Physics 9702/12 — May/June 2024

Background Concept

A physical quantity describes something measurable and is usually stated with a magnitude and a unit. Quantities come in two main types:

• Scalar: described completely by magnitude only (with a unit).
• Vector: requires magnitude and direction to be fully described.

Examples:

• Scalars: mass, time, temperature, energy, speed.
• Vectors: displacement, velocity, acceleration, force.

Understanding the Question

The question asks for the definition of a scalar quantity. You must choose which option matches “magnitude only” (i.e. no direction).

Approach

Recall the scalar definition and compare it against each option. Reject statements that describe vectors or that are physically impossible (e.g. “no unit” or “no magnitude”).

Step-by-Step Reasoning

• Option A: “represented as two perpendicular components” describes a vector (vectors can be resolved into components). Not a scalar.
• Option B: “does not require a unit” is false: scalars still have units (e.g. mass in kg, time in s).
• Option C: “without a direction” matches the defining feature of a scalar (magnitude only).
• Option D: “without a magnitude” cannot describe a meaningful physical quantity.

Therefore, the correct choice is C.

Key Takeaways

• A scalar has magnitude only.
• A vector has magnitude and direction and can be resolved into components.

Common Mistakes

• Choosing A because you remember “components” from mechanics: components are for vectors, not scalars.
• Thinking a scalar “does not need a unit”: almost all physical quantities, scalar or vector, require units.

Things to Be Careful About

• “No direction” is the key phrase for scalars.
• Both scalars and vectors have magnitudes and units; the difference is the direction.

Background Concept

For quantities with small uncertainties:

• If $W = AB$ or $W = \dfrac{A}{B}$ then the fractional (and hence percentage) uncertainties add:

$$\frac{\Delta W}{W} = \frac{\Delta A}{A} + \frac{\Delta B}{B}$$

• If a quantity is raised to a power, $W = A^n$, then the fractional uncertainty is multiplied by $|n|$:

$$\frac{\Delta W}{W} = |n|\frac{\Delta A}{A}$$

Constants (such as 2 in $2XY$) are treated as exact and do not contribute to the percentage uncertainty.

Understanding the Question

You are told:

• $X$ has a percentage uncertainty of $2\%$.
• $Y$ has a percentage uncertainty of $4\%$.
• A derived quantity $W$ (calculated using $X$ and $Y$) has a percentage uncertainty of $8\%$.

The task is to choose, from the given formulas, which one would produce $8\%$ uncertainty when uncertainties are combined correctly.

Approach

For each option, use:

• “add percentage uncertainties” for multiplication/division,
• “multiply the percentage uncertainty by the power” when a variable is squared.

Then compare the resulting percentage uncertainty with the required $8\%$.

Step-by-Step Reasoning

Option A: $W = XY$

$$\%\,\Delta W = 2\% + 4\% = 6\%$$

Not $8\%$.

Option B: $W = 2XY$
The factor 2 is a constant, so uncertainty is the same as for $XY$:

$$\%\,\Delta W = 2\% + 4\% = 6\%$$

Not $8\%$.

Option C: $W = \dfrac{X}{Y^2}$
$Y^2$ doubles the percentage uncertainty from $Y$:

$$\%\,\Delta (Y^2) = 2 \times 4\% = 8\%$$

Then for division, add uncertainties:

$$\%\,\Delta W = 2\% + 8\% = 10\%$$

Not $8\%$.

Option D: $W = \dfrac{Y}{X^2}$
$X^2$ doubles the percentage uncertainty from $X$:

$$\%\,\Delta (X^2) = 2 \times 2\% = 4\%$$

Then for division:

$$\%\,\Delta W = 4\% + 4\% = 8\%$$

This matches, so option D is correct.

Key Takeaways

• For products and quotients, percentage uncertainties add.
• For powers, multiply the percentage uncertainty by the power.
• Multiplying by an exact constant does not change percentage uncertainty.

Common Mistakes

• Treating $2XY$ as having a larger uncertainty because of the 2 (constants do not add uncertainty).
• Forgetting to multiply the uncertainty by 2 when a quantity is squared.
• Subtracting percentage uncertainties for division (they still add).

Things to Be Careful About

• Use absolute values of powers when applying the power rule (e.g. $X^{-2}$ still gives a factor of 2).
• These rules assume uncertainties are small and combine approximately (the standard A-Level method).
• Ensure you are combining percentage (or fractional) uncertainties, not absolute uncertainties, for multiplication/division.

