Physics 9702/13 — May/June 2025

Background Concept

A vector quantity has both magnitude and direction (e.g. displacement, velocity, acceleration, force, momentum).

A scalar quantity has magnitude only (e.g. mass, time, temperature, energy). Words like “increase/decrease”, “cooler”, or “later” may describe a change in a scalar, but they do not give a physical direction in space.

Understanding the Question

We must choose which option represents a vector quantity. So we look for an expression that includes:

• a numerical size with a unit (magnitude), and
• a spatial direction (e.g. up/down/left/right, at an angle, north, etc.).

Approach

Check each option for whether it contains both magnitude and direction.

Step-by-Step Reasoning

• A: $1\ \text{N}$ upwards. Newton ($\text{N}$) is the unit of force, and “upwards” gives a direction. Force is a vector, so this is a vector quantity.
• B: $2\ \text{kg}$ decrease. Kilogram is mass (a scalar). “Decrease” indicates change, not a direction in space.
• C: $3\ \text{K}$ cooler. Kelvin is temperature (a scalar). “Cooler” indicates a decrease in temperature.
• D: $4\ \text{s}$ later. Seconds are time (a scalar). “Later” is not a spatial direction.

Therefore, the only vector quantity given is option A.

Key Takeaways

• Vectors require magnitude + direction.
• Scalars can increase/decrease, but that does not make them vectors.
• Force (in newtons) is a standard example of a vector.

Common Mistakes

• Treating words like “decrease” or “later” as directions.
• Thinking any “change” implies a vector.
• Forgetting that temperature and mass are scalars.

Things to Be Careful About

• A vector direction must be a geometric direction (upwards, east, at $30^\circ$ to the horizontal, etc.), not merely “more/less”.
• Some quantities (like “velocity”) are vectors even if a direction is not explicitly written; but in this question you must pick the option that clearly includes direction in its statement.

Background Concept

For an electrical appliance, the rate of energy transfer (power) is related to the potential difference across it and the current through it by

$$P = VI$$

where $P$ is in watts (W), $V$ in volts (V) and $I$ in amperes (A). For domestic resistive heating appliances (kettles, toasters), the power is usually in the kilowatt range because heating water quickly needs a large energy transfer per second.

Understanding the Question

We are asked for a reasonable estimate of the current drawn by an electric kettle while it is operating. The options span from microamps to kiloamps, so the key skill is order-of-magnitude estimation.

Given information is not explicit, so we use typical mains voltage ($\approx 230\ \text{V}$) and typical kettle power ($\approx 2\text{–}3\ \text{kW}$).

Approach

1. Recall a typical kettle power rating (a few kilowatts).
2. Use $I = P/V$ with $V \approx 230\ \text{V}$.
3. Compare the resulting current with the options.

Step-by-Step Reasoning

Take a typical kettle: $P \approx 2.0\ \text{kW} = 2.0 \times 10^3\ \text{W}$.

Using $P = VI$:

$$I = \frac{P}{V} = \frac{2.0 \times 10^3}{230} \approx 8.7\ \text{A}$$

This is of order $10\ \text{A}$.

• $8\ \mu\text{A}$ and $8\ \text{mA}$ are far too small (typical of sensors / small electronics, not heaters).
• $8\ \text{kA}$ is far too large (would correspond to megawatts at mains voltage and would trip protection instantly).

So the reasonable estimate is $8\ \text{A}$, option C.

Key Takeaways

• Domestic heating appliances typically operate at kilowatt power levels.
• Use $P = VI$ to connect a power estimate and mains voltage to an estimated current.
• Correct order of magnitude here is amperes (around $10\ \text{A}$), not mA or kA.

Common Mistakes

• Confusing prefixes: mixing up $\text{mA}$ and $\text{A}$, or $\mu\text{A}$ and $\text{A}$.
• Forgetting that a kettle is a high-power device and assuming it draws “small electronics” currents.
• Using $P = I^2R$ without knowing $R$, leading to guesswork.

Things to Be Careful About

• Convert $\text{kW}$ to $\text{W}$ correctly: $1\ \text{kW} = 10^3\ \text{W}$.
• Use a sensible mains voltage (around $230\ \text{V}$ for many Cambridge contexts).
• Estimation questions reward the correct scale: here, a few to ~10 amperes is the realistic range.

Background Concept

Measurement errors are usually grouped into:

• Random errors: unpredictable variations between repeated readings (e.g. reaction time, reading a scale). They cause scatter about the mean value. If you repeat and average, positive and negative deviations tend to cancel, so the uncertainty in the mean decreases.
• Systematic errors: a consistent bias in the same direction (e.g. zero error, miscalibration). They shift all readings away from the true value by about the same amount. Repeating and averaging does not remove this bias.

Precision describes how close repeated readings are to each other (small scatter). Accuracy describes how close the result is to the true value (small systematic error and small random uncertainty overall).

Understanding the Question

You are given four statements about errors, accuracy and precision. You must choose the single statement that correctly matches the standard physics definitions.

Approach

Check each option against the definitions:

• Does it confuse accuracy with random error?
• Does it confuse precision with systematic error?
• Does it correctly describe how averaging affects random scatter?
• Does it correctly describe the effect of systematic error on the distribution of readings?

