Physics 9702/12 — May/June 2025

Background Concept

A physical quantity is something that can be measured and expressed using a number and a unit. For example, a length might be written as $2.3\ \text{m}$: here $2.3$ is the magnitude (numerical value) and $\text{m}$ is the unit.

Some physical quantities are scalars (magnitude only, e.g. mass, time, temperature), and some are vectors (magnitude and direction, e.g. displacement, velocity, force). Because not all quantities are vectors, direction is not a requirement for every physical quantity.

Prefixes (like milli-, kilo-) are optional ways of writing units and are not required.

Understanding the Question

The question asks for what all physical quantities must have, so we need a feature that is true for both scalars and vectors, in every case.

Approach

Use the definition of a physical quantity: it must be measurable and therefore expressible as a number with a unit. Then check each option for something that is always present.

Step-by-Step Reasoning

• Option A: “direction and magnitude” is only true for vectors, not for scalars like mass. So not all quantities.
• Option B: “direction and unit” again fails for scalars (no direction).
• Option C: “magnitude and a prefix” is false because prefixes are optional (e.g. $2\ \text{m}$ has no prefix).
• Option D: “magnitude and a unit” matches the definition of any physical quantity.

Therefore the correct option is D.

Key Takeaways

• Every physical quantity must be written as number + unit.
• Direction is only required for vector quantities.
• Prefixes are a choice, not a necessity.

Common Mistakes

• Thinking all physical quantities are vectors (so choosing an option involving direction).
• Confusing “unit” with “prefix”: prefixes modify units but are not always used.

Things to Be Careful About

• Words like “all” mean you must pick something universally true, not just true for many common examples.
• Remember the scalar/vector distinction: vectors have direction; scalars do not.

Background Concept

To convert units, rewrite every prefix and unit in terms of SI base units.

• Prefix: $\text{k} = 10^3$.
• Length: $1\ \text{mm} = 10^{-3}\ \text{m}$.
• A squared (or inverse-squared) unit must be converted using index laws: if $x = 10^a y$, then $x^{-2} = 10^{-2a} y^{-2}$.

Here $\text{N m}^{-2}$ is the SI unit of pressure/stress (pascal).

Understanding the Question

We are given a stress/pressure-like unit: $0.25\ \text{kN mm}^{-2}$.

We must express it in $\text{N m}^{-2}$. This means:

1. convert $\text{kN}$ to $\text{N}$,
2. convert $\text{mm}^{-2}$ to $\text{m}^{-2}$,
3. multiply the conversion factors.

Then choose the matching option.

Approach

Write

$$0.25\ \text{kN mm}^{-2} = 0.25\times (\text{kN})\times (\text{mm}^{-2})$$

Convert each bracket to SI:

• $\text{kN} \to \text{N}$ gives a factor of $10^3$.
• $\text{mm}^{-2} \to \text{m}^{-2}$ gives a factor of $10^6$ (because the millimetre is $10^{-3}$ m and the power is $-2$).
  Then multiply the factors and compare with the options.

Step-by-Step Reasoning

1. Convert the force unit:

$$1\ \text{kN} = 10^3\ \text{N}$$

So

$$0.25\ \text{kN mm}^{-2} = 0.25\times 10^3\ \text{N mm}^{-2}$$

1. Convert the area unit (carefully with the power $-2$):

$$1\ \text{mm} = 10^{-3}\ \text{m}$$

Raise both sides to the power $-2$:

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

1. Combine conversions:

$$0.25\times 10^3\ \text{N mm}^{-2} = 0.25\times 10^3\times 10^{6}\ \text{N m}^{-2}$$ $$= 0.25\times 10^{9}\ \text{N m}^{-2} = 2.5\times 10^{8}\ \text{N m}^{-2}$$

This is $250\ 000\ 000\ \text{N m}^{-2}$, which matches option D.

Key Takeaways

• Always convert prefixes first: $\text{k} = 10^3$.
• When converting $\text{mm}^{-2}$ to $\text{m}^{-2}$, the factor becomes $10^{6}$ (because $(10^{-3})^{-2} = 10^{6}$).
• Combine powers of ten systematically.

Common Mistakes

• Treating $\text{mm}^{-2}$ as if it converts with $10^{-3}$ instead of applying the power $-2$.
• Forgetting that the negative power flips the sign: $(10^{-3})^{-2}$ becomes $10^{+6}$, not $10^{-6}$.
• Converting $\text{kN}$ incorrectly (e.g. using $10^{-3}$).

Things to Be Careful About

• Apply index laws correctly: $(ab)^n = a^n b^n$ and $(10^a)^n = 10^{an}$.
• Keep units attached during conversion to avoid losing track of whether the factor should multiply or divide.
• Make sure the final unit is exactly $\text{N m}^{-2}$ before choosing the option.

