9709/42 May/June 2024 — Mathematics Solved Past Paper

Q13MMediumEnergy, Work and Power

A cyclist and bicycle have a total mass of $72\text{ kg}$ . The cyclist rides along a horizontal road against a total resistance force of $28\text{ N}$ .

Find the total work done by the cyclist to increase his speed from $8\text{ m s}^{-1}$ to $16\text{ m s}^{-1}$ while travelling a distance of $100$ metres.

DifficultyMedium

Worked solution

Approach

Use the work-energy principle: the work done by the cyclist equals the increase in kinetic energy plus the work done against the resistance force.

Working

Initial kinetic energy:

\frac{1}{2} \times 72 \times 8^2 = 2304\text{ J}

Final kinetic energy:

\frac{1}{2} \times 72 \times 16^2 = 9216\text{ J}

Increase in kinetic energy:

9216 - 2304 = 6912\text{ J}

Work done against resistance:

28 \times 100 = 2800\text{ J}

Work-energy equation:

\text{WD} - 2800 = 6912

Therefore:

\text{WD} = 6912 + 2800 = 9712\text{ J}

Answer

The total work done by the cyclist is $9712\text{ J}$ .

Final answer

9712 J

Detailed explanation

Walkthrough

This is a work-energy problem. The cyclist does work to increase the bicycle's kinetic energy, but some of that work is lost overcoming the resistance force. We therefore need three pieces: the initial kinetic energy, the final kinetic energy, and the work done against resistance.

First calculate the initial kinetic energy using $\frac{1}{2}mv^2$ with $m = 72$ and $v = 8$ :

\frac{1}{2} \times 72 \times 8^2 = 2304\text{ J}

Then calculate the final kinetic energy with $v = 16$ :

\frac{1}{2} \times 72 \times 16^2 = 9216\text{ J}

The increase in kinetic energy is the difference:

9216 - 2304 = 6912\text{ J}

Next, work done against resistance is force times distance:

28 \times 100 = 2800\text{ J}

The work-energy principle states that the net work done on the object equals its change in kinetic energy. Here the cyclist's work is positive, while resistance does negative work. So:

\text{WD} - 2800 = 6912

Solving gives:

\text{WD} = 6912 + 2800 = 9712\text{ J}

An alternative method uses suvat to find the acceleration and then Newton's second law to find the driving force, but the work-energy method is more direct.

Key Takeaways

Kinetic energy is $\frac{1}{2}mv^2$ .
Work done by a constant force is $Fd$ when force and displacement are in the same direction.
The work-energy principle connects net work to change in kinetic energy.
When a resistance force acts, the driving work must overcome both the resistance and the increase in kinetic energy.

Common Mistakes

Forgetting to include the work done against resistance.
Using the change in speed ( $16 - 8$ ) instead of the change in $v^2$ when finding the kinetic energy increase.
Sign errors: resistance work should be subtracted from the cyclist's work in the net work equation.
Using the total distance incorrectly, e.g. multiplying resistance by $v$ instead of distance.
Giving an unsupported final answer; the mark scheme requires the work-energy equation to be shown to earn the method mark.

Things to Be Careful About

The mass is $72\text{ kg}$ for the whole cyclist and bicycle; do not split it.
The speeds are in $\text{m s}^{-1}$ and distance in metres, so the work is in joules.
In the mark scheme, the final answer may be written as $9712\text{ J}$ ; $9710\text{ J}$ is condoned as a rounding, but the exact value is preferred.
The mark scheme says "Do not ISW", so if a candidate gives a correct value then writes incorrect subsequent working, the subsequent working should not be ignored.
Show the work-energy equation with all four terms to earn the method mark; an unsupported answer may not receive full credit.

Techniques used

calculate initial and final kinetic energiesapply the work-energy principleaccount for work done against resistance

Approach

Acceleration is the rate of change of velocity, so differentiate $v = 44t - 6t^2 - 36$ with respect to $t$ . Then require the acceleration to be positive and solve the resulting inequality, remembering that $t \ge 0$ .

Working

Differentiate $v$ :

a = \frac{dv}{dt} = 44 - 12t

For positive acceleration:

44 - 12t > 0

44 > 12t \quad \Rightarrow \quad t < \frac{11}{3}

Since $t \ge 0$ , the set of values is:

0 \le t < \frac{11}{3}

Answer

0 \le t < \frac{11}{3}

Final answer

0 <= t < 11/3

Detailed explanation

Walkthrough

The velocity function is $v = 44t - 6t^2 - 36$ . Since acceleration is the derivative of velocity with respect to time, differentiate each term:

\frac{d}{dt}(44t) = 44, \qquad \frac{d}{dt}(-6t^2) = -12t, \qquad \frac{d}{dt}(-36) = 0

So

a = \frac{dv}{dt} = 44 - 12t

We need this acceleration to be positive, so we solve

44 - 12t > 0

Rearranging gives

44 > 12t \quad \Rightarrow \quad t < \frac{11}{3}

Because $t$ represents time after the particle leaves $O$ , $t$ cannot be negative. Therefore the valid set is $0 \le t < \frac{11}{3}$ .

