Physics 9702/12 — February/March 2024

Background Concept

SI base quantities are fundamental physical quantities with base units that are not defined in terms of other SI units. Examples include mass (kg), time (s), electric current (A), and thermodynamic temperature (K).

A derived unit is formed from base units (e.g. newton, N, is derived from $\text{kg m s}^{-2}$).

Understanding the Question

You are given four rows, each pairing a physical quantity with a unit. The task is to identify the row where:

1. the quantity is an SI base quantity, and
2. the unit given is the correct SI base unit for that quantity.

Approach

List/check whether each quantity is a base quantity, then check whether the unit shown is the correct SI base unit (not just a commonly used unit).

Step-by-Step Reasoning

• A: current — A
  Electric current is an SI base quantity, and the ampere (A) is its SI base unit. So this row is correct.

• B: force — N
  Force is not a base quantity; it is derived from $F = ma$. Its unit newton is derived:

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

  So B is not a base quantity with a base unit.

• C: mass — g
  Mass is a base quantity, but its SI base unit is the kilogram (kg), not the gram (g). So C is incorrect.

• D: temperature — $^\circ\text{C}$
  Thermodynamic temperature is a base quantity, but its SI base unit is the kelvin (K). Degrees Celsius is accepted for use with SI, but it is not the SI base unit. So D is incorrect.

Therefore the correct option is A.

Key Takeaways

• Know the SI base quantities and their base units (e.g. current (\to) ampere, temperature (\to) kelvin, mass (\to) kilogram).
• Recognise derived quantities/units (e.g. force in newtons).

Common Mistakes

• Choosing D because $^\circ\text{C}$ is commonly used for temperature, forgetting the SI base unit is K.
• Choosing C because grams are commonly used, forgetting the SI base unit is kg.
• Treating force as a base quantity.

Things to Be Careful About

• The question asks for the base unit in the SI system, not merely a unit that is widely used.
• Unit symbols matter: kg is the base unit for mass, not g; K is the base unit for temperature, not $^\circ\text{C}$.

Background Concept

When forces act on an object along a straight line, the resultant (net) force is the algebraic sum of the forces, taking one direction as positive.

Newton's second law links resultant force to acceleration:

$$F_{\text{resultant}} = ma$$

So the acceleration magnitude is

$$a = \frac{|F_{\text{resultant}}|}{m}$$

Understanding the Question

A car of mass $850\ \text{kg}$ experiences two horizontal forces along the same straight line but in opposite directions: $1600\ \text{N}$ one way and $1200\ \text{N}$ the other way. You are asked for the magnitude of the acceleration.

Because the forces are horizontal and opposite, the resultant force is simply the difference in magnitudes.

Approach

1. Choose a positive direction (e.g. the direction of the $1600\ \text{N}$ force).
2. Compute the resultant force by subtracting the smaller opposing force from the larger.
3. Apply $a = F/m$.
4. Round appropriately and select the matching option.

Step-by-Step Reasoning

Take the direction of the $1600\ \text{N}$ force as positive. Then the opposing $1200\ \text{N}$ force is negative.

Resultant force:

$$F_{\text{resultant}} = 1600 - 1200 = 400\ \text{N}$$

Use Newton's second law:

$$a = \frac{F_{\text{resultant}}}{m} = \frac{400}{850}$$ $$a = 0.470588...\ \text{m s}^{-2} \approx 0.47\ \text{m s}^{-2}$$

This corresponds to option A.

Key Takeaways

• For collinear forces in opposite directions, the resultant is the difference between the larger and smaller force.
• Use $F = ma$ with the resultant force, not an individual force.
• The magnitude of acceleration is $|F|/m$.

Common Mistakes

• Adding the forces ($1600 + 1200$) instead of subtracting for opposite directions.
• Using the wrong mass unit (mass is already in $\text{kg}$, so no conversion is needed).
• Forgetting that the question asks for magnitude (so acceleration should be positive).

Things to Be Careful About

• Always decide whether forces are in the same or opposite directions before combining them.
• Keep units consistent: $\text{N}$, $\text{kg}$, giving $\text{m s}^{-2}$.
• In MCQs, round sensibly to match the given options (here $0.47\ \text{m s}^{-2}$).

Background Concept

An equation is homogeneous if both sides have the same base units (dimensions). For an oscillation period $T$, the base unit must be time, i.e. seconds (dimension $[T] = \text{s}$).

A spring constant $k$ is defined by Hooke’s law $F = kx$, so $k = F/x$. This lets us find the base units of $k$.

Understanding the Question

You are given a mass $m$ on a vertical spring of spring constant $k$, oscillating with period $T$. Four possible expressions are offered. You are not asked for the true formula, only which one has the correct units to be a time.

