9709/42 May/June 2025 — Mathematics Solved Past Paper

Q1Medium-EasyKinematics of Motion in a Straight LineEnergy, Work and Power

A crate is being pushed in a straight line along a horizontal surface by a force of magnitude $25\text{N}$ inclined at $20^\circ$ above the horizontal. The crate moves a distance of $12\text{m}$ in $8$ seconds with constant speed.

(a)

Find the constant speed of the crate.

1M Similar

(b)

Find the work done by the $25\text{N}$ force.

2M Similar

DifficultyMedium-Easy

Worked solution

Approach

Only the component of the $25\text{ N}$ force in the direction of motion does work. The force is $20^\circ$ above the horizontal and the crate moves horizontally, so the horizontal component is $25\cos 20^\circ$ . Multiply this by the displacement.

Working

W = Fd\cos\theta = 25\cos 20^\circ \times 12

Since $\cos 20^\circ \approx 0.93969$ ,

W \approx 25 \times 0.93969 \times 12 = 281.9077\ldots

So, to 3 significant figures,

W \approx 282\text{ J}

Answer

The work done is $282\text{ J}$ .

Final answer

282 J

Detailed explanation

Walkthrough

The force is inclined at $20^\circ$ above the horizontal, but the crate moves horizontally. Work done by a constant force is $W = Fd\cos\theta$ , where $\theta$ is the angle between the force and the displacement. Since the displacement is horizontal, the relevant component of the force is $25\cos 20^\circ$ . Multiplying this by the displacement $12\text{ m}$ gives

W = 25\cos 20^\circ \times 12.

Computing this gives $281.9077\ldots$ J, which is $282\text{ J}$ to 3 significant figures.

Key Takeaways

Only the component of force in the direction of motion does work.
The formula $W = Fd\cos\theta$ already accounts for that component.
Units of work are joules: $\text{N}\cdot\text{m} = \text{J}$ .

Common Mistakes

Using $25\sin 20^\circ \times 12$ : this is the vertical component and does not give the correct final answer for work done along the horizontal.
Multiplying $25 \times 12$ directly and ignoring the angle; this would overestimate the work.
Rounding too early; keep $281.9077\ldots$ for the next part.

Things to Be Careful About

The mark scheme requires the component multiplied by displacement to score the method mark, and the final answer must be $282\text{ J}$ (allow $281.9$ ). Ensure the calculator is in degree mode, not radians, because the angle is $20^\circ$ .

Techniques used

resolve the force into its horizontal componentmultiply the force component by displacementevaluate the work done

(c)

Find the power at which the $25\text{N}$ force is working.

1M Similar

DifficultyEasy

Worked solution

Approach

Power is the rate at which work is done: $P = \dfrac{W}{t}$ . Use the work from part (b) and divide by the time. Equivalently, $P = F_{\text{horizontal}} v$ , where $F_{\text{horizontal}} = 25\cos20^\circ$ and $v = 1.5$ .

Working

Using $W = 281.9077\ldots$ J and $t = 8$ s:

P = \frac{281.9077\ldots}{8} = 35.2385\ldots

So, to 3 significant figures,

P = 35.2\text{ W}

Answer

The power is $35.2\text{ W}$ .

Final answer

35.2 W

Detailed explanation

Walkthrough

Power is the rate of doing work, so divide the work from part (b) by the time taken. With $W = 281.9077\ldots\text{ J}$ and $t=8\text{ s}$ :

P = \frac{W}{t} = \frac{281.9077\ldots}{8} = 35.2385\ldots\text{ W}.

Therefore the power is $35.2\text{ W}$ to 3 significant figures. Alternatively, using $P = Fv$ with the horizontal component of force and the constant speed:

P = (25\cos20^\circ)(1.5) = 35.2385\ldots\text{ W}.

Both methods give the same answer.

Key Takeaways

Power equals work done per unit time.
Power can also be calculated as force component in the motion direction times speed.
Units of power are watts: $1\text{ W} = 1\text{ J s}^{-1}$ .

Common Mistakes

Using $P = 25 \times 1.5 = 37.5\text{ W}$ , ignoring the $20^\circ$ angle; the mark scheme awards B0 for this.
Forgetting to divide by 8 when using $P = W/t$ .
Giving the answer as $35.2\text{ kW}$ instead of $35.2\text{ W}$ ; the question asks for watts.
If following through from the rounded $282\text{ J}$ , a value of $35.3\text{ W}$ is allowed by the mark scheme, but it is better to keep unrounded values when possible.

Things to Be Careful About

Use the work value from part (b). If using $P = Fv$ , use $v = 1.5\text{ m s}^{-1}$ from part (a) and the horizontal component, not the full $25\text{ N}$ force. Keep the units as watts and give 3 significant figures.

Techniques used

divide work done by timeor multiply horizontal force component by speed

Q24MMediumMomentumEnergy, Work and Power

Two particles $P$ and $Q$ , of masses $0.2\text{kg}$ and $0.1\text{kg}$ respectively, are free to move in a straight line on a smooth horizontal plane. $P$ is projected towards $Q$ with speed $5\text{ms}^{-1}$ . At the same instant, $Q$ is projected away from $P$ with speed $2\text{ms}^{-1}$ . When $P$ collides with $Q$ , the particles coalesce.

Find the kinetic energy lost during the collision.

DifficultyMedium

Worked solution

Approach

The particles coalesce, so this is a perfectly inelastic collision. Use conservation of linear momentum to find the common speed $v$ after impact. Then calculate the total kinetic energy before and after the collision and subtract to find the energy lost.

Working

Let the common direction of motion be positive. Before the collision:

0.2(5) + 0.1(2) = (0.2 + 0.1)v

1.0 + 0.2 = 0.3v

1.2 = 0.3v

v = 4\text{ ms}^{-1}

Total kinetic energy before:

\text{KE}_{\text{before}} = \frac{1}{2}(0.2)(5^2) + \frac{1}{2}(0.1)(2^2)

= 2.5 + 0.2 = 2.7\text{ J}

Total kinetic energy after:

\text{KE}_{\text{after}} = \frac{1}{2}(0.3)(4^2) = \frac{1}{2}(0.3)(16) = 2.4\text{ J}

Kinetic energy lost:

2.7 - 2.4 = 0.3\text{ J}

Answer

0.3\text{ J}

Final answer

0.3 J

Detailed explanation

Walkthrough

This is a collision in which the two particles stick together, so it is a perfectly inelastic collision. On a smooth horizontal plane there are no external horizontal forces, so the total momentum of the two particles is conserved during the collision.

First choose a positive direction. Since $P$ is projected towards $Q$ and $Q$ is projected away from $P$ , both particles move in the same direction before impact. Therefore the total momentum before the collision is $0.2 \times 5 + 0.1 \times 2 = 1.2\text{ kg m s}^{-1}$ .

After coalescing, the combined mass is $0.2 + 0.1 = 0.3\text{ kg}$ and moves with common speed $v$ . Conservation of momentum gives $0.3v = 1.2$ , so $v = 4\text{ ms}^{-1}$ .

