Physics 9702/12 — February/March 2025

Background Concept

A scalar quantity has magnitude only (no direction). Examples include mass, time, energy and work.

A vector quantity has magnitude and direction, and must be added using vector rules. Examples include displacement, velocity, acceleration, force and momentum.

Understanding the Question

You are given four quantities (force, momentum, velocity, work) and asked which one is a scalar.

Approach

Identify which of the listed quantities requires a direction to be fully specified (vector) and which does not (scalar).

Step-by-Step Reasoning

• Force: described by both size and direction (e.g. 5 N to the right) (\Rightarrow) vector.
• Momentum: (\vec{p} = m\vec{v}), so it has the same direction as velocity (\Rightarrow) vector.
• Velocity: rate of change of displacement, so it includes direction of motion (\Rightarrow) vector.
• Work: defined by

$$W = \vec{F} \cdot \vec{s} = Fs\cos\theta$$

Work is the dot product of two vectors, which produces a scalar result (energy transferred). It can be positive or negative depending on (\theta), but it still has no direction.

Therefore the scalar quantity is work.

Key Takeaways

• Scalars: magnitude only.
• Vectors: magnitude + direction.
• Dot products of vectors (e.g. work) produce scalars.

Common Mistakes

• Thinking “negative work” implies a direction; it does not—negative is a sign indicating energy transfer opposite to the displacement direction, not a vector direction.
• Confusing speed (scalar) with velocity (vector).

Things to Be Careful About

• Any quantity written with an arrow (e.g. (\vec{F}, \vec{v}, \vec{p})) is a vector.
• Work and energy share the unit joule (J) and are both scalars.

Background Concept

Measurement errors are commonly grouped into:

• Systematic errors: a consistent bias that makes readings too large or too small each time (e.g. zero error, miscalibration). These affect accuracy (closeness to the true value).
• Random errors: unpredictable fluctuations between repeats (e.g. reaction time, small reading variations). These affect precision (spread of repeated readings).

Understanding the Question

You are asked for the effect of a systematic error on the measurement of a physical quantity. The options mention precision, range/spread, different repeated values, and being different from the correct value.

Approach

Recall what systematic error does to a set of measurements:

• It does not mainly increase scatter between repeats.
• It produces a consistent offset from the true value.
  So choose the option describing a measured value being different from the correct value.

Step-by-Step Reasoning

• With a systematic error (e.g. a ruler with a +2 mm zero error), every length measured will be too large by about 2 mm.
• Repeated measurements can still be tightly clustered (good precision), but the whole cluster is shifted away from the true value (poor accuracy).
• Therefore the key effect is: the measured value is different from the correct value.

So the correct option is D.

Key Takeaways

• Systematic error → affects accuracy (bias from true value).
• Random error → affects precision (scatter between repeats).

Common Mistakes

• Choosing options about “different values from each other” (that describes random error).
• Thinking systematic error “limits precision” (precision is about spread, not bias).

Things to Be Careful About

• In MCQs, phrases like “different from each other” and “range of values” usually indicate random error.
• “Different from the correct/true value” is the hallmark of a systematic error.

Background Concept

A constant resultant force produces a constant acceleration (Newton's second law):

$$F = ma$$

where $F$ is the resultant (net) force, $m$ is the mass, and $a$ is the acceleration.

From kinematics, the acceleration is the rate of change of velocity. On a velocity–time graph, this is the gradient:

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

Understanding the Question

The car experiences a constant resultant force of $300\ \text{N}$ for $5.0\ \text{s}$. The velocity–time graph (velocity in $\text{cm s}^{-1}$) shows the velocity increases linearly from $30\ \text{cm s}^{-1}$ at $t=0$ to $150\ \text{cm s}^{-1}$ at $t=5.0\ \text{s}$.

We need the car's mass, so we must find the acceleration from the graph and then use $m = F/a$.

Approach

1. Read the initial and final velocities from the graph and calculate the gradient $\Delta v / \Delta t$.
2. Convert the acceleration into SI units ($\text{m s}^{-2}$) because the force is in newtons.
3. Apply $m=F/a$ and match to the nearest option.