Background Concept

Air resistance (drag) is a force exerted by the air that opposes the direction of motion through the air. For typical speeds in air, the magnitude of drag increases with speed (often approximately proportional to $v$ or $v^2$).

The resultant (net) force on an object is the vector sum of all forces acting on it:

$$\vec{F}_{\text{resultant}} = \sum \vec{F}.$$

Weight $W = mg$ acts downward and is approximately constant during the motion.

Understanding the Question

A football has been kicked vertically upward. While it is moving upward:

• its velocity is upward,
• its speed is decreasing (it is slowing down as it rises),
• air resistance acts opposite to the motion (so it acts downward),
• weight acts downward.

The question asks how (i) the air resistance and (ii) the resultant force change as the ball rises.

Approach

1. Decide the direction of air resistance while the ball is moving upward.
2. Use the fact that the ball’s speed decreases as it goes up to infer how the magnitude of drag changes.
3. Add the forces (weight + drag) to see how the magnitude of the resultant force changes.

Step-by-Step Reasoning

• While the ball is moving upward, air resistance acts downward (opposes the upward velocity).
• As the ball rises, it slows down, so its speed $v$ decreases.
• Since drag increases with speed, a decreasing speed means the drag force decreases.
• The forces on the ball while it is rising are both downward:
  • weight $W = mg$ (constant),
  • drag $F_{\text{drag}}$ (decreasing).
• Therefore the resultant downward force is

$$F_{\text{resultant}} = W + F_{\text{drag}} \quad \text{(downward)}.$$

• As $F_{\text{drag}}$ decreases, the sum $W + F_{\text{drag}}$ decreases, so the resultant force decreases (still downward).

So: air resistance decreases; resultant force decreases → option A.

Key Takeaways

• Drag always acts opposite the direction of motion.
• If speed decreases, drag decreases.
• The resultant force is the vector sum; here weight and drag act in the same direction while the ball rises.

Common Mistakes

• Saying drag acts upward because the ball is moving upward (it acts opposite to motion, so it is downward here).
• Thinking the resultant force must decrease because the acceleration decreases; in fact the acceleration magnitude also decreases here because drag decreases, but you must justify it via forces.
• Forgetting to include weight when considering the resultant force.

Things to Be Careful About

• “Resultant force” means net force, not just air resistance.
• The direction matters: while moving up, both forces are downward; after the ball starts falling, drag reverses direction.
• The question asks for variation (increase/decrease), not the direction.

Background Concept

Key kinematics definitions:

• Velocity $v$ is the rate of change of displacement $s$:

$$v = \frac{\mathrm{d}s}{\mathrm{d}t}$$

• Speed is the rate of change of distance (distance is a scalar and does not include direction).
• Acceleration $a$ is the rate of change of velocity:

$$a = \frac{\mathrm{d}v}{\mathrm{d}t}$$

For a velocity–time graph:

• The gradient is $\frac{\Delta v}{\Delta t}$, which is acceleration.
• The area under the graph is $\int v\,\mathrm{d}t$, which gives displacement (signed area, so it can be negative if velocity is negative).

Understanding the Question

You are given four statements about:

• what gradients/areas on a $v$–$t$ graph represent, and
• definitions of velocity and acceleration.

You must choose the one statement that is NOT correct.

Approach

Check each option against:

1. the calculus/graph interpretation for a velocity–time graph (gradient and area), and
2. the precise definitions: velocity relates to displacement, speed relates to distance.

Step-by-Step Reasoning

• A: On a $v$–$t$ graph, gradient $= \frac{\Delta v}{\Delta t}$, which is acceleration $a$. So A is correct.
• B: By definition, acceleration is the rate of change of velocity, i.e. $a = \frac{\mathrm{d}v}{\mathrm{d}t}$. So B is correct.
• C: Displacement is found from velocity by integrating over time:

$$\Delta s = \int v\,\mathrm{d}t$$

This integral is the area under a $v$–$t$ graph (taking sign into account). So C is correct.

• D: Velocity is the rate of change of displacement, not distance. The rate of change of distance is speed. Therefore D is not correct.

So the incorrect statement is D.

Key Takeaways

• Gradient of a $v$–$t$ graph gives acceleration.
• Area under a $v$–$t$ graph gives displacement.
• Velocity is linked to displacement (vector); speed is linked to distance (scalar).

Common Mistakes

• Writing “velocity is rate of change of distance” (this is the definition of speed).
• Thinking the area under a $v$–$t$ graph gives distance without considering negative velocity (it gives signed displacement).

Things to Be Careful About

• Use the correct terms: distance vs displacement and speed vs velocity.
• Remember the sign convention: negative velocity contributes negative area and hence negative displacement.

