Physics 9702/12 — October/November 2025

Background Concept

Weight is the gravitational force on a mass.

$$W = mg$$

where $W$ is weight in newtons (N), $m$ is mass in kilograms (kg), and $g \approx 9.8\ \text{m s}^{-2}$ near Earth’s surface. For “reasonable estimates”, you mainly need the correct order of magnitude.

Understanding the Question

You are asked to choose a plausible value for the weight of an adult human from four powers-of-ten options. A typical adult has mass of order $10^{2}\ \text{kg}$ (more specifically around $70\ \text{kg}$), so their weight should be around $10^{3}\ \text{N}$.

Approach

1. Choose a sensible typical mass for an adult (e.g. $60$–$80\ \text{kg}$).
2. Multiply by $g \approx 10\ \text{m s}^{-2}$ for a quick estimate.
3. Compare the result with the given options.

Step-by-Step Reasoning

Take $m \approx 70\ \text{kg}$.

Using $g \approx 9.8\ \text{m s}^{-2}$:

$$W = mg \approx 70 \times 9.8 \approx 686\ \text{N} \approx 6.9 \times 10^{2}\ \text{N}$$

This is closest to $7 \times 10^{2}\ \text{N}$, which corresponds to option B.

(You could also do a faster estimate: $70 \times 10 \approx 700\ \text{N}$, giving the same option.)

Key Takeaways

• A typical adult mass is of order $10^{2}\ \text{kg}$.
• Weight near Earth is roughly $10\ \text{N}$ per kilogram, so an adult weighs of order $10^{3}\ \text{N}$.
• Order-of-magnitude reasoning quickly eliminates wildly unrealistic options.

Common Mistakes

• Confusing mass (kg) with weight (N) and choosing a number around $70$ instead of around $700$.
• Misplacing powers of ten (e.g. choosing $7 \times 10^{4}\ \text{N}$ which would imply a mass of several tonnes).

Things to Be Careful About

• Use $g \approx 9.8\ \text{m s}^{-2}$ (or $10\ \text{m s}^{-2}$ for estimates) and keep track that weight is in newtons.
• Check plausibility by reversing: $W = 7 \times 10^{2}\ \text{N}$ corresponds to $m \approx 70\ \text{kg}$, which is realistic.

Background Concept

Precision describes how closely repeated measurements of the same quantity agree with each other. High precision means small random uncertainty and therefore small scatter in repeated readings.

Accuracy describes how close a measurement (or the mean of many measurements) is to the accepted/true value. Accuracy is affected by systematic errors as well as random errors.

Instrument features like small scale divisions can help reduce reading uncertainty, but they do not guarantee high precision if the measurement technique is poor or the quantity fluctuates.

Understanding the Question

You are asked to choose which option best describes a set of data that has high precision. The key phrase is “set of data”: think about the spread of the repeated values rather than closeness to the accepted value.

Approach

Use the definition:

• High precision (\Rightarrow) repeated values are tightly grouped (small scatter).
  Then check which option states this most directly.

Step-by-Step Reasoning

• Option D: “repeats that are close to each other” directly describes low scatter in repeated readings (\Rightarrow) high precision.
• Option B: “close to the accepted value” describes accuracy, not precision.
• Option A: “equipment with small scale divisions” refers to resolution (smallest readable change). This may improve precision but does not necessarily describe the data set itself.
• Option C: “each value having a low uncertainty” is related, but precision is best identified by the agreement between repeats (scatter). In typical exam definitions, precision is described most clearly by the closeness of repeated readings, i.e. option D.

Therefore the best description of high precision is D.

Key Takeaways

• Precision (\neq) accuracy.
• High precision means repeated readings cluster closely (small random error).

Common Mistakes

• Choosing B because it sounds “good”: closeness to an accepted value is accuracy.
• Confusing instrument resolution (A) with the property of the resulting data set.

Things to Be Careful About

• A set of readings can be precise but inaccurate (tight cluster, wrong mean due to systematic error).
• The word precision in MCQs is most reliably linked to repeatability / small spread.

Background Concept

Many physics quantities are expressed in derived units built from SI base units. The newton is a derived unit:

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

To identify an unknown quantity from its units, rewrite everything in base units and simplify by cancelling powers.

Understanding the Question

You are given the unit

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

and asked which of the listed quantities (acceleration, force, mass, momentum) could have that unit.

Approach

1. Replace (\text{N}) with (\text{kg m s}^{-2}).
2. Multiply by (\text{s}^2) and (\text{m}^{-1}).
3. Cancel factors to find the remaining base unit.
4. Match that base unit to one of the options.

Step-by-Step Reasoning

Start with the given unit:

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

Substitute (\text{N} = \text{kg m s}^{-2}):

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

Now combine powers:

• (\text{s}^{-2}\times \text{s}^{2} = \text{s}^{0} = 1)
• (\text{m}^{1}\times \text{m}^{-1} = \text{m}^{0} = 1)

So the unit simplifies to:

$$\text{kg}$$

(\text{kg}) is the SI base unit of mass, so the correct option is C.