Next calculate the kinetic energy before the collision. Each particle contributes $\frac{1}{2}mv^2$ , so the total before is $\frac{1}{2}(0.2)(5^2) + \frac{1}{2}(0.1)(2^2) = 2.5 + 0.2 = 2.7\text{ J}$ .

After the collision the combined particle has kinetic energy $\frac{1}{2}(0.3)(4^2) = 2.4\text{ J}$ .

The kinetic energy lost is the difference $2.7 - 2.4 = 0.3\text{ J}$ .

Key Takeaways

In a collision where particles coalesce, momentum is conserved but kinetic energy is not.
The total momentum before and after must be written using the same positive direction.
Kinetic energy is always $\frac{1}{2}mv^2$ ; it is a scalar, so all terms are added.
The energy lost is found by subtracting the final total kinetic energy from the initial total kinetic energy.

Common Mistakes

Using $mg$ instead of $m$ in the momentum equation. The mark scheme allows at most M1A0M1A0 for this error.
Forgetting to include the kinetic energy of both particles before the collision.
Using the wrong mass after collision, such as $0.1\text{ kg}$ or $0.2\text{ kg}$ , instead of $0.3\text{ kg}$ .
Incorrectly subtracting in the wrong order; the mark scheme allows $-0.3$ but the physical answer should be stated as a positive loss.

Things to Be Careful About

Choose a positive direction and be consistent. Both velocities are in the same direction before impact, so both momentum terms have the same sign.
The mark scheme allows sign errors in the momentum equation for the method mark, but the final answer must be consistent.
Ensure all terms are dimensionally correct: mass in kg, speed in $\text{ms}^{-1}$ , energy in joules.
The phrase 'kinetic energy lost' means the initial kinetic energy minus the final kinetic energy.

Techniques used

apply conservation of linear momentum for a perfectly inelastic collisioncalculate total kinetic energy before collisioncalculate total kinetic energy after collisionfind the difference between initial and final kinetic energies

Approach

Identify the three forces acting on particle $P$ : its weight $mg$ acting vertically downwards, the tension $T$ in the string acting along the string towards the ceiling, and the applied force of $35\text{N}$ acting perpendicular to the string. Use the given lengths to find the trigonometric ratios of the angle $\theta$ between the string and the vertical. Resolve forces horizontally and vertically to form two equations, then solve for $T$ and $m$ .

Working

Let $\theta$ be the angle between the string and the vertical. From the diagram:

\cos\theta = \frac{2.4}{2.6} = \frac{12}{13}

The horizontal distance from $P$ to the point directly below the ceiling attachment point is:

\sqrt{2.6^2 - 2.4^2} = \sqrt{6.76 - 5.76} = \sqrt{1.0} = 1.0\text{ m}

So:

\sin\theta = \frac{1.0}{2.6} = \frac{5}{13}

The three forces at $P$ are:

Weight: $mg$ vertically downwards
Tension: $T$ along the string, directed up and to the left at angle $\theta$ to the vertical
Applied force: $35\text{N}$ perpendicular to the string, directed up and to the right

Since the $35\text{N}$ force is perpendicular to the string and the string makes angle $\theta$ with the vertical, the $35\text{N}$ force makes angle $\theta$ with the horizontal and angle $(90^\circ - \theta)$ with the vertical.

Resolving horizontally (taking right as positive):

35\cos\theta - T\sin\theta = 0

Substituting the trig ratios:

35 \times \frac{12}{13} = T \times \frac{5}{13}

\frac{420}{13} = \frac{5T}{13}

5T = 420

T = 84\text{ N}

Resolving vertically (taking upwards as positive):

T\cos\theta + 35\sin\theta - mg = 0

Substituting:

84 \times \frac{12}{13} + 35 \times \frac{5}{13} = mg

\frac{1008}{13} + \frac{175}{13} = mg

\frac{1183}{13} = mg

91 = mg

Using $g = 9.8\text{ m/s}^2$ :

m = \frac{91}{9.8} = 9.1\text{ kg}

Answer

m = 9.1, \quad T = 84

Final answer

m = 9.1, T = 84

Detailed explanation

Walkthrough

Step 1: Extract geometry from the diagram.
The particle $P$ is suspended from a ceiling by a string of length $2.6\text{m}$ , and the vertical distance from the ceiling to $P$ is $2.4\text{m}$ . Let $\theta$ be the angle between the string and the vertical. Using right-triangle trigonometry, $\cos\theta = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{2.4}{2.6} = \frac{12}{13}$ . The horizontal distance is $\sqrt{2.6^2 - 2.4^2} = 1.0\text{m}$ , so $\sin\theta = \frac{1.0}{2.6} = \frac{5}{13}$ . These clean fractions will make the algebra much simpler.

Step 2: Identify all forces.
There are three forces acting on $P$ :

Weight $mg$ acting straight down.
Tension $T$ acting along the string from $P$ toward the ceiling attachment point (up and to the left).
The applied $35\text{N}$ force acting perpendicular to the string, up and to the right.

Step 3: Resolve horizontally.
Taking right as positive, the horizontal component of the $35\text{N}$ force is $35\cos\theta$ (since the force makes angle $\theta$ with the horizontal, being perpendicular to a line at angle $\theta$ to the vertical). The horizontal component of $T$ is $T\sin\theta$ to the left. Setting the sum to zero:

35\cos\theta - T\sin\theta = 0

Substituting $\cos\theta = \frac{12}{13}$ and $\sin\theta = \frac{5}{13}$ :

35 \times \frac{12}{13} = T \times \frac{5}{13}

The denominators cancel, giving $5T = 420$ , so $T = 84\text{N}$ .

Step 4: Resolve vertically.
Taking upwards as positive, the vertical component of $T$ is $T\cos\theta$ (upwards), the vertical component of the $35\text{N}$ force is $35\sin\theta$ (upwards), and the weight is $mg$ (downwards). Setting the sum to zero:

T\cos\theta + 35\sin\theta = mg

Substituting $T = 84$ and the trig ratios:

84 \times \frac{12}{13} + 35 \times \frac{5}{13} = \frac{1008 + 175}{13} = \frac{1183}{13} = 91

So $mg = 91$ , and with $g = 9.8\text{ m/s}^2$ , $m = \frac{91}{9.8} = 9.1\text{ kg}$ .

Key Takeaways

When a particle is in equilibrium under three or more forces, resolving horizontally and vertically gives two independent equations that can be solved simultaneously.
If a force is perpendicular to a line at angle $\theta$ to the vertical, it makes angle $\theta$ with the horizontal — this swaps sin and cos for that force's components.
Using exact fractions (like $\frac{12}{13}$ and $\frac{5}{13}$ ) instead of decimal approximations keeps the arithmetic clean and avoids rounding errors.
Always verify the geometry: the string length, vertical drop, and horizontal offset must satisfy the Pythagorean theorem.