Background Concept

In an evacuated tube, air resistance is negligible, so the only significant force on the feather is its weight. This produces (approximately) constant downward acceleration equal to $g$.

For motion with constant acceleration $a$ from rest ($u=0$), the displacement after time $t$ is

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

So for free fall in vacuum ($a=g$), displacement is proportional to $t^2$.

Understanding the Question

The feather starts from rest at the top and takes total time $T$ to travel the whole tube length $L$.

You are asked: in half that time, $0.50T$, what fraction of the total distance $L$ has it fallen?

Key clue: it is not moving at constant speed; it is accelerating, so half the time is not half the distance.

Approach

Use the constant-acceleration displacement formula $s = \frac{1}{2}gt^2$.

1. Write an expression for $L$ in terms of $T$.
2. Write an expression for $s$ in terms of $0.50T$.
3. Divide to get $s/L$ (this cancels $g$).

Step-by-Step Reasoning

Full fall:

$$L = \frac{1}{2}gT^2$$

Half-time fall ($t=0.50T$):

$$s = \frac{1}{2}g(0.50T)^2$$

Compute the square:

$$(0.50T)^2 = 0.25T^2$$

So

$$s = \frac{1}{2}g(0.25T^2) = 0.25\left(\frac{1}{2}gT^2\right) = 0.25L$$

Hence the distance fallen in time $0.50T$ is $0.25L$, which corresponds to option B.

Key Takeaways

• In free fall from rest with constant acceleration, $s \propto t^2$.
• A fraction $k$ of the total time corresponds to a fraction $k^2$ of the total distance (when starting from rest and acceleration is constant).

Common Mistakes

• Choosing $0.50L$ by assuming constant speed (it is accelerating).
• Using $s = vt$ with $v$ taken as the final speed (speed is changing throughout).
• Forgetting to square the time fraction: $0.5^2 = 0.25$.

Things to Be Careful About

• The feather is in an evacuated tube, so you must model it like any object in free fall (constant $g$).
• The proportional method works because both distances are described by the same equation with the same $g$; $g$ cancels when comparing fractions.

Background Concept

On a velocity–time graph, the displacement over a time interval equals the area under the graph for that interval.

If the velocity changes uniformly (a straight line on the graph), the area under the line between two times forms a trapezium. The trapezium area is

$$A = \frac{1}{2}(\text{sum of parallel sides})\times(\text{separation})$$

Here, the parallel sides are the two velocities, and the separation is the time interval.

Understanding the Question

The brakes are applied from $t=1.0\,\text{s}$ to $t=4.0\,\text{s}$. The question asks for the distance travelled during braking, so we need the area under the $v$–$t$ graph only between these two times (not from $t=0$).

From the graph during braking:

• at $t=1.0\,\text{s}$, $v \approx 22\,\text{m s}^{-1}$
• at $t=4.0\,\text{s}$, $v \approx 8\,\text{m s}^{-1}$

Approach

1. Read the velocities at the start and end of braking from the graph.
2. Find the time interval: $\Delta t = 4.0 - 1.0$.
3. Since the graph segment is a straight line, use trapezium area (equivalently, average velocity (\times) time).
4. Match the computed distance to the options.

Step-by-Step Reasoning

Time interval while braking:

$$\Delta t = 4.0 - 1.0 = 3.0\,\text{s}$$

Trapezium area under the graph:

$$\text{distance} = \frac{1}{2}(v_1+v_2)\Delta t$$

Substitute $v_1 = 22\,\text{m s}^{-1}$ and $v_2 = 8\,\text{m s}^{-1}$:

$$\text{distance} = \frac{1}{2}(22+8)\times 3.0 = \frac{1}{2}(30)\times 3.0 = 15\times 3.0 = 45\,\text{m}$$

So the correct option is B.

Key Takeaways

• Displacement from a $v$–$t$ graph is found by area under the curve.
• A straight-line change in velocity over time gives a trapezium area:

$$\text{distance} = \text{average velocity} \times \text{time} = \frac{v_1+v_2}{2}\,\Delta t$$

Common Mistakes

• Using the wrong time interval (e.g. finding area from $0$ to $6\,\text{s}$ instead of $1$ to $4\,\text{s}$).
• Reading the velocities inaccurately (mixing up gridlines or using the wrong point).
• Calculating only a triangle area (which would correspond to velocity dropping to zero) rather than a trapezium.