Step-by-Step Reasoning

From the graph:

• Initial velocity: $v_0 = 30\ \text{cm s}^{-1}$
• Final velocity after $5.0\ \text{s}$: $v = 150\ \text{cm s}^{-1}$

So the change in velocity is:

$$\Delta v = 150 - 30 = 120\ \text{cm s}^{-1}$$

Acceleration is the gradient:

$$a = \frac{\Delta v}{\Delta t} = \frac{120\ \text{cm s}^{-1}}{5.0\ \text{s}} = 24\ \text{cm s}^{-2}$$

Convert to SI units:

$$24\ \text{cm s}^{-2} = 24 \times 10^{-2}\ \text{m s}^{-2} = 0.24\ \text{m s}^{-2}$$

Now use Newton's second law:

$$m = \frac{F}{a} = \frac{300\ \text{N}}{0.24\ \text{m s}^{-2}} = 1250\ \text{kg}$$

Rounded to the given options, $1250\ \text{kg} \approx 1300\ \text{kg}$, which corresponds to option C.

Key Takeaways

• The gradient of a velocity–time graph gives acceleration.
• Always convert to SI units before using $F=ma$ with $F$ in newtons.
• With MCQs, compute the value then choose the closest option consistent with sensible rounding.

Common Mistakes

• Using $v=150\ \text{cm s}^{-1}$ directly instead of the change in velocity $\Delta v$.
• Forgetting the unit conversion from $\text{cm s}^{-2}$ to $\text{m s}^{-2}$, giving a mass $100$ times too small/large.
• Taking the time as something other than $5.0\ \text{s}$ (e.g. misreading the axis scale).

Things to Be Careful About

• The graph does not start at zero velocity; you must use $150-30$, not just $150$.
• Conversion: $1\ \text{cm} = 10^{-2}\ \text{m}$, so accelerations in $\text{cm s}^{-2}$ must be multiplied by $10^{-2}$.
• Quote the final mass to a sensible number of significant figures consistent with the options (here, $\approx 1.3 \times 10^3\ \text{kg}$).

Background Concept

For motion with constant acceleration in a straight line, the SUVAT equations apply. One useful equation that links speed and distance without needing time is

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

where:

• $u$ is initial speed,
• $v$ is final speed,
• $a$ is constant acceleration,
• $s$ is displacement.

All quantities must be in SI units ($\text{m}$, $\text{s}$) so that $a$ comes out in $\text{m s}^{-2}$.

Understanding the Question

The aircraft:

• starts from rest, so $u = 0$,
• must reach take-off speed $v = 85\ \text{km h}^{-1}$,
• has at most $s = 1.20\ \text{km}$ runway distance.

“Minimum constant acceleration” means: to achieve the required final speed in the largest allowed distance (because for fixed $v$, larger $s$ gives smaller required $a$).

Approach

1. Convert $v$ and $s$ into SI units.
2. Use $v^2 = u^2 + 2as$ with $u=0$.
3. Rearrange to $a = \dfrac{v^2}{2s}$ and calculate.
4. Match the result to the nearest option.

Step-by-Step Reasoning

Convert the speed:

$$85\ \text{km h}^{-1} = 85\times \frac{1000\ \text{m}}{3600\ \text{s}} = 23.6\ \text{m s}^{-1}$$

Convert the distance:

$$1.20\ \text{km} = 1.20\times 10^3\ \text{m} = 1200\ \text{m}$$

Now apply the SUVAT equation:

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

With $u=0$:

$$v^2 = 2as \quad \Rightarrow \quad a = \frac{v^2}{2s}$$

Substitute:

$$a = \frac{(23.6)^2}{2(1200)} = \frac{556}{2400} = 0.232\ \text{m s}^{-2}$$

Rounding to 2 s.f. gives $0.23\ \text{m s}^{-2}$, which corresponds to option A.

Key Takeaways

• Use SUVAT equations only when acceleration is constant.
• Convert all values to SI units before substituting.
• For a required final speed, increasing the available distance reduces the required acceleration.

Common Mistakes

• Forgetting to convert $\text{km h}^{-1}$ to $\text{m s}^{-1}$ (this gives an answer too large by a factor of about $3.6^2$).
• Using $v=u+at$ (needs time, which is not given).
• Using $s=vt$ (only valid for constant speed, not accelerating motion).

Things to Be Careful About

• The distance is “no more than” $1.20\ \text{km}$: the minimum acceleration corresponds to using $s=1.20\ \text{km}$ exactly.
• Ensure $2s$ is used in the denominator in $a=\dfrac{v^2}{2s}$.
• Quote acceleration in $\text{m s}^{-2}$ and match to the given options.

Background Concept

For projectile motion with negligible air resistance:

• The only acceleration is gravitational acceleration $g$ vertically downward.
• Horizontal acceleration is zero, so the horizontal component of velocity stays constant.
• Vertical velocity changes uniformly: it decreases on the way up, becomes $0$ at the top, then increases in magnitude downward.

The speed is the magnitude of the velocity vector:

$$v = \sqrt{v_x^2 + v_y^2}$$

So even if one component becomes zero, the speed is not necessarily zero.