Common Mistakes

Using the wrong angle: resolving with $\theta$ as the angle to the horizontal instead of the vertical, which swaps sin and cos and gives incorrect equations.
Forgetting that the $35\text{N}$ force is perpendicular to the string, not to the vertical or horizontal — its components are $35\cos\theta$ horizontally and $35\sin\theta$ vertically, not the other way around.
Using decimal approximations like $\cos\theta = 0.923$ too early, which can lead to rounding errors that affect the final answer.
Not using $g = 9.8\text{ m/s}^2$ when converting from $mg$ to $m$ , giving an incorrect mass.

Things to Be Careful About

The mark scheme allows $T = 84.1$ and $m = 9.11$ if intermediate rounding is used, but the exact answers are $T = 84$ and $m = 9.1$ . Using exact fractions avoids this issue.
The mark scheme also accepts Lami's theorem as an alternative method, where $\frac{mg}{\sin 90^\circ} = \frac{35}{\sin(180^\circ - \theta)} = \frac{T}{\sin(90^\circ + \theta)}$ . This gives the same results but requires correctly identifying the angles between the force directions.
Ensure the direction of the $35\text{N}$ force is correctly interpreted: it acts perpendicular to the string and upwards, so its horizontal component is to the right and its vertical component is upwards.
The question asks for $m$ and $T$ in either order, so either $m = 9.1, T = 84$ or $T = 84, m = 9.1$ is acceptable.

Techniques used

extract geometry from diagram to find trig ratiosidentify all forces acting on a particle in equilibriumresolve forces horizontally and verticallysolve simultaneous equations for unknowns

Approach

The motion has three phases: uniform acceleration, constant speed, and uniform deceleration. Each phase corresponds to a line segment on the velocity-time graph. Calculate the key velocities and times, then sketch the three segments with correct labels.

Working

Phase 1: Acceleration (t = 0 to t = 30)

Initial speed $u = 15$ ms $^{-1}$ , acceleration $a = 0.4$ ms $^{-2}$ , time $t = 30$ s.

Speed at $t = 30$ :

v = u + at = 15 + 0.4 \times 30 = 27 \text{ ms}^{-1}

Phase 2: Constant speed (t = 30 to t = 30 + 3T)

Speed remains constant at $27$ ms $^{-1}$ .

Phase 3: Deceleration (t = 30 + 3T to t = 30 + 4T)

Deceleration $a = -0.2$ ms $^{-2}$ , duration $t = T$ s.

Speed at point $B$ ( $t = 30 + 4T$ ):

v = 27 + (-0.2)T = 27 - 0.2T \text{ ms}^{-1}

Since $T < 30$ , we have $0.2T < 6$ , so $27 - 0.2T > 21 > 0$ . The speed at $B$ is positive.

Answer

The velocity-time graph consists of three line segments:

From $(0, 15)$ to $(30, 27)$ with positive gradient $0.4$
From $(30, 27)$ to $(30 + 3T, 27)$ horizontal
From $(30 + 3T, 27)$ to $(30 + 4T, 27 - 0.2T)$ with negative gradient $-0.2$

Final answer

Three-line graph: (0, 15) to (30, 27), then horizontal to (30+3T, 27), then down to (30+4T, 27-0.2T)

Detailed explanation

Walkthrough

The question describes three distinct phases of motion. On a velocity-time graph, each phase is represented by a line segment whose gradient equals the acceleration.

Phase 1 (t = 0 to 30 s): The car accelerates uniformly at $0.4$ ms $^{-2}$ from $15$ ms $^{-1}$ . Using $v = u + at$ , the speed at $t = 30$ is $15 + 0.4 \times 30 = 27$ ms $^{-1}$ . This gives a straight line from $(0, 15)$ to $(30, 27)$ with gradient $0.4$ .

Phase 2 (t = 30 to 30 + 3T): The car moves at constant speed, so velocity does not change. This is a horizontal line at $v = 27$ from $t = 30$ to $t = 30 + 3T$ .

Phase 3 (t = 30 + 3T to 30 + 4T): The car decelerates at $0.2$ ms $^{-2}$ for $T$ seconds. The speed at $B$ is $27 - 0.2T$ . Since $T < 30$ , this speed is positive (greater than $21$ ms $^{-1}$ ), so the graph does not reach the $t$ -axis.

Key Takeaways

A velocity-time graph has three distinct segments for three-phase motion.
The gradient of each segment equals the acceleration during that phase.
Constant speed appears as a horizontal line.
Key points must be labelled with correct values or expressions.

Common Mistakes

Forgetting to label the key time values $30$ , $30 + 3T$ , and $30 + 4T$ on the horizontal axis.
Drawing the third segment reaching the $t$ -axis, implying the car stops at $B$ , which contradicts the problem.
Using incorrect initial velocity or acceleration values for the segments.

Things to Be Careful About

The third line segment must stop before the $t$ -axis since the speed at $B$ is positive.
All three time labels ( $30$ , $30 + 3T$ , $30 + 4T$ ) must be correctly placed on the horizontal axis.
The vertical axis must show $v = 15$ at $t = 0$ and $v = 27$ at $t = 30$ .

Techniques used

sketch velocity-time graphidentify key points and gradients

(b)

The distance from $A$ to $B$ is $2750\text{m}$ .

Find the value of $T$ .

6M Similar

DifficultyMedium-Hard

Worked solution

Approach

The distance from $A$ to $B$ equals the area under the velocity-time graph from $t = 0$ to $t = 30 + 4T$ . This area is the sum of three parts: a trapezium (acceleration phase), a rectangle (constant speed phase), and a trapezium (deceleration phase). Set this total area equal to $2750$ and solve for $T$ .

Working

Speed after 30 seconds:

v = 15 + 0.4 \times 30 = 27 \text{ ms}^{-1}

Speed at $B$ :

v_B = 27 + (-0.2)T = 27 - 0.2T \text{ ms}^{-1}

Area under the graph (distance from $A$ to $B$ ):

The area is split into three regions:

Region 1 (trapezium, $t = 0$ to $30$ ):

\text{Area}_1 = \frac{1}{2}(15 + 27) \times 30 = \frac{1}{2} \times 42 \times 30 = 630 \text{ m}

Region 2 (rectangle, $t = 30$ to $30 + 3T$ ):

\text{Area}_2 = 27 \times 3T = 81T \text{ m}

Region 3 (trapezium, $t = 30 + 3T$ to $30 + 4T$ ):

\text{Area}_3 = \frac{1}{2}\{(27 - 0.2T) + 27\} \times T = \frac{1}{2}(54 - 0.2T)T = 27T - 0.1T^2 \text{ m}

Total distance:

630 + 81T + 27T - 0.1T^2 = 2750

630 + 108T - 0.1T^2 = 2750

Rearranging:

0.1T^2 - 108T + 2120 = 0

Multiply by $10$ :

T^2 - 1080T + 21200 = 0

Using the quadratic formula:

T = \frac{1080 \pm \sqrt{1080^2 - 4 \times 21200}}{2}

T = \frac{1080 \pm \sqrt{1166400 - 84800}}{2}

T = \frac{1080 \pm \sqrt{1081600}}{2}

T = \frac{1080 \pm 1040}{2}

T = \frac{2120}{2} = 1060 \quad \text{or} \quad T = \frac{40}{2} = 20

Since $T < 30$ , we reject $T = 1060$ .