Key Takeaways

Acceleration is the rate of change of velocity: $a = \frac{dv}{dt}$ .
Differentiating a polynomial is done term by term.
The time domain $t \ge 0$ is important when giving a set of values for kinematics problems.

Common Mistakes

Using $a = \frac{v}{t}$ instead of differentiating $v$ ; this scores no method mark.
Solving $44 - 12t = 0$ and giving $t \le \frac{11}{3}$ instead of the strict inequality $t < \frac{11}{3}$ .
Including negative values of $t$ , which are not meaningful after leaving point $O$ .

Things to Be Careful About

The inequality must be strict because the question asks for positive acceleration, not zero acceleration.
If a lower limit is included, it must be $0$ : either $(0, \frac{11}{3})$ or $[0, \frac{11}{3})$ is acceptable.
The mark scheme allows equivalent decimal or fractional answers such as $3\frac{2}{3}$ or $3.67$ .

Techniques used

differentiate velocity with respect to timesolve a linear inequality for positive accelerationrestrict the solution to the time domain

(b)

Find the two values of $t$ at which $P$ returns to $O$ .

3M Similar

DifficultyMedium

Worked solution

Approach

Displacement is the integral of velocity. Integrate $v = 44t - 6t^2 - 36$ , impose the initial condition at $O$ that $s=0$ when $t=0$ , then set the displacement equal to zero and solve for $t$ . The root $t=0$ is the initial instant at $O$ , so exclude it.

Working

s = \int (44t - 6t^2 - 36)\,dt = 22t^2 - 2t^3 - 36t + c

Since $P$ starts at $O$ , $s=0$ when $t=0$ , so $c=0$ :

s = 22t^2 - 2t^3 - 36t

Returning to $O$ means $s=0$ :

22t^2 - 2t^3 - 36t = 0

Factorise:

-2t(t^2 - 11t + 18) = 0

-2t(t - 2)(t - 9) = 0

Thus $t = 0$ , $t = 2$ or $t = 9$ . Since $t=0$ is the starting time at $O$ , the two return times are:

t = 2, \quad t = 9

Answer

t = 2 \quad \text{and} \quad t = 9

Final answer

t = 2 and t = 9

Detailed explanation

Walkthrough

To find when the particle returns to $O$ , we need its displacement from $O$ as a function of time. Displacement is the integral of velocity, so integrate $v = 44t - 6t^2 - 36$ term by term:

s = \int (44t - 6t^2 - 36)\,dt = 22t^2 - 2t^3 - 36t + c

At $t=0$ , the particle is at $O$ , so $s=0$ . This gives $c=0$ , and therefore

s = 22t^2 - 2t^3 - 36t

Returning to $O$ means the displacement is zero again:

22t^2 - 2t^3 - 36t = 0

Factor out $-2t$ :

-2t(t^2 - 11t + 18) = 0

The quadratic factorises further:

-2t(t - 2)(t - 9) = 0

So the solutions are $t=0$ , $t=2$ , and $t=9$ . The value $t=0$ is simply the starting instant at $O$ , not a return to $O$ . Therefore the two required values are $t=2$ and $t=9$ .

Key Takeaways

Displacement is the integral of velocity with respect to time.
The constant of integration must be found using the initial condition at the starting point.
A particle returns to the origin when its displacement is zero, not when its velocity is zero.
Cubic equations can often be solved by factorisation after taking out a common factor.

Common Mistakes

Using $s = vt$ instead of integrating $v$ ; this is invalid because the velocity is not constant.
Forgetting the constant of integration. Here $c=0$ , but it must still be justified by $s=0$ at $t=0$ .
Trying to solve the cubic by expanding incorrectly, or missing the factor $t$ .
Including $t=0$ as one of the return times. It is the initial time at $O$ , not a return.

Things to Be Careful About

The expression $22t^2 - 2t^3 - 36t$ must be set equal to zero, not equated to the velocity.
The factorisation is $-2t(t-2)(t-9)$ , giving roots $0$ , $2$ , and $9$ .
Since the question asks for the two values at which $P$ returns to $O$ , only $t=2$ and $t=9$ should be given.
The units are seconds; the answer can be stated as $t=2$ and $t=9$ seconds.

Techniques used

integrate velocity to obtain displacementapply initial condition to find integration constantset displacement to zero and factorise cubicselect the non-initial roots

Q36MMediumForces and Equilibrium

Fig. 3

Four coplanar forces of magnitude $P\text{ N}$ , $10\text{ N}$ , $16\text{ N}$ and $2\text{ N}$ act at a point in the directions shown in the diagram. It is given that the forces are in equilibrium.

Find the values of $\theta$ and $P$ .

DifficultyMedium

Worked solution

Approach

Since the four forces are coplanar and in equilibrium, the vector sum of all forces must be zero. This means the sum of the horizontal components is zero and the sum of the vertical components is zero. We resolve the forces horizontally and vertically to form two equations in the two unknowns $P$ and $\theta$ .

