of motion.
4. Decide whether the acceleration is uniform by noting that drag depends on speed, and the speed is changing.

Step-by-Step Reasoning

1. At terminal velocity (before opening):
   $W = D$ so $F_{\text{res}} = 0$.

   Therefore:

   $$a = 0$$

   The skydiver falls downward at constant speed.

2. Just after the parachute opens fully:
   The parachute greatly increases the area, so for the same downward speed the drag force increases suddenly.

   So initially after opening:

   $$D > W$$

   Drag acts upward, weight acts downward, so the resultant force is upward.

3. Link to acceleration:
   Using $F_{\text{res}} = ma$, an upward resultant force means an upward acceleration.

   But the velocity is still downward at that moment, so the upward acceleration acts opposite to the motion. That means the skydiver’s downward speed decreases.

4. Uniform or non-uniform acceleration?
   As the skydiver slows down, their speed $v$ decreases, so the drag $D$ decreases (since $D$ depends on $v$).

   Therefore the resultant force $D - W$ decreases with time, so the acceleration changes with time. Hence the acceleration is non-uniform.

So the motion is: falling downward with decreasing speed and non-uniform acceleration → option B.

Key Takeaways

• Terminal velocity means $D = W$ and $a = 0$.
• After the parachute opens, drag becomes larger than weight, giving acceleration upward while motion is still downward, so speed decreases.
• Because drag depends on speed and the speed is changing, the acceleration is not constant (non-uniform).

Common Mistakes

• Saying the skydiver has “uniform deceleration”: drag changes as speed changes, so acceleration is not constant.
• Thinking acceleration must be downward because the skydiver is falling: acceleration depends on the resultant force, not the direction of motion.
• Confusing “slowing down” with “negative acceleration” without defining a sign convention; the safer statement is “acceleration is upward” or “opposite to the motion”.

Things to Be Careful About

• The question asks about the first few seconds after opening: initially $D$ is very large, but it quickly decreases as the skydiver slows.
• “Decreasing speed” does not mean the velocity is upward; the velocity is still downward until (much later) it could change direction (which does not happen here).
• “Non-uniform acceleration” here comes from the speed-dependence of drag, not from any change in weight.

Background Concept

In a collision on a straight line (1D), the total linear momentum is conserved provided external forces are negligible (an air track makes friction very small).

For a perfectly elastic collision, kinetic energy is also conserved. A very convenient equivalent statement is the coefficient of restitution:

$$e = \frac{\text{relative speed of separation}}{\text{relative speed of approach}}$$

For a perfectly elastic collision, $e=1$, so

$$\text{relative speed of separation} = \text{relative speed of approach}.$$

With objects $P$ and $Q$ in 1D, this becomes

$$v_Q - v_P = u_P - u_Q$$

provided you use a consistent sign convention.

Understanding the Question

Two gliders move towards each other:

• $m_P = 0.30\,\text{kg}$ with speed $1.2\,\text{m s}^{-1}$ to the right.
• $m_Q = 0.60\,\text{kg}$ with speed $1.8\,\text{m s}^{-1}$ to the left.

After a perfectly elastic head-on collision, we must find their speeds (magnitudes of velocities) and match them to options A–D.

Approach

1. Choose a positive direction (take right as positive).
2. Write the momentum conservation equation in that sign convention.
3. Write the elastic-collision condition using $e=1$ (relative speeds).
4. Solve the two simultaneous equations for $v_P$ and $v_Q$.
5. Convert to speeds by taking absolute values and compare to the options.

Step-by-Step Reasoning

1) Assign signs
Take right as positive.

• $u_P = +1.2\,\text{m s}^{-1}$.
• $u_Q = -1.8\,\text{m s}^{-1}$ (because it moves left).