So we test each option by checking whether it has units of seconds.

Approach

1. Write $[m]$ in base units: $\text{kg}$.
2. Find $[k]$ from $k = F/x$ using $[F] = \text{N} = \text{kg m s}^{-2}$.
3. Work out the units of $m/k$ and $k/m$.
4. Decide which option gives $\text{s}$ (not $\text{s}^2$, $\text{s}^{-1}$, etc.). Note that $2\pi$ is dimensionless and does not affect homogeneity.

Step-by-Step Reasoning

From $F = kx$,

$$[k] = \frac{[F]}{[x]} = \frac{\text{N}}{\text{m}}$$

Convert newtons to base units:

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

So

$$[k] = \frac{\text{kg m s}^{-2}}{\text{m}} = \text{kg s}^{-2}$$

Now compare the two possible ratios:

• For $k/m$:

$$\left[\frac{k}{m}\right] = \frac{\text{kg s}^{-2}}{\text{kg}} = \text{s}^{-2}$$

Taking a square root would give $\text{s}^{-1}$, not $\text{s}$.

• For $m/k$:

$$\left[\frac{m}{k}\right] = \frac{\text{kg}}{\text{kg s}^{-2}} = \text{s}^2$$

Taking a square root gives

$$\left[\sqrt{\frac{m}{k}}\right] = \text{s}$$

Therefore only

$$T = 2\pi \sqrt{\frac{m}{k}}$$

has units of time, so the homogeneous option is D.

Key Takeaways

• Homogeneity means matching base units on both sides.
• Use definitions of constants (here $k = F/x$) to find their units.
• $2\pi$ is dimensionless, so it never changes the units.
• Period must come out in seconds.

Common Mistakes

• Using $k = Fx$ instead of $k = F/x$, which gives the wrong units for $k$.
• Forgetting to convert $\text{N}$ into base units $\text{kg m s}^{-2}$.
• Assuming the correct physics formula without checking the dimensional requirement (the question only asks about homogeneity).

Things to Be Careful About

• Always simplify fully to base units (kg, m, s) before deciding.
• Remember that square roots halve powers: $\sqrt{\text{s}^2} = \text{s}$ but $\sqrt{\text{s}^{-2}} = \text{s}^{-1}$.
• Distinguish between $m/k$ and $k/m$; they are reciprocals with opposite powers of time.

Background Concept

Newton's second law links the resultant (net) force on an object to its acceleration:

$$F_{\text{resultant}} = ma$$

where:

• $F_{\text{resultant}}$ is the vector sum of all forces acting on the object,
• $m$ is the (constant) mass,
• $a$ is the acceleration.

If $m$ is fixed, then the resultant force is directly proportional to the acceleration. Therefore:

• constant $a$ (\Rightarrow) constant $F_{\text{resultant}}$,
• zero $a$ (\Rightarrow) zero $F_{\text{resultant}}$.

Understanding the Question

The object starts from rest at point $P$ and then moves away with uniform acceleration. The question asks what happens to the resultant force while it is moving.

Key information:

• mass is fixed (constant $m$),
• acceleration is uniform (constant $a$),
• it moves away from $P$ (so acceleration is not zero during the motion described).

Approach

1. Translate “uniform acceleration” into “constant acceleration”.
2. Use $F = ma$ with constant $m$ to infer how $F$ behaves.
3. Decide whether the force is zero or non-zero from whether $a$ is zero or non-zero.

Step-by-Step Reasoning

• “Uniform acceleration” means:

$$a = \text{constant}$$

• Apply Newton's second law:

$$F_{\text{resultant}} = ma$$

• Since $m$ is fixed and $a$ is constant, $F_{\text{resultant}}$ must be constant.
• The object is described as accelerating away from $P$, so $a \neq 0$ during the motion. Hence $F_{\text{resultant}}$ is constant but not zero.

Therefore the correct option is B.

Key Takeaways

• Uniform (constant) acceleration implies a constant resultant force only if mass is constant.
• Zero resultant force corresponds to zero acceleration (constant velocity, possibly at rest).

Common Mistakes

• Choosing D by thinking “moving at constant acceleration” might mean “moving steadily”: but steady motion is constant velocity, not uniform acceleration.
• Choosing A by confusing increasing velocity with increasing force: velocity can increase linearly in time even when force is constant.
• Choosing C: force proportional to displacement is the hallmark of simple harmonic motion (spring-like restoring force), not constant acceleration.