Background Concept

Density is defined as mass per unit volume:

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

When a quantity is calculated from measured values, its uncertainty depends on how those measurements combine.

For percentage uncertainties:

• For multiplication/division, percentage uncertainties add.
• If a measured quantity is raised to a power $n$, its percentage uncertainty is multiplied by $n$.

So if $V = L^3$, then:

$$\frac{\Delta V}{V} = 3\frac{\Delta L}{L}$$

Understanding the Question

You are given:

• mass $m = 975\ \text{g} \pm 10\ \text{g}$
• cube side length $L = 50\ \text{mm} \pm 1\ \text{mm}$

You must find:

1. the density of the cube
2. the percentage uncertainty in that density

The options all already suggest the density value is about $7.8\ \text{g cm}^{-3}$, so the discriminating step is the uncertainty.

Approach

1. Convert the side length into cm so that the final density can be in $\text{g cm}^{-3}$.
2. Calculate the volume $V = L^3$.
3. Calculate density $\rho = m/V$.
4. Find percentage uncertainty in mass.
5. Find percentage uncertainty in length, then multiply by 3 to get percentage uncertainty in volume.
6. Add the percentage uncertainties in mass and volume to get the percentage uncertainty in density.

Step-by-Step Reasoning

1) Unit conversion and volume

The side length is given in mm, but the density answers are in $\text{g cm}^{-3}$.

$$50\ \text{mm} = 5.0\ \text{cm}$$

Volume of a cube:

$$V = L^3 = (5.0\ \text{cm})^3 = 125\ \text{cm}^3$$

2) Density

$$\rho = \frac{m}{V} = \frac{975\ \text{g}}{125\ \text{cm}^3} = 7.8\ \text{g cm}^{-3}$$

(Indeed consistent with steel.)

3) Percentage uncertainty in density

Because $\rho = m / V$, percentage uncertainties in $m$ and $V$ add.

Mass percentage uncertainty:

$$\frac{\Delta m}{m} \times 100 = \frac{10}{975} \times 100 \approx 1.03\% \approx 1.0\%$$

Length percentage uncertainty:

$$\frac{\Delta L}{L} \times 100 = \frac{1}{50} \times 100 = 2.0\%$$

Volume depends on $L^3$, so multiply by 3:

$$\frac{\Delta V}{V} \times 100 = 3(2.0\%) = 6.0\%$$

Now add to get density percentage uncertainty:

$$\frac{\Delta \rho}{\rho} \times 100 \approx 1.0\% + 6.0\% = 7.0\%$$

This matches option B.

Key Takeaways

• Convert units early so your derived unit (here $\text{g cm}^{-3}$) comes out naturally.
• For powers: if $y = x^n$, then percentage uncertainty in $y$ is $n$ times percentage uncertainty in $x$.
• For products/quotients: add percentage uncertainties.

Common Mistakes

• Using $\Delta V/V = \Delta L/L$ instead of multiplying by 3 for $V = L^3$.
• Forgetting to convert $50\ \text{mm}$ to $5.0\ \text{cm}$ (leading to a volume in $\text{mm}^3$ and the wrong density unit).
• Combining uncertainties incorrectly (e.g. subtracting, or adding absolute uncertainties instead of percentage uncertainties for division).

Things to Be Careful About

• The uncertainty in length is relatively large compared to mass, and because volume depends on $L^3$, it dominates the final uncertainty.
• Rounding: keep a guard digit (e.g. $1.03\%$) and round at the end to match the option precision.
• Ensure the density unit is $\text{g cm}^{-3}$, not $\text{g mm}^{-3}$ or $\text{kg m}^{-3}$ unless explicitly asked.

Background Concept

For an equation to be homogeneous, the dimensions (base units) of the left-hand side (LHS) must match the dimensions of the right-hand side (RHS). Numerical constants such as $2\pi$ have no units and do not affect dimensional balance.

A useful way to do this quickly is:

• write each quantity in base dimensions (e.g. time $[T]$, length $[L]$),
• simplify the dimensions of any brackets,
• equate powers of each base dimension on both sides.

Understanding the Question

You are given

$$T = 2\pi \left( \frac{L}{g} \right)^n$$

where:

• $T$ is the time period, so it has dimension $[T]$,
• $L$ is a length, dimension $[L]$,
• $g$ is acceleration, dimension $[L\,T^{-2}]$.

The equation is stated to be homogeneous, so you must choose $n$ so that the RHS has the same dimensions as $T$.

Approach

1. Find the dimensions of $\dfrac{L}{g}$.
2. Raise that to the power $n$.
3. Set the resulting time power equal to $[T]^1$ and solve for $n$.
4. Match to the options.

Step-by-Step Reasoning

1. Dimensions of each quantity:

• $[T] = [T]$
• $[L] = [L]$
• $[g] = [L\,T^{-2}]$

1. Dimensions of the bracket:

$$\left[\frac{L}{g}\right] = \frac{[L]}{[L\,T^{-2}]} = [T^2]$$

So $\dfrac{L}{g}$ has dimensions of time-squared.