(As a quick check: acceleration has units (\text{m s}^{-2}), force is (\text{N}), and momentum is (\text{kg m s}^{-1}); none of these equal (\text{kg}).)

Key Takeaways

• Convert derived units (like (\text{N})) into SI base units to identify quantities.
• Use index laws to cancel units cleanly.
• If the final unit is (\text{kg}), the quantity is mass.

Common Mistakes

• Forgetting (\text{N} = \text{kg m s}^{-2}) and trying to guess from (\text{N}) directly.
• Cancelling incorrectly (e.g. thinking (\text{s}^{-2}\times \text{s}^{2} = \text{s}^{-4})).
• Confusing mass ((\text{kg})) with weight/force ((\text{N})).

Things to Be Careful About

• Treat units algebraically: keep track of indices on (\text{m}) and (\text{s}).
• Ensure the final simplified unit is a valid SI base unit combination.
• Don’t mix up momentum (\text{kg m s}^{-1}) with mass (\text{kg}): the extra (\text{m s}^{-1}) factor matters.

Background Concept

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

For powers, the key rule is:

• if $y \propto x^n$, then the percentage uncertainty in $y$ is $|n|$ times the percentage uncertainty in $x$.

This comes from how fractional (percentage) changes scale when you raise a value to a power. Constants (like 2) do not affect percentage uncertainty.

Understanding the Question

A ball is dropped from rest and falls a measured distance $s$ in a measured time $t$.

• uncertainty in $s$ is negligible (so we treat it as $0\%$)
• percentage uncertainty in $t$ is $4\%$
• we then compute $g$ assuming no air resistance.

The question asks: what percentage uncertainty should we assign to the calculated $g$?

Approach

1. Write the kinematics equation linking $s$, $t$, and $g$ for motion from rest.
2. Rearrange to make $g$ the subject.
3. Identify how $g$ depends on $t$ (the power of $t$).
4. Multiply the percentage uncertainty in $t$ by the magnitude of that power.

Step-by-Step Reasoning

From rest ($u=0$) with constant acceleration $g$:

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

Rearrange for $g$:

$$g = \frac{2s}{t^2}$$

So $g$ depends on $t$ as $g \propto t^{-2}$.

• The factor $2$ is a constant, so it does not change percentage uncertainty.
• The distance $s$ has negligible percentage uncertainty, so it contributes $0\%$.

Apply the power rule:

$$\%\,\Delta g = 2\times \%\,\Delta t = 2 \times 4\% = 8\%$$

Therefore the correct option is $8\%$.

Key Takeaways

• Use the correct kinematics equation: for a drop from rest, $s = \tfrac{1}{2}gt^2$.
• Rearranging often reveals a power law (here $g \propto t^{-2}$).
• For $x^n$, percentage uncertainty multiplies by $|n|$.

Common Mistakes

• Forgetting to square the time in $s=\tfrac{1}{2}gt^2$, leading to choosing $4\%$ instead of $8\%$.
• Adding uncertainties incorrectly (e.g. thinking $2s$ introduces an extra $2\%$ uncertainty).
• Using $s = vt$ (not valid here because speed is not constant).

Things to Be Careful About

• “Negligible” uncertainty means you ignore it in the propagation (treat as $0\%$), not that it is merely “small”.
• The negative power ($t^{-2}$) does not make uncertainty negative; you use the magnitude of the index.
• This is percentage (fractional) uncertainty, not absolute uncertainty.

Background Concept

For motion in a straight line with constant acceleration, the SUVAT equations relate displacement $d$, initial velocity $u$, final velocity $v$, acceleration $a$, and time $t$. One key relation that eliminates time is

$$v^{2} = u^{2} + 2ad$$

This is valid only when the acceleration is constant and $d$ is the displacement along the line of motion.

Understanding the Question

You are told a car starts with velocity $u$, accelerates at constant $a$ through a displacement $d$, and ends with velocity $v$. The question asks which option gives the expression for the initial velocity $u$ in terms of $v$, $a$, and $d$.

Approach

Choose the SUVAT equation that involves $u$, $v$, $a$, and $d$ but not $t$. Then rearrange it to make $u$ the subject. Finally, compare the rearranged expression with the four options.

Step-by-Step Reasoning

Start from the constant-acceleration relation:

$$v^{2} = u^{2} + 2ad$$

Rearrange to isolate $u^{2}$:

$$u^{2} = v^{2} - 2ad$$

Take the square root to get $u$:

$$u = \sqrt{v^{2} - 2ad}$$

(We take the positive root because $u$ is an initial speed/velocity magnitude in the same direction as the motion described.) This matches option B.