Understanding the Question

The object is launched at $20\ \text{m s}^{-1}$ at $45^\circ$ above the horizontal from ground level and lands back at the same level. We must choose the statement that correctly describes how its speed changes from launch, through the highest point, to just before it returns to the ground.

Key clues:

• “Air resistance is negligible” (\Rightarrow) mechanical energy is conserved and horizontal speed remains constant.
• “Immediately before it reaches the ground again” (\Rightarrow) compare speeds at the same height (launch height = landing height).

Approach

1. Resolve the initial velocity into horizontal and vertical components.
2. Use projectile-motion facts: $v_x$ constant; $v_y$ becomes zero at the top.
3. Find the minimum speed at the top from $v = \sqrt{v_x^2 + v_y^2}$.
4. Use symmetry / energy conservation to state the speed just before landing back at the original height.
5. Match this behaviour to the given options.

Step-by-Step Reasoning

1. Resolve the initial velocity:

$$v_x = 20\cos 45^\circ, \qquad v_y = 20\sin 45^\circ$$

1. With negligible air resistance:

• Horizontal acceleration $=0$, so $v_x$ stays constant throughout.
• Vertical velocity decreases on the way up because acceleration is $-g$.

1. At the highest point:

• Vertical component $v_y = 0$.
• Horizontal component still $v_x = 20\cos 45^\circ$.
  So the speed at the top is

$$v_{\min} = \sqrt{v_x^2 + 0^2} = v_x = 20\cos 45^\circ \approx 14\ \text{m s}^{-1}$$

This is greater than zero, so any option claiming the speed becomes zero is incorrect.

1. Just before reaching the ground again (back to the same height as launch):

• With no air resistance, mechanical energy is conserved, so at the same height the speed is the same as initially.
• Equivalently, projectile symmetry gives the same speed on return to the original level.
  Hence the speed becomes $20\ \text{m s}^{-1}$ again.

1. Therefore the speed:

• decreases from $20\ \text{m s}^{-1}$ to a value greater than zero (about $14\ \text{m s}^{-1}$),
• then increases back to $20\ \text{m s}^{-1}$.

This matches option A.

Key Takeaways

• In projectile motion (no air resistance), $v_x$ stays constant and only $v_y$ changes.
• The minimum speed occurs at the highest point and equals the horizontal component of the velocity.
• At the same height as launch, the speed is the same as the launch speed (energy conservation / symmetry).

Common Mistakes

• Confusing speed with vertical speed: the vertical component becomes zero at the top, but the overall speed does not.
• Thinking the object hits the ground faster than it was launched (true only if landing lower than launch, or with extra energy added).
• Assuming both velocity components decrease to zero at the top.

Things to Be Careful About

• The question asks for speed (a scalar), not velocity.
• “Immediately before it reaches the ground again” implies returning to the same height, so the speed matches the initial speed only because air resistance is negligible.
• If air resistance were present, the speed on return would be less than the initial speed.

Background Concept

Momentum $p$ is defined as

$$p = mv$$

where $m$ is mass and $v$ is velocity (a vector). The principle of conservation of momentum for a system states that the vector sum of the momenta of all objects in the system does not change, as long as the net external force on the system is zero. This follows from Newton's second law in momentum form:

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

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

Understanding the Question

You are asked to choose the correct statement of the conservation of momentum for a system. The key feature to look for is the condition under which momentum is conserved: it depends on external forces, not on whether the collision is elastic or inelastic.

Approach

Recall the correct conservation statement: “total momentum remains constant provided no net external force acts on the system.” Then compare this with the options.

Step-by-Step Reasoning

• Option A is wrong because kinetic energy is not always conserved (it is only conserved in elastic collisions).
• Option B is wrong because momentum is conserved in both elastic and inelastic collisions, provided external forces are negligible.
• Option C matches the correct condition: momentum is conserved provided that no external forces act (more precisely, no resultant external force).
• Option D is not a conservation principle; it is essentially the definition of momentum (and even then, it describes each object’s momentum, not a conservation law).

Therefore, the correct option is C.

Key Takeaways

• Momentum is a vector quantity: $p = mv$.
• Total momentum of a system is conserved if the net external force on the system is zero.
• Momentum conservation applies to elastic and inelastic interactions (condition: isolated system / negligible external forces).

Common Mistakes

• Thinking momentum is conserved only in elastic collisions (confusing momentum conservation with kinetic energy conservation).
• Forgetting that “no external forces” really means “no resultant external force” on the system.
• Choosing a statement that is just the definition of momentum rather than the conservation principle.