T = 20

Answer

T = 20

Final answer

T = 20

Detailed explanation

Walkthrough

The total distance from $A$ to $B$ is given as $2750$ m. On a velocity-time graph, the area under the curve equals the displacement. Since the motion is in one direction along a straight road, the area equals the distance.

Step 1: Find the speed after the acceleration phase.
Using $v = u + at$ with $u = 15$ , $a = 0.4$ , $t = 30$ :
$v = 15 + 0.4 \times 30 = 27$ ms $^{-1}$ .

Step 2: Find the speed at $B$ .
During the deceleration phase, $u = 27$ , $a = -0.2$ , $t = T$ :
$v_B = 27 - 0.2T$ .

Step 3: Calculate the area under the graph.
The graph has three regions:

A trapezium from $t = 0$ to $t = 30$ with parallel sides $15$ and $27$ , and width $30$ . Area $= \frac{1}{2}(15 + 27) \times 30 = 630$ .
A rectangle from $t = 30$ to $t = 30 + 3T$ with height $27$ and width $3T$ . Area $= 27 \times 3T = 81T$ .
A trapezium from $t = 30 + 3T$ to $t = 30 + 4T$ with parallel sides $27$ and $27 - 0.2T$ , and width $T$ . Area $= \frac{1}{2}(54 - 0.2T)T = 27T - 0.1T^2$ .

Step 4: Form and solve the equation.
Total area $= 630 + 81T + 27T - 0.1T^2 = 2750$ .
Rearranging gives $0.1T^2 - 108T + 2120 = 0$ , or $T^2 - 1080T + 21200 = 0$ .

Applying the quadratic formula: $T = \frac{1080 \pm \sqrt{1081600}}{2} = \frac{1080 \pm 1040}{2}$ .

This gives $T = 1060$ or $T = 20$ . Since the problem states $T < 30$ , we reject $T = 1060$ and accept $T = 20$ .

Key Takeaways

The area under a velocity-time graph gives the displacement (or distance, if motion is in one direction).
For multi-phase motion, split the area into simple geometric shapes.
Quadratic equations from kinematics often have two roots; always check against given constraints.

Common Mistakes

Forgetting that the area of a trapezium is $\frac{1}{2}(a + b)h$ and using the wrong formula.
Not checking the constraint $T < 30$ and accepting both roots.
Sign errors when rearranging the equation to standard quadratic form.

Things to Be Careful About

The constraint $T < 30$ must be used to reject $T = 1060$ .
Ensure all areas are in metres and times in seconds for consistency.
The equation $0.1T^2 - 108T + 2120 = 0$ must be correctly rearranged before applying the quadratic formula.

Techniques used

calculate speed using v = u + atfind area under velocity-time graphform quadratic equationsolve quadratic equation

(c)

The car continues its journey from $B$ , decelerating uniformly at $0.5\text{ms}^{-2}$ until it comes to rest at a point $C$ on the road.

Find the total distance from $A$ to $C$ .

3M Similar

DifficultyMedium-Easy

Worked solution

Approach

First, find the speed at point $B$ using the value of $T = 20$ from part (b). Then use $v^2 = u^2 + 2as$ for the motion from $B$ to $C$ (where the car comes to rest) to find the distance from $B$ to $C$ . Add this to the distance from $A$ to $B$ .

Working

Speed at $B$ :

v_B = 27 - 0.2 \times 20 = 27 - 4 = 23 \text{ ms}^{-1}

Distance from $B$ to $C$ :

For the motion from $B$ to $C$ : $v = 0$ , $u = 23$ , $a = -0.5$ , find $s$ .

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

0^2 = 23^2 + 2(-0.5)s

0 = 529 - s

s = 529 \text{ m}

Total distance from $A$ to $C$ :

\text{Distance } A \text{ to } C = 2750 + 529 = 3279 \text{ m}

Answer

3279 \text{ m}

Final answer

3279 m

Detailed explanation

Walkthrough

Step 1: Find the speed at $B$ .
From part (b), $T = 20$ . The speed at $B$ is:
$v_B = 27 - 0.2T = 27 - 0.2(20) = 27 - 4 = 23$ ms $^{-1}$ .

Step 2: Find the distance from $B$ to $C$ .
From $B$ to $C$ , the car decelerates at $0.5$ ms $^{-2}$ until it comes to rest. So $v = 0$ , $u = 23$ , $a = -0.5$ . We need $s$ .

Using $v^2 = u^2 + 2as$ :
$0 = 23^2 + 2(-0.5)s$
$0 = 529 - s$
$s = 529$ m.

Step 3: Find the total distance from $A$ to $C$ .
Distance from $A$ to $B$ is given as $2750$ m. Distance from $B$ to $C$ is $529$ m.
Total distance $= 2750 + 529 = 3279$ m.

Key Takeaways

The speed at the end of one phase becomes the initial speed for the next phase.
The equation $v^2 = u^2 + 2as$ is useful when time is not given or required.
Total distance is the sum of distances for each phase of the journey.

Common Mistakes

Using the wrong speed for the initial velocity in the $B$ to $C$ calculation.
Forgetting to add the distance from $A$ to $B$ to get the total distance from $A$ to $C$ .
Sign errors with the deceleration (should be negative in $v^2 = u^2 + 2as$ ).

Things to Be Careful About

Use the correct speed at $B$ ( $23$ ms $^{-1}$ ), not the speed after the first acceleration phase ( $27$ ms $^{-1}$ ).
The deceleration is $-0.5$ ms $^{-2}$ , so $a = -0.5$ in the formula.
The final answer may be accepted as $3280$ m (rounded), but $3279$ m is exact.

Techniques used

calculate speed at point Bapply v^2 = u^2 + 2as to find distancesum distances for total

Q5Medium-HardNewton's Laws of MotionForces and EquilibriumKinematics of Motion in a Straight Line

Fig. 5.1

One end of a light inextensible string is attached to a particle $A$ of mass $3\text{kg}$ . The other end of the string is attached to a particle $B$ of mass $4\text{kg}$ . Particle $A$ is in contact with a rough plane inclined at $30^\circ$ to the horizontal, and particle $B$ is in contact with a smooth horizontal plane. A second light inextensible string is attached to $B$ . The other end of this second string is attached to a particle $C$ of mass $5\text{kg}$ which hangs vertically.

Both strings are taut and pass over small smooth pulleys that are fixed at the ends of the horizontal plane. The part of the string from $A$ to the pulley is parallel to a line of greatest slope of the inclined plane, and $A$ , $B$ and $C$ are in the same vertical plane (see diagram).