1. Apply the power $n$:

$$\left[\left(\frac{L}{g}\right)^n\right] = [T^2]^n = [T^{2n}]$$

The factor $2\pi$ is dimensionless, so it does not change this.

1. Match LHS and RHS dimensions:

$$[T^1] = [T^{2n}] \Rightarrow 2n = 1 \Rightarrow n = \frac{1}{2}$$

So the correct option is C.

Key Takeaways

• Homogeneous equations must have matching dimensions on both sides.
• Acceleration has dimensions $[L\,T^{-2}]$.
• If $\dfrac{L}{g}$ has dimensions $[T^2]$, then the exponent must make it become $[T]$.

Common Mistakes

• Treating $2\pi$ as having units (it is dimensionless).
• Using $[g] = [T^{-2}]$ and forgetting the length dimension.
• Equating $n = 2$ because $T^2$ appears, instead of recognising that $(T^2)^n = T^{2n}$.

Things to Be Careful About

• When a bracket is raised to a power, the dimensions are raised to that power too: $(T^2)^n = T^{2n}$.
• Always cancel dimensions algebraically (e.g. the $[L]$ cancels in $[L]/[L\,T^{-2}]$).
• Make sure you equate the exponent of the time dimension correctly (here, $2n = 1$).

Background Concept

For any signal travelling at constant speed, the distance travelled is

$$\text{distance} = \text{speed} \times \text{time}.$$

Radio waves are electromagnetic waves, and in space (approximately vacuum) they travel at the speed of light,

$$c \approx 3.0 \times 10^8\ \text{m s}^{-1}.$$

If a pulse is sent out and then reflected back (an "echo" method), the measured time is the total time for the pulse to go to the planet and return. The one-way travel time is therefore half of the measured time.

Understanding the Question

A radio pulse is emitted from Earth, reflects off Jupiter, and is detected back on Earth. The time between sending and receiving is $3960\ \text{s}$.

That $3960\ \text{s}$ corresponds to travelling a distance of

$$2d$$

(where $d$ is the Earth-to-Jupiter distance). We must find $d$ and match it to one of the options in km.

Approach

1. Use $c = 3.0 \times 10^8\ \text{m s}^{-1}$ for the radio wave speed.
2. Halve the given time to get the one-way time.
3. Calculate $d = c \times (t/2)$.
4. Convert metres to kilometres and compare with the options.

Step-by-Step Reasoning

The total time is $t = 3960\ \text{s}$, so the one-way time is

$$\frac{t}{2} = \frac{3960}{2} = 1980\ \text{s}.$$

Now apply distance = speed × time:

$$d = c\left(\frac{t}{2}\right) = (3.0 \times 10^8)(1980).$$

Multiply $3.0 \times 1980 = 5940$, so

$$d = 5940 \times 10^8\ \text{m} = 5.94 \times 10^{11}\ \text{m}.$$

Convert to km using $1\ \text{km} = 10^3\ \text{m}$:

$$d = \frac{5.94 \times 10^{11}}{10^3} = 5.94 \times 10^8\ \text{km}.$$

This matches option A.

Key Takeaways

• In "radar"/echo distance problems, the measured time is for the out-and-back journey, so use $t/2$ for one-way distance.
• Electromagnetic waves travel at $c \approx 3.0 \times 10^8\ \text{m s}^{-1}$ in vacuum.
• Convert units carefully: dividing by $10^3$ changes metres to kilometres.

Common Mistakes

• Using $d = ct$ without halving the time, giving a distance twice as large (would lead to option B).
• Converting units the wrong way (multiplying by $10^3$ instead of dividing when going from m to km).
• Using $3.0 \times 10^5\ \text{m s}^{-1}$ (confusing with km s$^{-1}$ or other speeds).

Things to Be Careful About

• The question says the pulse reflects and is detected again on Earth: that explicitly indicates a round trip.
• Keep track of powers of ten when converting $5.94 \times 10^{11}\ \text{m}$ to km.
• Options are in km, so ensure your final comparison is also in km.

Background Concept

A velocity–time graph tells you two key things:

• The gradient of the graph is the acceleration:

$$a = \frac{\Delta v}{\Delta t}$$

• The area under the graph between two times is the displacement (and the distance travelled if the velocity stays positive):

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

For straight-line sections, areas can be found using triangles and rectangles.

Understanding the Question

You are given a $v$–$t$ graph:

• from $t=0$ to $2\ \text{s}$, $v$ increases linearly from $0$ to $12\ \text{m s}^{-1}$,
• from $t=2$ to $6\ \text{s}$, $v$ is constant at $12\ \text{m s}^{-1}$,
• from $t=6$ to $10\ \text{s}$, $v$ decreases linearly back to $0$.