Things to Be Careful About

• Ensure the distance is only for the braking period: $t=1.0$ to $t=4.0\,\text{s}$.
• Use correct units: velocity in $\text{m s}^{-1}$, time in $\text{s}$, so area gives $\text{m}$.
• For a straight segment, trapezium area is valid; if the graph were curved, you would need a different method (e.g. counting squares/estimation).

Background Concept

In a collision, two key ideas are used:

• Linear momentum: $p = mv$ (a vector). For a system of objects, the total momentum is the vector sum of each momentum.
• Conservation of momentum: if the resultant external force on the system is zero (or negligible), then the total momentum of the system remains constant.

Kinetic energy is

$$E_k = \frac{1}{2}mv^2$$

• In an elastic collision, total kinetic energy is conserved.
• In an inelastic collision, total kinetic energy is not conserved; some is transformed into internal energy (deformation, heating) and/or sound. So total kinetic energy decreases.

Understanding the Question

Two satellites collide inelastically in deep space. “Deep space” indicates there are negligible external forces (e.g. no significant gravitational forces or drag acting during the collision time), so the pair of satellites can be treated as an isolated system.

The question asks what happens to:

• total kinetic energy
• total momentum

Approach

1. Decide whether the system is isolated: deep space implies external forces are negligible.
2. Use that to decide whether momentum is conserved.
3. Use the fact it is explicitly stated to be inelastic to decide what happens to kinetic energy.
4. Match this pair of statements to the options.

Step-by-Step Reasoning

1. Momentum: For the two satellites taken together as one system, during the collision the forces they exert on each other are internal forces. With negligible external resultant force, conservation of momentum applies, so total momentum is conserved.

2. Kinetic energy: The collision is stated to be inelastic, meaning kinetic energy is converted into other forms (e.g. deformation of the satellites, heating). Therefore the total kinetic energy after the collision is less than before: it is reduced.

So the correct combination is: total kinetic energy reduced, total momentum conserved, which is option C.

Key Takeaways

• Momentum is conserved in collisions when the system has negligible external resultant force.
• Inelastic collisions do not conserve kinetic energy; total kinetic energy decreases.
• “Deep space” is a clue that external forces can be neglected.

Common Mistakes

• Thinking kinetic energy is always conserved in collisions (it is only conserved in elastic collisions).
• Thinking momentum is reduced because objects “lose speed”; momentum is still conserved if there is no external resultant force.
• Forgetting momentum is a vector (direction matters), though this question only asks about conservation qualitatively.

Things to Be Careful About

• Momentum conservation depends on external forces, not on whether the collision is elastic or inelastic.
• Kinetic energy conservation depends on whether the collision is elastic; “inelastic” directly implies kinetic energy is not conserved.

Background Concept

Momentum is defined as

$$p = mv$$

where $p$ is momentum (in $\text{kg m s}^{-1}$), $m$ is mass (in $\text{kg}$) and $v$ is speed/velocity (in $\text{m s}^{-1}$). For an estimate question, you choose reasonable typical values and aim for the correct order of magnitude (power of ten) rather than an exact number.

Understanding the Question

We need an order-of-magnitude estimate of the momentum of a typical family car moving at $25\ \text{km h}^{-1}$. The answer options differ by factors of 10, so the key is:

• convert $25\ \text{km h}^{-1}$ to $\text{m s}^{-1}$,
• use a sensible car mass (about $10^3\ \text{kg}$),
• compute $p = mv$ and match to the nearest option.

Approach

1. Estimate a typical family car mass: about $1000$ to $1500\ \text{kg}$ (i.e. about $10^3\ \text{kg}$).
2. Convert $25\ \text{km h}^{-1}$ into $\text{m s}^{-1}$.
3. Multiply $m$ and $v$ to get momentum and round to the nearest power of ten.

Step-by-Step Reasoning

1. Mass estimate: a family car is typically around $1.0 \times 10^3\ \text{kg}$.

2. Speed conversion:

$$v = 25\ \text{km h}^{-1} = 25 \times \frac{1000\ \text{m}}{3600\ \text{s}} \approx 6.94\ \text{m s}^{-1} \approx 7\ \text{m s}^{-1}$$

1. Momentum:

$$p = mv \approx (1.0 \times 10^3)(7) = 7 \times 10^3\ \text{kg m s}^{-1}$$

This is closest in order of magnitude to $1 \times 10^4\ \text{kg m s}^{-1}$.

So the correct option is A.

Key Takeaways

• Use $p = mv$ with SI units.
• A typical car mass is about $10^3\ \text{kg}$.
• $25\ \text{km h}^{-1}$ is about $7\ \text{m s}^{-1}$.
• Estimation MCQs usually test order of magnitude (powers of ten).