2) Conservation of momentum
Total initial momentum equals total final momentum:

$$m_Pu_P + m_Qu_Q = m_Pv_P + m_Qv_Q.$$

Substitute values:

$$0.30(1.2) + 0.60(-1.8) = 0.30v_P + 0.60v_Q.$$

Compute the left-hand side:

$$0.36 - 1.08 = -0.72$$

so

$$-0.72 = 0.30v_P + 0.60v_Q.$$

3) Elastic collision condition ($e=1$)
Relative approach speed is $u_P - u_Q$ (with the chosen signs):

$$u_P - u_Q = 1.2 - (-1.8) = 3.0.$$

So relative separation speed must also be $3.0$:

$$v_Q - v_P = 3.0.$$

Rearrange:

$$v_Q = v_P + 3.0.$$

4) Solve simultaneously
Substitute $v_Q = v_P + 3.0$ into momentum:

$$-0.72 = 0.30v_P + 0.60(v_P+3.0).$$

Expand:

$$-0.72 = 0.30v_P + 0.60v_P + 1.8 = 0.90v_P + 1.8.$$

So

$$0.90v_P = -2.52 \Rightarrow v_P = -2.8\,\text{m s}^{-1}.$$

Then

$$v_Q = v_P + 3.0 = -2.8 + 3.0 = +0.2\,\text{m s}^{-1}.$$

5) Convert to speeds and match option
Speeds are magnitudes:

• Speed of $P$ is $|v_P| = 2.8\,\text{m s}^{-1}$.
• Speed of $Q$ is $|v_Q| = 0.2\,\text{m s}^{-1}$.

This matches option C.

Key Takeaways

• In 1D collisions, pick a sign convention first; velocities towards the negative direction must be negative.
• Use momentum conservation plus either kinetic energy conservation or the $e=1$ relative-speed condition for perfectly elastic collisions.
• Final speeds are magnitudes of the final velocities.

Common Mistakes

• Treating both initial velocities as positive because the question gives “speeds”. You must convert speeds to signed velocities.
• Using $v_P - v_Q = u_P - u_Q$ (wrong order): the correct elastic condition is $v_Q - v_P = u_P - u_Q$ for consistent signs.
• Forgetting the question asks for speeds, so quoting $v_P=-2.8\,\text{m s}^{-1}$ as the speed.

Things to Be Careful About

• The heavier glider does not necessarily reverse direction; here $Q$ ends up moving right slowly.
• Keep units consistent (all are already in SI units here).
• When matching to multiple-choice options, compare speeds (positive values) not signed velocities.

Background Concept

A couple consists of two forces that are:

• equal in magnitude,
• opposite in direction,
• parallel, and
• separated by a perpendicular distance.

Because the forces are equal and opposite, their resultant (net) force is zero. However, because they act along different lines (they are separated), they produce a non-zero turning effect (moment).

The moment of a couple is

$$\tau = Fd$$

where $F$ is the magnitude of one of the forces and $d$ is the perpendicular separation between their lines of action.

Understanding the Question

The options ask what a couple can produce:

• translational motion (requires a non-zero resultant force),
• rotational motion (requires a non-zero resultant moment/torque).

So we decide whether a couple has net force, net moment, or both.

Approach

1. Use the definition of a couple to evaluate the net force.
2. Use the separated lines of action to evaluate the net moment.
3. Match these conclusions to the statements A–D.

Step-by-Step Reasoning

• In a couple, the two forces have equal magnitude and opposite direction.
  Therefore the vector sum of forces is

$$\sum \vec{F} = 0$$

So the object does not accelerate translationally due to the couple alone.

• Even though the net force is zero, the forces act at different positions. Each force produces a moment about (almost) any point, and these moments act in the same rotational sense, so they add.

  Hence

$$\sum \tau \neq 0$$

So a couple produces rotation.

Therefore, the correct statement is: it acts to produce rotation only.

Key Takeaways

• Translational motion needs $\sum \vec{F} \neq 0$.
• Rotational motion needs $\sum \tau \neq 0$.
• A couple gives $\sum \vec{F} = 0$ but $\sum \tau \neq 0$, so it causes rotation without translation.