Working

Horizontal resolution:
Taking forces to the right as positive, the rightward component is $10\cos 25^\circ$ . The leftward components are $2\cos 40^\circ$ (from the 2 N force) and $16\sin\theta$ (from the 16 N force, noting that $\theta$ is measured from the vertical).

\begin{aligned} 10\cos 25^\circ &= 2\cos 40^\circ + 16\sin\theta \\ 16\sin\theta &= 10\cos 25^\circ - 2\cos 40^4 \\ \sin\theta &= \frac{10\cos 25^\circ - 2\cos 40^\circ}{16} \end{aligned}

Calculating the numerical values:

\sin\theta = \frac{9.06307787 - 1.532088886}{16} = \frac{7.530988984}{16} \approx 0.4706868115

Solving for $\theta$ :

\theta = \sin^{-1}(0.4706868115) \approx 28.0789^\circ

Rounding to 3 significant figures:

\theta = 28.1^\circ

Vertical resolution:
Taking forces upwards as positive, the upward force is $P$ . The downward components are $10\sin 25^\circ$ (from the 10 N force), $16\cos\theta$ (from the 16 N force), and $2\sin 40^\circ$ (from the 2 N force).

P = 10\sin 25^\circ + 16\cos\theta + 2\sin 40^\circ

Substituting the unrounded value $\theta = 28.0789^\circ$ :

P = 10\sin 25^\circ + 16\cos(28.0789^\circ) + 2\sin 40^\circ

Calculating the numerical values:

P = 4.226182617 + 14.46418098 + 1.285575219 \approx 19.6285636

Rounding to 3 significant figures:

P = 19.6 \text{ N}

Answer

\theta = 28.1^\circ, \quad P = 19.6 \text{ N}

Final answer

theta = 28.1 degrees, P = 19.6 N

Detailed explanation

Walkthrough

The problem states that four coplanar forces act at a point and are in equilibrium. The fundamental principle here is that for a system of forces to be in equilibrium, the resultant force in any direction must be zero. We choose two perpendicular directions (horizontal and vertical) to resolve the forces.

Step 1: Horizontal Resolution
We look at all forces and their horizontal components. Taking the right direction as positive:

The 10 N force is at $25^\circ$ below the positive x-axis, so its horizontal component is $10\cos 25^\circ$ to the right.
The 2 N force is at $40^\circ$ below the negative x-axis, so its horizontal component is $2\cos 40^\circ$ to the left.
The 16 N force is at $\theta$ to the negative y-axis. Since the angle is measured from the vertical, the horizontal component is $16\sin\theta$ to the left.

Setting the sum of rightward forces equal to the sum of leftward forces:

10\cos 25^\circ = 2\cos 40^\circ + 16\sin\theta

We rearrange this to isolate $\sin\theta$ :

\sin\theta = \frac{10\cos 25^\circ - 2\cos 40^\circ}{16}

Evaluating this gives $\sin\theta \approx 0.470687$ , so $\theta \approx 28.1^\circ$ .

Step 2: Vertical Resolution
Now we consider the vertical components. Taking the upward direction as positive:

The force $P$ acts vertically upwards.
The 10 N force has a vertical component of $10\sin 25^\circ$ downwards.
The 16 N force has a vertical component of $16\cos\theta$ downwards (since $\theta$ is from the vertical).
The 2 N force has a vertical component of $2\sin 40^\circ$ downwards.

Setting the upward force equal to the sum of downward forces:

P = 10\sin 25^\circ + 16\cos\theta + 2\sin 40^\circ

Substituting the value of $\theta$ found in Step 1 (using the unrounded value for accuracy) gives $P \approx 19.6$ N.

Key Takeaways

Equilibrium condition: For coplanar forces in equilibrium, $\sum F_x = 0$ and $\sum F_y = 0$ . This provides two independent equations for two unknowns.
Resolving forces: Always identify the reference axis for each angle. If an angle is given to the horizontal, use cosine for the horizontal component and sine for the vertical. If given to the vertical, use sine for the horizontal and cosine for the vertical.
Accuracy in intermediate steps: When solving for one unknown to substitute into another equation, always use the unrounded value to avoid compounding rounding errors.

Common Mistakes

Mixing up sine and cosine: A very common error is using $\cos\theta$ for the horizontal component of the 16 N force. Since $\theta$ is measured from the vertical (negative y-axis), the horizontal component must be $16\sin\theta$ .
Sign errors: Forgetting that forces pointing left or down are negative when setting up the equilibrium equations. It is often safer to write $\sum \text{rightward} = \sum \text{leftward}$ and $\sum \text{upward} = \sum \text{downward}$ to avoid sign confusion.
Rounding too early: Calculating $\theta$ and rounding to $28.1^\circ$ before substituting into the vertical equation can lead to a slightly incorrect value for $P$ (e.g., $19.6$ might still be correct, but in other cases it could push the answer to the next rounding boundary).