Common Mistakes

• Forgetting to convert $\text{km h}^{-1}$ to $\text{m s}^{-1}$ (leads to a value too large by a factor of $3.6$).
• Using an unrealistic car mass (e.g. $10^2\ \text{kg}$ or $10^5\ \text{kg}$).
• Choosing an option based on the unconverted speed number (e.g. multiplying $1000 \times 25$ without unit conversion).

Things to Be Careful About

• The conversion factor: $1\ \text{km h}^{-1} = \frac{1000}{3600}\ \text{m s}^{-1} \approx 0.278\ \text{m s}^{-1}$.
• The options are spaced by decades (powers of ten), so rounding appropriately is expected.
• Momentum has units $\text{kg m s}^{-1}$ (do not write $\text{kg m s}$ or $\text{N s}$ unless you are consistent and know they are equivalent).

Background Concept

Velocity is a vector, so direction matters. The change in velocity is defined by

$$\Delta v = v - u$$

where $u$ is the initial velocity and $v$ is the final velocity. Because velocity is a vector, subtraction must be done with a consistent sign convention (e.g. right is positive, left is negative). The sign of $\Delta v$ then tells you the direction of the change.

Understanding the Question

The ball initially moves to the right at $8\ \text{m s}^{-1}$ and after bouncing off the wall it moves to the left at $3\ \text{m s}^{-1}$. You are asked for the change in velocity during the collision (not the speed and not the final velocity).

Approach

1. Choose a positive direction (take right as +).
2. Write $u$ and $v$ with signs.
3. Use $\Delta v = v-u$.
4. Convert the sign of $\Delta v$ into “to the left/right” and select the matching option.

Step-by-Step Reasoning

Take right as positive.

• Initial velocity to the right:

u = u = +8\ \text{m s}^{-1}$$ - Final velocity to the left:

v = -3\ \text{m s}^{-1}$$

Now calculate the change:

$$\Delta v = v-u = (-3) - (+8) = -11\ \text{m s}^{-1}$$

The negative sign means the change in velocity is towards the left, with magnitude $11\ \text{m s}^{-1}$.

So the correct option is C.

Key Takeaways

• Use $\Delta v = v-u$ for change in velocity.
• Treat velocity as a vector: assign signs for direction.
• A negative $\Delta v$ (with right positive) means the change is to the left.

Common Mistakes

• Adding speeds: $8 + 3 = 11$ but then stating “to the right” without sign reasoning.
• Using $\Delta v = u - v$ (wrong order), which flips the direction.
• Treating “left” as a separate label rather than using a sign convention.

Things to Be Careful About

• Always define which direction is positive before substituting values.
• The answer is the change in velocity, not the final velocity.
• A result like $-11\ \text{m s}^{-1}$ must be converted into a magnitude and direction: $11\ \text{m s}^{-1}$ to the left.

Background Concept

This situation has two distinct stages:

1. Collision (very short time): external forces like weight act, but the collision time is so small that the impulse due to external forces is negligible compared with the impulse between pellet and block. So total momentum is conserved during the impact.

For a perfectly inelastic collision (objects stick together),

$$\text{initial momentum} = \text{final momentum}$$

1. Rise after collision: after the pellet is embedded, the combined mass moves upward and slows under gravity until its speed becomes zero at the top. Neglecting air resistance, mechanical energy is conserved during this rise:

$$\frac{1}{2}mv^2 = mgh$$

so

$$h = \frac{v^2}{2g}$$

Note the mass cancels, so the height depends only on the speed just after the collision.

Understanding the Question

A $5.0\text{ g}$ pellet travels upward at $200\text{ m s}^{-1}$ and embeds in a stationary $95\text{ g}$ clay block. You are asked for the maximum height the block (with pellet) rises above its initial position.

Key clues:

• “embeds itself” (\Rightarrow) perfectly inelastic collision (momentum conserved, kinetic energy not conserved in the collision).
• “rise freely after impact” (\Rightarrow) after collision, treat it like a projectile moving under gravity.