The options describe acceleration or distances in certain time intervals. We check each against what the graph actually implies.

Approach

Use quick checks:

1. For any claim about accelerating, compute the gradient $\Delta v/\Delta t$ on the relevant section.
2. For any claim about distance travelled, compute the area under the graph over the relevant time interval.
3. Also watch for statements about stopping or reversing: that would require $v=0$ for a time (stopping) or negative velocity (reversing).

Step-by-Step Reasoning

Option A: “accelerates for $2\ \text{s}$, then stops for $4\ \text{s}$ and then reverses.”

• It does accelerate for the first $2\ \text{s}$.
• But from $2$ to $6\ \text{s}$ the graph is horizontal at $v=12\ \text{m s}^{-1}$, so it is moving at constant speed, not stopped.
• The velocity never goes negative, so it never reverses.
  So A is false.

Option B: “accelerates at $12\ \text{m s}^{-2}$ for $2\ \text{s}$.”
Acceleration in the first section is the gradient:

$$a = \frac{12-0}{2-0} = 6\ \text{m s}^{-2}$$

So B is false.

Option C: “travels a distance of $36\ \text{m}$ in the first $4\ \text{s}$.”
Distance in first $4\ \text{s}$ is area from $t=0$ to $t=4$.

• From $0$ to $2\ \text{s}$: triangle area

$$\frac{1}{2}\times 2 \times 12 = 12\ \text{m}$$

• From $2$ to $4\ \text{s}$: rectangle area

$$2 \times 12 = 24\ \text{m}$$

Total:

$$12 + 24 = 36\ \text{m}$$

So C is correct.

Option D: “travels a distance of $48\ \text{m}$ in the last $4\ \text{s}$.”
Last $4\ \text{s}$ means $t=6$ to $t=10$, which is a triangle down from $12$ to $0$:

$$\text{area} = \frac{1}{2} \times 4 \times 12 = 24\ \text{m}$$

So D is false.

Therefore the correct statement is C.

Key Takeaways

• Gradient of a $v$–$t$ graph gives acceleration.
• Area under a $v$–$t$ graph gives displacement; if $v \ge 0$ it also equals distance travelled.
• Constant non-zero velocity is motion at steady speed, not stopping.

Common Mistakes

• Using the height of the graph ($v$) as if it were acceleration, instead of the slope.
• Forgetting to split the area into shapes (triangle + rectangle) when the graph changes form.
• Confusing last 4 seconds ($6$ to $10\ \text{s}$) with from $0$ to $4\ \text{s}$.

Things to Be Careful About

• Time intervals: always identify the correct start and end times from the graph.
• Units: area under $v$–$t$ has units $(\text{m s}^{-1})\times \text{s} = \text{m}$.
• “Reverses” requires negative velocity (graph below the time axis), which is not present here.

Background Concept

When forces on an object are constant, the acceleration is constant and is found from Newton’s second law:

$$\sum F = ma$$

Weight acts downward with magnitude $mg$. If an additional upward force acts, the resultant force is the vector sum (taking care with signs). Once $a$ is known, the velocity after time $t$ follows from constant-acceleration kinematics:

$$v = u + at$$

“Speed” means the magnitude of the velocity (always positive).

Understanding the Question

The object is already moving downward at $60\ \text{m s}^{-1}$ when a new constant force of $50\ \text{N}$ upward starts acting.

Given:

• $m = 1.0\ \text{kg}$
• initial velocity at that instant: $u = 60\ \text{m s}^{-1}$ downward
• forces after that instant: weight $mg$ downward and $50\ \text{N}$ upward
• time after force starts: $t = 2.0\ \text{s}$

Find the speed after $2.0\ \text{s}$.

Approach

1. Choose a sign convention (e.g. downward positive).
2. Compute resultant force: $F = mg - 50$.
3. Use $a = F/m$.
4. Use $v = u + at$ to get the velocity after $2.0\ \text{s}$.
5. Convert to speed by taking the magnitude.

Step-by-Step Reasoning

Take downward as positive.

1) Resultant force

• Weight: $+mg = +(1.0)(9.8) = +9.8\ \text{N}$
• Upward force: $-50\ \text{N}$

So

$$F = 9.8 - 50 = -40.2\ \text{N}$$

Negative means the resultant force is actually upward.

2) Acceleration

$$a = \frac{F}{m} = \frac{-40.2}{1.0} = -40.2\ \text{m s}^{-2}$$

So the acceleration is upward (opposite to the initial downward motion), meaning the object is slowing down while moving downward.

3) Velocity after 2.0 s

$$v = u + at = 60 + (-40.2)(2.0) = -20.4\ \text{m s}^{-1}$$

The negative sign shows the object has reversed direction and is moving upward. The speed is therefore

$$|v| = 20.4\ \text{m s}^{-1} \approx 20\ \text{m s}^{-1}$$

So the correct option is A.