Step-by-Step Reasoning

• Option A: Accuracy does not “always” imply small random error; you could have small random scatter but a large systematic offset (precise but inaccurate), or large random error but the mean happens to be close to true by chance. So A is false.
• Option B: Precision is about small random scatter, not about systematic error. A set can be very precise and have either small or large systematic error. So B is false.
• Option C: Random errors vary from reading to reading. Taking many readings and calculating the mean reduces the effect of random fluctuations (uncertainty in the mean decreases). This is correct.
• Option D: A systematic error does not create a “random set spread out about the true value”; it shifts readings away from the true value (they may still be tightly clustered, just around the wrong value). So D is false.

Hence the correct option is C.

Key Takeaways

• Random error (\rightarrow) scatter; reduced by repeats and averaging.
• Systematic error (\rightarrow) consistent bias; not reduced by repeats.
• Precision relates to random scatter; accuracy relates to closeness to the true value.

Common Mistakes

• Saying “accurate means small random error” (accuracy is about closeness to true value, strongly affected by systematic error).
• Saying “precise means small systematic error” (precision is mainly about random scatter).
• Claiming averaging removes systematic error (it does not).

Things to Be Careful About

• In exam wording, “always” is a strong claim: accuracy/precision do not guarantee a particular type of error is small.
• Averaging improves the estimate of the mean for random error, but the best method still requires checking for zero errors/calibration to address systematic error.

Background Concept

In any physically correct equation, the units on the left-hand side must match the units on the right-hand side (homogeneity of units). This lets you find the units of an unknown constant by treating each quantity as a unit “symbol” and rearranging.

Here, the equation is a form of Fourier’s law of heat conduction:

$$\frac{Q}{t} = \frac{kA\Delta T}{L}$$

where $Q/t$ is the rate of energy transfer (power), $A$ is area, $L$ is thickness/length, and $\Delta T$ is a temperature difference. In SI, temperature differences are measured in kelvin (K). (A change of $1\ ^\circ\text{C}$ equals a change of $1\ \text{K}$ in size, but the SI base unit is K.)

Understanding the Question

You are given a relationship linking $Q/t$ to $k$, geometry ($A$, $L$), and temperature difference $(T_1-T_2)$. The task is to find the SI base units of $k$ and choose the correct option.

Known SI units:

• $Q$ is energy in joules (J)
• $t$ in seconds (s)
• $A$ in $\text{m}^2$
• $L$ in m
• $(T_1-T_2)$ in K

Approach

1. Rearrange the given equation to make $k$ the subject.
2. Replace each quantity with its SI units.
3. Reduce everything to SI base units and compare with the options (noting that K is the SI temperature unit).

Step-by-Step Reasoning

Start with

$$\frac{Q}{t} = \frac{kA(T_1-T_2)}{L}$$

Rearrange:

$$k = \frac{Q}{t}\frac{L}{A(T_1-T_2)}$$

Now work out units:

• $Q/t$ is energy per time = power.

$$\left[\frac{Q}{t}\right] = \text{W} = \text{J s}^{-1}$$

Convert joule to base units:

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

So

$$\text{W} = \text{J s}^{-1} = \text{kg m}^2\text{ s}^{-3}$$

Other units:

• $[L]=\text{m}$
• $[A]=\text{m}^2$
• $[T_1-T_2]=\text{K}$

Substitute into $[k]$:

$$[k] = \frac{\text{kg m}^2\text{ s}^{-3}\times \text{m}}{\text{m}^2\times \text{K}}$$

Cancel one factor of m:

$$[k] = \text{kg m s}^{-3}\text{K}^{-1}$$

This matches option B.

Key Takeaways

• Use unit homogeneity to find the units of a constant.
• Convert derived units carefully: $\text{J} = \text{kg m}^2\text{ s}^{-2}$ and $\text{W} = \text{kg m}^2\text{ s}^{-3}$.
• In SI, temperature is measured in kelvin, so the unit should be $\text{K}^{-1}$.

Common Mistakes

• Using $^\circ\text{C}^{-1}$ in the final unit even though the question asks for SI base units.
• Forgetting that $Q/t$ is power and not energy.
• Converting $\text{J}$ incorrectly (e.g. missing a factor of m).

Things to Be Careful About

• A temperature difference has the same numerical size in $^\circ\text{C}$ and K, but the SI base unit is still K, so choose $\text{K}^{-1}$.
• Track powers of metres: the factor $A/L$ contributes $\text{m}^2/\text{m} = \text{m}$, which must be cancelled correctly when isolating $k$.

Background Concept

On a velocity–time graph, the displacement over a time interval is the signed area under the graph:

• If $v$ is positive (graph above the time axis), the area contributes a positive displacement.
• If $v$ is negative (graph below the time axis), the area contributes a negative displacement (the object moves in the opposite direction).

Mathematically,

$$\text{displacement} = \int v\,\mathrm{d}t$$

For straight-line segments, you can find the integral by simple geometry (triangles/trapezia).