Things to Be Careful About

• In exam wording, “no external forces act” is usually accepted as shorthand for “no net external force”; small external forces (e.g. friction) would break exact conservation.
• Momentum conservation is about the total (vector) momentum of the system, not the momentum of each object separately.

Background Concept

Momentum is a vector quantity:

$$\vec{p} = m\vec{v}$$

In an isolated system (no external resultant force), total momentum is conserved:

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

If objects stick together, the collision is perfectly inelastic, but momentum is still conserved.

Often we only need one component of momentum (e.g. along the original direction of P). Because momentum is a vector, we can conserve momentum separately in perpendicular directions by resolving into components.

Understanding the Question

P has mass $6.0\ \text{kg}$ and moves horizontally to the right at $3.0\ \text{m s}^{-1}$. Q has mass $2.0\ \text{kg}$ and moves at $4.2\ \text{m s}^{-1}$ at $35^\circ$ to P’s path, as shown (Q is moving downwards and to the left toward P).

They collide and stick together. The question asks for the magnitude of the component of the final momentum in P’s original direction (take this as the +x direction).

Because momentum is conserved, this final x-component equals the initial total x-component.

Approach

1. Choose axes with +x along P’s original direction.
2. Calculate P’s momentum in the x-direction.
3. Resolve Q’s velocity into an x-component and hence find Q’s x-momentum (with the correct sign).
4. Add the x-components to get the total (and therefore final) momentum component along x.

Step-by-Step Reasoning

1) Momentum of P in the x-direction
P moves entirely along +x, so

$$p_{Px} = mv = 6.0 \times 3.0 = 18\ \text{kg m s}^{-1}$$

2) x-component of Q’s momentum
Q’s speed is $4.2\ \text{m s}^{-1}$ at $35^\circ$ to the horizontal, but from the diagram it is directed to the left. So the x-component of Q’s velocity has magnitude

$$v_{Qx} = 4.2\cos 35^\circ$$

but it is negative (opposite +x). Hence

$$p_{Qx} = m v_{Qx} = 2.0 \times (-4.2\cos 35^\circ)$$

Numerically,

$$2.0 \times 4.2\cos 35^\circ \approx 6.9\ \text{kg m s}^{-1}$$

so

$$p_{Qx} \approx -6.9\ \text{kg m s}^{-1}$$

3) Total x-momentum (and final x-momentum)
Since the system is isolated,

$$p_{x,\text{final}} = p_{x,\text{initial}} = p_{Px} + p_{Qx}$$

So

$$p_{x,\text{final}} = 18 + (-6.9) = 11.1\ \text{kg m s}^{-1} \approx 11\ \text{kg m s}^{-1}$$

This matches option B.

Key Takeaways

• Momentum is a vector; conserve it using components.
• In an isolated system, total momentum is conserved even in perfectly inelastic collisions.
• Use $p_x = mv\cos\theta$ for the component along a chosen axis, and include the correct sign from the direction.

Common Mistakes

• Ignoring direction/sign and adding $18 + 6.9$ to get $25\ \text{kg m s}^{-1}$ (option D).
• Using $\sin 35^\circ$ instead of $\cos 35^\circ$ for the horizontal component.
• Calculating final velocity first (unnecessary here since only momentum component is asked).

Things to Be Careful About

• The phrase “component in the original direction of P” means along P’s original line of motion (the x-axis).
• Always decide from the diagram whether Q’s x-component is in the same direction as P or opposite; here it is opposite.
• Quote the momentum component with correct unit $\text{kg m s}^{-1}$ and appropriate rounding to match the options.

Background Concept

The gravitational force on a mass near Earth is its weight:

$$W = mg$$

where $m$ is the mass and $g$ is the gravitational field strength. This force depends on the gravitational field, not on whether the object is accelerating.

When an object is in contact with a surface (e.g. standing on a floor), there is also a normal contact force (reaction) $R$ from the surface. This is often what people informally call the "apparent weight" (what a scale reads), and it does change when there is acceleration.

Understanding the Question

At rest on the launchpad, the astronaut experiences gravitational force $F = mg$.
The spacecraft then accelerates upwards with acceleration $0.2g$.
The question asks specifically for the gravitational force on the astronaut during the launch.

Approach

Identify what is meant by gravitational force: it is weight $mg$, which stays the same as long as $g$ is unchanged.
Check the common trap: using Newton’s second law to find the resultant force and confusing that with the gravitational force.

Step-by-Step Reasoning

The gravitational force is always

$$W = mg$$

Even when the spacecraft accelerates upward, the gravitational field is (by the question’s assumption) unchanged, so $W$ remains $mg$.