The system is released from rest. In the subsequent motion, $C$ moves vertically downwards with acceleration $2\text{ms}^{-2}$ , and neither $A$ nor $B$ reach a pulley.

(a)

Find the tensions in each of the strings.

3M Similar

DifficultyMedium-Easy

Worked solution

Approach

Apply Newton's second law to particles C and B separately. Since the strings are inextensible and taut, all particles share the same acceleration magnitude of $2\text{ ms}^{-2}$ . Particle C accelerates downwards, while particles A and B accelerate in their respective directions of motion.

Working

For particle C (mass $5\text{ kg}$ ), moving vertically downwards with acceleration $2\text{ ms}^{-2}$ :

\begin{aligned} 5g - T_{BC} &= 5 \times 2 \\ 50 - T_{BC} &= 10 \\ T_{BC} &= 40 \text{ N} \end{aligned}

For particle B (mass $4\text{ kg}$ ), moving horizontally with acceleration $2\text{ ms}^{-2}$ . The horizontal plane is smooth, so there is no friction acting on B:

\begin{aligned} T_{BC} - T_{AB} &= 4 \times 2 \\ 40 - T_{AB} &= 8 \\ T_{AB} &= 32 \text{ N} \end{aligned}

Answer

T_{BC} = 40 \text{ N}, \quad T_{AB} = 32 \text{ N}

Final answer

T_BC = 40 N, T_AB = 32 N

Detailed explanation

Walkthrough

Step 1: Analyse particle C.
Particle C has mass $5\text{ kg}$ and accelerates vertically downwards at $2\text{ ms}^{-2}$ . Taking the downward direction as positive, the forces acting on C are its weight $5g$ downwards and the tension $T_{BC}$ upwards. Applying Newton's second law ( $F = ma$ ):

5g - T_{BC} = 5 \times 2

Using $g = 10\text{ ms}^{-2}$ , this gives $50 - T_{BC} = 10$ , so $T_{BC} = 40\text{ N}$ .

Step 2: Analyse particle B.
Particle B has mass $4\text{ kg}$ and accelerates horizontally at $2\text{ ms}^{-2}$ towards the second pulley. The horizontal plane is smooth, so there is no friction. The forces in the horizontal direction are $T_{BC}$ pulling B to the right and $T_{AB}$ pulling B to the left. Applying Newton's second law:

T_{BC} - T_{AB} = 4 \times 2

Substituting $T_{BC} = 40\text{ N}$ : $40 - T_{AB} = 8$ , giving $T_{AB} = 32\text{ N}$ .

Key Takeaways

When particles are connected by inextensible strings, they all share the same acceleration magnitude.
Newton's second law can be applied to each particle independently, using the correct direction of acceleration for each.
Smooth surfaces contribute no frictional force, simplifying the equations of motion.

Common Mistakes

Using the wrong mass in Newton's second law for a given particle.
Getting the sign wrong in the equation of motion (e.g., writing $T_{AB} - T_{BC} = 4 \times 2$ instead of $T_{BC} - T_{AB} = 4 \times 2$ ). The mark scheme allows sign errors as long as the correct mass is used.
Forgetting that the horizontal plane is smooth, and incorrectly adding a friction term for particle B.

Things to Be Careful About

Ensure consistent use of $g = 10\text{ ms}^{-2}$ throughout the question.
The tension $T_{BC}$ is the same on both sides of the smooth pulley at the end of the horizontal plane.
Particle B is on a smooth horizontal plane, so no friction acts on it — only the two tensions and the vertical weight/normal reaction (which are irrelevant for horizontal motion).

Techniques used

apply Newton's second law to particle Capply Newton's second law to particle Bsolve simultaneous equations for tensions

(b)

Find the coefficient of friction between $A$ and the inclined plane.

4M Similar

DifficultyMedium

Worked solution

Approach

Apply Newton's second law to particle A, resolving forces into components parallel and perpendicular to the inclined plane. Use the perpendicular equation to find the normal reaction $R$ , then use the parallel equation to find the friction force $F$ . Finally, apply $F = \mu R$ to find the coefficient of friction.

Working

Particle A (mass $3\text{ kg}$ ) moves up the $30^\circ$ inclined plane with acceleration $2\text{ ms}^{-2}$ . The forces acting on A are:

Tension $T_{AB} = 32\text{ N}$ up the plane
Weight $3g = 30\text{ N}$ acting vertically downwards
Normal reaction $R$ perpendicular to the plane
Friction $F$ acting down the plane (opposing the motion)

Resolving perpendicular to the plane:

\begin{aligned} R &= 3g\cos 30^\circ \\ &= 3 \times 10 \times \frac{\sqrt{3}}{2} \\ &= 15\sqrt{3} \text{ N} \end{aligned}

Resolving parallel to the plane (up the plane is positive):

\begin{aligned} T_{AB} - F - 3g\sin 30^\circ &= 3 \times 2 \\ 32 - F - 3 \times 10 \times \frac{1}{2} &= 6 \\ 32 - F - 15 &= 6 \\ F &= 11 \text{ N} \end{aligned}

Applying the friction model $F = \mu R$ :

\begin{aligned} 11 &= \mu \times 15\sqrt{3} \\ \mu &= \frac{11}{15\sqrt{3}} \\ &= \frac{11\sqrt{3}}{45} \\ &\approx 0.423 \end{aligned}

Answer

\mu = \frac{11\sqrt{3}}{45} \approx 0.423

Final answer

μ = 11√3/45 ≈ 0.423

Detailed explanation

Walkthrough

Step 1: Identify forces on particle A.
Particle A is on a rough inclined plane at $30^\circ$ to the horizontal. The forces are: tension $T_{AB} = 32\text{ N}$ pulling up the plane, weight $3g = 30\text{ N}$ acting vertically down, normal reaction $R$ perpendicular to the plane, and friction $F$ acting down the plane (since A is moving up).

Step 2: Resolve perpendicular to the plane.
There is no acceleration perpendicular to the plane, so the forces balance:

R = 3g\cos 30^\circ = 30 \times \frac{\sqrt{3}}{2} = 15\sqrt{3} \text{ N}

Step 3: Resolve parallel to the plane.
Taking up the plane as positive, and using $a = 2\text{ ms}^{-2}$ :

T_{AB} - F - 3g\sin 30^\circ = 3a

Substituting known values:

32 - F - 15 = 6 \implies F = 11 \text{ N}

Step 4: Apply $F = \mu R$ .

\mu = \frac{F}{R} = \frac{11}{15\sqrt{3}} = \frac{11\sqrt{3}}{45} \approx 0.423

Key Takeaways

On an inclined plane, always resolve forces parallel and perpendicular to the plane.
The component of weight perpendicular to the plane is $mg\cos\theta$ , and parallel to the plane is $mg\sin\theta$ .
Friction opposes the direction of motion (or intended motion) and is given by $F = \mu R$ at limiting equilibrium.