Things to Be Careful About

Angle references: Pay close attention to how angles are defined in the diagram. The 2 N force is at $40^\circ$ to the horizontal (negative x-axis), while the 16 N force is at $\theta$ to the vertical (negative y-axis). This difference dictates whether sine or cosine is used for each component.
Units: Ensure the final answer for $P$ includes the unit of force (Newtons, N), while $\theta$ is an angle (degrees).
Mark scheme guidance: The mark scheme allows sign errors in the initial resolution equation (M1) as long as the correct number of relevant terms are present, but the final numerical answers must be correct to 3 significant figures.

Techniques used

resolve forces horizontallyresolve forces verticallyapply equilibrium conditionsolve for unknowns using trigonometry

Approach

Since the car moves at constant speed, acceleration is $0$ , so the resultant force along the slope is zero. Resolve forces parallel to the slope: driving force $DF$ must balance the resistance plus the component of the car’s weight down the slope. Then use $P = DF \times v$ .

Working

Along the slope:

DF - 480 - mg\sin\alpha = 0

With $m = 1400\text{ kg}$ , $g = 10\text{ m s}^{-2}$ and $\sin\alpha = 0.12$ :

DF = 480 + 1400 \times 10 \times 0.12

DF = 480 + 1680 = 2160\text{ N}

Therefore the power is

P = DF \times v = 2160 \times 24

P = 51840\text{ W} = 51.84\text{ kW}

Answer

The power of the car’s engine is $51840\text{ W}$ (or $51.84\text{ kW}$ ).

Final answer

51840 W

Detailed explanation

Walkthrough

Constant speed means the acceleration is zero. Newton’s second law therefore tells us that the net force along the direction of motion is zero.

On the hill, three forces act in the direction parallel to the slope: the driving force forwards, the resistance backwards, and the component of the car’s weight down the slope. To resolve the weight, multiply $mg$ by $\sin\alpha$ , because the component along a slope inclined at angle $\alpha$ is $mg\sin\alpha$ .

Set the forward driving force equal to the sum of the two backward forces:

DF = 480 + 1400 \times 10 \times 0.12 = 2160\text{ N}

Finally, the power of the engine is the rate at which the driving force does work, given by $P = DF \times v$ when the car is moving at constant speed in the direction of the force.

Key Takeaways

On an incline, the component of weight along the slope is $mg\sin\alpha$ .
At constant speed the resultant force is zero.
Power supplied by an engine equals driving force times speed for motion in the line of the force.

Common Mistakes

Using $mg\cos\alpha$ instead of $mg\sin\alpha$ for the component along the slope.
Forgetting the resistance term when setting up the balance of forces.
Using $v = 24$ with the wrong force, or forgetting that power uses the driving force, not the net force.
If acceleration is zero, treating the car as if it still has a nonzero resultant force.

Things to Be Careful About

The mark scheme expects a clear statement of Newton’s second law with three terms in the balance, so show all three forces. Also, if the final answer is given in kW, include the unit; otherwise a bare number may not be accepted.

Techniques used

resolve forces along an inclined planeapply the equilibrium condition for constant speedcalculate engine power using P = Fv

(b)

The car now moves at a constant speed on a straight level road.

Given that its engine is working at $54\text{ kW}$ , find this speed.

3M Similar

DifficultyMedium

Worked solution

Approach

On a level road the car moves at constant speed, so acceleration is zero and the driving force equals the resistance. Express the driving force in terms of $v$ using $P = DF \times v$ , equate it to $kv^2$ , and solve the resulting cubic equation.

Working

Let the speed be $v\text{ m s}^{-1}$ . The resistance is

F_R = kv^2 = \frac{5}{6}v^2

At constant speed on level ground,

DF = F_R = \frac{5}{6}v^2

Using $P = DF \times v$ with $P = 54\text{ kW} = 54000\text{ W}$ :

54000 = \frac{5}{6}v^2 \times v = \frac{5}{6}v^3

v^3 = 54000 \times \frac{6}{5} = 64800

v = \sqrt[3]{64800} = 40.1659...

Hence the speed is $40.2\text{ m s}^{-1}$ .

Answer

v = 40.2\text{ m s}^{-1}

Final answer

40.2 m s^-1

Detailed explanation

Walkthrough

On a level road the car’s weight is perpendicular to the motion, so it does not have a component along the road. Since the car moves at constant speed, there is no acceleration. Newton’s second law along the horizontal direction gives:

DF - \frac{5}{6}v^2 = 0

so the driving force equals the resistance.

The engine power is related to force and speed by $P = DF \times v$ . Substituting $DF = \frac{5}{6}v^2$ :

54000 = \frac{5}{6}v^3

Solve this cubic by multiplying by $\frac{6}{5}$ and taking the cube root.

Key Takeaways

On level ground at constant speed, the driving force is exactly balanced by the resistance.
$P = Fv$ links engine power, driving force and speed.
When power is constant but resistance depends on $v^2$ , the speed satisfies a cubic equation.