If we apply Newton’s second law to see where option A comes from:

Upward resultant force on astronaut:

$$R - mg = ma$$

with $a = 0.2g$:

$$R - mg = m(0.2g)$$ $$R = 1.2mg$$

So $1.2mg$ is the normal reaction / apparent weight, not the gravitational force.

Therefore the gravitational force is $mg$ → option B.

Key Takeaways

• Gravitational force near Earth is $W = mg$.
• Acceleration affects the contact force (apparent weight), not the gravitational force.
• Use Newton’s second law carefully: it relates to the resultant force, not any single force.

Common Mistakes

• Choosing $1.2mg$ by calculating the reaction force and calling it the gravitational force.
• Thinking weight becomes $m(g+a)$ whenever there is upward acceleration.
• Answering $0.8mg$ by subtracting the acceleration effect from $g$ (confusing with apparent weight in downward-accelerating situations).

Things to Be Careful About

• The question says gravitational force, not "force on the astronaut" or "apparent weight".
• Only if $g$ itself changes significantly (e.g. large altitude change) would the gravitational force change; that is not implied here.

Background Concept

The weight of an object acts through its centre of gravity (centre of mass in a uniform gravitational field). For a uniform plank, the mass is evenly distributed, so its centre of gravity is at the geometrical midpoint.

A moment (torque) about a pivot is given by

$$\tau = F \times d$$

where $F$ is the force and $d$ is the perpendicular distance from the pivot to the line of action of the force. If a force’s line of action passes through the pivot, then $d=0$ and the moment is zero.

In equilibrium:

• net moment about any point is zero (clockwise moments = anticlockwise moments), and
• net force is zero.

Understanding the Question

A uniform plank is horizontal and balanced on a pivot at its midpoint. Two children sit on opposite sides. You are asked which statement about the weight of the plank is correct.

Key facts:

• “Uniform plank” (\Rightarrow) weight acts at plank midpoint.
• “Pivot at its midpoint” (\Rightarrow) that weight produces no turning effect about the pivot.

Approach

1. Locate where the plank’s weight acts (centre of gravity).
2. Decide whether that force produces a clockwise, anticlockwise, or zero moment about the pivot.
3. Check the other options for common misconceptions (e.g. confusing plank weight with total reaction at pivot).

Step-by-Step Reasoning

1. Because the plank is uniform, its centre of gravity is at its midpoint.
2. The pivot is also at the midpoint, so the line of action of the plank’s weight passes through the pivot.
3. Therefore the perpendicular distance $d$ from pivot to the weight’s line of action is $0$, so

$$\tau_{\text{plank}} = W_{\text{plank}} \times 0 = 0.$$

So the plank’s weight causes no clockwise or anticlockwise moment.

Evaluating the options:

• A True: weight can be considered to act at the midpoint.
• B/C False: the plank’s weight produces zero moment about the midpoint pivot.
• D False: the pivot reaction equals the total downward force (weights of both children + plank), not just the plank’s weight.

Key Takeaways

• A uniform object’s weight acts at its geometrical centre.
• A force whose line of action passes through the pivot produces zero moment.
• The support force at a pivot balances the total weight in vertical equilibrium.

Common Mistakes

• Saying the plank’s weight causes a clockwise/anticlockwise moment without noticing it acts at the pivot.
• Confusing “force from the pivot” with “weight of the plank”; the pivot supports all loads.
• Thinking the weight acts where the label is drawn or where the children sit, rather than at the centre of gravity.

Things to Be Careful About

• Moments depend on perpendicular distance to the line of action, not just “left” or “right”.
• Even though the plank has mass, if its weight acts exactly at the pivot, it does not affect the moment balance (but it still affects the vertical force balance).

Background Concept

Archimedes' principle states that the upthrust on a submerged object equals the weight of the fluid displaced. For an object fully submerged in a liquid, the volume of liquid displaced equals the volume of the object. Upthrust $U = \rho_{\text{liquid}} V g$, where $\rho_{\text{liquid}}$ is the density of the liquid, $V$ is the volume of the object, and $g$ is the acceleration due to gravity. The weight of the object is $W = \rho_{\text{object}} V g$. The ratio of upthrust to weight is therefore $U/W = \rho_{\text{liquid}} / \rho_{\text{object}}$.

Understanding the Question

The question asks which single change would double the value of the ratio $U/W$. The ratio depends only on the densities of the liquid and the object, not on the volume. We need to evaluate each option to see which one doubles $\rho_{\text{liquid}} / \rho_{\text{object}}$.

