Physics 9702/11 — October/November 2025

Background Concept

A scalar quantity has magnitude only (no direction), e.g. mass, time, energy.

A vector quantity has both magnitude and direction and must be added using vector rules (or resolved into components), e.g. displacement, velocity, acceleration, force.

A quick test is: if the quantity can point in different directions (e.g. left/right, up/down) and reversing direction changes the meaning/sign of the quantity, it is a vector.

Understanding the Question

You are given four physical quantities (acceleration, charge, kinetic energy, wavelength) and must decide which are scalars and which are vectors, then choose the option row A–D that matches.

Approach

1. Decide scalar/vector for each quantity using the definition (direction or no direction).
2. Compare your list with the four rows in the table.

Step-by-Step Reasoning

• Acceleration: defined as rate of change of velocity. Since velocity is a vector, acceleration must also have direction. Therefore acceleration is a vector.

• Charge: an amount of electric charge does not have a direction; it may be positive or negative, but that is not a spatial direction. Therefore charge is a scalar.

• Kinetic energy: given by

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

Energy has magnitude only and no direction. Therefore kinetic energy is a scalar.

• Wavelength: it is a length (distance between successive wavefronts/peaks) and is not a directed quantity. Therefore wavelength is a scalar.

So the pattern is: acceleration vector; charge scalar; kinetic energy scalar; wavelength scalar, which corresponds to row D.

Key Takeaways

• Vectors require direction (e.g. acceleration, velocity, force).
• Energies and lengths are scalars (e.g. kinetic energy, wavelength).
• “Positive/negative” for charge is not the same as a direction in space.

Common Mistakes

• Thinking charge is a vector because electric field is a vector (field is vector; charge is scalar).
• Thinking wavelength is a vector because waves can travel in a direction (the wave propagation direction can be represented by a vector, but the wavelength is just a scalar distance).
• Confusing velocity (vector) with speed (scalar), then extending that confusion to acceleration.

Things to Be Careful About

• A quantity can be negative and still be a scalar (e.g. charge), because the sign does not indicate a spatial direction.
• When in doubt, ask whether you could meaningfully add two such quantities using vector addition (head-to-tail) or whether ordinary algebra is sufficient; scalars add normally, vectors require components/direction.

Background Concept

When a calculated quantity depends on measured quantities with uncertainties, we estimate the uncertainty in the result by propagating uncertainties.

For multiplication/division, fractional (percentage) uncertainties add:

• If $Q = ab/c$, then

$$\frac{\Delta Q}{Q} = \frac{\Delta a}{a} + \frac{\Delta b}{b} + \frac{\Delta c}{c}.$$

For powers, the fractional uncertainty is multiplied by the power:

• If $Q = a^n$, then

$$\frac{\Delta Q}{Q} = |n|\frac{\Delta a}{a}.$$

Here, $X$ is proportional to $L$ and inversely proportional to $T^2$, so $T$ contributes twice its fractional uncertainty.

Understanding the Question

You are given:

• $L = 6.8 \pm 0.1\ \text{cm}$
• $T = 2.42 \pm 0.08\ \text{s}$

and the relationship:

$$X = \frac{4\pi^2 L}{T^2}.$$

You must calculate:

1. the numerical value of $X$
2. the uncertainty in $X$

Then choose which option matches both the value and the uncertainty.

Approach

1. Substitute the central values of $L$ and $T$ into the formula to find $X$.
2. Use fractional uncertainty propagation:

$$\frac{\Delta X}{X} = \frac{\Delta L}{L} + 2\frac{\Delta T}{T}.$$

1. Convert the fractional uncertainty into an absolute uncertainty using $\Delta X = X(\Delta X/X)$.
2. Round the uncertainty sensibly (usually to 1 s.f. unless it begins with 1 or 2), then round $X$ to match.

Step-by-Step Reasoning

1) Calculate $X$

$$X = \frac{4\pi^2 L}{T^2}.$$

Compute $T^2$:

$$T^2 = (2.42)^2 = 5.86\ \text{s}^2\ \text{(3 s.f.)}.$$

Compute numerator $4\pi^2 L$:

$$4\pi^2 \approx 39.48,$$

so

$$4\pi^2 L = 39.48 \times 6.8 = 268.5.$$

Then

$$X = \frac{268.5}{5.86} = 45.8\ \text{cm s}^{-2}.$$

(Units: $L$ is in cm and $T^2$ is in $\text{s}^2$, so $X$ has units $\text{cm s}^{-2}$.)