Things to Be Careful About

• “Uniform acceleration” means constant acceleration, not “constant speed”.
• The question asks about the resultant force (net force), not individual forces that might change while still producing a constant net force.
• Mass is stated fixed; if mass were changing (e.g. rocket), constant acceleration would not necessarily imply constant force.

Background Concept

A projectile’s launch velocity is a vector. If it makes an angle $\theta$ to the horizontal, its components are

$$v_x = v\cos\theta, \qquad v_y = v\sin\theta$$

A useful relation comes from dividing these:

$$\tan\theta = \frac{v_y}{v_x} \;\;\Rightarrow\;\; v_y = v_x\tan\theta$$

This is often quickest when one component (here, horizontal) and the angle are given.

Understanding the Question

You are told the projectile is launched at $25^\circ$ above the horizontal, and its horizontal component of velocity at launch is $13\ \text{m s}^{-1}$. You must find the vertical component at launch, and then choose the matching option.

Approach

Use the component relation

$$v_y = v_x\tan\theta$$

with $v_x = 13\ \text{m s}^{-1}$ and $\theta = 25^\circ$.

Step-by-Step Reasoning

Start from the standard component equations:

$$v_x = v\cos 25^\circ, \qquad v_y = v\sin 25^\circ$$

You are given $v_x = 13$, so $v\cos 25^\circ = 13$.

Now express $v_y$ in terms of $v\cos 25^\circ$:

$$v_y = v\sin 25^\circ = v\cos 25^\circ \tan 25^\circ$$

Substitute $v\cos 25^\circ = 13$:

$$v_y = 13\tan 25^\circ$$

Calculate:

$$v_y = 13 \times 0.466 \approx 6.06\ \text{m s}^{-1} \approx 6.1\ \text{m s}^{-1}$$

This corresponds to option B.

Key Takeaways

• Velocity is a vector; resolve it using $\cos$ for adjacent (horizontal) and $\sin$ for opposite (vertical).
• If $v_x$ and the launch angle are known, the fastest method is $v_y = v_x\tan\theta$.

Common Mistakes

• Using $v_y = v_x\cos\theta$ or $v_y = v_x\sin\theta$ (mixing up trig ratios).
• Using $\tan\theta = v_x/v_y$ instead of $v_y/v_x$.
• Forgetting to round to match the given options.

Things to Be Careful About

• The given $13\ \text{m s}^{-1}$ is a component, not the total speed.
• Check angle is to the horizontal (so horizontal uses $\cos$, vertical uses $\sin$).
• Keep units as $\text{m s}^{-1}$ throughout.

Background Concept

A velocity–time graph contains two key pieces of information:

• The gradient gives acceleration.
• The (signed) area between the graph and the time axis gives displacement.

Mathematically, displacement is the integral of velocity:

$$s = \int v\,dt$$

On a graph, this becomes a geometric area calculation. Regions above the time axis (positive velocity) contribute positive displacement; regions below the axis (negative velocity) contribute negative displacement.

Understanding the Question

The ball starts from rest at $t=0$ at position $X$. The given $v$-$t$ graph shows:

• From $0$ to $1.0\,\text{s}$: velocity increases linearly from $0$ to $+10\,\text{m s}^{-1}$ (falling).
• At $1.0\,\text{s}$: the ball bounces and its velocity instantaneously changes to $-5\,\text{m s}^{-1}$ (rebound upward, hence negative on this graph).
• From $1.0$ to $1.5\,\text{s}$: velocity increases from $-5$ to $0\,\text{m s}^{-1}$ (still moving upward but slowing to rest at the top).

We are asked for the total displacement from $X$ at $t=1.5\,\text{s}$, i.e. net change in position, not total distance travelled.

Approach

Compute the signed area under the graph from $t=0$ to $1.5\,\text{s}$:

1. Split the graph into simple shapes (triangles).
2. Find each area using $\tfrac{1}{2} \times \text{base} \times \text{height}$.
3. Apply a negative sign to the part below the time axis.
4. Add areas to get net displacement.

Step-by-Step Reasoning

1) From $0$ to $1.0\,\text{s}$
This section is a triangle above the axis with base $1.0\,\text{s}$ and height $10\,\text{m s}^{-1}$.

$$A_1 = \tfrac{1}{2} \times 1.0 \times 10 = 5.0\,\text{m}$$

This is positive displacement (velocity is positive).