Approach

1. Use conservation of momentum to find the common upward speed (v) immediately after impact.
2. Convert that kinetic energy into gravitational potential energy to find (h):

$$\frac{1}{2}(M+m)v^2 = (M+m)gh$$

Step-by-Step Reasoning

1. Convert grams to kilograms:

$$m = 5.0\text{ g} = 0.0050\text{ kg},\quad M = 95\text{ g} = 0.095\text{ kg}$$

Total mass after sticking:

$$M+m = 0.100\text{ kg}$$

1. Momentum before impact:

• Pellet momentum: (mu = 0.0050\times 200)
• Block momentum: (0) (stationary)
  So total initial momentum is

$$p = 0.0050\times 200 = 1.0\text{ kg m s}^{-1}$$

1. Momentum after impact:
   They move together at speed (v):

$$(M+m)v = 0.100\,v$$

Conservation of momentum gives

$$1.0 = 0.100\,v \Rightarrow v = 10\text{ m s}^{-1}$$

1. Find rise height using energy (or SUVAT). At the top, final speed is (0). Using energy:

$$\frac{1}{2}(M+m)v^2 = (M+m)gh$$

Cancel ((M+m)):

$$\frac{1}{2}v^2 = gh \Rightarrow h = \frac{v^2}{2g}$$

Substitute (v=10\text{ m s}^{-1}) and (g=9.81\text{ m s}^{-2}):

$$h = \frac{100}{19.62} = 5.10\text{ m}$$

This corresponds to option A.

Key Takeaways

• For an inelastic collision, conserve momentum, not kinetic energy.
• After the collision, use energy conservation (or kinematics) for the upward motion under gravity.
• The height depends on the post-collision speed, with (h = v^2/(2g)).

Common Mistakes

• Conserving kinetic energy during the collision (not valid when the pellet embeds).
• Forgetting to convert grams to kilograms.
• Using the pellet mass alone when calculating the rise, instead of the combined mass (though it cancels if you do the energy step correctly).
• Using (h = v^2/g) instead of (h = v^2/(2g)).

Things to Be Careful About

• Momentum conservation applies during the impact because the collision time is very small; it does not mean momentum is conserved during the rise.
• Use a consistent value for (g) (typically (9.81\text{ m s}^{-2}) or (9.8\text{ m s}^{-2})); rounding should still clearly match option A.
• The question asks for height above its initial position (the block’s starting level), which is exactly what the energy method gives here.

Background Concept

A couple consists of two forces that:

• are equal in magnitude,
• act in opposite directions,
• are parallel,
• act along different (separated) lines of action.

Because the forces are equal and opposite, the resultant force is zero (so the object does not accelerate linearly). However, because the lines of action are separated, the forces produce a turning effect (a moment/torque).

The moment of a couple is

$$\tau = Fd$$

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

Understanding the Question

You are shown four diagrams (A–D) of a single object with two vertical forces acting at different positions.

You must choose which diagram represents a couple.

So we must look for: same size forces, opposite directions, separated.

Approach

For each option:

1. Check if the forces are equal in magnitude.
2. Check if they are opposite in direction.
3. Check if their lines of action are separated (not collinear), so a turning effect exists.

Only a diagram that satisfies all three is a couple.

Step-by-Step Reasoning

• A: both forces are $6\,\text{N}$ but both act downwards. Resultant force is not zero, so not a couple.
• B: one force is $6\,\text{N}$ down and the other is $6\,\text{N}$ up, and they act on opposite sides (separated). Resultant force is zero, but there is a turning effect: this is a couple.

• C: forces are both downwards and also unequal ($12\,\text{N}$ and $6\,\text{N}$). Not a couple.
• D: forces are opposite in direction but unequal ($12\,\text{N}$ and $6\,\text{N}$). Resultant force is not zero, so not a couple.

Therefore the correct option is B.

Key Takeaways

• A couple requires equal and opposite forces with separated lines of action.
• Couples cause rotation without translation (zero net force, non-zero moment).

Common Mistakes

• Choosing two forces that are separated but in the same direction (gives a net force, not a couple).
• Choosing opposite forces that are not equal (net force not zero).
• Forgetting that the forces must be parallel and not acting along the same line.

Things to Be Careful About

• “Opposite sides” alone is not enough: you must check equal magnitude and opposite direction.
• A couple always has zero resultant force; if the forces are not equal and opposite, it cannot be a couple.

Background Concept

For a rigid body in equilibrium:

1. The resultant force is zero.
2. The resultant moment (torque) about any point is zero.

The moment of a force about a point is

$$\tau = F r \sin\theta$$

where $r$ is the distance from the pivot to the point of application of the force, and $\theta$ is the angle between the force direction and the line from the pivot to the point of application. Only the component of the force perpendicular to the rod contributes to the turning effect.

Understanding the Question

A uniform rod $XY$ is hinged at $Y$ to a vertical wall. The rod makes $60^\circ$ to the wall (so the rod is at $60^\circ$ to the vertical). The rod’s weight is $3.0\text{ N}$ acting vertically downward at its midpoint.