2) Calculate fractional uncertainty in $X$

Because $X \propto L$ and $X \propto T^{-2}$:

$$\frac{\Delta X}{X} = \frac{\Delta L}{L} + 2\frac{\Delta T}{T}.$$

Calculate each term:

$$\frac{\Delta L}{L} = \frac{0.1}{6.8} = 0.0147,$$ $$\frac{\Delta T}{T} = \frac{0.08}{2.42} = 0.0331.$$

So

$$\frac{\Delta X}{X} = 0.0147 + 2(0.0331) = 0.0809 \approx 8.1\%.$$

3) Convert to absolute uncertainty

$$\Delta X = 0.0809 \times 45.8 = 3.7\ \text{cm s}^{-2} \approx 4\ \text{cm s}^{-2}.$$

So the result is

$$X \approx 46 \pm 4\ \text{cm s}^{-2},$$

which matches option D.

Key Takeaways

• For products/quotients, add fractional uncertainties.
• For a power $n$, multiply the fractional uncertainty by $|n|$.
• After finding $\Delta X$, round the uncertainty appropriately and round the value to the same decimal place/significance.

Common Mistakes

• Forgetting the factor of 2 from $T^2$ (using $\Delta T/T$ instead of $2\Delta T/T$).
• Subtracting uncertainties for division instead of adding fractional uncertainties.
• Giving an over-precise uncertainty (e.g. $\pm 3.7$) when the options and convention expect $\pm 4$.
• Rounding $X$ inconsistently with the uncertainty.

Things to Be Careful About

• Use fractional uncertainty rules (not absolute) for expressions involving multiplication/division.
• Keep enough significant figures during intermediate steps to avoid rounding error, then round at the end.
• Ensure units follow from the formula: here $\text{cm s}^{-2}$, not $\text{m s}^{-2}$.

Background Concept

SI base units are built from the base quantities (kg, m, s, A, K, mol, cd). Many physical units are derived from definitions, and to express a derived unit in SI base units you repeatedly substitute definitions until only base units remain.

Power is defined as the rate of transfer of energy:

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

So $1\ \text{W} = 1\ \text{J s}^{-1}$. The joule is also derived from work done $W = Fd$, giving $1\ \text{J} = 1\ \text{N m}$, and the newton comes from $F = ma$, giving $1\ \text{N} = 1\ \text{kg m s}^{-2}$.

Understanding the Question

You are asked to identify which option expresses the watt using only SI base units. The options include combinations like $\text{J s}^{-1}$ and $\text{N m s}^{-1}$, which are correct forms for watts but are not yet in base units because $\text{J}$ and $\text{N}$ are derived units. We must reduce everything to kg, m, and s.

Approach

Start from a known definition of the watt:

• $\text{W} = \text{J s}^{-1}$ (power is energy per time).

Then convert $\text{J}$ into $\text{N m}$, and convert $\text{N}$ into $\text{kg m s}^{-2}$. Finally, simplify the powers of metres and seconds.

Step-by-Step Reasoning

From the definition of power:

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

Convert joule to newton-metre:

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

So

$$\text{W} = \frac{\text{N m}}{\text{s}} = \text{N m s}^{-1}$$

Now convert newton into base units using $F = ma$:

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

Substitute into the expression for watt:

$$\text{W} = (\text{kg m s}^{-2})\,\text{m}\,\text{s}^{-1} = \text{kg m}^2\text{ s}^{-3}$$

This matches option B.

Key Takeaways

• $\text{W} = \text{J s}^{-1}$ and $\text{J} = \text{N m}$ are useful stepping stones, but not SI base units.
• Reduce derived units step-by-step until only base units remain.
• For watts, the SI base-unit form is $\text{kg m}^2\text{ s}^{-3}$.

Common Mistakes

• Stopping at $\text{J s}^{-1}$ or $\text{N m s}^{-1}$ and calling them “SI base units” (they still contain derived units).
• Incorrectly recalling $\text{N}$ as $\text{kg m s}^{-1}$ (missing one power of $\text{s}$).
• Losing track of indices when dividing by time (e.g. writing $\text{s}^{-2}$ instead of $\text{s}^{-3}$).

Things to Be Careful About

• The phrase “SI base units” means only base-unit symbols are allowed.
• When multiplying/dividing units, treat indices like algebra: dividing by $\text{s}$ subtracts 1 from the power of $\text{s}$.
• Ensure the final expression is simplified (combine the two metres into $\text{m}^2$).

Background Concept

Forces are vectors, so to combine them you must add them using vector addition (not simple arithmetic addition unless they act in the same direction). If two vectors of magnitudes $A$ and $B$ have an angle $\theta$ between them, the magnitude $R$ of their resultant is found from the cosine rule:

$$R^2 = A^2 + B^2 + 2AB\cos\theta$$

This comes directly from constructing the triangle (or parallelogram) formed by the two vectors.

Understanding the Question

Two equal forces of $6.0\,\text{N}$ act from the same point, with an angle of $60^\circ$ between their lines of action. The task is to find the magnitude (size) of the single force that would have the same effect (the resultant).