2) From $1.0$ to $1.5\,\text{s}$
This section is a triangle below the axis. The speed decreases from $5\,\text{m s}^{-1}$ to $0$ over $0.50\,\text{s}$.
Magnitude of area:

$$\text{area magnitude} = \tfrac{1}{2} \times 0.50 \times 5.0 = 1.25\,\text{m}$$

Because the graph is below the axis, this contributes negative displacement:

$$A_2 = -1.25\,\text{m}$$

3) Net displacement at $1.5\,\text{s}$

$$s = A_1 + A_2 = 5.0 - 1.25 = 3.75\,\text{m}$$

So the ball is still $3.75\,\text{m}$ in the positive displacement direction from its starting point at $t=1.5\,\text{s}$.

Key Takeaways

• Displacement over a time interval equals the signed area under a velocity–time graph.
• After a bounce, the velocity can change sign; the area below the axis must be subtracted.
• The question asks for displacement (net change), not distance (sum of magnitudes).

Common Mistakes

• Adding magnitudes of both triangular areas (that would give distance travelled, not displacement).
• Forgetting the negative sign for the region below the time axis.
• Using $v \times t$ with the final velocity instead of using the triangular area (velocity is changing, so you must use the area of the shape).

Things to Be Careful About

• Check whether the graph uses the convention that downward is positive (as here) or upward is positive; either way, the area method still works as long as you keep the signs consistent.
• Use the correct base widths: $0$ to $1.0\,\text{s}$ and $1.0$ to $1.5\,\text{s}$.
• Units: $(\text{m s}^{-1})\times(\text{s})=\text{m}$, so the area is directly a displacement in metres.

Background Concept

Acceleration is defined as the rate of change of velocity. As velocity is a vector quantity (it has both magnitude and direction), acceleration is also a vector. Mathematically, for motion in a straight line, the average acceleration is given by $a = \frac{\Delta v}{\Delta t}$, where $\Delta v$ is the change in velocity and $\Delta t$ is the time taken. The instantaneous acceleration is the limit of this as $\Delta t$ approaches zero.

Understanding the Question

The question asks for the correct definition of acceleration from four options. It is a multiple-choice question testing the precise wording of a fundamental concept in kinematics.

Approach

Compare each option to the standard definition of acceleration:

• Option A: "change in velocity per unit time" — this directly matches the definition.
• Option B: "rate of change of speed per unit time" — speed is a scalar; acceleration involves velocity.
• Option C: "rate of change of velocity per unit time" — this is redundant because "rate of change" already means "per unit time".
• Option D: "resultant force per unit mass" — this describes acceleration via Newton's second law, not a definition.

Step-by-Step Reasoning

1. Recall the formal definition: Acceleration is the rate of change of velocity, or equivalently, the change in velocity divided by the time interval.
2. Option A uses "change in velocity per unit time", which is a correct and concise statement of that definition.
3. Option B incorrectly uses "speed" instead of "velocity". Since speed does not include direction, it is not the same as velocity.
4. Option C is incorrect because it says "rate of change ... per unit time", which is tautological (rate of change already includes per unit time). The phrase "rate of change of velocity" alone would be correct, but adding "per unit time" makes it wrong.
5. Option D is a statement of Newton's second law ($F = ma$ rearranged to $a = F/m$), which relates acceleration to force, but it is not the definition of acceleration.

Key Takeaways

• Acceleration is defined as the rate of change of velocity (vector).
• It is important to distinguish between velocity and speed.
• Definitions in physics must be precise; redundant phrases or substitution of scalar for vector quantities can change the meaning.

Common Mistakes

• Confusing acceleration with the rate of change of speed (e.g., from option B).
• Accepting option C because "rate of change of velocity" is familiar; but the extra "per unit time" makes it incorrect.
• Using Newton's second law as a definition instead of the kinematic definition.

Things to Be Careful About

• The term "rate of change" already implies division by time, so do not add "per unit time" after it.
• Acceleration is a vector; definitions must refer to velocity, not speed.
• In multiple-choice questions, read each option carefully for such subtle differences.

Background Concept

In a one-dimensional collision:

• Momentum is conserved (always, if no external resultant impulse):

$$mu_1 + mu_2 = mv_1 + mv_2$$

• If the collision is perfectly elastic, kinetic energy is also conserved:

$$\frac{1}{2}mu_1^2 + \frac{1}{2}mu_2^2 = \frac{1}{2}mv_1^2 + \frac{1}{2}mv_2^2$$

For a head-on perfectly elastic collision between two equal masses, these two conservation laws together imply a very useful result:

• the two objects exchange velocities after the collision.

Understanding the Question

Ball $P$ and ball $Q$ have equal mass and move along the same straight line towards each other.

• $P$ initially moves to the right at $1.30\ \text{m s}^{-1}$.
• $Q$ initially moves to the left at $0.50\ \text{m s}^{-1}$.
  The collision is perfectly elastic and momentum is conserved, so we must pick the option that matches the correct final velocities.