Key Takeaways

• For constant acceleration, use SUVAT equations; to avoid time, use $v^{2} = u^{2} + 2ad$.
• Solving for an initial speed usually produces a square root.
• Always check that the form is dimensionally consistent (here, everything under the square root has units of $\text{m}^2\text{s}^{-2}$).

Common Mistakes

• Using $v = u + at$ and then trying to eliminate $t$ without also using $d = ut + \tfrac{1}{2}at^{2}$.
• Forgetting the factor of $2$ in $2ad$.
• Writing $u = v - ad$ (confusing with $v = u + at$ and mixing up $d$ with $t$).

Things to Be Careful About

• $d$ is displacement, not time, so expressions like $v - ad$ cannot be correct dimensionally.
• When taking square roots, remember $u$ is a velocity; the algebra gives $u = \pm\sqrt{v^{2}-2ad}$, but the physically appropriate sign depends on the direction chosen.

Background Concept

For motion with constant acceleration (including constant deceleration), the kinematic (SUVAT) equations apply. A particularly useful one when time is not involved is

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

where:

• $u$ is the initial speed,
• $v$ is the final speed,
• $a$ is the constant acceleration (negative for deceleration in the chosen positive direction),
• $s$ is the displacement along the line of motion.

Understanding the Question

The bicycle slows down uniformly from $8\ \text{m s}^{-1}$ to $6\ \text{m s}^{-1}$ while travelling $7\ \text{m}$. That tells us the (constant) deceleration. The question then asks for the additional distance travelled starting from $6\ \text{m s}^{-1}$ until the speed becomes $0$.

Approach

1. Use $v^2 = u^2 + 2as$ for the first braking segment to calculate $a$.
2. Use the same equation again for the second segment ($u = 6\ \text{m s}^{-1}$, $v = 0$) to find the stopping distance from $6\ \text{m s}^{-1}$.

Step-by-Step Reasoning

1) Find the constant acceleration
For the first part: $u = 8$, $v = 6$, $s = 7$.

$$v^2 = u^2 + 2as$$ $$6^2 = 8^2 + 2a(7)$$ $$36 = 64 + 14a$$ $$14a = -28 \Rightarrow a = -2.0\ \text{m s}^{-2}$$

The negative sign shows it is a deceleration.

2) Find further distance to rest
Now: $u = 6$, $v = 0$, with the same $a = -2.0\ \text{m s}^{-2}$.

$$0^2 = 6^2 + 2(-2.0)s$$ $$0 = 36 - 4s$$ $$4s = 36 \Rightarrow s = 9.0\ \text{m}$$

So the bicycle travels a further $9\ \text{m}$ before stopping, which corresponds to option B.

Key Takeaways

• “Uniform deceleration” means acceleration is constant (negative in the direction of motion).
• When time is not given, $v^2 = u^2 + 2as$ is usually the quickest SUVAT equation.
• With constant acceleration, you can determine $a$ from one interval and reuse it for another.

Common Mistakes

• Using $v = u + at$ (needs time, which is not given).
• Forgetting to square the speeds in $v^2 = u^2 + 2as$.
• Losing the negative sign for $a$ and then getting a negative distance.
• Adding the initial $7\ \text{m}$ to the requested “further distance” (the question asks only the additional distance after reaching $6\ \text{m s}^{-1}$).

Things to Be Careful About

• Keep a consistent sign convention: take the direction of motion as positive so deceleration gives $a < 0$.
• Distinguish between the total braking distance from $8\ \text{m s}^{-1}$ to rest and the further distance from $6\ \text{m s}^{-1}$ to rest.
• Quote the final distance with a sensible number of significant figures (here it is exactly $9.0\ \text{m}$ from the given numbers).

Background Concept

For one-dimensional motion, velocity is the rate of change of displacement:

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

So the displacement over a time interval is

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

On a velocity–time graph, this integral corresponds to the area between the graph and the time axis, taking account of sign:

• above the axis: $v>0$ so area contributes positively to displacement,
• below the axis: $v<0$ so area contributes negatively to displacement.

Understanding the Question

The graph has two labelled regions:

• area 1 is above the time axis (positive velocity),
• area 2 is below the time axis (negative velocity).

The question asks for an expression for the total displacement, meaning the net change in position, not the total distance travelled.

Approach

Add the areas algebraically (signed areas):

• include $+\text{area 1}$,
• include $-\text{area 2}$ because that section has negative velocity.

Then match this expression to the given options.

Step-by-Step Reasoning

1. Displacement equals the signed area under the $v$–$t$ graph:

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

1. The region labelled area 1 lies where $v>0$, so it contributes

$$+\text{area 1}$$

1. The region labelled area 2 lies where $v<0$. Even if the diagram labels the geometric area as “area 2” (a positive number), its contribution to displacement is negative:

$$-\text{area 2}$$

1. Therefore the total displacement is

$$\Delta s = \text{area 1} - \text{area 2}$$

This corresponds to option A.