Common Mistakes

Mixing up $\sin$ and $\cos$ when resolving the weight component. Remember: perpendicular to the plane uses $\cos\theta$ , parallel uses $\sin\theta$ .
Forgetting that friction acts down the plane when the particle moves up the plane.
Not using the correct value of $T_{AB} = 32\text{ N}$ from part (a).
Arithmetic errors when computing $3g\sin 30^\circ = 15$ or $3g\cos 30^\circ = 15\sqrt{3}$ .

Things to Be Careful About

The friction force $F$ is found to be $11\text{ N}$ , which is less than the maximum possible friction $\mu R$ . The question states the system is in motion, so we use $F = \mu R$ (the particle is on the rough plane and moving, so limiting friction applies).
Always rationalise the denominator in the final answer: $\frac{11}{15\sqrt{3}} = \frac{11\sqrt{3}}{45}$ .
The mark scheme accepts $\mu = 0.423$ to 3 significant figures.

Techniques used

resolve forces parallel to inclined planeresolve forces perpendicular to inclined planeapply Newton's second law on inclined planeuse F = μR to find coefficient of friction

(c)

When the system has been in motion for $1.5\text{s}$ , the string attached to $A$ breaks.

Find the total distance that $A$ travels up the plane from the instant that the system is released from rest to the instant that $A$ comes to instantaneous rest.

5M Similar

DifficultyMedium-Hard

Worked solution

Approach

The motion of A occurs in two phases:

Phase 1: The system is in motion for $1.5\text{ s}$ with constant acceleration $2\text{ ms}^{-2}$ . Use suvat equations to find the distance travelled and the speed at $t = 1.5\text{ s}$ .
Phase 2: After the string breaks, A continues up the plane but decelerates due to friction and the component of weight. Apply Newton's second law to find the new acceleration, then use suvat equations to find the additional distance until A comes to rest.

The total distance is the sum of the distances from both phases.

Working

Phase 1: System in motion ( $t = 0$ to $t = 1.5\text{ s}$ )

Initial velocity $u = 0$ , acceleration $a = 2\text{ ms}^{-2}$ , time $t = 1.5\text{ s}$ .

Distance travelled:

\begin{aligned} s_1 &= ut + \frac{1}{2}at^2 \\ &= 0 + \frac{1}{2} \times 2 \times 1.5^2 \\ &= 2.25 \text{ m} \end{aligned}

Speed at $t = 1.5\text{ s}$ :

\begin{aligned} v &= u + at \\ &= 0 + 2 \times 1.5 \\ &= 3 \text{ ms}^{-1} \end{aligned}

Phase 2: After the string breaks

The tension $T_{AB}$ is removed. Particle A continues moving up the plane but now decelerates. The forces acting on A parallel to the plane are:

Friction $F = 11\text{ N}$ down the plane
Component of weight $3g\sin 30^\circ = 15\text{ N}$ down the plane

Taking up the plane as positive, apply Newton's second law:

\begin{aligned} -F - 3g\sin 30^\circ &= 3a \\ -11 - 15 &= 3a \\ a &= -\frac{26}{3} \text{ ms}^{-2} \end{aligned}

Now use $v^2 = u^2 + 2as$ with $v = 0$ , $u = 3\text{ ms}^{-1}$ , and $a = -\frac{26}{3}\text{ ms}^{-2}$ :

\begin{aligned} 0 &= 3^2 + 2 \left(-\frac{26}{3}\right) s_2 \\ 0 &= 9 - \frac{52}{3} s_2 \\ \frac{52}{3} s_2 &= 9 \\ s_2 &= \frac{27}{52} \text{ m} \end{aligned}

Total distance:

\begin{aligned} s_{\text{total}} &= s_1 + s_2 \\ &= 2.25 + \frac{27}{52} \\ &= \frac{9}{4} + \frac{27}{52} \\ &= \frac{117}{52} + \frac{27}{52} \\ &= \frac{144}{52} \\ &= \frac{36}{13} \\ &\approx 2.77 \text{ m} \end{aligned}

Answer

s_{\text{total}} = \frac{36}{13} \approx 2.77 \text{ m}

Final answer

36/13 m ≈ 2.77 m

Detailed explanation

Walkthrough

Phase 1: System in motion (0 to 1.5 s)

Particle A starts from rest ( $u = 0$ ) and accelerates at $2\text{ ms}^{-2}$ up the plane. Using the suvat equation $s = ut + \frac{1}{2}at^2$ :

s_1 = 0 + \frac{1}{2} \times 2 \times (1.5)^2 = 2.25 \text{ m}

The speed at $t = 1.5\text{ s}$ is found using $v = u + at$ :

v = 0 + 2 \times 1.5 = 3 \text{ ms}^{-1}

Phase 2: After the string breaks

When the string attached to A breaks, the tension $T_{AB} = 32\text{ N}$ disappears. Particle A is still moving up the plane at $3\text{ ms}^{-1}$ , but now the only forces parallel to the plane are friction ( $F = 11\text{ N}$ , down the plane) and the component of weight ( $3g\sin 30^\circ = 15\text{ N}$ , down the plane). Both act to decelerate A.

Applying Newton's second law (up the plane positive):

-11 - 15 = 3a \implies a = -\frac{26}{3} \text{ ms}^{-2}

Now use $v^2 = u^2 + 2as$ with $v = 0$ (A comes to instantaneous rest):

0 = 9 + 2\left(-\frac{26}{3}\right)s_2 \implies s_2 = \frac{27}{52} \text{ m}

Total distance:

s_{\text{total}} = 2.25 + \frac{27}{52} = \frac{36}{13} \approx 2.77 \text{ m}

Key Takeaways

When a system's configuration changes (e.g., a string breaks), the motion must be split into phases with different accelerations.
The speed at the end of one phase becomes the initial speed for the next phase.
After the string breaks, friction and the weight component both act to decelerate the particle up the plane.
Always check that the acceleration is negative (deceleration) when the particle is slowing down.

Common Mistakes

Forgetting to add the distances from both phases — only finding $s_2$ and stopping.
Using the wrong acceleration in Phase 2. The acceleration is NOT $-2\text{ ms}^{-2}$ ; it must be recalculated using Newton's second law with the new forces.
Sign errors in the equation of motion for Phase 2. Both friction and the weight component act down the plane, so both are negative when up the plane is positive.
Arithmetic errors when adding $2.25 + \frac{27}{52}$ . Converting to a common denominator helps: $\frac{9}{4} + \frac{27}{52} = \frac{117 + 27}{52} = \frac{144}{52} = \frac{36}{13}$ .
Using $g = 9.8$ instead of $g = 10$ (the mark scheme uses $g = 10$ ).