Given:

• $F_1 = 6.0\,\text{N}$
• $F_2 = 6.0\,\text{N}$
• angle between them $\theta = 60^\circ$

Unknown:

• resultant magnitude $R$

Approach

Treat the two forces as sides of a vector triangle with included angle $60^\circ$. Use the cosine rule to find $R$. Finally, match the numerical result to the closest option provided.

Step-by-Step Reasoning

Let $A = B = 6.0\,\text{N}$ and $\theta = 60^\circ$.

Apply the cosine rule:

$$R^2 = A^2 + B^2 + 2AB\cos\theta$$

Substitute values:

$$R^2 = 6.0^2 + 6.0^2 + 2(6.0)(6.0)\cos 60^\circ$$

Compute:

• $6.0^2 = 36$
• $\cos 60^\circ = 0.5$

So:

$$R^2 = 36 + 36 + 72(0.5) = 36 + 36 + 36 = 108$$

Then:

$$R = \sqrt{108} = 10.4\,\text{N}$$

The options are given to about 2 s.f., so $10.4\,\text{N}$ corresponds to $10\,\text{N}$, which is option C.

Key Takeaways

• Forces add as vectors, not scalars.
• For two vectors with known angle between them, the cosine rule gives the resultant magnitude:

$$R^2 = A^2 + B^2 + 2AB\cos\theta$$

• Always round sensibly to match the precision of the options.

Common Mistakes

• Adding magnitudes directly: $6.0 + 6.0 = 12\,\text{N}$ (only valid if angle is $0^\circ$).
• Using Pythagoras: $R = \sqrt{6^2 + 6^2}$ (only valid if angle is $90^\circ$).
• Using $\cos 60^\circ = \sqrt{3}/2$ (that is $\sin 60^\circ$).

Things to Be Careful About

• Ensure the angle used in the cosine rule is the included angle between the two forces (here $60^\circ$).
• In MCQs, your calculated value may not exactly match an option; choose the nearest appropriate rounding (here $10.4\,\text{N} \to 10\,\text{N}$).

Background Concept

For motion in a straight line:

• Constant speed: displacement $s = vt$.
• Uniform (constant) acceleration: displacement

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

where $u$ is initial speed, $a$ is acceleration, and $t$ is time from the chosen start.

To find when one object “overtakes” another (same direction), you compare their displacements from the same origin: overtaking occurs when their positions (displacements) are equal at the same time.

Understanding the Question

At $t=0$:

• The goods train passes the station (take station as $s=0$) at constant speed $10\,\text{m s}^{-1}$.
• The express train is initially at rest at the station, but starts moving at $t=0$ with uniform acceleration $0.5\,\text{m s}^{-2}$.

We need the time $t$ when the express train’s displacement from the station first becomes equal to the goods train’s displacement from the station.

Approach

1. Write an expression for the goods train displacement $s_g(t)$ using constant speed.
2. Write an expression for the express train displacement $s_e(t)$ using $s = ut + \tfrac{1}{2}at^2$ with $u=0$.
3. Set $s_e(t)=s_g(t)$ and solve for $t$. Ignore the trivial solution $t=0$ because that is just when they start together.
4. Match the value to the given options.

Step-by-Step Reasoning

Goods train moves at constant speed $v=10\,\text{m s}^{-1}$:

$$s_g = vt = 10t$$

Express train starts from rest, so $u=0$, and has $a=0.5\,\text{m s}^{-2}$:

$$s_e = ut + \frac{1}{2}at^2 = 0 + \frac{1}{2}(0.5)t^2 = 0.25t^2$$

At the overtaking point, positions are equal:

$$0.25t^2 = 10t$$

Bring all terms to one side and factor:

$$0.25t^2 - 10t = 0$$ $$t(0.25t - 10) = 0$$

So $t=0$ (the instant they are together at the start) or

$$0.25t = 10 \Rightarrow t = 40\,\text{s}$$

Therefore the correct option is D.

Key Takeaways

• Use the same origin and the same time reference for both objects.
• Overtaking in 1D means equal displacement at the same time.
• Constant speed uses $s=vt$; uniform acceleration uses $s=ut+\tfrac{1}{2}at^2$.

Common Mistakes

• Using $v = u + at$ for the express train and setting its speed equal to $10\,\text{m s}^{-1}$ (that only finds when speeds match, not when it catches up).
• Forgetting that the express train starts from rest ($u=0$).
• Not discarding the trivial solution $t=0$.

Things to Be Careful About

• Both trains start at the same position at $t=0$, so the displacement equality equation will always have $t=0$ as a solution.
• Keep units consistent (here everything is already in SI).
• Because one displacement is proportional to $t$ and the other to $t^2$, there is exactly one non-zero catch-up time in this situation.