Key Takeaways

• Displacement from a $v$–$t$ graph is the signed area under the graph.
• Sections below the time axis (negative velocity) subtract from the total displacement.
• Distance travelled would be $\text{area 1} + \text{area 2}$, but that is not asked here.

Common Mistakes

• Adding $\text{area 1} + \text{area 2}$ (this gives distance, not displacement).
• Using “area 2 − area 1” (wrong sign convention for the labelled regions).
• Thinking that any area must be positive and forgetting that velocity can be negative.

Things to Be Careful About

• The labels “area 1” and “area 2” usually represent geometric areas (positive magnitudes); you must apply the sign from the velocity.
• Always check whether the question asks for displacement (signed) or distance travelled (always positive).

Background Concept

Weight is a force. It is the gravitational force acting on a mass in a gravitational field. Near the Earth’s surface its magnitude is given by

$$W = mg$$

where $m$ is the mass and $g$ is the gravitational field strength (also the free-fall acceleration), about $9.81\ \text{m s}^{-2}$.

Mass is not a force: it is a property of the object (a measure of inertia, i.e. resistance to acceleration).

Understanding the Question

You are asked which statement correctly describes the weight of an object. The options mix up force definitions (Newton’s laws), terminal velocity ideas, and inertia. The correct choice should match the definition of weight as a gravitational force.

Approach

Identify the definition of weight, then compare each option:

• If it describes a gravitational force, it’s correct.
• If it describes inertia (mass) or a different force situation (terminal velocity balance), it’s not the definition of weight.

Step-by-Step Reasoning

• Option C: “the force acting on the object due to a gravitational field” matches the definition of weight exactly. So C is correct.

Why the others are wrong:

• A: “mass multiplied by its acceleration” is $F = ma$, the resultant (net) force producing acceleration $a$, not specifically weight. Weight would be $mg$, where $g$ is the gravitational field strength, not “its acceleration” in general.
• B: At terminal velocity, acceleration is zero so resultant force is zero. Weight is not zero; it is balanced by drag. So weight is not “equal to the resultant force” in that case.
• D: “property that resists change in motion” is inertia (mass), not weight.

Key Takeaways

• Weight is a gravitational force: $W = mg$.
• Mass is a property (inertia), not a force.
• At terminal velocity, forces balance: resultant force $=0$ but weight is still present.

Common Mistakes

• Confusing $F = ma$ with weight: not every force is weight.
• Thinking terminal velocity means weight becomes zero; actually drag equals weight.
• Using “inertia” as a description of weight (inertia describes mass).

Things to Be Careful About

• Weight depends on gravitational field strength $g$ (varies by location); mass does not.
• Terminal velocity means $a=0$ and net force $=0$, but individual forces (weight, drag) are non-zero and equal in magnitude.

Background Concept

Momentum $p$ is defined by

$$p = mv$$

The principle of conservation of momentum states:

• If the resultant external force on a system is zero (i.e. the system is isolated), then the total momentum of the system does not change.

This follows from Newton’s second law in momentum form:

$$F_{\text{external}} = \frac{\Delta p_{\text{system}}}{\Delta t}$$

So if $F_{\text{external}} = 0$, then $\Delta p_{\text{system}} = 0$ and the total momentum is constant.

Understanding the Question

You are asked to identify which option is a correct statement of conservation of momentum.

Key phrase to look for: isolated system / no external force implies total momentum stays constant.

Approach

Scan the options and select the one that:

1. Refers to a system (not a single body),
2. Imposes the condition of being isolated (or equivalent wording), and
3. Concludes that momentum is constant (total momentum before = total momentum after).

Step-by-Step Reasoning

• A: Mentions an elastic collision. Momentum is conserved in both elastic and inelastic collisions (provided no external resultant force), so restricting to elastic is not the general principle statement.
• B: This is the definition of momentum, not the conservation principle.
• C: This is Newton’s second law in momentum form, not the conservation principle.
• D: “The momentum of an isolated system is constant.” This is exactly the conservation principle.

Therefore the correct choice is D.

Key Takeaways

• Conservation of momentum applies to a system, not necessarily to each object separately.
• The condition is no resultant external force (an isolated system).
• The conclusion is total momentum remains constant.

Common Mistakes

• Choosing A because collisions are mentioned: conservation is not limited to elastic collisions.
• Choosing C: confusing Newton’s second law with conservation of momentum.
• Forgetting the key condition: momentum is conserved only when external resultant force on the system is zero.

Things to Be Careful About

• “Isolated” means external forces are zero or cancel in resultant; internal forces between objects do not change the total momentum of the system.
• Momentum is a vector; conservation applies to total momentum in each direction separately (e.g. horizontal and vertical components).