Approach

Use a sign convention (right positive), then apply the equal-mass perfectly elastic collision rule: swap the velocities. As a quick check, confirm that both total momentum and total kinetic energy are unchanged.

Step-by-Step Reasoning

1. Choose right as positive.

$$u_P = +1.30\ \text{m s}^{-1},\qquad u_Q = -0.50\ \text{m s}^{-1}$$

1. For equal masses in a perfectly elastic head-on collision, the velocities exchange:

$$v_P = u_Q = -0.50\ \text{m s}^{-1}$$ $$v_Q = u_P = +1.30\ \text{m s}^{-1}$$

This means $P$ rebounds left at $0.50\ \text{m s}^{-1}$, and $Q$ moves right at $1.30\ \text{m s}^{-1}$.

1. Check with momentum conservation (optional but reassuring):
   Initial momentum:

$$p_i = m(1.30) + m(-0.50) = 0.80m$$

Final momentum:

$$p_f = m(-0.50) + m(1.30) = 0.80m$$

Matches.

1. Check with kinetic energy conservation (also optional):
   Initial:

$$E_{k,i} = \frac{1}{2}m(1.30^2 + 0.50^2)$$

Final uses the same two speeds, so $E_{k,f} = E_{k,i}$.

So the correct diagram is the one showing $P$ moving left at $0.50\ \text{m s}^{-1}$ and $Q$ moving right at $1.30\ \text{m s}^{-1}$, which is Option D.

Key Takeaways

• In 1D, momentum is conserved in collisions (if external impulse is negligible).
• Perfectly elastic means both momentum and kinetic energy are conserved.
• For equal masses in a head-on perfectly elastic collision, the objects exchange velocities.

Common Mistakes

• Getting signs wrong (e.g. taking both velocities as positive even though one is to the left).
• Thinking the objects must move off together with a common speed (that corresponds to a perfectly inelastic collision, not elastic).
• Using momentum conservation alone: in an elastic collision you need the elastic condition as well (or use the known equal-mass result).

Things to Be Careful About

• Always define a positive direction before substituting into equations.
• Distinguish between speed (always positive) and velocity (can be negative in 1D).
• The “exchange of velocities” shortcut is valid only for equal masses and a head-on perfectly elastic collision.

Background Concept

For vertical motion near Earth with air resistance negligible, the only significant force on the ball after it is struck is its weight. This produces a constant acceleration of magnitude $g \approx 9.81\ \text{m s}^{-2}$ downward.

For constant acceleration, the SUVAT kinematics equations apply. The most direct one here relates speeds and displacement:

$$v^2 = u^2 + 2as$$

where $u$ is initial speed, $v$ is final speed, $a$ is acceleration, and $s$ is displacement in the chosen positive direction.

Understanding the Question

The ball is hit downwards with speed $2.4\ \text{m s}^{-1}$ from a height of $0.90\ \text{m}$ above the ground. It then falls the $0.90\ \text{m}$ to the ground with constant downward acceleration $g$.

We are asked for the speed on impact (magnitude of velocity) when it reaches the ground.

Approach

Choose downward as the positive direction so that $a = +g$ and the displacement to the ground is $s = +0.90\ \text{m}$. Use

$$v^2 = u^2 + 2as$$

because time is not needed and not given.

Step-by-Step Reasoning

Initial speed (downwards):

$$u = 2.4\ \text{m s}^{-1}$$

Acceleration (downwards):

$$a = g = 9.81\ \text{m s}^{-2}$$

Displacement to the ground (downwards):

$$s = 0.90\ \text{m}$$

Substitute into $v^2 = u^2 + 2as$:

$$v^2 = (2.4)^2 + 2(9.81)(0.90)$$

Compute:

$$v^2 = 5.76 + 17.658 = 23.418$$

So

$$v = \sqrt{23.418} = 4.84\ \text{m s}^{-1}$$

Matching to the options, $4.84\ \text{m s}^{-1}$ rounds to $4.8\ \text{m s}^{-1}$, which is option B.

Key Takeaways

• With negligible air resistance, vertical motion is constant acceleration with $a = g$ downward.
• When time is not required, $v^2 = u^2 + 2as$ is often the quickest SUVAT equation.
• “Speed” means the magnitude, so take the positive square root.

Common Mistakes

• Using $s = 0.90\ \text{m}$ but taking $a = -9.81\ \text{m s}^{-2}$ without a consistent sign convention.
• Forgetting to square the initial speed (using $u$ instead of $u^2$).
• Choosing the wrong SUVAT equation and introducing an unnecessary unknown time.