Background Concept

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

• The gradient gives acceleration:

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

• The area under the graph (taking account of sign) gives displacement:

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

For motion under gravity (free fall), the acceleration is constant and equal to $g$ downward. On a $v$–$t$ graph this appears as a straight line with constant gradient.

A bounce involves a collision with the ground. If the ball leaves the ground with a smaller speed than it arrived, kinetic energy has been lost (to heat, sound, deformation), so the collision is inelastic.

Understanding the Question

The graph shows the ball released, speeding up as it falls, then colliding with the ground and bouncing repeatedly with decreasing speeds.

Points $P,Q,R,S$ lie on the velocity–time graph:

• $P$ is at the start (release).
• $Q$ is just before the first impact.
• A vertical jump to $R$ represents the rapid change in velocity during the collision.
• $S$ is later during the next fall.

We must decide which statement (A–D) is not correct.

Approach

Check each option using standard $v$–$t$ interpretations:

1. Area under a segment → displacement (height fallen/rise).
2. Vertical change at impact + reduced speed → inelastic.
3. Gradient of straight-line flight segments → gravitational acceleration.
4. Identify where the maximum upward velocity occurs (most negative velocity if downward is taken as positive).

Step-by-Step Reasoning

A: Segment $PQ$ is the motion from release to just before first impact. The displacement from release to impact equals the initial height. On a $v$–$t$ graph, displacement is area under the curve, so area under $PQ$ represents that height (with the sign depending on the chosen positive direction). So A is correct.

B: At impact, the velocity changes sign (direction reverses) and the magnitude after the bounce is smaller than before (successive peaks shrink). That means kinetic energy is not conserved in the collision → inelastic. So B is correct.

C: Segment $RS$ is straight, meaning constant acceleration. During flight the only significant acceleration is due to gravity, so the gradient corresponds to $g$ (with sign depending on axis convention). So C is correct.

D: At $P$ the ball is released from rest, so $v=0$ at $P$. The maximum upward speed occurs immediately after a bounce, at the point where the velocity is most upward (largest magnitude upward), which is at $R$ (most negative on this graph). Therefore D is the statement that is not correct.

Key Takeaways

• On a $v$–$t$ graph: gradient → acceleration, area → displacement.
• Free-fall motion gives straight-line segments with gradient $\pm g$.
• Reduced rebound speed indicates an inelastic collision.
• “Maximum upward velocity” is at the most upward-directed point after the bounce, not at the initial release from rest.

Common Mistakes

• Thinking the area under a $v$–$t$ graph gives distance regardless of sign (it gives displacement; distance would use area under $|v|$).
• Saying the collision is elastic just because the ball bounces (elastic would require same speed before and after impact).
• Identifying $P$ as a maximum because it is a labelled point, ignoring that the ball is released from rest.

Things to Be Careful About

• The sign convention: in this graph, downward velocity appears positive (since the line from $P$ to $Q$ goes upward while the ball is falling). Upward velocities are therefore negative.
• “Maximum upward velocity” refers to the most negative velocity value here (largest magnitude upward), occurring just after the bounce (at $R$).

Background Concept

In projectile motion with negligible air resistance, the horizontal and vertical motions are independent:

• Horizontally there is no acceleration, so the horizontal velocity remains constant at its initial value $u$.
• Vertically the object accelerates downward at $g$, like free fall.

Key equations:

• Vertical displacement (starting with zero vertical velocity, because it is projected horizontally):

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

• Horizontal displacement:

$$x = ut$$

So the horizontal range is proportional to the time of flight.

Understanding the Question

The first object falls a vertical distance $y$ and travels horizontally a distance $x$. That tells us the time of flight $t$ (from vertical motion) and hence $x$ (from horizontal motion).

A second object is launched horizontally with the same initial horizontal speed $u$, but it falls a vertical distance $4y$ before it lands. We must find its horizontal distance in terms of $x$.

Approach

1. Use the vertical free-fall equation $y = \tfrac{1}{2}gt^2$ to find how the time of flight scales when the vertical distance changes from $y$ to $4y$.
2. Use $x = ut$ and the fact that $u$ is the same for both launches to scale the horizontal distance in the same ratio as the times.

Step-by-Step Reasoning

For the first object (vertical motion):

$$y = \frac{1}{2}gt^2 \;\Rightarrow\; t = \sqrt{\frac{2y}{g}}$$

Horizontal motion has constant speed $u$, so

$$x = ut$$

For the second object, the vertical distance is $4y$:

$$4y = \frac{1}{2}gt_2^2 \;\Rightarrow\; t_2 = \sqrt{\frac{8y}{g}}$$

Compare $t_2$ with $t$:

$$\frac{t_2}{t} = \frac{\sqrt{8y/g}}{\sqrt{2y/g}} = \sqrt{\frac{8}{2}} = \sqrt{4} = 2$$

So the second object is in the air for twice as long. Since the horizontal speed is unchanged ($u$ is the same and there is no horizontal acceleration), the horizontal distance doubles:

$$x_2 = ut_2 = u(2t) = 2(ut) = 2x$$

Therefore the correct option is $2x$.