Understanding the Question

You are given a piecewise straight-line $v$–$t$ graph starting at $(0,0)$. The question states that at $t=0$ the displacement is zero, so the displacement at $t=20\ \text{s}$ is just the net signed area between the graph and the time axis from $0$ to $20\ \text{s}$.

Key read-offs from the graph:

• At $t=8\ \text{s}$, $v=+3\ \text{m s}^{-1}$.
• At $t=16\ \text{s}$, $v=0$.
• From $16$ to $24\ \text{s}$, $v$ decreases linearly to $-3\ \text{m s}^{-1}$, so at $t=20\ \text{s}$ (halfway in time) $v=-1.5\ \text{m s}^{-1}$.

Approach

1. Split the interval $0$ to $20\ \text{s}$ into regions where the graph shape is simple.
2. Compute areas of triangles for $0\to 8$ and $8\to 16$.
3. Compute the signed area from $16\to 20$ as a trapezium (since velocity changes linearly).
4. Add the areas, remembering the negative sign for the part below the axis.

Step-by-Step Reasoning

1) Area from $0$ to $8\ \text{s}$
This is a triangle with base $8\ \text{s}$ and height $3\ \text{m s}^{-1}$.

$$A_1 = \frac{1}{2}\times 8 \times 3 = 12\ \text{m}$$

2) Area from $8$ to $16\ \text{s}$
Another triangle above the axis with base $8\ \text{s}$ and height $3\ \text{m s}^{-1}$.

$$A_2 = \frac{1}{2}\times 8 \times 3 = 12\ \text{m}$$

So by $t=16\ \text{s}$,

$$A_1 + A_2 = 24\ \text{m}$$

3) Area from $16$ to $20\ \text{s}$
From $16$ to $24\ \text{s}$ the graph goes linearly from $0$ to $-3\ \text{m s}^{-1}$. At $t=20\ \text{s}$ (4 seconds after 16, i.e. halfway to 24), velocity is halfway to $-3$:

$$v(20)= -1.5\ \text{m s}^{-1}$$

The region from $16$ to $20\ \text{s}$ is a trapezium with parallel sides $0$ and $-1.5$ and width $4$.

$$A_3 = \frac{1}{2}(0 + (-1.5))\times 4 = -3\ \text{m}$$

Negative sign indicates displacement in the negative direction.

4) Total displacement at $20\ \text{s}$

$$\text{displacement} = 24 + (-3) = 21\ \text{m}$$

This matches option B.

Key Takeaways

• Displacement from a $v$–$t$ graph is the signed area under the curve.
• Break complicated shapes into triangles/trapezia.
• When the graph is below the time axis, the area is negative.

Common Mistakes

• Adding areas below the axis as positive (forgetting that negative velocity gives negative displacement).
• Using the gradient (acceleration) instead of the area.
• Using total distance travelled instead of displacement (distance would add the magnitude of the negative part).

Things to Be Careful About

• Read values at the correct times: here $t=20\ \text{s}$ lies on the downward sloping section, so you must interpolate to get $v(20)=-1.5\ \text{m s}^{-1}$.
• Keep units consistent: $(\text{m s}^{-1})\times (\text{s})=\text{m}$ for area under a $v$–$t$ graph.
• Use the signed area because the question asks for displacement, not distance.

Background Concept

In a vacuum, a falling object experiences negligible air resistance, so it undergoes free fall with (approximately) constant acceleration equal to the gravitational field strength, $g \approx 9.8\ \text{m s}^{-2}$. For motion with constant acceleration, the displacement $s$ after time $t$ is related to initial velocity $u$ by

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

For a drop “from rest”, $u = 0$, and for free fall $a = g$.

Understanding the Question

A feather is dropped from rest in a vertical evacuated tube, so we treat it as free fall. The museum wants the time of fall to be close to $0.5\ \text{s}$. We must find the tube length (the distance fallen in that time) and choose the closest option.

Given:

• $u = 0$
• $t = 0.50\ \text{s}$
• $a = g = 9.8\ \text{m s}^{-2}$

Unknown:

• $s$ (the required tube length)

Approach

Use the constant-acceleration displacement equation for motion from rest:

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

Compute $s$ for $t = 0.50\ \text{s}$ and then compare with the listed tube lengths.

Step-by-Step Reasoning

Start from

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

With $u=0$ and $a=g$:

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

Substitute $g=9.8\ \text{m s}^{-2}$ and $t=0.50\ \text{s}$:

$$s = \frac{1}{2} \times 9.8 \times (0.50)^2$$

Calculate $(0.50)^2 = 0.25$:

$$s = 0.5 \times 9.8 \times 0.25 = 1.225\ \text{m}$$

This is $\approx 1.2\ \text{m}$, which corresponds to option A.

Key Takeaways

• In vacuum, falling motion is modelled as constant acceleration $g$.
• For a drop from rest, displacement after time $t$ is $s = \tfrac{1}{2}gt^2$.
• Always compare your computed value to the options (round appropriately).