Things to Be Careful About

• Keep directions consistent: pick up or down as positive and stick to it.
• Take the square root at the end to get $v$ from $v^2$.
• Round to match the options (here $4.84\to 4.8\ \text{m s}^{-1}$).

Background Concept

When an object moves at constant velocity, its acceleration is zero, so by Newton’s second law the resultant force is zero.

For a body moving through a fluid, typical vertical forces are:

• weight $W = mg$ (downwards)
• upthrust $U$ (upwards)
• drag/resistive force $D$ (opposes the motion; here upwards because the ball moves down)

Upthrust is given by Archimedes’ principle:

$$U = \rho g V$$

where $\rho$ is the fluid density and $V$ is the volume of fluid displaced. If the object is fully submerged and the fluid density is uniform, then neither $\rho$ nor $V$ changes with depth, so $U$ does not change with depth.

Understanding the Question

A ball is falling through a liquid at constant speed (so it has reached terminal speed). The liquid density is stated to be uniform. We must choose which of the given statements about how the forces compare or vary with depth is correct.

Approach

1. Use “constant speed” to infer $a=0$ and hence the forces must balance.
2. Use $U = \rho gV$ and the fact that $\rho$ is uniform (and $V$ is fixed for the ball) to decide whether upthrust changes with depth.
3. Check each option against these facts.

Step-by-Step Reasoning

• Constant speed means acceleration is zero:

$$\sum F = ma = 0$$

• Taking downward as positive, weight acts down and upthrust and drag act up:

$$W - (U + D) = 0 \quad \Rightarrow \quad W = U + D$$

So the weight is equal to (not greater than) the sum of drag and upthrust.

• Upthrust:

$$U = \rho g V$$

Given the liquid has uniform density $\rho$ and the ball’s volume $V$ is constant, $U$ is constant with depth.

• Therefore:
  • A is false: drag depends mainly on speed (and fluid properties); since speed is constant, there is no reason for drag to increase with depth here.
  • B is false: it incorrectly adds forces without considering direction (it would imply drag balances both weight and upthrust).
  • C is true: upthrust is constant with depth in a uniform-density liquid.
  • D is false: at constant speed, $W = U + D$, not greater.

Hence the correct statement is C.

Key Takeaways

• Constant speed $\Rightarrow a=0 \Rightarrow$ resultant force is zero.
• At terminal speed for a falling object in a fluid: $W = U + D$.
• In a uniform-density fluid, upthrust $U = \rho g V$ is independent of depth (for a fully submerged object of fixed volume).

Common Mistakes

• Forgetting that “constant speed” implies zero resultant force.
• Treating forces as scalars and writing incorrect sums like $D = W + U$ (ignoring directions).
• Thinking upthrust must increase with depth because pressure increases with depth: although pressure does increase, the pressure difference between top and bottom of the object in a uniform-density fluid depends on $\rho g$ and the object’s size, not on absolute depth.

Things to Be Careful About

• Drag force depends on speed (e.g. proportional to $v$ or $v^2$ in common models); if $v$ is constant, $D$ is constant.
• The conclusion that upthrust is constant assumes the ball is not compressing and remains fully submerged with constant displaced volume.
• Pressure increases with depth, but upthrust depends on the pressure difference across the object, not the absolute pressure.

Background Concept

Weight is the gravitational force on a mass:

$$W = mg$$

where $m$ is mass (in $\text{kg}$) and $g \approx 9.81\ \text{m s}^{-2}$. The unit of weight is the newton (N), and prefixes are useful here: $1\ \text{MN} = 10^6\ \text{N}$.

When identical cubes are stacked to form a larger cube, the number of small cubes is found from a scale factor. If the large cube is $k$ times longer along each edge, then the total number of small cubes is $k^3$.

Understanding the Question

Each small cube has mass $1.0\ \text{kg}$ and side length $5.0\ \text{cm}$. They are stacked to form a larger cube of side $2.0\ \text{m}$. The task is to find the weight of the entire large cube, i.e. find its total mass and multiply by $g$.

Approach

1. Convert the small cube side length to metres so the ratio with $2.0\ \text{m}$ is consistent.
2. Find how many small cubes fit along one edge of the large cube.
3. Cube that number to get the total number of small cubes.
4. Multiply by $1.0\ \text{kg}$ to get total mass.
5. Use $W = mg$ and express the result in kN/MN to match the options.