Approach

Write the ratio in terms of densities. Then consider each option: how does it affect $\rho_{\text{liquid}}$ and $\rho_{\text{object}}$? The volume of the object does not affect the ratio, so options that only change volume (B and D) leave the ratio unchanged. Option A doubles $\rho_{\text{liquid}}$, doubling the ratio. Option C doubles $\rho_{\text{object}}$, halving the ratio. Thus A is correct.

Step-by-Step Reasoning

1. Write expressions: $U = \rho_{\text{liquid}} V g$, $W = \rho_{\text{object}} V g$.
2. Ratio $R = U/W = (\rho_{\text{liquid}} V g) / (\rho_{\text{object}} V g) = \rho_{\text{liquid}} / \rho_{\text{object}}$.
3. For $R$ to double, $\rho_{\text{liquid}}$ must double or $\rho_{\text{object}}$ must halve.
4. Evaluate options:
   • A: $\rho_{\text{liquid}}$ doubles → $R$ doubles.
   • B: $V$ halves, but $R$ independent of $V$ → $R$ unchanged.
   • C: $\rho_{\text{object}}$ doubles → $R$ halves.
   • D: $V$ doubles, but $R$ independent of $V$ → $R$ unchanged.
5. Therefore, only option A doubles the ratio.

Key Takeaways

• Upthrust depends on the density of the fluid and the volume of the object submerged.
• Weight depends on the density and volume of the object.
• For a fully submerged object, the ratio of upthrust to weight simplifies to the ratio of densities.
• Changing the volume of the object does not affect this ratio.

Common Mistakes

• Thinking that changing the volume of the object changes the ratio. Since both weight and upthrust are proportional to volume, the volume cancels out.
• Confusing weight/upthrust with upthrust/weight. The question asks for upthrust/weight, so doubling the liquid density doubles the ratio.
• Misapplying Archimedes' principle: forgetting that upthrust equals weight of displaced fluid, not weight of the object.

Things to Be Careful About

• Ensure the ratio is correctly identified: upthrust/weight, not weight/upthrust.
• The volume of the liquid itself is irrelevant; only the volume of the object matters.
• The object must be fully submerged for the displaced volume to equal the object's volume.

Background Concept

A pin-jointed frame made of light (negligible weight) straight rods often forms two-force members: if a rod is only pinned at its ends and has no other loads, the force it exerts at each end must act along the rod’s length (either tension or compression).

For a pin joint in equilibrium, the vector sum of forces is zero:

$$\sum F_x = 0, \qquad \sum F_y = 0.$$

When a force acts along a rod at an angle $\theta$ to the horizontal, its components are:

$$F_x = F\cos\theta, \qquad F_y = F\sin\theta.$$

Understanding the Question

The sign has weight $75\,\text{N}$ acting downward at point $Y$. Two rods meet at $Y$:

• rod $XY$ is horizontal,
• rod $YZ$ makes $30^\circ$ to the horizontal and goes up-left to the wall.

You are asked for the horizontal force exerted by the wall on rod $XY$, i.e. the horizontal hinge reaction at $X$ (which equals the axial force in rod $XY$ because the rod is a two-force member).

Approach

1. Consider equilibrium of the joint at $Y$ (the only place where an external load, $75\,\text{N}$, is applied).
2. Assume forces from rods act along their lengths: one horizontal (from $XY$) and one at $30^\circ$ (from $YZ$).
3. Use vertical equilibrium to find the force in $YZ$.
4. Use horizontal equilibrium to find the force in $XY$, which is the horizontal force transmitted to the wall at $X$.

Step-by-Step Reasoning

At joint $Y$, there are three forces:

• weight $W = 75\,\text{N}$ downward,
• force from rod $YZ$ along the rod (up-left at $30^\circ$), magnitude $T_{YZ}$,
• force from rod $XY$ along the rod (horizontal), magnitude $F_{XY}$.

Vertical equilibrium (upward components balance the weight):

$$T_{YZ}\sin 30^\circ = 75.$$

Since $\sin 30^\circ = 0.5$,

$$T_{YZ} = \frac{75}{0.5} = 150\,\text{N}.$$

Horizontal equilibrium (rightward and leftward components balance):
The horizontal component of $T_{YZ}$ is leftward with magnitude $T_{YZ}\cos 30^\circ$, so the horizontal force in rod $XY$ must be equal in magnitude:

$$F_{XY} = T_{YZ}\cos 30^\circ = 150\cos 30^\circ.$$

With $\cos 30^\circ = 0.866$,

$$F_{XY} = 150 \times 0.866 \approx 1.30\times 10^2\,\text{N}.$$

So the horizontal force exerted by the wall on rod $XY$ has magnitude $130\,\text{N}$ (option C).