Key Takeaways

• Always use the resultant force to find acceleration: $a = \sum F / m$.
• A force opposite to the direction of motion produces acceleration that can slow, stop, and reverse the motion.
• “Speed” is the magnitude of velocity, so it is always positive.

Common Mistakes

• Adding forces as scalars without direction: using $50 + mg$ instead of $50 - mg$ for the resultant.
• Forgetting that speed is positive: getting $v=-20.4\ \text{m s}^{-1}$ and choosing a negative-speed option (none exist).
• Assuming the object keeps accelerating downward because it is falling: after the force starts, the net force is upward, so acceleration is upward.

Things to Be Careful About

• Sign convention: pick one direction as positive and stick to it throughout.
• Value of $g$: many MCQs assume $g = 10\ \text{m s}^{-2}$, giving exactly $20\ \text{m s}^{-1}$; using $9.8$ gives $20.4\ \text{m s}^{-1}$ which still rounds to option A.
• Units: ensure $F$ in $\text{N}$, $m$ in $\text{kg}$, so $a$ comes out in $\text{m s}^{-2}$.

Background Concept

Projectile motion (with air resistance negligible) is motion under the influence of gravity only. The key idea is that the motion can be split into two independent perpendicular components:

• Horizontal (x-direction): no force (\Rightarrow) no acceleration (a_x = 0) (\Rightarrow) horizontal velocity (v_x) is constant.
• Vertical (y-direction): the only force is weight (mg) downward (\Rightarrow) constant acceleration (a_y = -g) (taking upward as positive).

A constant horizontal velocity combined with a constant vertical acceleration produces a parabolic (curved) trajectory.

Understanding the Question

A stone is thrown upward at an angle, and it follows a curved path. Air resistance is negligible, so we ignore drag. The question asks which statement about the stone’s horizontal and vertical motion explains the curved shape.

So we must choose the option that correctly describes the accelerations/velocities in each direction.

Approach

1. Identify forces acting on the stone during flight.
2. Convert forces into accelerations (Newton’s second law qualitatively).
3. State what that implies for horizontal and vertical components of velocity.
4. Match to the correct option.

Step-by-Step Reasoning

1. After the stone is released, the only significant force is its weight (mg) acting vertically downward.
2. Therefore:
   • Horizontally: no force (\Rightarrow) (a_x = 0) (\Rightarrow) (v_x) is constant.
   • Vertically: constant force (mg) downward (\Rightarrow) (a_y) is constant downward, equal to (-g).
3. This exactly matches option B: constant horizontal velocity and constant vertical acceleration.

Why this gives the curved shape: as time passes, the stone continues moving sideways at steady rate, while its vertical velocity changes (reduces to zero at the top then becomes downward), so the direction of motion continuously changes, producing the arc.

Key Takeaways

• With negligible air resistance, projectiles have no horizontal acceleration and constant vertical acceleration (-g).
• The horizontal and vertical motions are independent.
• The curved path results from combining constant horizontal velocity with changing vertical velocity.

Common Mistakes

• Thinking there is a horizontal acceleration because the path is curved (curvature here comes from vertical acceleration, not horizontal).
• Mixing up velocity and acceleration (e.g. saying “constant vertical velocity” even though gravity must change vertical velocity).
• Thinking acceleration changes direction part way through; under gravity alone the acceleration is always downward.

Things to Be Careful About

• “Air resistance negligible” is the crucial clue: it removes any horizontal force and any speed-dependent drag forces.
• Vertical acceleration is constant throughout the flight (upwards motion does not mean acceleration is upwards).
• The vertical velocity changes sign at the top, but the vertical acceleration does not.

Background Concept

A force is the rate of change of momentum:

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

For a rocket, the engine throws mass (exhaust) backwards. If the exhaust leaves with speed $v$ relative to the rocket and mass is ejected at a rate $\dot{m}$ (in $\text{kg s}^{-1}$), then the momentum carried away each second is $\dot{m}v$.

By Newton’s third law, the rocket experiences an equal and opposite force (thrust) forwards of magnitude:

$$F = \dot{m}v$$

This is essentially “momentum per second”.

Understanding the Question

You are told:

• exhaust mass ejected per second: $90\ \text{kg s}^{-1}$
• exhaust speed relative to the rocket: $190\ \text{m s}^{-1}$

You must find the force on the rocket due to this ejected gas, and then choose the closest option in kN.

Approach

1. Treat $90\ \text{kg}$ per second as the mass flow rate $\dot{m}$.
2. Use $F = \dot{m}v$ (rate of change of momentum).
3. Convert the answer from N to kN and match to an option.