Common Mistakes

• Thinking a couple must cause translation because “forces are acting”: the forces cancel in pairs, so the net force is zero.
• Confusing a single force causing a moment (about some pivot) with a couple: a single force generally gives both net force and a moment, whereas a couple has zero net force.

Things to Be Careful About

• The key feature is separation of lines of action: equal and opposite forces at the same line of action would cancel both force and moment.
• A couple’s moment $\tau = Fd$ is independent of the choice of pivot point (unlike the moment of a single force).

Background Concept

If an object is in equilibrium, both translational and rotational equilibrium apply. For a particle / for the condition of no net translation, the key statement is:

$$\sum \vec{F} = \vec{0}$$

For three forces acting in equilibrium, this means that if you draw the three force vectors head-to-tail in sequence, they must form a closed triangle (a “vector triangle”). Each side of the triangle represents one force, with its arrow showing the force direction.

Tension forces always act along the string, pulling away from the object towards the support.

Understanding the Question

A picture frame is hanging symmetrically from two segments of string that meet at a pin above. The frame is at rest, so it is in equilibrium.

So the frame has exactly three forces:

• its weight $W$ acting vertically downward through its centre of mass,
• a tension at the left attachment point directed along the left string up towards the pin,
• a tension at the right attachment point directed along the right string up towards the pin.

The question asks which option shows the correct vector triangle for these three forces.

Approach

1. Draw (or imagine) a free-body diagram of the frame and mark the directions of the three forces.
2. Use $\sum \vec{F} = \vec{0}$: the three vectors must close to form a triangle.
3. Decide what each side must look like:
   • $W$ must be vertical and downward.
   • the two tensions must both point upward (one up-left, one up-right), because each string pulls the frame towards the pin.
4. Compare with the options and select the only triangle with these directions and with head-to-tail closure.

Step-by-Step Reasoning

1. The weight acts straight down: draw $W$ vertically downward.
2. The left string slopes up towards the pin, so the tension on the frame at the left corner acts along the string towards the pin: up and towards the middle (up-right).
3. Similarly, the right tension acts up and towards the middle (up-left).

Now form the vector triangle (head-to-tail):

• Place one tension vector.
• From its head, place the other tension vector.
• The resultant of the two tensions is vertically upward (horizontal components cancel by symmetry).
• To make the sum zero, the third vector (weight) must be vertically downward, closing the triangle.

So the correct option must show:

• a downward vertical vector for $W$,
• two upward sloping vectors for the tensions.

Among C and D (the only ones with a downward vertical arrow), only D has both sloping arrows directed upwards in the correct sense to close the triangle.

Key Takeaways

• Equilibrium implies $\sum \vec{F} = \vec{0}$.
• For three forces in equilibrium, the vectors form a closed head-to-tail triangle.
• Tension acts along the string, pulling the object towards the support.

Common Mistakes

• Drawing the tension direction the wrong way (tension does not push; it pulls along the string).
• Using an upward weight vector (weight must be vertically downward).
• Drawing vectors tail-to-tail and calling it a “triangle” (the triangle must be head-to-tail to represent addition).

Things to Be Careful About

• The vector triangle is about directions and closure, not about where forces are applied on the object.
• Ensure the vertical side corresponds to $W$ (downward), not to the resultant of tensions.
• In a symmetric situation the tensions are equal, but you don’t need lengths here—only correct directions and closure to match the options.

Background Concept

Pressure is defined as force per unit area:

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

For an object resting on a surface, the force it exerts on the surface due to its weight is essentially its weight $W = mg$ (assuming it is at rest and there are no extra vertical forces).

Density links mass and volume:

$$\rho = \frac{m}{V} \quad \Rightarrow \quad m = \rho V$$

For a cuboid, the volume is the product of its three side lengths. The contact area depends on which face is on the ground. Since $p = F/A$, for a fixed force $F$ the pressure is largest when the contact area $A$ is smallest.