At the free end $X$, a force $F$ acts at $30^\circ$ to the rod. The rod is in equilibrium, so clockwise moments about the hinge must balance anticlockwise moments.

Approach

Choose the hinge point $Y$ as the moment-taking point because the (unknown) hinge reaction forces act through $Y$ and therefore produce zero moment. Then:

• Write the moment of $F$ about $Y$ using $r=L$ and angle $30^\circ$.
• Write the moment of the weight about $Y$ using $r=L/2$ and the angle between the rod and the weight direction (vertical), which is $60^\circ$.
• Equate the two moments and solve for $F$.

Step-by-Step Reasoning

Let the rod length be $L$.

1) Moment due to $F$ about $Y$
The force is applied at $X$, a distance $L$ from $Y$. The angle between the rod (the line $YX$) and $F$ is $30^\circ$, so the perpendicular component is $F\sin 30^\circ$.

$$\tau_F = F L \sin 30^\circ$$

2) Moment due to the weight about $Y$
Because the rod is uniform, the weight acts at the midpoint, distance $L/2$ from $Y$.
The weight acts vertically downward, and the rod makes $60^\circ$ with the vertical, so the angle between $r$ (along the rod) and the weight is $60^\circ$.

$$\tau_W = 3.0 \left(\frac{L}{2}\right) \sin 60^\circ$$

3) Equilibrium condition (principle of moments)
Clockwise moment = anticlockwise moment (magnitudes equal):

$$F L \sin 30^\circ = 3.0 \left(\frac{L}{2}\right) \sin 60^\circ$$

Cancel $L$ and use $\sin 30^\circ = 0.5$:

$$F \times 0.5 = 3.0 \times \frac{1}{2} \sin 60^\circ$$

Both sides contain a factor $1/2$, so

$$F = 3.0\sin 60^\circ = 3.0 \times 0.866 \approx 2.6\text{ N}$$

So the correct option is C.

Key Takeaways

• Taking moments about the hinge removes unknown hinge forces from the calculation.
• Use $\tau = Fr\sin\theta$: the angle is between the force and the line from pivot to point of action.
• For a uniform rod, weight acts at the midpoint.

Common Mistakes

• Using $\cos$ instead of $\sin$ for the torque factor.
• Using the wrong lever arm for the weight (using $L$ instead of $L/2$).
• Using the wrong angle for the weight: it is the angle between the rod and the vertical (the weight’s direction), i.e. $60^\circ$.
• Trying to balance forces rather than moments (force balance alone is not sufficient to find $F$ here).

Things to Be Careful About

• Always identify the pivot and measure $r$ from that pivot.
• The hinge reaction may have both horizontal and vertical components, but both pass through the pivot so their moments about $Y$ are zero.
• Check which angle is given: “rod at $60^\circ$ to the wall” means $60^\circ$ to the vertical (since the wall is vertical).

Background Concept

Density $\rho$ is defined as mass per unit volume:

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

For a substance of uniform density, mass can be found from

$$m = \rho V$$

When two liquids are mixed (and we are told to assume the volumes add with no change), the total mass is the sum of the masses of each liquid and the total volume is the sum of their volumes:

$$m_{\text{total}} = m_1 + m_2, \qquad V_{\text{total}} = V_1 + V_2$$

So the mixture density is

$$\rho_{\text{mix}} = \frac{m_{\text{total}}}{V_{\text{total}}}$$

Understanding the Question

You start with $40\ \text{cm}^3$ of glycerine (density $1.3\ \text{g cm}^{-3}$). You then add some volume of water (density $1.0\ \text{g cm}^{-3}$) so that the final mixture has density $1.1\ \text{g cm}^{-3}$. The unknown is the volume of water that must be added.

The key instruction is: total volume does not change on mixing. That means the final volume is simply $40 + V_w$.

Approach

1. Convert the given volumes and densities into masses using $m = \rho V$.
2. Add masses to get total mass, and add volumes to get total volume.
3. Set the required mixture density $1.1\ \text{g cm}^{-3}$ equal to $m_{\text{total}}/V_{\text{total}}$.
4. Solve the resulting equation for $V_w$ and choose the matching option.