Step-by-Step Reasoning

Mass flow rate:

$$\dot{m} = 90\ \text{kg s}^{-1}$$

Exhaust speed (relative to rocket):

$$v = 190\ \text{m s}^{-1}$$

Thrust magnitude:

$$F = \dot{m}v = 90 \times 190 = 17100\ \text{N}$$

Convert to kilonewtons ($1\ \text{kN} = 1000\ \text{N}$):

$$F = \frac{17100}{1000} = 17.1\ \text{kN}$$

The closest option is $17\ \text{kN}$, which is B.

Key Takeaways

• Use $F = \Delta p / \Delta t$ for situations involving momentum change.
• For steady exhaust: $F = \dot{m}v$.
• Always check unit conversions (N to kN here).

Common Mistakes

• Multiplying by $g$ unnecessarily (this is not a weight question).
• Using the rocket’s speed relative to the ground (not given and not needed); the question gives speed relative to the rocket.
• Converting to kN incorrectly (e.g. dividing by $100$ instead of $1000$).

Things to Be Careful About

• The velocity given is relative to the rocket, which is exactly what you need for thrust in this simplified model.
• Quote the final choice as the nearest option; $17.1\ \text{kN}$ rounds to $17\ \text{kN}$.

Background Concept

A collision between two objects (treated as an isolated system) always conserves total linear momentum.

The key distinction is whether kinetic energy is conserved:

• Elastic collision: total kinetic energy is conserved.
• Inelastic collision: total kinetic energy decreases (some is transferred to internal energy such as heat, sound, deformation).

A useful equivalent statement for a 1D elastic collision is that the coefficient of restitution $e = 1$:

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

So for an elastic collision, the relative speed of approach equals the relative speed of separation.

Understanding the Question

You are given four statements about a collision and asked which one does not describe an elastic collision between two objects.

So you should check each option against the properties of an elastic collision (momentum conserved, kinetic energy conserved, relative speed approach = relative speed separation).

Approach

Use the definition:

1. Identify which statements are true for elastic collisions.
2. The odd one out (false for elastic collisions) is the answer.

Step-by-Step Reasoning

• A: “Relative speed of approach equals relative speed of separation.” This is true for a perfectly elastic collision ($e = 1$). So A describes an elastic collision.
• B: “Total kinetic energy is conserved.” This is the defining property of an elastic collision. So B describes an elastic collision.
• C: “Total kinetic energy is reduced.” This contradicts the definition of elastic; it describes an inelastic collision. So C does not describe an elastic collision.
• D: “Total linear momentum is conserved.” Momentum is conserved for collisions in an isolated system (elastic or inelastic), so it is consistent with an elastic collision.

Therefore the statement that does not describe an elastic collision is C.

Key Takeaways

• Elastic collision: momentum conserved and kinetic energy conserved.
• Inelastic collision: momentum conserved but kinetic energy decreases.
• Relative speed of approach = relative speed of separation is a hallmark of a perfectly elastic collision.

Common Mistakes

• Thinking momentum conservation distinguishes elastic from inelastic (it does not; momentum is conserved in both, if external forces are negligible).
• Mixing up “kinetic energy conserved” with “total energy conserved” (total energy is always conserved, but kinetic energy may convert to other forms).

Things to Be Careful About

• The question asks for the statement that does not describe an elastic collision (so you are looking for the inelastic feature).
• “Kinetic energy reduced” is specifically the indicator of an inelastic collision, not an elastic one.

Background Concept

Newton’s third law states that when two bodies interact, they exert forces on each other that are:

• equal in magnitude,
• opposite in direction,
• the same type of force,
• and act on different bodies.

A key point is that third-law pairs do not act on the same object, so they cannot directly cancel each other on one object.

Understanding the Question

The back wheel exerts a force on the road directed backward (this is the wheel pushing on the road at the contact patch). The question asks: what force is equal and opposite to that particular force?

So we must find the force exerted by the other body in that interaction (the road) on the wheel.

Approach

1. Identify the two bodies in the interaction: (i) back wheel, (ii) road.
2. Use Newton’s third law: force of wheel on road backward implies force of road on wheel forward.
3. Match that description to the options.

Step-by-Step Reasoning

• The wheel and road interact via contact forces (static friction in this case).
• Given: “backward push of the back wheel on the road” = force by wheel on road, direction backward.
• Third law pair must be: force by road on wheel, equal magnitude, opposite direction (forward).
• Option B says “the forward push of the road on the back wheel”, which matches exactly.

Why the other options are not third-law pairs:

• A and C involve different interactions (pedals/chain), not the road-wheel interaction.
• D is the resultant resistive force on the cyclist/bike; it is not the reaction to the wheel’s force on the road, and it acts on a different body/system.

Key Takeaways

• Newton’s third law pairs are equal and opposite forces acting on different bodies.
• To find the pair, name both interacting bodies and swap “force of X on Y” to “force of Y on X” with opposite direction.

Common Mistakes

• Choosing a force that balances another on the same object (that relates to Newton’s first/second law, not the third law).
• Picking a force from a different interaction (e.g. chain tension) because it seems “related” to motion.