Background Concept

In any collision on a frictionless horizontal surface, there is no external horizontal impulse on the two-ball system. Therefore the total momentum vector of the system is conserved.

Momentum is a vector quantity, so conservation means:

$$\vec{p}_{\text{before}} = \vec{p}_{\text{after}}$$

When two objects stick together (a perfectly inelastic collision), they move off with a common velocity, but the total momentum is still the vector sum of the individual momenta just before impact.

Understanding the Question

You are shown two momentum vectors (one of magnitude $4.0\ \text{N s}$ horizontally to the right, and one of magnitude $8.0\ \text{N s}$ directed up-and-left). After collision the balls stick together, so the momentum after the collision must equal the vector sum of these two given momenta.

The task is to choose which option (A–D) matches the direction (and sensible relative length) of the resultant momentum.

Approach

1. Use conservation of momentum: resultant momentum after collision is $\vec{p}_1 + \vec{p}_2$.
2. Add the vectors either by a tip-to-tail sketch or by considering components:
   • vertical component comes only from the $8.0\ \text{N s}$ vector, so the resultant must point upward;
   • horizontal component is the rightward $4.0\ \text{N s}$ combined with a leftward component from the $8.0\ \text{N s}$ vector.
3. Decide whether the net horizontal component is left or right, and whether the resultant should be long or short compared with $8.0\ \text{N s}$.

Step-by-Step Reasoning

• The $4.0\ \text{N s}$ momentum is purely to the right, so it contributes no vertical component.
• The $8.0\ \text{N s}$ momentum points up-left, so it contributes:
  • a positive vertical component (upward);
  • a negative horizontal component (to the left).

Therefore the resultant must have an upward component (so options pointing downward, C and D, are impossible).

Now compare horizontal components: adding a rightward $4.0\ \text{N s}$ vector will partially cancel the leftward component of the $8.0\ \text{N s}$ vector. From the diagram, the $8.0\ \text{N s}$ vector is not almost horizontal, so its upward component is large and its leftward component is not tiny; thus the resultant should still be slightly to the left of vertical (upward with a small left tilt), not strongly up-left.

Also, since you are adding a vector that points partly opposite in the horizontal direction, the magnitude of the resultant should be less than $8.0\ \text{N s}$ (because there is some cancellation in the horizontal components).

Among the remaining options A and B:

• A is a strongly up-left, long vector (too much leftward component and/or too large magnitude).
• B is mostly upward with slight leftward tilt and a moderate length, consistent with partial cancellation.

So the correct diagram is B.

Key Takeaways

• Momentum is conserved as a vector in collisions when external forces are negligible.
• For 2D momentum problems, use vector addition (tip-to-tail) or resolve into components.
• In a perfectly inelastic collision, objects stick together but total momentum still equals the vector sum before impact.

Common Mistakes

• Treating momentum as a scalar and adding $4.0 + 8.0$ to get $12.0\ \text{N s}$.
• Choosing a downward-pointing option even though one momentum clearly has an upward component and nothing provides a downward component.
• Assuming the resultant must point in the direction of the larger momentum vector without considering component cancellation.

Things to Be Careful About

• Always conserve momentum vectorially (direction matters).
• Check vertical and horizontal components separately to eliminate impossible options quickly.
• The resultant magnitude can be smaller than the larger vector when there is partial opposition in one component (cancellation).

Background Concept

For vertical motion of a falling skydiver, the main forces are:

• weight $mg$ downward (approximately constant),
• drag force $D$ upward (depends on speed and effective area).

Newton’s second law in the downward direction gives

$$mg - D = ma$$

where $a$ is the (downward) acceleration.

A velocity–time graph links directly to acceleration:

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

So the gradient of the $v$–$t$ graph is the acceleration (sign shows direction; magnitude gives how large the acceleration is).

At terminal velocity, speed is constant so $a=0$ and the resultant force is zero, hence $D = mg$.

Understanding the Question

The graph shows:

• initially, a horizontal line at a large speed: the skydiver is falling at constant (terminal) velocity,
• at time $t_2$ the parachute opens and the velocity quickly decreases (the graph slopes down steeply),
• later the graph levels off to a new lower horizontal line, and at time $t_3$ she again has constant (terminal) velocity.

We must decide which of the three statements about acceleration and drag force at $t_1$, $t_2$, and $t_3$ are correct.

Approach

1. Use the slope (gradient) of the $v$–$t$ graph to compare the magnitude of acceleration at the marked times.
2. Use “horizontal section = constant velocity = $a=0$” to infer force balance ($D = mg$) at $t_1$ and $t_3$.
3. Use $mg - D = ma$ to infer when drag is largest (largest upward resultant corresponds to largest deceleration).

Step-by-Step Reasoning

1. Statement 1: maximum magnitude of acceleration at $t_2$

Acceleration magnitude is $|a| = \left|\mathrm{d}v/\mathrm{d}t\right|$, i.e. the steepness of the graph.