Common Mistakes

• Using $s = gt^2$ (missing the factor $\tfrac{1}{2}$), which would give $2.45\ \text{m}$ and wrongly suggest option B.
• Forgetting that “from rest” means $u=0$.
• Using $g = 10\ \text{m s}^{-2}$ without checking closeness (it would give $1.25\ \text{m}$, still closest to A, but accuracy matters).

Things to Be Careful About

• Keep units consistent: $g$ in $\text{m s}^{-2}$, $t$ in $\text{s}$ gives $s$ in $\text{m}$.
• Squaring the time is essential: a small change in $t$ changes $s$ significantly because of $t^2$.
• Choose the closest option; here $1.225\ \text{m}$ rounds naturally to $1.2\ \text{m}$.

Background Concept

For projectile motion with negligible air resistance:

• The horizontal and vertical motions are independent.
• The vertical acceleration is constant and equal to $g$ downward for any projectile, regardless of how it is launched.
• Mechanical energy is conserved: loss of gravitational potential energy becomes gain in kinetic energy.

A very powerful result (when air resistance is negligible) is that the speed at a given height depends only on:

• the initial speed, and
• the change in height,
  not on the launch direction.

Understanding the Question

Two coins leave the same point at the edge of a horizontal table:

• Coin X is launched horizontally with speed $u$.
• Coin Y is launched with the same speed $u$ but at $30^{\circ}$ above the horizontal.

Both land on the horizontal ground below the table. We must choose which statement A–D is correct.

Key given information:

• Same initial speed $u$.
• Same initial height above the ground (same table).
• Air resistance negligible.

Approach

Check each option using the physics of projectiles:

• Use conservation of mechanical energy to compare the impact speeds.
• Use vertical-motion ideas to compare vertical distance travelled, time of flight, and vertical acceleration.

Energy is the quickest way to decide option A.

Step-by-Step Reasoning

Option A: same impact speed?

Let the table height above the ground be $h$.

Initial mechanical energy for each coin (taking ground as zero potential):

• Kinetic: $\frac{1}{2}mu^2$
• Potential: $mgh$

At impact, potential energy is $0$ and kinetic is $\frac{1}{2}mv^2$.

Conservation of energy gives

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

so

$$v^2 = u^2 + 2gh$$

This expression contains no launch angle, so both coins have the same impact speed. Hence A is correct.

Why the others are wrong (briefly)

B (same vertical distance): Coin X falls straight down a vertical distance $h$. Coin Y first rises then falls to the ground, so its total vertical distance travelled is greater than $h$.

C (Y hits ground before X): Coin Y has an initial upward vertical component $u\sin 30^{\circ}$, so it spends extra time going up and then coming back down; it therefore lands after coin X.

D (Y has smaller vertical acceleration): In projectile motion without air resistance, vertical acceleration is always $g$ downward for both coins, so it is the same, not smaller.

Key Takeaways

• With negligible air resistance, the impact speed depends only on initial speed and vertical drop: $v^2 = u^2 + 2g\Delta h$.
• Launch angle changes the components of velocity and the time of flight, but not the gravitational acceleration (always $g$ downward).

Common Mistakes

• Thinking a projectile launched upward must hit the ground faster/slower in speed because it “uses up” speed going up (it changes vertical component during flight, but total energy is conserved).
• Saying vertical acceleration depends on initial velocity or direction; it does not (ignoring air resistance).
• Interpreting “vertical distance travelled” as “vertical displacement” (here the distance for coin Y includes up and down segments).

Things to Be Careful About

• Conservation of energy applies because air resistance is stated negligible; with air resistance, impact speeds would differ.
• Distinguish between speed (magnitude of velocity) and velocity components; the components at impact differ, but the resultant speed is the same.
• Time to hit the ground depends on vertical motion (initial vertical component), not on horizontal speed.

Background Concept

Terminal velocity occurs when an object falls through a fluid and the forces balance so that its acceleration becomes zero. The main forces are:

• weight $W = mg$ (downwards)
• drag force $D$ (upwards, increases with speed)

Newton's second law gives

$$F_{\text{res}} = ma.$$

At terminal velocity, the speed is constant, so $a = 0$ and therefore $F_{\text{res}}=0$. This also implies the forces are balanced, so (taking upwards as positive)

$$D - W = 0 \quad \Rightarrow \quad D = W.$$

Understanding the Question

Two spheres are dropped from the same height and both reach terminal velocity. They have equal volumes, but one sphere has a larger density (so it has a larger mass and weight). We must decide which statement (A–D) is correct while both spheres are at terminal velocity.

Approach

Use the defining feature of terminal velocity: constant velocity implies zero acceleration, so the resultant force must be zero for each sphere. Then quickly check each option using:

• resultant force at terminal velocity
• relationship between weight and density when volume is fixed.

Step-by-Step Reasoning

1. At terminal velocity, each sphere moves with constant velocity, so

$$a = 0.$$

1. From Newton's second law,

$$F_{\text{res}} = ma = 0.$$

So the resultant force on each sphere is zero; therefore the resultant forces are equal. This matches option B.

1. Check why others are wrong:

• A (drag forces equal): at terminal velocity, $D=W$ for each sphere. With equal volume but different density, masses are different:

$$m = \rho V.$$

So the denser sphere has larger $m$ and larger $W=mg$, hence larger drag at terminal velocity. So drag forces are not equal.