Step-by-Step Reasoning

Small cube side:

$$5.0\ \text{cm} = 0.050\ \text{m}$$

Number of cubes along one edge of the large cube:

$$N_{\text{edge}} = \frac{2.0}{0.050} = 40$$

Because it is a cube, the stacking happens in three dimensions, so total number of cubes is:

$$N = 40^3 = 64000 = 6.4 \times 10^4$$

Total mass:

$$m = N \times 1.0\ \text{kg} = 6.4 \times 10^4\ \text{kg}$$

Weight:

$$W = mg = (6.4 \times 10^4)(9.81) \approx 6.28 \times 10^5\ \text{N}$$

Convert to MN:

$$6.28 \times 10^5\ \text{N} = 0.628\ \text{MN} \approx 0.63\ \text{MN}$$

So the closest option is C.

Key Takeaways

• Use a linear scale factor and cube it to count small cubes in a large cube.
• Find total mass by multiplying number of cubes by mass per cube.
• Weight is found from $W = mg$, and unit prefixes (kN, MN) must be handled carefully.

Common Mistakes

• Forgetting to convert $5.0\ \text{cm}$ to metres, leading to the wrong scale factor.
• Using $40^2$ instead of $40^3$ (counting layers but not the full 3D stack).
• Treating $g$ as $10$ without checking whether the rounding changes the option choice significantly.
• Incorrect conversion between N, kN, MN, and GN.

Things to Be Careful About

• Ratios must use the same unit: $2.0\ \text{m}$ with $0.050\ \text{m}$.
• The final weight is about $6 \times 10^5\ \text{N}$; this is $\sim 10^6\ \text{N}$ order of magnitude, so MN (not kN or GN) is the sensible prefix.
• Rounding: $0.628\ \text{MN}$ rounds to $0.63\ \text{MN}$, matching the option precisely.

Background Concept

For a liquid of density $\rho$ at rest, the pressure increase with depth $h$ is

$$\Delta p = \rho g h$$

where $g$ is the gravitational field strength. If there are layers of different liquids, the total pressure increase from the top to the bottom is the sum of the pressure increases across each layer:

$$\Delta p_{\text{total}} = \rho_1 g h_1 + \rho_2 g h_2 + \cdots$$

This works because pressure is continuous at the boundary between the liquids, and each layer contributes according to its own density and thickness.

Understanding the Question

A vertical borehole is $2000\ \text{m}$ deep. The top part is oil of density $830\ \text{kg m}^{-3}$ with unknown depth $x$, and the rest is water of density $1000\ \text{kg m}^{-3}$ with depth $(2000-x)$.

The pressure at the bottom due to the liquids (so the hydrostatic/gauge pressure from the columns of oil and water) is given as $17.5\ \text{MPa} = 17.5\times 10^6\ \text{Pa}$. We must find $x$ and then choose the correct option.

Approach

1. Write an expression for the pressure at the bottom as the sum of the hydrostatic pressures from the oil column and the water column.
2. Substitute the given densities and depths ($x$ for oil and $2000-x$ for water).
3. Solve the resulting linear equation for $x$.
4. Compare with the options.

Step-by-Step Reasoning

Pressure contribution from oil (depth $x$):

$$p_{\text{oil}} = \rho_{\text{oil}} g x = 830 g x$$

Pressure contribution from water (depth $2000-x$):

$$p_{\text{water}} = \rho_{\text{water}} g (2000-x) = 1000 g (2000-x)$$

Total pressure due to liquids at the bottom:

$$17.5\times 10^6 = 830gx + 1000g(2000-x)$$

Expand and simplify inside the brackets:

$$830gx + 1000g(2000-x)=g\left(830x + 2.00\times 10^6 -1000x\right)=g\left(2.00\times 10^6 -170x\right)$$

So

$$17.5\times 10^6 = g(2.00\times 10^6 -170x)$$

Divide by $g$ (taking $g=9.81\ \text{m s}^{-2}$):

$$\frac{17.5\times 10^6}{9.81}=2.00\times 10^6 -170x$$

Numerically,

$$1.78\times 10^6 \approx 2.00\times 10^6 -170x$$

Rearrange:

$$170x = 2.00\times 10^6 - 1.78\times 10^6 = 2.16\times 10^5$$ $$x \approx \frac{2.16\times 10^5}{170} \approx 1.27\times 10^3\ \text{m}$$

So the oil depth is about $1270\ \text{m}$, which corresponds to option D.

Key Takeaways

• Hydrostatic pressure increases with depth according to $\Delta p = \rho g h$.
• For multiple liquid layers, add the contributions: $\Delta p_{\text{total}} = \sum \rho_i g h_i$.
• Convert MPa to Pa correctly: $1\ \text{MPa} = 10^6\ \text{Pa}$.