Key Takeaways

• In projectile motion (no air resistance), time of flight is determined by vertical motion only.
• If vertical drop increases by a factor of 4, time of flight increases by a factor of $\sqrt{4}=2$.
• Horizontal distance is proportional to time of flight when horizontal speed is constant.

Common Mistakes

• Assuming distance scales directly with vertical distance (choosing $4x$) instead of with the square root via $y \propto t^2$.
• Forgetting that the initial vertical velocity is zero for a horizontal projection and incorrectly using $y = ut + \tfrac{1}{2}gt^2$ with a non-zero $u$ vertically.
• Treating the speed as constant in 2D and trying to use total speed instead of separating components.

Things to Be Careful About

• The independence of components only holds when air resistance is negligible (as stated).
• Make sure you use the vertical displacement equation with $u_y = 0$.
• Recognise scaling: $y \propto t^2$ means multiplying $y$ by 4 multiplies $t$ by 2, not by 4.

Background Concept

When an object falls through air, two main vertical forces act:

• Weight $W = mg$ acting downward (approximately constant).
• Air resistance (drag) $F_\text{drag}$ acting upward (increases with speed).

Newton's second law states that the resultant force is related to acceleration:

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

So the direction of the acceleration is the same as the direction of the resultant force. On a velocity–time graph, the acceleration at any point is the gradient of the graph.

Understanding the Question

The sphere starts from rest and its velocity increases and then levels off (approaches terminal velocity). Point $P$ is on the rising part of the $v$–$t$ curve, before it becomes horizontal.

You must choose the force diagram that matches the situation at point $P$.

Approach

1. Use the slope (gradient) of the velocity–time graph at $P$ to decide whether the acceleration is zero or not.
2. Use $F = ma$ to decide whether the resultant force is upward, downward, or zero.
3. Translate that into a comparison of force arrow lengths: compare $W$ (down) with $F_\text{drag}$ (up).

Step-by-Step Reasoning

• At point $P$, the graph is still rising, so its gradient is not zero. That means the sphere is still speeding up downward, so $a$ is downward.
• Since $a$ is downward, the resultant force must be downward.
• The vertical forces are $W$ downward and $F_\text{drag}$ upward, so a downward resultant means:

$$W - F_\text{drag} > 0 \quad \Rightarrow \quad W > F_\text{drag}$$

So the correct force diagram must show both forces, with the weight arrow longer than the air resistance arrow.

That corresponds to option C.

Key Takeaways

• Gradient of a $v$–$t$ graph gives acceleration.
• Non-zero acceleration implies a non-zero resultant force in the same direction.
• For a falling object before terminal velocity: $W > F_\text{drag}$.

Common Mistakes

• Choosing equal arrows (terminal velocity) even though the graph at $P$ is not horizontal.
• Thinking that because the sphere is moving downward, the resultant force must be downward (motion direction is not the same as resultant force direction).
• Omitting drag entirely (there is drag whenever the object moves through air).

Things to Be Careful About

• Terminal velocity occurs only when the $v$–$t$ graph becomes horizontal (gradient $=0$), meaning $W = F_\text{drag}$.
• At early times drag is smaller because speed is smaller, so the weight arrow should be longer than the drag arrow, but not necessarily by a huge amount depending on where $P$ is.

Background Concept

Linear momentum is $\vec{p} = m\vec{v}$ (a vector). For a system of particles,

$$\vec{F}_{\text{external}} = \frac{\mathrm{d}\vec{P}}{\mathrm{d}t}$$

where $\vec{P}$ is the total momentum of the system. If the resultant external force on the system is zero (an isolated system), then

$$\frac{\mathrm{d}\vec{P}}{\mathrm{d}t} = 0 \;\Rightarrow\; \vec{P} = \text{constant}$$

So total linear momentum is conserved in any interaction (including any type of collision: elastic or inelastic) provided the system is isolated.

Understanding the Question

The question asks which collision situation guarantees momentum conservation in all cases. The key word is always: we need the condition that makes total momentum conserved regardless of masses, speeds, or whether the objects stick together.

Approach

Use the criterion: momentum is conserved if and only if the system is isolated (no net external force / external impulse). Then match this condition to the options.