• C (velocities equal): the denser sphere needs a larger drag to balance its larger weight, which generally requires a higher speed (drag increases with speed), so terminal speeds are not equal.
• D (weights equal): equal volume but different density means different mass, so weights are not equal.

Key Takeaways

• Terminal velocity means $a=0$ and hence resultant force is zero.
• At terminal velocity, drag equals weight: $D=W$.
• For fixed volume, weight depends on density via $m=\rho V$.

Common Mistakes

• Thinking “terminal velocity” means “maximum velocity” but not applying $a=0$.
• Claiming drag forces are equal because the spheres have the same size; drag depends on speed and must match each sphere’s weight at terminal velocity.
• Forgetting that equal volume does not imply equal mass when densities differ.

Things to Be Careful About

• The question asks about the situation while both are at terminal velocity, not during the initial acceleration.
• “Resultant forces are equal” here means both are zero, not that the individual forces (weight or drag) are equal.
• Use consistent sign convention if writing $D-W=0$ or $W-D=0$; the conclusion $F_{\text{res}}=0$ is the key point.

Background Concept

A force is defined (Newton's second law) as the time rate of change of momentum of a body.

Momentum is

$$p = mv$$

and the resultant force is

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

If the mass is constant, then

$$F = \frac{\mathrm{d}(mv)}{\mathrm{d}t} = m\frac{\mathrm{d}v}{\mathrm{d}t} = ma$$

So $F=ma$ is a useful special case, but the more general defining form is $F = \mathrm{d}p/\mathrm{d}t$.

Understanding the Question

You are asked which option gives the definition of force. The options offer different equations involving force:

• $F=ma$ (only true when mass is constant)
• $F = \Delta p/\Delta t$ (general definition)
• $F = W/s$ (related to work done)
• $F = \tau / d_\perp$ (rearrangement of the definition of moment)

The correct definition must be the general law, not a rearranged result from another definition.

Approach

Choose the statement that matches Newton's second law in its defining (momentum) form. Then check the others are either special cases or definitions of other quantities (work, moment).

Step-by-Step Reasoning

• Newton's second law states that the resultant force is the rate of change of momentum:

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

This is exactly option B.

• Option A uses $F = ma$, which follows from $F=\mathrm{d}p/\mathrm{d}t$ only if $m$ is constant; it is not the defining statement in full generality.

• Option C rearranges the work done relation $W = Fs$ (for constant force in the direction of motion). This is not a definition of force.

• Option D rearranges the moment relation $\tau = F d_\perp$ (definition of moment), again not a definition of force.

Therefore the correct choice is B.

Key Takeaways

• The defining statement of force is

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

• $F=ma$ is a special case for constant mass.
• Equations involving work or moment do not define force; they define work and moment.

Common Mistakes

• Choosing $F=ma$ as the definition without recognising it assumes constant mass.
• Confusing a rearranged formula (e.g. $F=W/s$) with a definition.
• Forgetting that moment $\tau$ is defined by $\tau = F d_\perp$, so $F=\tau/d_\perp$ is not a force definition.

Things to Be Careful About

• In A-Level mark schemes, the most general and accepted defining form is $F = \mathrm{d}p/\mathrm{d}t$.
• Conditions hidden in formulas: $W=Fs$ needs constant force and motion along the force; $F=ma$ needs constant mass.
• The question asks for a definition, so prefer the fundamental law over derived/special-case relationships.

Background Concept

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

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

• If $F_{\text{resultant}} = 0$, then $a = 0$.
• Zero acceleration means the velocity is constant (could be zero, i.e. at rest, or a constant non-zero velocity in a straight line).

Also, in momentum form:

$$F_{\text{resultant}} = \frac{\Delta p}{\Delta t}$$

So if $F_{\text{resultant}} = 0$, momentum does not change.

Understanding the Question

The diagram shows four coplanar forces acting through the object:

• $12\ \text{N}$ upwards and $12\ \text{N}$ downwards
• $20\ \text{N}$ to the left and $20\ \text{N}$ to the right

You are asked which statement (A–D) could describe the object's motion.

Approach

1. Find the resultant force by cancelling equal and opposite force pairs.
2. Use $F_{\text{resultant}} = ma$ (or $F = \Delta p / \Delta t$) to deduce what happens to acceleration (and therefore velocity and momentum).
3. Choose the option consistent with that conclusion.

Step-by-Step Reasoning

• Vertical direction: the upward force is $12\ \text{N}$ and the downward force is $12\ \text{N}$, so the resultant vertical force is $0\ \text{N}$.
• Horizontal direction: the rightward force is $20\ \text{N}$ and the leftward force is $20\ \text{N}$, so the resultant horizontal force is $0\ \text{N}$.

Therefore the overall resultant force is:

$$F_{\text{resultant}} = 0$$

From Newton's second law:

$$a = \frac{F_{\text{resultant}}}{m} = 0$$

So the object is not accelerating; its velocity is constant. That means it could be moving in a straight line at constant speed (or be stationary).