Common Mistakes

• Using a single density for the full $2000\ \text{m}$ instead of treating oil and water separately.
• Forgetting to convert $17.5\ \text{MPa}$ to $\text{Pa}$.
• Writing the water depth as $x$ instead of $(2000-x)$.
• Using $g=10\ \text{m s}^{-2}$ here would not match the options; this question effectively requires $g\approx 9.8$ or $9.81\ \text{m s}^{-2}$.

Things to Be Careful About

• The phrase “pressure due to the liquids” indicates hydrostatic (gauge) pressure from the columns, not including atmospheric pressure.
• Keep units consistent: $\rho$ in $\text{kg m}^{-3}$, $h$ in $\text{m}$, giving pressure in $\text{Pa}$.
• When you simplify, check the sign: since water is denser than oil, increasing the oil depth should reduce the bottom pressure (because it replaces denser water), which is consistent with the term $(2.00\times 10^6 -170x)$ decreasing as $x$ increases.

Background Concept

The moment (torque) of a force about a pivot is

$$M = F \times r_\perp$$

where $r_\perp$ is the perpendicular distance from the pivot to the line of action of the force. For a uniform rod pivoted at one end, its weight $W=mg$ acts vertically downward at its centre of mass, which is at the midpoint of the rod.

A key idea: as the rod rotates, the force $W$ stays vertical, but the perpendicular distance from the pivot to the vertical line of action changes with angle.

Understanding the Question

You are told a rod is pivoted at one end and rotated in a vertical plane. The angle $\theta$ is measured between the rod and the horizontal, increasing from $0^\circ$ (rod horizontal) to $180^\circ$ (rod horizontal but pointing the opposite way). The question asks which graph of moment $M$ (due to the rod’s weight only) against $\theta$ is correct.

So we need how the lever arm of the weight changes with $\theta$, including the change of sign (direction) of the turning effect.

Approach

1. Take the rod length as $L$ and weight as $W$.
2. Locate the centre of mass: at distance $L/2$ from the pivot along the rod.
3. Write the perpendicular distance from pivot to the vertical line of action of $W$ in terms of $\theta$.
4. Use $M = Wr_\perp$ and decide the sign of $M$ for $\theta$ from $0^\circ$ to $180^\circ$.
5. Match the resulting shape to the given options.

Step-by-Step Reasoning

Let the pivot be at the origin. When the rod makes angle $\theta$ to the horizontal, the midpoint has horizontal coordinate

$$x = \frac{L}{2}\cos\theta$$

The weight acts vertically downward through this point. A vertical line has constant $x$, so the perpendicular distance from the pivot (origin) to that vertical line is simply $|x|$.

So the magnitude of the moment is

$$|M| = W\left|\frac{L}{2}\cos\theta\right|$$

But the question’s graphs include negative values, so we consider direction:

• At $\theta = 0^\circ$, the rod points to the right; the weight acts downward at $x=+L/2$, producing (say) an anticlockwise moment taken as positive. So $M$ is positive and maximum.
• At $\theta = 90^\circ$, the rod is vertical; the weight’s line of action passes through the pivot ($x=0$), so the moment is zero.
• At $\theta = 180^\circ$, the rod points to the left; the midpoint is at $x=-L/2$ and the moment reverses direction compared with $\theta=0^\circ$, so $M$ is negative with the same magnitude.

This is captured by

$$M = W\left(\frac{L}{2}\cos\theta\right)$$

So $M$ follows a cosine variation: smooth curve from positive at $0^\circ$, through zero at $90^\circ$, to negative at $180^\circ$.

Among the options, this corresponds to the smooth curve crossing zero at $90^\circ$ (not a straight line), i.e. option D.

Key Takeaways

• Moment depends on the perpendicular distance to the line of action, not the distance along the rod.
• For a uniform rod pivoted at one end, the centre of mass is at $L/2$.
• As the rod rotates, the lever arm of the vertical weight is the horizontal component: proportional to $\cos\theta$.
• Moment changes sign when the line of action moves from one side of the pivot to the other.

Common Mistakes

• Using $\sin\theta$ instead of $\cos\theta$ (mixing up which component is perpendicular to the vertical force).
• Assuming the moment is always positive (ignoring the reversal of turning direction after $90^\circ$).
• Thinking the moment is maximum at $90^\circ$; actually it is zero there because the line of action passes through the pivot.
• Choosing the straight-line graph (option C) rather than recognising the trigonometric dependence.

Things to Be Careful About

• Always identify the line of action of the force (here vertical) and measure the perpendicular distance to that line.
• Decide and keep a consistent sign convention for moments; the sign change is an essential clue for selecting the correct graph.
• The functional form here is smooth (cosine), so any graph with sharp corners or linear behaviour is unlikely.