• At $t_1$ the graph is horizontal, so gradient $=0$ and $|a|=0$.
• At $t_2$ the graph begins its steep downward drop; this is where the magnitude of the negative gradient is greatest (the strongest deceleration) before it later becomes less steep as a new terminal speed is approached.

So statement 1 is correct.

1. Statement 2: drag at $t_1$ equals drag at $t_3$

At $t_1$ and $t_3$ the velocity is constant (both are on horizontal parts of the graph), so $a=0$ at each time.

Using Newton’s second law:

$$mg - D = m(0) = 0 \quad \Rightarrow \quad D = mg$$

Thus $D(t_1) = mg$ and $D(t_3) = mg$, so they are equal.

So statement 2 is correct.

1. Statement 3: drag force is maximum at $t_1$

At $t_1$, we already found $D = mg$.

Just after the parachute opens (around $t_2$), the skydiver is slowing down rapidly. That means the acceleration is upward (or, taking downward as positive, $a$ is large and negative). From

$$mg - D = ma$$

a large negative $a$ requires $mg - D$ to be large and negative, i.e. $D$ must be greater than $mg$.
So drag becomes larger than its terminal value $mg$ immediately after opening the parachute.

Therefore drag cannot be maximum at $t_1$; it is larger around $t_2$.

So statement 3 is false.

Conclusion: statements 1 and 2 only are correct (\Rightarrow) option A.

Key Takeaways

• The gradient of a $v$–$t$ graph is acceleration; the steepest slope gives the greatest $|a|$.
• Horizontal sections on a $v$–$t$ graph mean $a=0$ and hence resultant force $=0$.
• At terminal velocity, drag equals weight ($D=mg$), even if the terminal speed changes.
• A rapid decrease in speed after opening a parachute implies drag temporarily exceeds weight.

Common Mistakes

• Thinking drag must be different at $t_1$ and $t_3$ because the speeds are different; at terminal velocity the drag still equals $mg$.
• Confusing “maximum velocity” with “maximum acceleration”; acceleration depends on slope, not on the size of $v$.
• Assuming drag is maximum at the highest speed; after the parachute opens, drag can increase sharply due to increased area even while speed decreases.

Things to Be Careful About

• Use magnitude of acceleration: compare steepness, not sign.
• The moment the parachute opens, the drag changes due to increased cross-sectional area; the graph’s steepest region corresponds to largest resultant upward force.
• Always tie force statements to Newton’s second law and whether $a=0$ (terminal) or not.

Background Concept

In a one-dimensional elastic collision, both momentum and kinetic energy are conserved. A very useful consequence (derived from those two conservation laws) is:

relative speed of approach = relative speed of separation

If we take rightward velocities as positive, then:

• relative speed of approach is $u_P - u_Q$ (before collision)
• relative speed of separation is $v_Q - v_P$ (after collision)

So for an elastic collision:

$$u_P - u_Q = v_Q - v_P$$

This rule works regardless of the masses.

Understanding the Question

You are told the blocks move towards each other on a frictionless surface and collide elastically.

From the diagram after the collision:

• block $P$ moves left at $70\ \text{cm s}^{-1}$, so $v_P = -70\ \text{cm s}^{-1}$
• block $Q$ moves right at $20\ \text{cm s}^{-1}$, so $v_Q = +20\ \text{cm s}^{-1}$

The question asks which option could be the velocities before collision.

Approach

1. Choose a sign convention (right positive).
2. Calculate the relative speed of separation $v_Q - v_P$ from the given after-collision velocities.
3. Use $u_P - u_Q = v_Q - v_P$ for an elastic collision.
4. Check each option’s $u_P - u_Q$ and select the one that matches.

Step-by-Step Reasoning

With right as positive:

After collision:

$$v_P = -70\ \text{cm s}^{-1},\quad v_Q = +20\ \text{cm s}^{-1}$$

Relative speed of separation:

$$v_Q - v_P = 20 - (-70) = 90\ \text{cm s}^{-1}$$

Elastic condition:

$$u_P - u_Q = 90\ \text{cm s}^{-1}$$

Now test the options:

• A: $u_P - u_Q = 10 - (-60) = 70$ (not $90$)
• B: $u_P - u_Q = 50 - 0 = 50$ (not $90$)
• C: $u_P - u_Q = (-20) - (70) = -90$ (wrong sign and also they are not moving towards each other)
• D: $u_P - u_Q = 60 - (-30) = 90$ (matches)

So option D is possible.

Key Takeaways

• In a 1D elastic collision: $u_P - u_Q = v_Q - v_P$.
• Use a clear sign convention; velocities in opposite directions must have opposite signs.
• For MCQs, a quick relative-speed check can identify the correct row without needing masses.