Common Mistakes

Using $v$ instead of $v^3$ in the power equation.
Using $kv^2$ as the driving force directly, without multiplying by $v$ for power.
Forgetting to convert $54\text{ kW}$ to $54000\text{ W}$ .
Taking a square root instead of a cube root.

Things to Be Careful About

The mark scheme allows follow-through on the value of $k$ from part (a), so if you used a different $k$ , still set up $54000 = kv^3$ and solve. The final speed must be positive; the cubic has one real positive root here. Include units in the final answer.

Techniques used

equate driving force and resistance at constant speeduse P = Fv to relate power and speedsolve a cubic equation

Q58MMediumForces and Equilibrium

Fig. 5

A particle of mass $0.8\text{ kg}$ lies on a rough plane which is inclined at an angle of $28^\circ$ to the horizontal. The particle is kept in equilibrium by a force of magnitude $T\text{ N}$ . This force acts at an angle of $35^\circ$ above a line of greatest slope of the plane (see diagram). The coefficient of friction between the particle and the plane is $0.2$ .

Find the least and greatest possible values of $T$ .

DifficultyMedium

Worked solution

Approach

Resolve the forces acting on the particle perpendicular and parallel to the inclined plane. Since the particle is in equilibrium with friction potentially acting in either direction along the plane, consider the two cases of limiting equilibrium: impending motion up the plane (minimum $T$ ) and impending motion down the plane (maximum $T$ ). Use the limiting friction condition $F = \mu R$ to solve for $T$ in each case.

Working

Resolving perpendicular to the plane:
The forces perpendicular to the plane are the normal reaction $R$ (upwards), the component of $T$ perpendicular to the plane ( $T\sin 35^\circ$ , upwards), and the component of weight perpendicular to the plane ( $mg\cos 28^\circ$ , downwards). Equating forces perpendicular to the plane:

\begin{aligned} R + T\sin 35^\circ &= 0.8g\cos 28^\circ \\ R + 0.573576T &= 0.8 \times 9.8 \times \cos 28^\circ \\ R + 0.573576T &= 7.06358 \end{aligned}

Resolving parallel to the plane:
The forces parallel to the plane are the component of $T$ parallel to the plane ( $T\cos 35^\circ$ , up the plane), the component of weight parallel to the plane ( $mg\sin 28^\circ$ , down the plane), and the frictional force $F$ .

Case 1: Impending motion down the plane (greatest $T$ )
Friction acts up the plane to oppose the motion. Equating forces parallel to the plane:

\begin{aligned} T\cos 35^\circ &= 0.8g\sin 28^\circ + F \\ 0.819152T &= 0.8 \times 9.8 \times \sin 28^\circ + F \\ 0.819152T &= 3.75577 + F \end{aligned}

Case 2: Impending motion up the plane (least $T$ )
Friction acts down the plane to oppose the motion. Equating forces parallel to the plane:

\begin{aligned} T\cos 35^\circ &= 0.8g\sin 28^\circ - F \\ 0.819152T &= 3.75577 - F \end{aligned}

Using limiting friction $F = 0.2R$ :
Substitute $R = 7.06358 - 0.573576T$ into $F = 0.2R$ :

F = 0.2(7.06358 - 0.573576T) = 1.41272 - 0.114715T

For Case 1 (greatest $T$ ):

\begin{aligned} 0.819152T &= 3.75577 + 1.41272 - 0.114715T \\ 0.819152T + 0.114715T &= 5.16849 \\ 0.933867T &= 5.16849 \\ T &= 5.53450 \end{aligned}

For Case 2 (least $T$ ):

\begin{aligned} 0.819152T &= 3.75577 - (1.41272 - 0.114715T) \\ 0.819152T &= 2.34305 + 0.114715T \\ 0.819152T - 0.114715T &= 2.34305 \\ 0.704437T &= 2.34305 \\ T &= 3.32614 \end{aligned}

Answer

The least possible value of $T$ is $3.33$ N and the greatest possible value is $5.53$ N.

Final answer

Least T = 3.33 N, Greatest T = 5.53 N

Detailed explanation

Walkthrough

First, we resolve the forces perpendicular to the inclined plane to find an expression for the normal reaction $R$ in terms of $T$ . The forces are the normal reaction $R$ acting upwards perpendicular to the plane, the perpendicular component of the applied force $T\sin 35^\circ$ also acting upwards, and the perpendicular component of the weight $mg\cos 28^\circ$ acting downwards. Setting the sum of upward forces equal to the downward force gives $R + T\sin 35^\circ = 0.8g\cos 28^\circ$ .

Next, we resolve the forces parallel to the plane. The parallel component of the applied force is $T\cos 35^\circ$ acting up the plane, and the parallel component of the weight is $mg\sin 28^\circ$ acting down the plane. Friction $F$ acts parallel to the plane to oppose impending motion.

Since we want the least and greatest values of $T$ , we consider the two limiting cases:

Greatest $T$ : The particle is on the verge of sliding up the plane. Friction acts down the plane. The equation is $T\cos 35^\circ = mg\sin 28^\circ + F$ .
Least $T$ : The particle is on the verge of sliding down the plane. Friction acts up the plane. The equation is $T\cos 35^\circ = mg\sin 28^\circ - F$ .