Understanding the Question

You are given a metal cuboid with density $2700\ \text{kg m}^{-3}$ and dimensions $1.0\ \text{m}$, $2.0\ \text{m}$, and $4.0\ \text{m}$. The cuboid can rest on any face, so the contact area can be $1.0\times 2.0$, $1.0\times 4.0$, or $2.0\times 4.0$ (each occurs twice among the six faces).

You must find the largest pressure due to weight, so you need:

1. the weight (same in all orientations),
2. the smallest possible contact area.

Approach

1. Calculate the cuboid volume $V$ from its dimensions.
2. Use $m = \rho V$ to find the mass.
3. Use $W = mg$ to find the weight.
4. List the possible face areas and choose the smallest.
5. Compute $p_{\max} = W/A_{\min}$ and match to the closest option.

Step-by-Step Reasoning

1. Volume

$$V = (1.0)(2.0)(4.0) = 8.0\ \text{m}^3$$

1. Mass from density

$$m = \rho V = 2700 \times 8.0 = 21600\ \text{kg}$$

1. Weight (taking $g \approx 9.81\ \text{m s}^{-2}$)

$$W = mg = 21600 \times 9.81 \approx 2.12 \times 10^5\ \text{N}$$

1. Smallest contact area
   Possible face areas:

$$1.0 \times 2.0 = 2.0\ \text{m}^2,\quad 1.0 \times 4.0 = 4.0\ \text{m}^2,\quad 2.0 \times 4.0 = 8.0\ \text{m}^2$$

So

$$A_{\min} = 2.0\ \text{m}^2$$

1. Maximum pressure

$$p_{\max} = \frac{W}{A_{\min}} = \frac{2.12 \times 10^5}{2.0} = 1.06 \times 10^5\ \text{Pa}$$

Convert to kPa:

$$1.06 \times 10^5\ \text{Pa} = 106\ \text{kPa} \approx 110\ \text{kPa}$$

This corresponds to option D.

Key Takeaways

• Use $p = F/A$ for pressure.
• For an object at rest on a surface, $F$ is its weight $W = mg$.
• For a fixed weight, maximum pressure occurs with the minimum contact area.
• Use $m = \rho V$ to connect density to weight.

Common Mistakes

• Using the largest face area instead of the smallest (this gives the minimum pressure).
• Forgetting to convert from Pa to kPa (or converting incorrectly: $1\ \text{kPa} = 10^3\ \text{Pa}$).
• Calculating volume incorrectly (e.g. adding dimensions instead of multiplying).
• Using density as if it were weight density (mixing up $\rho$ with $\rho g$).

Things to Be Careful About

• Keep units consistent: density is in $\text{kg m}^{-3}$ so volume must be in $\text{m}^3$.
• The cuboid has six faces but only three distinct areas (each area occurs twice).
• Your final option should match the nearest given value; $106\ \text{kPa}$ rounds to $110\ \text{kPa}$ in the options.

Background Concept

A floating object is in vertical equilibrium, so the resultant vertical force is zero. The two key forces are:

• Weight of the object: $W = mg = \rho_i V_{\text{total}} g$
• Upthrust (buoyant force): equal to the weight of displaced fluid, $U = \rho_w V_{\text{displaced}} g$

For an object that floats (not accelerating up or down):

$$U = W$$

This immediately links the fraction submerged to the ratio of densities.

Understanding the Question

We have a cylindrical iceberg of total height $H$ and uniform cross-sectional area $A$. A portion of height $h$ is above the sea-water surface, so the submerged height is $H - h$.

We are given the densities: ice $\rho_i$ and sea water $\rho_w$, and we must choose the expression for $h$.

Approach

1. Let the submerged height be $x$ (so $x = H-h$).
2. Write expressions for weight of the iceberg and the upthrust in terms of densities and volumes.
3. Use equilibrium $\rho_w g V_{\text{sub}} = \rho_i g V_{\text{total}}$.
4. Solve for $x$ and then calculate $h = H-x$.

Step-by-Step Reasoning

Let cross-sectional area be $A$.