Now compare to options:

• A (accelerating): false because $a=0$.
• B (moving in a straight line): true for constant velocity motion.
• C (decreasing speed): would require acceleration opposite to motion, so $F_{\text{resultant}} \neq 0$.
• D (increasing momentum): would require $\Delta p/\Delta t > 0$, so $F_{\text{resultant}} \neq 0$.

Hence option B.

Key Takeaways

• Equal and opposite forces give zero resultant force.
• Zero resultant force implies zero acceleration.
• With zero acceleration, velocity is constant (straight-line motion at constant speed, or rest).
• Increasing/decreasing speed or momentum requires a non-zero resultant force.

Common Mistakes

• Thinking “forces are present so it must accelerate”: acceleration depends on the resultant force, not on the existence of forces.
• Choosing D because forces look “large”: momentum only changes if the resultant force is not zero.
• Assuming the object must be stationary: it can also move with constant velocity.

Things to Be Careful About

• Resultant force is a vector sum: cancel forces separately in perpendicular directions.
• Options C and D both imply a change in velocity/momentum, which requires $F_{\text{resultant}} \neq 0$.
• The diagram also shows no resultant turning effect (pairs act symmetrically), but for this question the key point is the zero resultant force and hence no linear acceleration.

Background Concept

In an isolated system (no external forces), the total momentum vector is conserved:

$$\vec{p}_{\text{initial}} = \vec{p}_{\text{final}}$$

Momentum is a vector quantity, so conservation must hold separately in perpendicular directions (e.g. $x$ and $y$):

$$\sum p_x\; \text{conserved} \qquad \text{and} \qquad \sum p_y\; \text{conserved}$$

Here the collision occurs in a vacuum and the question states that no other particles are involved, so the two nuclei form an isolated system during the collision.

Understanding the Question

One nucleus is moving horizontally towards a stationary nucleus. After collision, exactly two nuclei move away along the shown paths.

Because the incoming nucleus moves purely horizontally, the initial total momentum has:

• a positive $x$-component (to the right)
• zero $y$-component (no up/down component)

So any physically possible outgoing paths must allow the two final momentum vectors to add to a resultant that points purely to the right.

Approach

Use momentum conservation in the vertical direction:

• compute (qualitatively) the sign of the final $p_y$ components from each diagram
• reject any diagram where the final total $p_y$ cannot equal the initial total $p_y = 0$

This avoids needing any masses or speeds.

Step-by-Step Reasoning

1. Initial momentum:

   • stationary target nucleus has $\vec{p} = \vec{0}$
   • incoming nucleus has momentum to the right only

   Therefore:

$$\left(\sum p_y\right)_{\text{initial}} = 0$$

1. Apply conservation of momentum in the $y$ direction:

$$\left(\sum p_y\right)_{\text{final}} = 0$$

1. Inspect each option qualitatively:
   • A: one nucleus goes up-right and the other down-right, so one has $p_y > 0$ and the other has $p_y < 0$. These can cancel (with suitable speeds), so A can be possible.
   • B: similarly one goes slightly up-right and one slightly down-right, so $p_y$ can cancel. Possible.
   • C: both outgoing paths are down-right, so both nuclei have $p_y < 0$.

Then the total final $y$-momentum is negative:

$$\left(\sum p_y\right)_{\text{final}} = p_{y1} + p_{y2} < 0$$

which cannot equal $0$. So C is not possible.

• D: one nucleus goes up-right ($p_y>0$), the other down-left ($p_y<0$). The $y$-components can cancel, and the total $x$-momentum can still be to the right if the rightward $p_x$ is larger in magnitude than the leftward $p_x$. So D can be possible.

Therefore, the only impossible diagram is C.

Key Takeaways

• Momentum is conserved as a vector in an isolated collision.
• If the initial momentum has zero component in some direction (here $y$), the final total momentum component in that direction must also be zero.
• Two vectors that both have negative $y$-components cannot sum to give zero $y$-component.

Common Mistakes

• Treating momentum as a scalar and only considering “total speed” rather than direction.
• Forgetting that conservation must hold in both $x$ and $y$ directions separately.
• Assuming any pair of outgoing directions is possible without checking the initial transverse momentum.

Things to Be Careful About

• The statement “No other particles are involved” is crucial: it means there are only two final momentum vectors to add.
• The target nucleus is initially stationary, so the initial momentum is entirely from the incoming nucleus.
• A diagram with one particle going backwards (like D) can still be possible as long as the other particle has enough forward momentum to keep the total $x$-momentum positive.

Background Concept

The moment (turning effect) of a force about a point is

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

where $F$ is the force and $d$ is the perpendicular distance from the pivot to the force’s line of action.

For equilibrium (no angular acceleration) about a chosen pivot,

$$\sum \text{clockwise moments} = \sum \text{anticlockwise moments}$$

A very useful trick is to choose the pivot at a point where an unknown force acts (e.g. a contact force), because its moment about that point is zero.

Understanding the Question

The bar of length $50\ \text{cm} = 0.50\ \text{m}$ lies on the ground. A $400\ \text{N}$ block is attached at the centre of the bar (so its weight acts at $0.25\ \text{m}$ from end $P$). The bar itself weighs $200\ \text{N}$, acting at its centre of gravity, a distance $x$ from end $P$ (this is what we want).