Things to Be Careful About

• Constant speed means net force on the cyclist/bike is zero, but that is separate from identifying a third-law pair.
• The third-law partner of “wheel on road” must involve the road and the wheel only, with directions reversed.

Background Concept

When an object falls in air, two main forces act:

• weight $W = mg$ downward (approximately constant)
• air resistance / drag $F_D$ upward (increases as speed increases)

Newton’s second law gives

$$\sum F = ma$$

Taking downward as positive:

$$mg - F_D = ma$$

At the start, $v=0$ so $F_D \approx 0$ and the acceleration is close to $g$. As the stone speeds up, $F_D$ increases, reducing the resultant force and hence reducing $a$. Eventually, at terminal velocity, $F_D = mg$ so the resultant force is zero and $a=0$.

Understanding the Question

You are asked which graph of acceleration $a$ against time $t$ could represent a stone released from rest that falls a long distance in air.
Key clues:

• “released from rest” (\Rightarrow) initially no drag
• “falls a long distance” (\Rightarrow) it has time to approach terminal velocity, so acceleration should approach zero at large $t$

Approach

Decide the behaviour of $a$ at:

1. $t=0$ (immediately after release)
2. as $t$ increases (as speed increases and drag grows)
3. at large $t$ (terminal velocity)
   Then choose the option whose curve matches: starting near $g$ and decreasing asymptotically towards $0$.

Step-by-Step Reasoning

1. At $t=0$, the stone is at rest ($v=0$). Drag depends on speed, so initially $F_D \approx 0$.

   Hence

$$mg - 0 = ma \Rightarrow a \approx g$$

So the graph must start at a positive value (about $g$), not at zero.

1. As the stone falls, its speed increases, so $F_D$ increases.

$$mg - F_D = ma$$

As $F_D$ gets larger, $mg - F_D$ gets smaller, so $a$ decreases with time.

1. After a long time, terminal velocity is approached. At this point $F_D = mg$.

$$mg - F_D = 0 \Rightarrow a = 0$$

So $a$ tends towards zero, usually by flattening off (approaching zero asymptotically) rather than dropping linearly.

1. Comparing to the options:

• A starts at $a=0$ and rises, which is not correct.
• B decreases linearly to $0$; the early-time drag increase is not linear with time, so this is not the best match.
• C is constant $a$ (vacuum case), not in air for a long fall.
• D starts positive and decreases quickly at first, then levels off towards $0$, matching terminal-velocity behaviour.

Therefore the correct graph is D.

Key Takeaways

• In air, acceleration is not constant: drag increases with speed.
• Immediately after release, $a \approx g$ because drag is negligible.
• As speed increases, acceleration decreases.
• At terminal velocity, resultant force is zero so $a=0$.

Common Mistakes

• Choosing a constant-acceleration graph (like C), which would only apply if air resistance is negligible.
• Choosing a graph that starts at $a=0$ (like A); acceleration is not zero at release.
• Assuming acceleration decreases linearly with time (like B); drag typically depends on speed (often (\propto v) or (\propto v^2)), so the approach to zero is curved.

Things to Be Careful About

• The graph is for acceleration, not velocity: terminal velocity corresponds to $a=0$, not $v=0$.
• The question says “falls a long distance”, so you should expect terminal behaviour (acceleration approaching zero) to be shown.
• The sign convention: the options show positive acceleration; interpret that as taking downward as positive.

Background Concept

Momentum is a vector quantity defined by

$$\vec p = m\vec v$$

If the resultant external force in some direction is zero, then momentum in that direction is conserved:

$$\sum \vec p_{\text{before}} = \sum \vec p_{\text{after}}$$

Kinetic energy is

$$E_k = \frac12 mv^2$$

In a collision where objects stick together (perfectly inelastic collision), momentum is conserved (in the direction(s) with no external impulse), but kinetic energy is not conserved because some is transferred into internal energy (deformation, sound, heating).

Understanding the Question

An empty cart moves horizontally at constant velocity on a track, and resistive forces are negligible. A rock is dropped vertically into it and stays inside. The question asks how the momentum and kinetic energy of the combined cart+rock (after the rock lands and they move together) compare to those of the cart alone (before the rock was added).

Key idea: the rock initially has no horizontal velocity, so it brings in no horizontal momentum, but it does increase the mass of the moving system once it is inside the cart.

Approach

1. Consider only the horizontal direction: there is no significant external horizontal force, so horizontal momentum is conserved during the short collision as the rock is captured.
2. Use momentum conservation to infer that the final total horizontal momentum equals the cart’s initial horizontal momentum.
3. Use the fact that the mass increases but momentum stays the same to deduce the speed decreases.
4. Compute/compare kinetic energies using $E_k = \tfrac12 mv^2$ (or show algebraically that $E_{k,f} < E_{k,i}$).