Step-by-Step Reasoning

• Option A: Not always true because if the two-object system is not isolated (e.g. friction with the ground, external push), external impulse changes total momentum.
• Option B: Equal and opposite velocities do not guarantee isolation; external forces could still act. Also, this condition is about initial momentum, not whether it stays constant.
• Option C: If the two objects form an isolated system, $\vec{F}_{\text{external}} = 0$, so total momentum is conserved throughout the collision.
• Option D: Same mass does not guarantee isolation; external forces can still change total momentum.

Therefore, the only situation where momentum is always conserved is an isolated system.

Key Takeaways

• Momentum conservation requires zero resultant external force (or zero external impulse) on the system.
• Details like equal/opposite velocities or equal masses are irrelevant unless the system is isolated.
• Momentum conservation applies to both elastic and inelastic collisions when isolation holds.

Common Mistakes

• Assuming momentum is conserved in all collisions without checking for external forces.
• Confusing “initial total momentum happens to be zero” (e.g. equal and opposite velocities) with “momentum is conserved”.
• Thinking equal masses ensures conservation (it does not).

Things to Be Careful About

• The system must include all interacting objects; leaving out the environment (e.g. Earth when friction acts) can make a non-isolated situation look isolated.
• “Isolated” means no net external force; internal forces during the collision do not affect total momentum because they occur in equal and opposite pairs.

Background Concept

For a head-on (one-dimensional) collision with no external forces, total momentum is conserved:

$$m_1u_1 + m_2u_2 = m_1v_1 + m_2v_2$$

An elastic collision also conserves kinetic energy. A very useful equivalent statement (for 1D elastic collisions) is that the relative speed of separation equals the relative speed of approach:

$$v_2 - v_1 = u_1 - u_2$$

This comes from combining momentum conservation with kinetic energy conservation, and it avoids having to write out the kinetic-energy equation.

Understanding the Question

• Object 1: mass $m$, initial velocity $u_1 = +9.0\ \text{m s}^{-1}$.
• Object 2: mass $2m$, initially at rest so $u_2 = 0$.
• After collision, both move with velocities $v_1$ (for mass $m$) and $v_2$ (for mass $2m$).
• We must find $v_2$.

The collision is head-on, so we treat it as 1D with a sign convention.

Approach

1. Write conservation of momentum for the two objects.
2. Use the elastic condition $v_2 - v_1 = u_1 - u_2$.
3. Solve the two simultaneous equations to obtain $v_2$, then match it to the options.

Step-by-Step Reasoning

1. Choose the initial direction of the mass $m$ as positive.

2. Momentum before collision:

$$p_{\text{before}} = m(9.0) + (2m)(0) = 9m$$

1. Momentum after collision:

$$p_{\text{after}} = mv_1 + 2mv_2$$

1. Apply momentum conservation ($p_{\text{before}}=p_{\text{after}}$):

$$9m = mv_1 + 2mv_2 \Rightarrow 9 = v_1 + 2v_2$$

1. Apply the 1D elastic collision relative-speed condition:

$$v_2 - v_1 = u_1 - u_2 = 9.0 - 0 = 9.0$$

1. Solve simultaneously. From $v_2 - v_1 = 9$,

$$v_1 = v_2 - 9$$

Substitute into $9 = v_1 + 2v_2$:

$$9 = (v_2 - 9) + 2v_2 = 3v_2 - 9$$ $$18 = 3v_2 \Rightarrow v_2 = 6.0\ \text{m s}^{-1}$$

So the object of mass $2m$ moves forward at $+6.0\ \text{m s}^{-1}$, which corresponds to option D.

Key Takeaways

• With no external forces, total momentum is conserved.
• In 1D elastic collisions, you can use either kinetic energy conservation or the shortcut $v_2 - v_1 = u_1 - u_2$.
• Keep a consistent sign convention for velocities.

Common Mistakes

• Using momentum conservation alone (not enough information for two unknown final velocities).
• Getting the sign wrong in the elastic condition (writing $v_1 - v_2 = u_1 - u_2$ without thinking about the order).
• Assuming the heavier object must move more slowly without calculation (here it does move more slowly than $9.0$, but the exact value matters).

Things to Be Careful About

• The “relative speed” equation is specific to 1D elastic collisions; it is not valid for inelastic collisions.
• Negative options would indicate the $2m$ mass rebounds backwards; check algebra and sign convention if you obtain a negative value.
• Quote the velocity with the unit $\text{m s}^{-1}$ and match it to the given options.

Background Concept

Resultant force is linked to momentum by Newton's second law in momentum form:

$$F = \frac{\mathrm{d}p}{\mathrm{d}t}$$

where:

• $p$ is momentum ($p = mv$), unit $\text{kg m s}^{-1}$,
• $t$ is time (s),
• $F$ is the resultant (net) force, unit N.