Common Mistakes

• Forgetting to assign negative signs to leftward velocities.
• Using $v_P - v_Q$ instead of $v_Q - v_P$ (mixing up the order for “separation” vs “approach”).
• Choosing an option where the blocks are not moving towards each other before the collision.

Things to Be Careful About

• The elastic-collision relative-speed rule is about differences in velocity, so the order matters.
• Keep units consistent (all are in $\text{cm s}^{-1}$ here, so no conversion is needed).
• “Towards each other” means initially $P$ must be to the left moving right, and $Q$ to the right moving left (so typically $u_P>0$ and $u_Q<0$).

Background Concept

A couple is a pair of forces that:

• are equal in magnitude,
• act in opposite directions,
• are parallel, and
• act along different lines of action (so there is a separation).

Key consequences:

• The resultant force is zero (so there is no linear acceleration of the centre of mass).
• The pair produces a non-zero moment (torque), causing rotation.

The moment of a couple is

$$\tau = Fd$$

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

Understanding the Question

You are shown four force diagrams (A–D) acting on a bar. You must choose which diagram represents a couple.

So you need to identify the diagram where the forces satisfy the definition above: equal, opposite, parallel, and separated.

Approach

For each option, check:

1. Are there two forces?
2. Are they equal and opposite?
3. Are they parallel?
4. Are their lines of action separated (not the same line)?

Only if all are true do you have a couple.

Step-by-Step Reasoning

• Option A: One force $F$ upwards at one end and one force $F$ downwards at the other end.

  • Equal magnitudes: yes ($F$ and $F$)
  • Opposite directions: yes (up vs down)
  • Parallel: yes (both vertical)
  • Different lines of action: yes (left end vs right end)
    Therefore this is a couple (zero resultant force but a turning effect).

• Option B: Two equal upward forces.

  • Not opposite directions, so resultant force is not zero.
    Not a couple.

• Option C: Forces are not parallel (one is angled), so they cannot form a couple.

• Option D: Only one force is present, so cannot be a couple.

Hence the correct diagram is A.

Key Takeaways

• A couple requires two forces: equal, opposite, parallel, separated.
• A couple gives rotation without translation (net force $=0$, net moment $\neq 0$).

Common Mistakes

• Choosing two equal forces in the same direction (like B): that gives a resultant force, not a pure turning effect.
• Forgetting the parallel requirement (rejecting C).
• Thinking a single off-centre force (like D) is a couple: a single force can produce a moment about a point, but it is not a couple.

Things to Be Careful About

• The separation must be the perpendicular distance between lines of action, not simply the distance between where arrows touch the bar.
• A couple’s moment does not depend on the choice of pivot: it is a “pure torque” with zero resultant force.

Background Concept

For an object in rotational equilibrium, the net moment (torque) about any point is zero.

• Moment of a force about a point: $\text{moment} = F \times \text{perpendicular distance}$.
• In equilibrium: sum of clockwise moments $=$ sum of anticlockwise moments.

A uniform bar has its centre of gravity at its midpoint, i.e. at distance $L/2$ from either end. The weight $Mg$ acts vertically downward through this midpoint.

Understanding the Question

A uniform bar (mass $M$, length $L$) rests on a pivot located a distance $x$ from the left end X. An object of mass $m$ is placed somewhere on the bar so that the bar does not rotate.

The question asks for an expression for $y$, the distance of the object from the pivot, in terms of $M$, $m$, $L$, and $x$.

Approach

1. Mark where the bar’s weight acts: at the centre, $L/2$ from end X.
2. Find the distance between the pivot and the bar’s centre of gravity: $(L/2 - x)$ (with the pivot measured from the same end X).
3. Take moments about the pivot (so the pivot’s reaction force produces no moment).
4. Set clockwise moment equal to anticlockwise moment and solve for $y$.

Step-by-Step Reasoning

1. The bar’s weight is $Mg$ acting at its midpoint, at position $L/2$ from end X.
2. The pivot is at position $x$ from end X.
3. So the perpendicular distance from the pivot to the bar’s weight is:

$$\left(\frac{L}{2} - x\right)$$

1. The object’s weight is $mg$ acting at distance $y$ from the pivot.
2. For equilibrium (no rotation), moments about the pivot balance:

$$mg\,y = Mg\left(\frac{L}{2}-x\right)$$

1. Cancel $g$ and solve for $y$:

$$y = \frac{M}{m}\left(\frac{L}{2}-x\right)=\frac{M}{2m}(L-2x)$$

This matches option C.

Key Takeaways

• A uniform bar’s weight acts at its midpoint.
• Taking moments about the pivot eliminates the unknown pivot reaction.
• In equilibrium: clockwise moments equal anticlockwise moments.

Common Mistakes

• Using $L-x$ instead of $(L/2-x)$ (forgetting the weight acts at the centre, not at the end).
• Taking moments about an endpoint, introducing an extra unknown reaction force and complicating the calculation.
• Forgetting that $g$ cancels (so the final expression for $y$ should not contain $g$).