Key Takeaways

• For a light pin-jointed rod with no other loads, forces at its ends act along the rod.
• For equilibrium at a joint: resolve forces into components and set $\sum F_x = 0$ and $\sum F_y = 0$.
• A single angled support often provides the vertical component needed to balance weight.

Common Mistakes

• Using $\cos 30^\circ$ instead of $\sin 30^\circ$ for the vertical balance (vertical component is $F\sin\theta$).
• Forgetting that the force in $YZ$ is not $75\,\text{N}$; it must be larger because only its vertical component supports the weight.
• Treating the $0.50\,\text{m}$ length as necessary here; equilibrium at joint $Y$ does not require taking moments.

Things to Be Careful About

• Be consistent about the angle: the rod is at $30^\circ$ to the horizontal, so vertical component uses $\sin 30^\circ$.
• The question asks for the horizontal force from the wall on rod $XY$: its magnitude equals the axial force in $XY$ for this two-force member.
• Round sensibly: $150\cos 30^\circ = 129.9\,\text{N}$ rounds to $130\,\text{N}$, matching the option.

Background Concept

When quantities are multiplied or divided, their percentage (or fractional) uncertainties add.

• If $Q = AB$, then (approximately) $\Delta Q\% = \Delta A\% + \Delta B\%$.
• If $Q = \dfrac{A}{B}$, then $\Delta Q\% = \Delta A\% + \Delta B\%$.
• If $Q = A^n$, then $\Delta Q\% = |n|\,\Delta A\%$.

These rules are used for “uncertainty in a derived quantity” when the uncertainties are small and independent.

Understanding the Question

You are given uncertainties in:

• height $h = 0.20\ \text{m} \pm 2\%$,
• internal diameter $d = 0.05\ \text{m} \pm 3\%$,
• mass $m = 0.36\ \text{kg} \pm 10\%$.

The density is calculated from

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

where the liquid’s volume $V$ is the volume of a cylinder (beaker), using the measured $h$ and $d$. The question asks for the percentage uncertainty in $\rho$.

Approach

1. Write $V$ in terms of $d$ and $h$.
2. Substitute into $\rho = m/V$ to see how $\rho$ depends on $m$, $d$, and $h$.
3. Add percentage uncertainties, remembering that $d$ is squared in the volume.

Step-by-Step Reasoning

1. Cylinder volume using diameter $d$:

$$V = \pi r^2 h, \quad r = \frac{d}{2}$$

So

$$V = \pi \left(\frac{d}{2}\right)^2 h = \frac{\pi d^2 h}{4}$$

1. Density:

$$\rho = \frac{m}{V} = \frac{m}{\pi d^2 h/4}$$

Ignoring constants ($\pi$ and $4$ have no uncertainty), the dependence is:

$$\rho \propto \frac{m}{d^2 h}$$

1. Combine percentage uncertainties:

• from $m$: $10\%$
• from $h$ in the denominator: $2\%$
• from $d^2$ in the denominator: $2 \times 3\% = 6\%$

Total:

$$\Delta \rho\% = 10\% + 6\% + 2\% = 18\%$$

So the correct option is D.

Key Takeaways

• For $\rho = m/V$, uncertainty in $\rho$ depends on uncertainties in both $m$ and the measured dimensions used to find $V$.
• For a cylinder, $V \propto d^2 h$, so diameter uncertainty contributes twice.
• For products/quotients, add percentage uncertainties.

Common Mistakes

• Forgetting that $d$ is squared, and using $3\%$ instead of $6\%$.
• Subtracting uncertainties because a variable is in the denominator (you still add percentage uncertainties).
• Using radius uncertainty without converting correctly from diameter (here $r = d/2$; the factor $1/2$ does not affect percentage uncertainty, but mixing $r$ and $d$ can lead to errors).

Things to Be Careful About

• Use the relationship $V = \frac{\pi d^2 h}{4}$ (not $\pi d h$).
• Constants like $\pi$ and $4$ have no uncertainty, so they do not contribute.
• Apply the power rule carefully: if a quantity is squared, its percentage uncertainty doubles.

Background Concept

For a rigid body in static equilibrium:

1. Resultant force is zero (vertical equilibrium here):

$$\sum F_y = 0$$

1. Resultant moment about any point is zero:

$$\sum \tau = 0$$

The plank is uniform, so its weight acts at its centre of gravity, i.e. at the midpoint (here, $2.0\ \text{m}$ from either end).

Understanding the Question

A uniform plank $XY$ of length $4.0\ \text{m}$ has weight $300\ \text{N}$ acting at its centre. It is supported at both ends, giving upward reactions at $X$ and at $Y$.