At limiting equilibrium, the frictional force is $F = \mu R = 0.2R$ . We substitute the expression for $R$ from the perpendicular resolution into this friction equation, giving $F = 0.2(7.06358 - 0.573576T)$ . Substituting this into both parallel resolution equations allows us to solve for $T$ in each case, yielding $T = 5.53$ N for the greatest value and $T = 3.33$ N for the least value.

Key Takeaways

Resolving forces perpendicular and parallel to an inclined plane is a fundamental technique in equilibrium problems.
When finding the range of a force for equilibrium on a rough surface, always consider both directions of impending motion (up and down the plane) to account for friction acting in opposite directions.
The limiting friction condition $F = \mu R$ must be applied at the boundary of equilibrium.

Common Mistakes

Forgetting to consider both cases of impending motion (only finding one value of $T$ instead of the least and greatest).
Resolving incorrectly by mixing up sine and cosine components for the angle of the applied force $T$ relative to the plane.
Using the wrong angle for the weight components (must use $28^\circ$ , the angle of the plane to the horizontal).
Not substituting $F = \mu R$ correctly or making algebraic errors when solving the simultaneous equations.

Things to Be Careful About

Ensure $g = 9.8$ is used consistently throughout the calculation.
Keep intermediate values to at least 4-5 decimal places to avoid rounding errors in the final answer.
The final answers should be given to 3 significant figures as per standard practice (3.33 and 5.53).

Techniques used

resolve forces perpendicular to the planeresolve forces parallel to the planeapply limiting friction condition F = μRsolve simultaneous equations

Approach

Take the direction from $X$ towards $Z$ as positive. For each collision the particles coalesce, so they have a common velocity afterwards. Since the track is smooth and horizontal, horizontal momentum is conserved in each direct impact.

Working

First collision: $A$ (mass $5\text{ kg}$ , speed $6\text{ m s}^{-1}$ ) hits stationary $B$ (mass $1\text{ kg}$ ) and they form $D$ .

\begin{aligned} 5 \times 6 &= (5+1)v_D \\ 30 &= 6v_D \\ v_D &= 5\ \text{m s}^{-1} \end{aligned}

Second collision: $D$ (mass $6\text{ kg}$ , speed $5\text{ m s}^{-1}$ ) moves towards $C$ , while $C$ (mass $2\text{ kg}$ ) moves towards $D$ with speed $v$ , so in the positive $X\to Z$ direction its velocity is $-v$ .

\begin{aligned} 6(5) - 2v &= (5+1+2)v_E \\ 30 - 2v &= 8v_E \\ v_E &= \frac{30-2v}{8} = \frac{15-v}{4} \end{aligned}

Answer

v_E = \frac{15-v}{4}\ \text{m s}^{-1}

Final answer

Speed of E is (15 - v)/4 m/s

Detailed explanation

Walkthrough

Here two separate impacts occur. In an impact where the particles coalesce, the final object has mass equal to the sum of the two masses and moves with one common velocity. No external horizontal force acts during the impact because the track is smooth and horizontal, so the total horizontal momentum is conserved.

For the first impact, only $A$ is moving before the impact; $B$ is initially at rest. Therefore

5 \times 6 = (5+1)v_D,

which gives $v_D = 5\ \text{m s}^{-1}$ .

For the second impact, choose positive direction from $X$ to $Z$ . The combined particle $D$ moves in this direction, so its momentum is positive. Particle $C$ , however, is projected from $Z$ towards $Y$ , which is the negative direction, so its velocity is $-v$ and its momentum is $-2v$ . Setting total momentum before equal to total momentum after gives

6v_D - 2v = 8v_E.

Substituting $v_D = 5$ and solving gives $v_E = \frac{15-v}{4}$ , as required.

Key Takeaways

In a coalescing collision the two masses stick together, so the final mass is the sum of the masses.
Momentum is conserved in a direct impact on a smooth horizontal track.
A velocity in the opposite direction to the chosen positive direction must be given a negative sign.

Common Mistakes

Forgetting that $C$ is moving towards $B$ , not away from it, and writing its velocity as $+v$ .
Using the mass of $B$ twice, or forgetting to add the masses of $A$ and $B$ when forming $D$ .
Treating the two collisions as if they could be combined into one collision.

Things to Be Careful About

This is a show that result: the final answer must be written in terms of $v$ and all working must be shown.
The speed $v_E$ must be positive; the formula $\frac{15-v}{4}$ is consistent with $E$ moving towards $Z$ only when $v<15$ .

Techniques used

apply conservation of momentum to a coalescing impactassign signed velocities for opposite directionssolve for the combined particle's velocity

(b)

The total loss of kinetic energy of the system due to the two collisions is $63\text{ J}$ .

Use the result from (a) to show that $v = 3$ .