Step-by-Step Reasoning

Mass of glycerine:

$$m_g = \rho_g V_g = (1.3\ \text{g cm}^{-3})(40\ \text{cm}^3) = 52\ \text{g}$$

Let the volume of water added be $V_w\ \text{cm}^3$. The mass of that water is

$$m_w = \rho_w V_w = (1.0\ \text{g cm}^{-3})V_w = V_w\ \text{g}$$

Total mass and total volume of mixture:

$$m_{\text{total}} = 52 + V_w\ \text{g}$$ $$V_{\text{total}} = 40 + V_w\ \text{cm}^3$$

Required density is $1.1\ \text{g cm}^{-3}$:

$$\rho_{\text{mix}} = \frac{52+V_w}{40+V_w} = 1.1$$

Solve:

$$52+V_w = 1.1(40+V_w) = 44 + 1.1V_w$$ $$52-44 = 1.1V_w - V_w = 0.1V_w$$ $$8 = 0.1V_w \Rightarrow V_w = 80\ \text{cm}^3$$

This corresponds to option D.

A quick sense-check: the target density $1.1$ is much closer to water (1.0) than to glycerine (1.3), so you need to add a relatively large volume of water compared with $40\ \text{cm}^3$ of glycerine. $80\ \text{cm}^3$ is consistent with that.

Key Takeaways

• Use $m = \rho V$ to turn densities and volumes into masses.
• For mixtures with no volume change: add masses and add volumes separately.
• Set $\rho_{\text{mix}} = m_{\text{total}}/V_{\text{total}}$ and solve for the unknown.

Common Mistakes

• Adding densities directly (e.g. averaging $1.0$ and $1.3$) instead of using mass and volume.
• Forgetting the instruction that volumes add (some mixtures contract/expand, but here it is explicitly excluded).
• Rearrangement error when solving $\frac{52+V_w}{40+V_w} = 1.1$ (especially mishandling $1.1V_w$).

Things to Be Careful About

• Keep units consistent: here $\text{g cm}^{-3}$ with $\text{cm}^3$ naturally gives grams.
• When clearing a fraction, multiply both sides by the full denominator $(40+V_w)$.
• The mixture density must lie between $1.0$ and $1.3\ \text{g cm}^{-3}$; if your algebra gives a negative volume or a density outside this range, something has gone wrong.

Background Concept

A resistive (drag) force that is proportional to the square of speed has the form

$$F = kv^2$$

where $k$ is a constant for that car under those conditions.

To maintain a steady speed, the engine must provide a driving force equal in magnitude to the resistive force (resultant force $=0$). The useful output power needed at that speed is

$$P = Fv$$

where $F$ is the force the engine must supply at that speed.

Understanding the Question

You are told:

• at $v = 20.0\ \text{m s}^{-1}$, the resistive force is $800\ \text{N}$,
• resistive force varies as $v^2$,
• find the useful output power needed to keep the car moving steadily at $40.0\ \text{m s}^{-1}$.

So you must (1) scale the force from $20.0$ to $40.0\ \text{m s}^{-1}$, then (2) use $P = Fv$.

Approach

1. Use the proportionality $F \propto v^2$ to find the new force at $40.0\ \text{m s}^{-1}$.
2. Since speed is steady, the engine’s useful force output equals this resistive force.
3. Compute power using $P = Fv$ and match to the given options.

Step-by-Step Reasoning

Because

$$F \propto v^2$$

then the ratio of forces at two speeds is

$$\frac{F_2}{F_1} = \left(\frac{v_2}{v_1}\right)^2.$$

With $F_1 = 800\ \text{N}$ at $v_1 = 20.0\ \text{m s}^{-1}$ and $v_2 = 40.0\ \text{m s}^{-1}$:

$$F_2 = 800\left(\frac{40.0}{20.0}\right)^2 = 800 \times 4 = 3200\ \text{N}.$$

At steady speed, the engine must supply $F = 3200\ \text{N}$.

Now use

$$P = Fv$$

with $v = 40.0\ \text{m s}^{-1}$:

$$P = 3200 \times 40.0 = 1.28 \times 10^5\ \text{W} = 128\ \text{kW}.$$

This corresponds to option C.

Key Takeaways

• If $F \propto v^2$, doubling $v$ makes $F$ four times larger.
• Power to maintain constant speed against a resistive force is $P = Fv$.
• With quadratic drag, power increases very rapidly with speed (here by a factor of $8$ when speed doubles).

Common Mistakes

• Scaling the force linearly with $v$ instead of with $v^2$ (getting $1600\ \text{N}$).
• Forgetting that power depends on both force and speed, and using $P=F$ or $P=\frac{F}{v}$.
• Converting to kW incorrectly (missing the factor of $10^3$).

Things to Be Careful About

• Use the ratio method carefully: square the speed ratio.
• Keep units consistent: $\text{N} \times \text{m s}^{-1} = \text{W}$.
• “Useful output power” here means the mechanical power delivered to the wheels to balance resistive forces at steady speed (not the chemical power from fuel).