A child of weight $600\ \text{N}$ stands either at $X$ or at $Y$. We want the upward reaction at the left support ($X$), called $F$, in each case, then choose the option that matches both values.

Approach

For each position of the child:

1. Use moments about $X$ to find the reaction at $Y$ (because the reaction at $X$ has zero moment about $X$).
2. Use vertical force balance to get $F$ from

$$F + R_Y = 900\ \text{N}.$$

Step-by-Step Reasoning

Let $F$ be the upward reaction at $X$ and $R_Y$ the upward reaction at $Y$.

Total downward force is always

$$300 + 600 = 900\ \text{N}.$$

1) Child stands at $X$

Take moments about $X$.

• Child is at $X$, so its moment about $X$ is $0$.
• Plank weight $300\ \text{N}$ acts at $2.0\ \text{m}$ from $X$.
• Reaction at $Y$ acts at $4.0\ \text{m}$ from $X$.

So

$$R_Y(4.0) = 300(2.0)$$ $$R_Y = \frac{600}{4.0} = 150\ \text{N}.$$

Vertical equilibrium:

$$F = 900 - 150 = 750\ \text{N}.$$

2) Child stands at $Y$

Take moments about $X$.

• Child weight $600\ \text{N}$ acts at $4.0\ \text{m}$ from $X$.
• Plank weight contributes the same $300(2.0)$.

So

$$R_Y(4.0) = 300(2.0) + 600(4.0)$$ $$R_Y = \frac{600 + 2400}{4.0} = 750\ \text{N}.$$

Vertical equilibrium:

$$F = 900 - 750 = 150\ \text{N}.$$

This corresponds to option D.

Key Takeaways

• For a body in equilibrium: use $\sum F = 0$ and $\sum \tau = 0$.
• Taking moments about a support is efficient because it removes that support’s reaction from the equation.
• A uniform object’s weight acts at its midpoint.

Common Mistakes

• Using the full plank length ($4.0\ \text{m}$) as the distance for the plank’s own weight instead of $2.0\ \text{m}$.
• Forgetting to include the plank’s weight when the child’s weight is present.
• Mixing up which reaction you are solving for when taking moments.

Things to Be Careful About

• Distances in moments must be perpendicular distances to the line of action (here they are horizontal distances along the plank).
• Keep a consistent sign convention for clockwise/anticlockwise moments.
• Remember that when you pivot about $X$, the moment due to $F$ is zero (because its line of action passes through the pivot).

Background Concept

Power $P$ is the rate of doing work:

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

If a constant force $F$ acts along the direction of motion and produces a displacement $s$, then the work done is

$$W = Fs$$

Combining these gives

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

To turn $\dfrac{s}{t}$ into something recognisable, we use the definition of (constant) velocity:

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

or equivalently $s = vt$.

Understanding the Question

You are told the final equation is

$$\text{power} = \text{force} \times \text{velocity}$$

The question asks which one of the listed relationships (A–D) is used in deriving this result. So we look for the missing step that connects displacement $s$ and time $t$ to velocity $v$.

Approach

1. Start from the standard definition $P = W/t$.
2. Replace $W$ using $W = Fs$ (force along displacement).
3. Recognise that you then need a relationship to rewrite $s/t$ in terms of $v$. That relationship is $s = vt$.

Step-by-Step Reasoning

Start with

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

Substitute $W = Fs$:

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

Rearrange to highlight $s/t$:

$$P = F\left(\frac{s}{t}\right)$$

Now use the kinematics definition of velocity:

$$v = \frac{s}{t} \quad \Rightarrow \quad s = vt$$

So $\dfrac{s}{t}=v$ and therefore

$$P = Fv$$

Hence the required relationship is displacement $= \text{velocity} \times \text{time}$ (Option A).

Key Takeaways

• $P = W/t$ and $W = Fs$ are the starting definitions.
• To reach $P = Fv$, you must convert $s/t$ into $v$ using $s = vt$.

Common Mistakes

• Choosing $F = ma$ (B): not needed because the derivation does not involve acceleration.
• Choosing $p = mv$ (C): momentum is irrelevant here.
• Choosing $v = at$ (D): only applies for acceleration from rest and still doesn’t directly link displacement and time to velocity for this derivation.

Things to Be Careful About

• $W = Fs$ requires the force component parallel to displacement; more generally $W = Fs\cos\theta$.
• Don’t confuse $v = s/t$ (average velocity for a displacement over time) with the constant-acceleration formulae; the derivation assumes the simple rate form is appropriate.