3M Similar

DifficultyMedium

Worked solution

Approach

The total kinetic energy after both collisions is the kinetic energy of the single final particle $E$ , because all mass has coalesced. The loss in kinetic energy is

\text{initial KE} - \text{final KE} = 63\ \text{J}.

Use the speed of $E$ obtained in part (a), then solve the resulting quadratic in $v$ .

Working

Initial kinetic energy of the system:

\text{KE}_i = \frac12(5)(6^2) + \frac12(2)(v^2) = 90 + v^2

Final kinetic energy, when the single particle $E$ has mass $5+1+2 = 8\text{ kg}$ and speed $\frac{15-v}{4}$ :

\text{KE}_f = \frac12(8)\left(\frac{15-v}{4}\right)^2 = \frac{(15-v)^2}{4}

Loss:

\begin{aligned} 90 + v^2 - \frac{(15-v)^2}{4} &= 63 \\ 360 + 4v^2 - 225 + 30v - v^2 &= 252 \\ 135 + 3v^2 + 30v &= 252 \\ 3v^2 + 30v - 117 &= 0 \\ v^2 + 10v - 39 &= 0 \\ (v+13)(v-3) &= 0 \end{aligned}

Since $v$ is a speed, $v>0$ , so $v=3$ .

Answer

v=3

Final answer

v = 3

Detailed explanation

Walkthrough

Before any collision, the moving particles are $A$ and $C$ , so the initial kinetic energy has two terms: $\frac12(5)(6^2)$ for $A$ and $\frac12(2)(v^2)$ for $C$ . After both collisions every particle has coalesced into $E$ , so the final kinetic energy is just the kinetic energy of one particle of mass $8\text{ kg}$ moving at the speed from part (a).

Subtract final KE from initial KE and set the loss equal to $63\text{ J}$ . Clearing the fraction gives a quadratic equation. Factorising gives $v=3$ or $v=-13$ . Since $v$ is a speed and $C$ is projected towards $B$ , only $v=3$ is physically possible.

Key Takeaways

Kinetic energy is not conserved in a perfectly inelastic (coalescing) collision; use conservation of momentum to find speeds, then calculate energy loss.
The total kinetic energy after all collisions is that of the single final combined object.
A negative speed is not physically valid.

Common Mistakes

Forgetting the second initial kinetic energy term for particle $C$ .
Using $v_E = v$ in the final kinetic energy, or failing to use the result from part (a).
Setting the loss equal to $-63$ or not clearing the fraction correctly.
Giving $v=-13$ as a valid solution without discarding it.

Things to Be Careful About

This is also a show that result. The mark scheme requires the algebraic solving of the quadratic to be shown; just writing the quadratic and the answer is not enough.
When clearing the fraction, multiply every term by $4$ .

Techniques used

calculate the total kinetic energy before collisionscalculate the kinetic energy of the final combined particleset the kinetic-energy loss equal to the given valuesolve a quadratic equation and discard the non-physical root

(c)

It is given that the distance $XY$ is $36\text{ m}$ and the distance $YZ$ is $98\text{ m}$ .

5M

(i)

Find the time between the two collisions.

4M Similar

DifficultyMedium

Worked solution

Approach

First find how long $A$ takes to travel $XY=36\text{ m}$ at $6\text{ m s}^{-1}$ . During this time $C$ also moves, so find how far $C$ is from $Y$ when the first collision occurs. After the first collision, $D$ and $C$ move towards each other; use their combined approach speed to find the time until the second collision.

Working

Time for $A$ to reach $B$ :

t_1 = \frac{36}{6} = 6\ \text{s}

In these $6\text{ s}$ , $C$ moves distance $3 \times 6 = 18\text{ m}$ from $Z$ . Therefore the distance from $C$ to $B$ (at $Y$ ) is

BC = 98 - 3(6) = 80\ \text{m}

After the first collision, $D$ has speed $5\text{ m s}^{-1}$ towards $C$ , and $C$ has speed $3\text{ m s}^{-1}$ towards $D$ . If the time between the collisions is $t$ seconds, the sum of the distances they travel before meeting is the gap of $80\text{ m}$ :

5t + 3t = 80 \quad\Rightarrow\quad 8t = 80 \quad\Rightarrow\quad t = 10\ \text{s}

Answer

The time between the two collisions is $10$ seconds.

Final answer

10 s

Detailed explanation

Walkthrough

Assume all particles move with constant speeds, because after projection or coalescence there is no horizontal force on a smooth track. $A$ takes $6$ seconds to travel $36\text{ m}$ . At this moment $C$ has also been moving for $6$ seconds at $3\text{ m s}^{-1}$ , so it has covered $18\text{ m}$ from $Z$ . Since $YZ=98\text{ m}$ , the gap between $C$ and the point $Y$ is $80\text{ m}$ .

After $A$ and $B$ coalesce, $D$ starts from $Y$ at $5\text{ m s}^{-1}$ towards $C$ , while $C$ continues at $3\text{ m s}^{-1}$ towards $D$ . Together they close the $80\text{ m}$ gap at $5+3=8\text{ m s}^{-1}$ , so the time is $10$ seconds.