Things to Be Careful About

• Distances must be measured from the same reference (here, end X) before subtracting.
• The sign/direction is handled by choosing clockwise vs anticlockwise consistently; the final distance $y$ is a magnitude.
• If $x > L/2$, the bar’s centre of gravity would lie on the other side of the pivot, changing which side the mass must be placed on; the algebraic form still comes from the same moment-balance idea.

Background Concept

Density $\rho$ measures how much mass is contained in a given volume. It is defined by

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

where $m$ is mass and $V$ is volume. Since density is a ratio, changing both $m$ and $V$ can have competing effects.

Understanding the Question

You are told that both mass and volume are changed, and you must choose the pair of changes that will definitely (i.e. for any amounts of change) make the density larger.

So we need a combination where the ratio $m/V$ must increase, not just might increase.

Approach

Use $\rho = m/V$ and consider how the ratio changes:

• increasing the numerator ($m$) tends to increase $\rho$,
• increasing the denominator ($V$) tends to decrease $\rho$,
• decreasing the numerator tends to decrease $\rho$,
• decreasing the denominator tends to increase $\rho$.

Pick the option where both changes act in the direction of increasing the ratio.

Step-by-Step Reasoning

Start with

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

Check each option qualitatively:

• A: $m$ decreases (would reduce $\rho$) and $V$ decreases (would increase $\rho$). Competing effects: not guaranteed.
• B: $m$ decreases and $V$ increases. Both changes reduce $m/V$, so density must decrease, not increase.
• C: $m$ increases and $V$ decreases. Both changes increase $m/V$, so density must increase.
• D: $m$ increases (increases $\rho$) and $V$ increases (decreases $\rho$). Competing effects: not guaranteed.

Therefore the only combination that must increase density is option C.

Key Takeaways

• Density is a ratio $\rho = m/V$.
• To guarantee a ratio increases, increase the numerator and decrease the denominator.

Common Mistakes

• Thinking “mass decreases and volume decreases” must increase density; it depends on which decreases more.
• Choosing “mass increases and volume increases” because mass increases; the volume increase can outweigh it.
• Forgetting density is proportional to $m$ but inversely proportional to $V$.

Things to Be Careful About

• The word “must” means the result must be true for any sizes of change, not just for some particular numerical example.
• When both numerator and denominator change in the same direction (both up or both down), the effect on a ratio is ambiguous without numbers.

Background Concept

For a body fully submerged in a fluid, the upthrust (buoyant force) equals the weight of the fluid displaced (Archimedes’ principle). Quantitatively,

$$U = \rho g V$$

where:

• $\rho$ is the density of the liquid,
• $g$ is the gravitational field strength (acceleration of free fall),
• $V$ is the volume of liquid displaced (for full submersion, this is the object’s volume).

So, for the same object in the same liquid, the only factor that changes from planet to planet is $g$.

Understanding the Question

You are told the object is fully submerged on Earth and on Mars in the same liquid, and that the liquid density and object volume are unchanged. You are asked for the ratio $U_M/U_E$ given $g_M = 3.7\ \text{m s}^{-2}$. This is a direct comparison of buoyant forces under different $g$.

Approach

1. Write $U = \rho g V$ for Earth and Mars.
2. Take the ratio $U_M/U_E$ so that $\rho$ and $V$ cancel.
3. Substitute $g_M$ and the standard Earth value $g_E \approx 9.8\ \text{m s}^{-2}$.
4. Match the numerical result to the given options.

Step-by-Step Reasoning

For Mars:

$$U_M = \rho g_M V$$

For Earth:

$$U_E = \rho g_E V$$

Now form the ratio:

$$\frac{U_M}{U_E} = \frac{\rho g_M V}{\rho g_E V}$$

Cancel common factors $\rho$ and $V$ (because they are the same in both situations):

$$\frac{U_M}{U_E} = \frac{g_M}{g_E}$$

Substitute values:

$$\frac{U_M}{U_E} = \frac{3.7}{9.8} = 0.377... \approx 0.38$$

So the correct option is A.

Key Takeaways

• Upthrust for a submerged object is $U = \rho g V$.
• When comparing the same object in the same liquid, $U \propto g$.
• Ratios are powerful because shared quantities cancel.

Common Mistakes

• Assuming upthrust is independent of gravity and choosing $1.0$.
• Using $g_M$ alone without dividing by Earth’s $g$.
• Using the object’s weight ($mg$) instead of buoyant force ($\rho g V$).

Things to Be Careful About

• The question says same liquid density and same object volume, so only $g$ changes.
• Use a sensible Earth value $g_E \approx 9.8\ \text{m s}^{-2}$ (or $9.81\ \text{m s}^{-2}$; both round to the same option).
• Round to match the options given (here $0.38$).