Things to Be Careful About

The friction force $F = 11\text{ N}$ remains the same in Phase 2 because the normal reaction $R = 15\sqrt{3}\text{ N}$ hasn't changed, and $\mu$ is the same. The coefficient of friction doesn't change when the string breaks.
Particle A comes to instantaneous rest, so $v = 0$ in Phase 2. Do not assume it then slides back down — the question only asks for the distance up the plane until rest.
The acceleration in Phase 2 is $-\frac{26}{3} \approx -8.67\text{ ms}^{-2}$ , which is a large deceleration. This is because both friction and the weight component oppose the motion.
The mark scheme accepts the answer as $\frac{36}{13}$ or $2.769230789\ldots$ or $2.77$ to 3 significant figures.

Techniques used

use suvat equation to find distance in first phasefind speed at instant string breaksapply Newton's second law after string breaksuse suvat equation for second phase of motionadd distances from both phases

Q6MediumKinematics of Motion in a Straight Line

A particle $P$ moves in a straight line and passes through the point $A$ at time $t = 0$ . The velocity $v\text{ms}^{-1}$ of $P$ at time $t$ seconds is given by

v = (2t + 1)^{\frac{3}{2}} - 2t^2, \text{ where } 0 \le t \le 3.

(a)

Find the maximum velocity of $P$ in the interval $0 \le t \le 3$ .

5M Similar

DifficultyMedium

Worked solution

Approach

Differentiate $v$ with respect to $t$ to locate stationary points, then choose the value of $t$ in $0 \le t \le 3$ that gives the greatest velocity.

Working

Differentiate using the chain rule:

\frac{dv}{dt} = 3(2t+1)^{1/2} - 4t

Set $\frac{dv}{dt} = 0$ :

3(2t+1)^{1/2} - 4t = 0

3\sqrt{2t+1} = 4t

Square both sides:

9(2t+1) = 16t^2

16t^2 - 18t - 9 = 0

Solve:

t = \frac{18 \pm \sqrt{18^2 - 4(16)(-9)}}{2(16)} = \frac{18 \pm 30}{32}

So $t = \frac{48}{32} = \frac{3}{2}$ or $t = -\frac{12}{32} = -\frac{3}{8}$ .

Since $0 \le t \le 3$ , the relevant stationary point is $t = \frac{3}{2}$ .

At this time,

v = (2\cdot\frac{3}{2} + 1)^{3/2} - 2\left(\frac{3}{2}\right)^2 = 4^{3/2} - \frac{9}{2} = 8 - \frac{9}{2} = \frac{7}{2} = 3.5

Checking endpoints: $v(0)=1$ and $v(3)=7\sqrt7 - 18 \approx 0.52$ , both less than $3.5$ . Hence this is the maximum.

Answer

\text{Maximum velocity} = 3.5\ \text{m s}^{-1}

Final answer

3.5 m/s

Detailed explanation

Walkthrough

We are given $v=(2t+1)^{3/2}-2t^2$ for $0\le t\le3$ . To find the maximum velocity, we need the point where the derivative changes from positive to negative: a stationary point. Differentiating with the chain rule gives $3\sqrt{2t+1}-4t$ . Setting this to zero isolates the square root, and squaring both sides removes it to produce a quadratic. Solving gives $t=1.5$ and $t=-0.375$ . Only $t=1.5$ lies in the interval. Substituting back gives $v=3.5$ . Checking the endpoint velocities shows this is larger than $v(0)=1$ and $v(3)\approx0.52$ , so it is the maximum.

Key Takeaways

The maximum or minimum of a velocity function is found by differentiating and solving $\frac{dv}{dt}=0$ .
The chain rule is needed when differentiating expressions of the form $(2t+1)^{3/2}$ .
After solving an equation with a square root, reject roots that are outside the given domain or that do not satisfy the original equation.

Common Mistakes

Using $a=\frac{v}{t}$ instead of differentiating is not accepted.
Forgetting the factor $2$ from differentiating $2t+1$ under the chain rule, giving $\frac{3}{2}(2t+1)^{1/2}$ instead of $3(2t+1)^{1/2}$ .
Squaring incorrectly or failing to produce the three-term quadratic $16t^2-18t-9=0$ .
Not rejecting $t=-\frac{3}{8}$ because it is outside $0\le t\le3$ .

Things to Be Careful About

When an equation has been squared, check that any candidate root satisfies the original equation and lies in the given interval.
The mark scheme requires the simplified maximum value $3.5\ \text{m s}^{-1}$ as the final answer.
If verifying $t=1.5$ , you must show that it makes the derivative zero, not just guess it from a graph or table.

Techniques used

differentiate a composite velocity function using the chain ruleset the derivative equal to zero to locate a stationary pointsolve the resulting quadratic equation for timecheck the stationary value against the endpoints to confirm the maximum

(b)

It is given that in the interval $0 \le t \le 3$ the velocity of $P$ is always positive.

Find the distance of $P$ from $A$ at the instant when $P$ is moving at this maximum velocity.

4M Similar

DifficultyMedium

Worked solution

Approach

Because $v>0$ on the interval, the distance from $A$ equals the displacement $\int_0^T v\,dt$ . Integrate $v$ and evaluate the definite integral between $t=0$ and $t=\frac{3}{2}$ .

Working

Integrating:

s(t)=\int\left[(2t+1)^{3/2}-2t^2\right]dt =\frac{(2t+1)^{5/2}}{5}-\frac{2t^3}{3}+c

At $t=\frac{3}{2}$ :

s\left(\frac{3}{2}\right) =\frac{4^{5/2}}{5}-\frac{2\left(\frac{3}{2}\right)^3}{3} =\frac{32}{5}-\frac{9}{4} =\frac{83}{20}

At $t=0$ :

s(0)=\frac{1^{5/2}}{5}-0=\frac{1}{5}

Therefore the distance from $A$ is

\frac{83}{20}-\frac{1}{5} =\frac{83}{20}-\frac{4}{20} =\frac{79}{20} =3.95\ \text{m}

Final answer

79/20 m (3.95 m)

Detailed explanation

Walkthrough

The particle starts at $A$ at $t=0$ . Since $v$ is always positive on $0\le t\le3$ , the distance travelled is the same as the displacement from $A$ . The displacement is the definite integral of velocity. We integrate using the reverse of the chain rule: the power increases by $1$ and we divide by the new power and by the derivative of $2t+1$ , giving denominator $5$ . At $t=1.5$ , $F(t)=\frac{83}{20}$ . At $t=0$ , $F(0)=\frac15$ . The distance is the difference

\frac{83}{20}-\frac{1}{5}=\frac{79}{20}=3.95\ \text{m}.

Key Takeaways

The definite integral of velocity from $t=0$ to $t=T$ gives displacement; if velocity is positive, distance equals displacement.
The reverse chain rule is used to integrate composite functions such as $(2t+1)^{3/2}$ .
Definite integration uses $F(b)-F(a)$ , so the arbitrary constant cancels.

Common Mistakes

Using $s=vt$ is invalid because the velocity is not constant.
Forgetting to use $t=1.5$ from part (a); using an arbitrary time.
Making arithmetic errors when computing $4^{5/2}$ or $\frac{2\left(\frac{3}{2}\right)^3}{3}$ .
Forgetting to subtract $F(0)$ , or incorrectly handling the constant of integration.