An upward force of $360\ \text{N}$ is applied at end $Q$. It is “just sufficient to lift end $Q$”, meaning the contact force (reaction) at $Q$ has just fallen to zero and the bar is about to pivot about $P$.

Approach

At the instant $Q$ just lifts:

1. Treat $P$ as the pivot point.
2. Set the anticlockwise moment from the applied upward force at $Q$ equal to the clockwise moments from the two weights (block and bar).
3. Solve for $x$.

Step-by-Step Reasoning

1. Distances from pivot $P$:

   • End $Q$ is $0.50\ \text{m}$ from $P$.
   • The block acts at the centre: $0.25\ \text{m}$ from $P$.
   • The bar’s weight acts at its centre of gravity: $x$ from $P$.

2. Take moments about $P$ (reaction at $P$ gives zero moment because it acts through the pivot).

   • Anticlockwise moment from the upward force at $Q$:

$$360 \times 0.50 = 180\ \text{N m}$$

• Clockwise moments from the weights:

$$400 \times 0.25 = 100\ \text{N m}$$

and

$$200 \times x$$

1. Equate clockwise and anticlockwise moments:

$$360 \times 0.50 = 400 \times 0.25 + 200x$$ $$180 = 100 + 200x$$ $$80 = 200x \Rightarrow x = 0.40\ \text{m}$$

So the centre of gravity of the bar is $0.40\ \text{m}$ from end $P$.

Key Takeaways

• “Just lifts” implies the reaction at the lifted end is zero and the bar pivots about the other end.
• Use the principle of moments about the pivot to avoid unknown reactions.
• Convert cm to m before calculating moments so units are consistent.

Common Mistakes

• Taking moments about the wrong point (e.g. about the centre instead of about $P$ at the tipping point).
• Forgetting that the block is at $0.25\ \text{m}$ from $P$ (centre of a $0.50\ \text{m}$ bar).
• Leaving lengths in cm, leading to moments in \text{N cm} and numerical confusion.

Things to Be Careful About

• The applied force is upward at $Q$, so it produces the opposite sense of rotation to the downward weights.
• The distance for each moment must be the perpendicular distance to the force line of action (here all forces are vertical, so the perpendicular distance is just the horizontal distance along the bar).
• The answer requested is distance from the bar’s centre of gravity to end $P$ (this is $x$, not the distance to $Q$).

Background Concept

Density, symbol $\rho$, describes how much mass is contained in a given volume. It is defined by the relationship

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

where $m$ is mass and $V$ is volume. The SI unit is $\text{kg m}^{-3}$ (kilograms per cubic metre).

Understanding the Question

The question asks for the definition of density in words. You must choose the option that matches $\rho = m/V$.

Approach

1. Recall the formula for density: $\rho = m/V$.
2. Translate this into words: “mass per unit volume”.
3. Match that wording to the options given.

Step-by-Step Reasoning

From the definition,

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

This reads as “density equals mass divided by volume”, i.e. mass per unit volume.

• Option D states “mass per unit volume”, which is exactly $m/V$.
• Option C (“mass per cubic metre”) is a specific unit description ($\text{kg}$ per $\text{m}^3$), not the general definition.
• Option A (“mass of one cubic metre”) is missing the idea of “per unit volume” as a ratio definition.
• Option B (“mass of a unit volume”) is close in meaning, but the standard definition is expressed as “mass per unit volume”, matching D.

Therefore the correct choice is D.

Key Takeaways

• Density is defined by $\rho = m/V$.
• In words: density is mass per unit volume.
• Definitions are best matched to the general ratio form, not to a particular unit size (like $1\,\text{m}^3$).

Common Mistakes

• Choosing “mass per cubic metre” as a definition: that describes the SI unit rather than stating the general relationship.
• Confusing “mass of one cubic metre” with the definition: that is a special case ($V=1\,\text{m}^3$), not the definition itself.

Things to Be Careful About

• A definition should work for any volume, so it should be phrased as “per unit volume” (a ratio), not tied to a specific volume such as $1\,\text{m}^3$.
• Remember the SI unit follows from the definition: since $m$ is in $\text{kg}$ and $V$ in $\text{m}^3$, $\rho$ is in $\text{kg m}^{-3}$.

Background Concept

For an object in equilibrium under three forces, the resultant force is zero:

$$\vec{F}_1 + \vec{F}_2 + \vec{F}_3 = \vec{0}$$

Graphically, this means that if you draw the three force vectors head-to-tail in any order (without changing their directions), they must form a closed triangle (a “vector triangle”).

A block on a rough slope typically has three forces:

• Weight $W = mg$ vertically downward.
• Normal contact force $N$ perpendicular to the plane.
• Friction $F$ parallel to the plane, opposing the direction the block would otherwise slide.

Understanding the Question

The block is at rest on a slope rising to the right. You must choose which of the four given vector triangles correctly represents the directions of the three forces on the block in equilibrium.