On a momentum–time graph, the gradient (slope) equals $\mathrm{d}p/\mathrm{d}t$, so it directly gives the resultant force.

Understanding the Question

You are given a straight-line graph of momentum $p$ against time $t$ for a motorcycle. The question asks for the resultant force on the motorcycle.

Since force equals the rate of change of momentum, you just need the gradient of the line.

Approach

1. Use $F = \Delta p / \Delta t$ (valid here because the graph is a straight line, so the gradient is constant).
2. Read two clear points on the line (the endpoints are easiest).
3. Calculate the gradient and match it to one of the options.

Step-by-Step Reasoning

From the graph:

• At $t = 0\ \text{s}$, $p = 0\ \text{kg m s}^{-1}$.
• At $t = 10\ \text{s}$, $p = 5000\ \text{kg m s}^{-1}$.

So:

$$\Delta p = 5000 - 0 = 5000\ \text{kg m s}^{-1}$$ $$\Delta t = 10 - 0 = 10\ \text{s}$$

Resultant force is the gradient:

$$F = \frac{\Delta p}{\Delta t} = \frac{5000}{10} = 500\ \text{N}$$

This corresponds to option A.

Key Takeaways

• $F = \mathrm{d}p/\mathrm{d}t$.
• On a $p$–$t$ graph, the gradient equals the resultant force.
• Use two well-separated points on a straight line to reduce reading error.

Common Mistakes

• Using the area under the graph (that would be relevant for a force–time graph giving impulse, not a momentum–time graph).
• Reading the final momentum value as the force (wrong units: $\text{kg m s}^{-1}$ is not N).
• Mixing up axes and calculating $\Delta t / \Delta p$ instead of $\Delta p / \Delta t$.

Things to Be Careful About

• Check units: $\text{kg m s}^{-1}$ divided by $\text{s}$ gives $\text{kg m s}^{-2} = \text{N}$.
• Make sure you take the change in momentum over the change in time using consistent endpoints (here $0$ to $10\ \text{s}$).

Background Concept

For straight-line motion, Newton's second law states that the resultant (net) force on an object equals mass (m) times acceleration (a):

$$\sum F = ma$$

Weight (W) is the gravitational force on the object:

$$W = mg$$

where (g) is the acceleration of free fall.

Understanding the Question

A rocket is moving vertically upwards just as it leaves the ground. The engine provides an upward force (thrust) (T). The rocket's weight (W) acts downward. The rocket's initial acceleration is upward with magnitude (a). We are asked to find an expression for the initial upward force due to the engine, in terms of (W), (a), and (g).

Approach

1. Choose a positive direction (upward is most convenient).
2. Write Newton's second law for vertical forces: thrust upward minus weight downward equals (ma).
3. Replace (m) using (W = mg) so the final expression is in terms of (W), (a), (g).
4. Match the resulting expression to one of the options.

Step-by-Step Reasoning

Take upward as positive.

Forces on the rocket:

• Upward thrust: (T)
• Downward weight: (W)

So the resultant upward force is (T - W). Apply Newton’s second law:

$$T - W = ma$$

But we are not given (m); we are given (W). Use (W = mg):

$$m = \frac{W}{g}$$

Substitute into (T - W = ma):

$$T - W = \left(\frac{W}{g}\right)a$$

Rearrange for (T):

$$T = \frac{Wa}{g} + W$$

This matches option A.

Key Takeaways

• Always write (\sum F = ma) in the direction of acceleration.
• Weight links mass to (g): (m = W/g).
• For upward acceleration, thrust must exceed weight, so the answer should be (> W).

Common Mistakes

• Writing (T = ma - W) (sign error): weight opposes the upward acceleration so it must be added when solving for (T).
• Using (W) as if it were mass (mixing units): (W) is in newtons, not kilograms.
• Forgetting to convert from (m) to (W/g), leaving the answer in terms of (m).

Things to Be Careful About

• Direction/sign convention: define upward as positive before writing the force equation.
• Sense check: if (a = 0), then (T = W) (just hovering at constant velocity initially). The expression (T = \frac{Wa}{g} + W) gives this correctly.
• Units: (\frac{Wa}{g}) has units (\text{N}\cdot\text{m s}^{-2} / \text{m s}^{-2} = \text{N}), so it can be added to (W).

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

$$\tau = F \times d_\perp$$

where $d_\perp$ is the perpendicular distance from the point to the line of action of the force.

Taking moments about the hinge is powerful because the (unknown) hinge reaction forces pass through the hinge, so their moments about the hinge are zero and they do not appear in the equation.

Understanding the Question

A uniform rod (so its weight acts at its midpoint) is hinged at end $X$ and held in equilibrium by a force $F$ applied at the other end.