Things to Be Careful About

The mark scheme accepts $\frac{79}{20}\ \text{m}$ or $3.95\ \text{m}$ only if the full integration method is shown.
If using an indefinite integral, remember the constant of integration; it cancels when evaluating $F\left(\frac{3}{2}\right)-F(0)$ . For reference, the constant would be $c=-0.2$ if $s(0)=0$ is imposed.
Since $v>0$ , there is no need to consider absolute values or changes of direction.

Techniques used

integrate velocity with respect to time using the reverse chain ruleevaluate the displacement between the initial time and the maximum-velocity timeuse the positivity of velocity to equate distance with displacementsimplify the resulting fractional arithmetic

Q76MMediumEnergy, Work and PowerForces and Equilibrium

A particle $P$ of mass $3\text{kg}$ is projected with a speed of $8\text{ms}^{-1}$ up a line of greatest slope of a rough plane inclined at $30^\circ$ to the horizontal. $P$ is projected from a point $A$ on the plane and comes to instantaneous rest at a point $B$ on the plane. $P$ then slides back down the plane. The coefficient of friction between $P$ and the plane is $\frac{1}{12}\sqrt{3}$ .

Using an energy method throughout, find the speed of $P$ at the instant it returns to $A$ .

DifficultyMedium

Worked solution

Approach

Use the work-energy principle. Between $A$ and $B$ the particle's initial kinetic energy is converted into gravitational potential energy and work done against friction. On the return journey the gravitational energy is converted back into kinetic energy, while friction again removes energy. Treating the round trip lets us compare the final kinetic energy with the initial kinetic energy directly.

Working

Let $d$ be the distance $AB$ and take $g=10\,\text{m s}^{-2}$ .

Resolving perpendicular to the plane, the normal reaction is

R=3g\cos 30^\circ.

The friction force, which opposes motion in either direction, has magnitude

F=\mu R=\frac{\sqrt{3}}{12}\times 3g\cos 30^\circ = \frac{3g}{8}.

With $g=10$ ,

F=\frac{3\times 10}{8}=3.75\,\text{N}.

The initial kinetic energy at $A$ is

\frac12 \times 3 \times 8^2 = 96\,\text{J}.

For the motion $A\to B$ , the gain in gravitational potential energy is

3g\,d\sin 30^\circ = 3\times 10 \times d \times \frac12 = 15d\,\text{J},

and the work done against friction is

Fd = 3.75d\,\text{J}.

The work-energy principle for $A\to B$ gives

96 = 15d + 3.75d.

Thus

96 = 18.75d,

so

d = \frac{96}{18.75}=5.12\,\text{m}.

The work done against friction on the way up is therefore

Fd = 3.75\times 5.12 = 19.2\,\text{J}.

For the round trip $A\to B\to A$ , the particle starts and finishes at the same height, so the net gravitational potential energy change is zero. The only net loss of energy is the total work done against friction, which is $19.2\,\text{J}$ on the way up and the same on the way down. If the required speed is $v$ , then

\frac12 \times 3 \times v^2 = \frac12 \times 3 \times 8^2 - 2(19.2).

So

1.5v^2 = 96 - 38.4 = 57.6,

and therefore

v^2 = 38.4,

v = \sqrt{38.4} = 6.196773\ldots

Answer

The speed of $P$ at the instant it returns to $A$ is

v = \frac{8\sqrt{15}}{5} = 6.20\,\text{m s}^{-1}.

Final answer

6.20 m s^-1

Detailed explanation

Walkthrough

Start by naming the unknown distance $AB$ as $d$ . The energy method needs every form of energy involved. At $A$ , the particle has only kinetic energy $\frac12 \times 3 \times 8^2$ . As it moves up the plane to $B$ , it rises through a vertical height $d\sin 30^\circ$ ; this appears as gravitational potential energy $mgd\sin 30^\circ$ . At the same time, friction does work against the motion, removing energy from the particle. The frictional force is found from $F=\mu R$ . The normal reaction $R$ is perpendicular to the plane, and since there is no acceleration perpendicular to the plane, $R=mg\cos 30^\circ$ . Substituting the coefficient of friction gives $F=3g/8$ .

The work-energy equation for the upward half is therefore: initial kinetic energy equals gravitational potential energy gained plus work done against friction. This gives one equation in $d$ only, so $d$ can be found. Once $d$ is known, the work done against friction on the upward journey is also known.

For the return journey, the particle starts from rest at $B$ , slides down to $A$ , and regains kinetic energy as it loses gravitational potential energy. Because $A$ and the end of the whole journey are at the same height, a neat way to finish is to look at the complete round trip. The particle starts and finishes at the same height, so the net change in gravitational potential energy is zero. The only net energy loss is the work done against friction on both halves, which is twice the work on one half. Therefore the final kinetic energy is the initial kinetic energy minus the total friction work. Solving gives the required speed; only the positive square root is meaningful.

Key Takeaways

The work-energy principle states that the total work done by non-gravitational forces equals the change in mechanical energy. Here friction, a non-gravitational force, removes mechanical energy. The gravitational potential energy of a particle on an inclined plane is $mgh$ , where $h$ is the vertical rise, not the distance along the plane. The normal reaction on an inclined plane is $R=mg\cos\theta$ , and the frictional force is $F=\mu R$ . Because friction always opposes motion, it does work on both the upward and the downward halves of the journey. For a round trip returning to the same height, gravitational potential energy cancels, leaving only the total work against friction as the energy loss.

Common Mistakes

A common error is to use Newton's second law instead of an energy method; the question explicitly requires an energy method throughout, so a Newton's-law solution loses the main method marks. Mixing up sine and cosine is another frequent mistake: the vertical height uses $\sin 30^\circ$ , while the normal reaction and friction use $\cos 30^\circ$ . Some candidates forget that friction acts on the way down as well as on the way up, so the total friction work over the whole trip is doubled. Sign errors in the energy equation are also common: on the way up, both the gain in gravitational potential energy and the work against friction are positive terms on the right-hand side.

Things to Be Careful About

Use $g=10\,\text{m s}^{-2}$ , as is standard in this syllabus. Every term in the energy equation must be dimensionally correct; in particular, the weight term must include the component $mg\sin 30^\circ$ for the vertical rise. The final answer must be positive, and the exact value is $\frac{8\sqrt{15}}{5}\approx 6.19677\,\text{m s}^{-1}$ , which rounds to $6.20\,\text{m s}^{-1}$ . Writing an unsupported final value may lose the final accuracy mark, so all working should be shown. If a special case using Newton's second law is considered, it can only recover limited special-case marks and is not the intended full solution.

Techniques used

resolve forces into components parallel and perpendicular to the planeapply the work-energy principlecalculate work done against frictionaccount for energy loss over a round trip

Mathematics 9709/42 — May/June 2025

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