So you need:

1. The correct directions of $W$, $N$, and $F$.
2. A triangle where these three vectors, drawn head-to-tail, close.

Approach

1. Decide the direction of each force from the geometry of the slope.
2. Eliminate options with an incorrect direction for any force.
3. For the remaining option(s), check that the friction direction is consistent (up the slope) and the three vectors can form a closed triangle.

Step-by-Step Reasoning

1. Weight $W$ is always vertically downward.

   • So the weight vector in the triangle must point straight down. Options with an upward vertical vector are incorrect.

2. Normal reaction $N$ is perpendicular to the slope.

   • Since the slope rises to the right, the perpendicular outward normal points upwards and to the left.
   • So the normal vector should be a slanted vector pointing up-left.

3. Friction $F$ acts parallel to the slope and opposes the tendency to slide.

   • With the plane rising to the right, “down the slope” is towards the left.
   • Without friction, the block would tend to slide down the slope (left), so friction must act up the slope (to the right).

4. Comparing with the options:

   • Eliminate any triangle where the vertical force points up (not weight).
   • Eliminate any triangle where the sloping force points down-right/down-left (wrong for the normal here).
   • Between the remaining ones, choose the one where the third force points to the right (friction up the slope).

This matches option C.

Key Takeaways

• For equilibrium under three forces, the vectors form a closed triangle.
• On an incline: $W$ is vertical, $N$ is perpendicular to the plane, friction is parallel to the plane and opposite to impending motion.

Common Mistakes

• Drawing friction down the slope (it must oppose the tendency to slide).
• Taking the normal force to be vertical (it is perpendicular to the surface, not to the ground).
• Forgetting that a force triangle must be drawn head-to-tail and must close for equilibrium.

Things to Be Careful About

• Directions matter, not where the vectors are drawn: in a vector triangle you may translate a vector, but you must not rotate it.
• The slope rising to the right means the component of weight down the slope is to the left, so friction is to the right (up the slope).

Background Concept

In a liquid at rest, the (gauge) pressure due to the liquid above a point is

$$p = \rho g h$$

where:

• $\rho$ is the density of the liquid,
• $g$ is gravitational field strength,
• $h$ is the vertical depth below the free surface.

For the same liquid and same $g$, pressure is directly proportional to depth: $p \propto h$.

Understanding the Question

The cylinder contains $80\ \text{cm}^3$ of liquid, so the free surface is at the $80\ \text{cm}^3$ mark.

We are told the pressure due to the liquid at the $60\ \text{cm}^3$ mark is $1200\ \text{Pa}$. That pressure is caused only by the liquid column between the surface (at $80$) and that mark (at $60$).

We want the pressure due to the liquid at the base (the bottom of the cylinder, i.e. below the $0\ \text{cm}^3$ mark), which corresponds to the full depth of the liquid from $80$ down to $0$.

Approach

Because $p = \rho g h$, we can compare pressures using a ratio of depths:

$$\frac{p_1}{p_2} = \frac{h_1}{h_2}$$

We are not given actual heights, only volume marks. In a measuring cylinder (assumed uniform cross-sectional area $A$), volume and height are related by $V = Ah$, so height is proportional to volume.

Thus, the depth between two marks is proportional to the difference in their volume readings.

Step-by-Step Reasoning

Let the cylinder have constant cross-sectional area $A$.

1. Depth below the surface at the $60\ \text{cm}^3$ mark:

• Surface at $80\ \text{cm}^3$.
• Volume of liquid above the $60\ \text{cm}^3$ mark is $80 - 60 = 20\ \text{cm}^3$.

Convert that volume difference to height:

$$h_{60} = \frac{V}{A} = \frac{20}{A}$$

(We do not need $A$ numerically.)

1. Depth below the surface at the base:

• Volume of liquid above the base corresponds to $80 - 0 = 80\ \text{cm}^3$.

So

$$h_{\text{base}} = \frac{80}{A}$$

1. Use proportionality of pressure to depth:

$$\frac{p_{\text{base}}}{p_{60}} = \frac{h_{\text{base}}}{h_{60}} = \frac{80/A}{20/A} = 4$$

1. Calculate base pressure:

$$p_{\text{base}} = 4 \times 1200 = 4800\ \text{Pa}$$

So the correct option is D.

Key Takeaways

• Hydrostatic pressure in a liquid at rest increases linearly with depth: $p \propto h$.
• In a uniform measuring cylinder, volume markings are proportional to height, so you can use volume differences to compare depths.
• Pressure at a point depends on the liquid column above that point (difference between the surface mark and the point’s mark).

Common Mistakes

• Using $60\ \text{cm}^3$ as the depth below the surface instead of the correct difference $80-60$.
• Assuming pressure is proportional to the volume below the point rather than the liquid above it.
• Forgetting that the surface is at $80\ \text{cm}^3$ (since that is the amount of liquid in the cylinder).

Things to Be Careful About

• The method relies on the cylinder having constant cross-sectional area; that is implied by a standard measuring cylinder with uniform scale.
• The question asks for “pressure due to the liquid” (gauge pressure), not including atmospheric pressure.
• Keep the proportional reasoning clear: use differences in readings to represent the depth of liquid above the point.