Step-by-Step Reasoning

Let the cart have mass $m_c$ and initial horizontal speed $v$. Let the rock have mass $m_r$.

Before the rock lands:

• Cart horizontal momentum: $p_i = m_c v$.
• Rock horizontal momentum: $0$ (it is dropped vertically).
  So total initial horizontal momentum is $m_c v$.

After the rock lands and moves with the cart, they share a common horizontal speed $v'$:

$$p_f = (m_c + m_r)v'$$

Because resistive forces are negligible, the external horizontal impulse is negligible, so

$$(m_c+m_r)v' = m_c v$$

This immediately shows the final momentum is the same as the initial:

$$p_f = m_c v = p_i$$

Also,

$$v' = \frac{m_c}{m_c+m_r}v$$

so $v' < v$.

Now compare kinetic energies.

Initial kinetic energy (cart only):

$$E_{k,i} = \frac12 m_c v^2$$

Final kinetic energy (cart + rock):

$$E_{k,f} = \frac12 (m_c+m_r)v'^2$$

Substitute for $v'$:

$$E_{k,f} = \frac12 (m_c+m_r)\left(\frac{m_c}{m_c+m_r}v\right)^2$$ $$E_{k,f} = \frac12 \frac{m_c^2}{m_c+m_r}v^2$$

Since $m_c+m_r > m_c$, then

$$\frac{m_c^2}{m_c+m_r} < m_c$$

so

$$E_{k,f} < \frac12 m_c v^2 = E_{k,i}$$

Therefore: same momentum, smaller kinetic energy.

Key Takeaways

• Conserve momentum only in directions where external impulse is negligible (here: horizontal).
• A dropped object can change the mass of a moving system without adding horizontal momentum.
• When objects stick together, kinetic energy decreases even though momentum is conserved.

Common Mistakes

• Conserving kinetic energy for a sticking collision (it is not conserved in an inelastic capture).
• Forgetting momentum is a vector and incorrectly including the rock’s vertical momentum when considering horizontal momentum conservation.
• Assuming the speed stays the same because “no resistive forces”: the collision itself provides an internal interaction that changes the cart’s speed.

Things to Be Careful About

• Use horizontal momentum only: the track provides external vertical forces (normal reaction) so vertical momentum is not conserved for the cart+rock system.
• The correct comparison is between the cart-only state before and the combined cart+rock after: the total mass changes, so you must be consistent about which mass is used in $p=mv$ and $E_k=\tfrac12 mv^2$.

Background Concept

For an object submerged in a fluid, the fluid exerts an upward buoyant force called the upthrust. By Archimedes' principle, the upthrust equals the weight of fluid displaced:

$$U = \rho gV_{\text{displaced}}$$

If the block is fully submerged and has volume $V$, then $V_{\text{displaced}} = V$, so $U = \rho gV$.

When an object is held stationary, it is in equilibrium, so the resultant force is zero. In one dimension (vertical), this means the sum of upward forces equals the sum of downward forces.

Understanding the Question

The wooden block is submerged and tied by a string to the bottom of the container, so the string must be preventing the block from rising (wood is less dense than water, so it tends to float).

Forces on the block:

• weight $mg$ downward,
• upthrust $\rho gV$ upward,
• tension $T$ from the string (on the block, the string pulls downward).

The question asks for the magnitude of the force due to the tension, i.e. $T$.

Approach

1. Write the upthrust using Archimedes' principle.
2. Write the vertical equilibrium condition (resultant force $=0$).
3. Rearrange to solve for the tension $T$.
4. Match the expression to the options.

Step-by-Step Reasoning

Upthrust (block fully submerged, volume $V$):

$$U = \rho gV$$

Choose upward as positive. Then the forces in the vertical direction are:

• upward: $+U$
• downward: $-mg$ and $-T$

Equilibrium gives:

$$U - mg - T = 0$$

Substitute $U = \rho gV$:

$$\rho gV - mg - T = 0$$

Solve for $T$:

$$T = \rho gV - mg$$

This corresponds to option D.

Key Takeaways

• For a fully submerged object, upthrust is $\rho gV$.
• A string attached to the bottom provides a downward tension on a floating object.
• Equilibrium means the net force is zero: upward forces = downward forces.

Common Mistakes

• Using $T = mg - \rho gV$ (sign error from not considering which way the tension acts).
• Assuming tension acts upward: a string can only pull, so the string attached below the block pulls the block downward.
• Using the wrong displaced volume (here the block is submerged so displaced volume is $V$).

Things to Be Careful About

• Direction matters: decide a sign convention and stick to it.
• The question asks for the force due to tension on the block (not the force the block exerts on the string, though the magnitudes are equal by Newton's third law).
• The expression $\rho gV - mg$ is only positive if the block is less dense than water; that is consistent with needing a string to hold it down.