• Total volume of iceberg:

$$V_{\text{total}} = AH$$

• Submerged volume (displaced water volume):

$$V_{\text{sub}} = Ax$$

Weight of iceberg:

$$W = \rho_i (AH) g$$

Upthrust:

$$U = \rho_w (Ax) g$$

Floating equilibrium gives:

$$\rho_w (Ax) g = \rho_i (AH) g$$

Cancel $A$ and $g$:

\rho_w x = \rho_i H n$$ So

x = \frac{\rho_i}{\rho_w}H

But $h$ is the height above water:

h = H - x = H - \frac{\rho_i}{\rho_w}H = \left(1-\frac{\rho_i}{\rho_w}\right)H

This matches option **A**. ## Key Takeaways - For a floating object: $\rho_w V_{\text{displaced}} = \rho_{\text{object}} V_{\text{object}}$. - The fraction submerged is $\dfrac{V_{\text{displaced}}}{V_{\text{object}}} = \dfrac{\rho_{\text{object}}}{\rho_w}$. - The height above the surface is total height minus submerged height. ## Common Mistakes - Using $\rho_i/\rho_w$ as the *height above water* instead of the *submerged fraction*. - Writing $U = \rho_i V_{\text{sub}} g$ (wrong fluid density in the upthrust). - Choosing an expression that makes $h$ negative (e.g. option B) even though the iceberg is clearly partly above water. ## Things to Be Careful About - Upthrust depends on the density of the surrounding fluid ($\rho_w$), not the object. - The displaced volume is the submerged volume (here $Ax$), not the total volume. - A quick reasonableness check: since typically $\rho_i < \rho_w$, we expect $0<h<H$; option A satisfies this.

Background Concept

Work done on an object by a force is the energy transferred when a force causes a displacement.

For a constant force, the work done is

$$W = F s \cos\theta$$

where:

• $F$ is the magnitude of the force,
• $s$ is the displacement,
• $\theta$ is the angle between the force and the displacement.

So, if there is no displacement, the work done is zero. Also, if the force is always perpendicular to the displacement ($\theta = 90^\circ$), then $\cos\theta = 0$ and the work done is zero.

Understanding the Question

We must pick the situation where energy is transferred to the object by a force doing work, i.e. where there is a displacement in the direction of a force acting on the object.

Approach

Check each option for:

1. whether the object moves (non-zero displacement), and
2. whether there is a force component along the motion.

Step-by-Step Reasoning

A: Object slides at constant velocity on a horizontal frictionless surface in a vacuum.

• No resistive forces; weight and normal reaction are vertical and cancel.
• No horizontal force is needed to maintain constant velocity.
• Hence no force component along the displacement, so no work is done on the object.

B: Held at arm’s length at fixed height.

• Displacement is zero.
• So $s=0$ and $W=0$.

C: Pushed up a frictionless ramp.

• The object moves up the ramp, so $s \neq 0$.
• The applied push has a component along the ramp in the direction of motion.
• Therefore $W = Fs\cos\theta > 0$ and work is done on the object.
• This work appears mainly as an increase in gravitational potential energy ($\Delta E_p = mg\Delta h$).

D: Stationary object floats partially submerged.

• Displacement is zero.
• So $W=0$.

Therefore the correct option is C.

Key Takeaways

• Work is done only when a force causes displacement along (or partly along) its direction.
• Stationary situations (no displacement) imply zero work done on the object.
• Moving an object upward typically involves work against weight, increasing gravitational potential energy.

Common Mistakes

• Thinking that “a force exists” automatically means work is done (e.g. weight and upthrust exist in D, but no displacement so $W=0$).
• Thinking that constant velocity implies work must be done (in A, with no friction/drag, no driving force is needed).
• Confusing work done by a person with work done on the object (in B the person may get tired, but physically the object has no displacement so no work is done on it).

Things to Be Careful About

• Work depends on the component of force along the displacement: if the force is perpendicular to the motion, work is zero.
• Always check whether the object’s position changes; “fixed height” or “stationary” means $s=0$ immediately.