• Weight $W = 28\text{ N}$ acts vertically downward at the midpoint.
• The rod makes an angle of $40^\circ$ to the horizontal.
• The force $F$ is perpendicular to the rod.

We are asked for the magnitude of $F$.

Approach

1. Take moments about the hinge at $X$.
2. Write the anticlockwise moment from $F$.
3. Write the clockwise moment from the weight using the perpendicular distance from hinge to the weight's line of action.
4. Equate clockwise and anticlockwise moments and solve for $F$.

Step-by-Step Reasoning

Let the rod length be $L$.

1) Moment from $F$ about hinge
$F$ acts at the far end and is perpendicular to the rod. The perpendicular distance from hinge to the line of action of $F$ is therefore the full length $L$.

$$\tau_F = F \times L$$

2) Moment from the weight about hinge
The weight acts at the midpoint, so its point of application is at distance $L/2$ along the rod from the hinge.

But the perpendicular distance to the vertical line of action of the weight is the horizontal distance of the midpoint from the hinge.

Since the rod is at $40^\circ$ to the horizontal, the horizontal component of $L/2$ is:

$$d_\perp = \frac{L}{2}\cos 40^\circ$$

So the clockwise moment of the weight is

$$\tau_W = 28 \times \frac{L}{2}\cos 40^\circ$$

3) Equilibrium of moments

$$F L = 28\left(\frac{L}{2}\cos 40^\circ\right)$$

Cancel $L$:

$$F = 14\cos 40^\circ$$

Numerically:

$$F = 14(0.766) = 10.7\text{ N} \approx 11\text{ N}$$

So the correct option is B.

Key Takeaways

• In equilibrium, clockwise moment = anticlockwise moment about any chosen point.
• Choose the hinge/pivot for moments to eliminate unknown reaction forces.
• Always use the perpendicular distance to the line of action (here: horizontal distance for a vertical weight).

Common Mistakes

• Using $\sin 40^\circ$ instead of $\cos 40^\circ$ for the perpendicular distance (the moment arm is horizontal, not vertical).
• Taking distance along the rod ($L/2$) as the perpendicular distance (it is not unless the force is perpendicular to the rod).
• Forgetting that the rod is uniform, so weight acts at the midpoint.

Things to Be Careful About

• Check which distance is perpendicular to each force’s line of action.
• Don’t include hinge forces in the moment equation about the hinge (they produce zero moment about that point).
• Round to match the options: $10.7\text{ N}$ rounds to $11\text{ N}$.

Background Concept

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

$$\tau = F \times r_\perp$$

where $r_\perp$ is the perpendicular distance from the pivot to the line of action of the force.

A couple is a pair of equal and opposite forces whose lines of action are separated. A couple produces rotation without a resultant force. The turning effect of a couple can be found either by adding the moments of the two forces about any point, or by using

$$\tau_{\text{couple}} = F \times s$$

where $s$ is the perpendicular separation between the lines of action of the two forces.

Understanding the Question

Two forces, each of magnitude $F$, act on a rod in opposite directions. Each is applied at a distance $d$ from the pivot $P$ (one on the left, one on the right). We are asked for the torque about $P$.

Key point: although the forces are opposite, they act on opposite sides of the pivot, so their moments about $P$ act in the same rotational sense.

Approach

Calculate the moment of each force about $P$ using $\tau = Fd$ (since the force is perpendicular to the rod and the distance to the pivot is $d$). Then add the two moments because both turn the rod the same way.

(Equivalently, note that the separation between lines of action is $2d$, so the couple torque is $F(2d)=2Fd$.)

Step-by-Step Reasoning

• Left force: magnitude $F$, perpendicular distance from pivot is $d$.

$$\tau_1 = Fd$$

• Right force: also magnitude $F$, also at distance $d$ from the pivot. Even though it acts downward, because it is on the opposite side of the pivot it produces rotation in the same sense as the left force.

$$\tau_2 = Fd$$

• Total torque about $P$:

$$\tau = \tau_1 + \tau_2 = Fd + Fd = 2Fd$$

So the correct option is C.

Key Takeaways

• A couple has zero resultant force but a non-zero turning effect.
• About the midpoint pivot here, each force provides moment $Fd$, and the two moments add to $2Fd$.
• Couple torque can also be found as $F \times$ (separation between the lines of action).

Common Mistakes

• Choosing $0$ because the forces are equal and opposite (that only refers to resultant force, not torque).
• Taking the torque as $Fd$ (counting only one force and forgetting the second moment).
• Using $4Fd$ by incorrectly multiplying both force and distance factors (as in option D).

Things to Be Careful About

• Always check the direction of rotation about the pivot: forces on opposite sides of the pivot can create moments in the same sense even if the forces are opposite.
• The perpendicular separation between lines of action is $2d$, not $d$.