Key Takeaways

Distance = speed $\times$ time can be used for constant-speed motion.
To find when two moving objects meet, add their speeds when they travel towards each other.
Position at the moment of the first collision is found by subtracting the distance $C$ has covered from $YZ$ .

Common Mistakes

Using $YZ=98$ as the initial gap between $D$ and $C$ without subtracting $C$ 's motion during the first $6$ seconds.
Treating the relative speed as $5-3$ instead of $5+3$ , since the two bodies move towards each other.
Using $v=3$ for $D$ 's speed instead of $v_D=5$ from part (a).

Things to Be Careful About

The mark scheme allows follow-through from the $6\text{ s}$ found in the first step.
The $80\text{ m}$ gap is the distance between the particles at the instant of the first collision, not the original $YZ$ distance.

Techniques used

find the time for a constant-speed journeytrack the distance moved by the second particleuse relative speed of approach to find collision time

(ii)

Find the time between the instant that $A$ is projected from $X$ and the instant that $E$ reaches $Z$ .

1M Similar

DifficultyMedium-Easy

Worked solution

Approach

After both collisions, $E$ moves towards $Z$ with the speed found in part (a). The point at which the second collision occurs is exactly the point that $C$ has reached after moving from $Z$ for the whole time up to that collision. So find that distance, divide by $E$ 's speed, and add all three stage times.

Working

From part (b), $v=3$ , so from part (a)

v_E = \frac{15-3}{4} = 3\ \text{m s}^{-1}

The second collision occurs $6+10=16\text{ s}$ after $A$ is projected. During that time $C$ has travelled from $Z$ a distance

3 \times 16 = 48\ \text{m}.

So $E$ takes $48/3 = 16\text{ s}$ to reach $Z$ . Total time:

6 + 10 + 16 = 32\ \text{s}

Answer

The time is $32$ seconds.

Final answer

32 s

Detailed explanation

Walkthrough

The only new part is the journey of $E$ from the second collision point to $Z$ . The second collision happens at the position that $C$ has reached. Since $C$ starts at $Z$ and moves for $16$ seconds before the second collision, that point is $48\text{ m}$ from $Z$ . $E$ has speed $3\text{ m s}^{-1}$ , so $E$ takes $16$ seconds to reach $Z$ . Add the $6$ seconds for $A$ to reach $B$ and the $10$ seconds between collisions, giving $6+10+16=32$ seconds.

Key Takeaways

The final stage time is found by dividing the remaining distance to $Z$ by the speed of $E$ .
The remaining distance can be found from the total distance $C$ has travelled from $Z$ up to the second collision.
Total journey time is the sum of the times of the separate stages.

Common Mistakes

Forgetting to add all three stage times together.
Using the whole distance $YZ=98\text{ m}$ as the distance $E$ must travel after the second collision.
Forgetting that $v=3$ from part (b), so using an incorrect speed for $E$ .

Things to Be Careful About

The second collision occurs $16$ seconds after $A$ is projected, not $10$ seconds after.
All speeds are constant, so $\text{time} = \text{distance}/\text{speed}$ is valid for the final stage.

Techniques used

use the speed of the final particlecompute the distance remaining to the endpointadd the times of the separate stages

Q7MediumNewton's Laws of MotionEnergy, Work and Power

Fig. 7

Two particles $P$ and $Q$ of masses $2.5\text{ kg}$ and $0.5\text{ kg}$ respectively are connected by a light inextensible string that passes over a small smooth pulley fixed at the top of a plane inclined at an angle of $30^\circ$ to the horizontal. Particle $P$ is on the plane and $Q$ hangs below the pulley such that the level of $Q$ is $2\text{ m}$ below the level of $P$ (see diagram).

Particle $P$ is released from rest with the string taut and slides down the plane. The plane is rough with coefficient of friction $0.2$ between the plane and $P$ .

(a)

Find the acceleration of $P$ .

5M Similar

DifficultyMedium

Worked solution

Approach

Apply Newton's second law to each particle separately. For P, resolve the weight along and perpendicular to the plane; friction acts up the plane because P slides down. For Q, the only forces are its weight and the tension. The string is inextensible, so both particles share the same magnitude of acceleration $a$ . Add the two equations to eliminate the tension $T$ and solve for $a$ .

Working

For particle $P$ on the inclined plane:

Perpendicular to the plane (no acceleration), the normal reaction balances the component of the weight:

R = 2.5g\cos 30°

Friction opposes P's motion down the plane, so it acts up the plane:

F = \mu R = 0.2 \times 2.5g\cos 30° = 0.5g\cos 30°

Along the plane (taking down-the-plane as positive for P, since P accelerates in that direction):

2.5g\sin 30° - T - F = 2.5a \quad\text{...(i)}

For particle $Q$ (taking upward as positive, since Q accelerates upward when P slides down):

T - 0.5g = 0.5a \quad\text{...(ii)}

Adding equations (i) and (ii) eliminates $T$ :

2.5g\sin 30° - F - 0.5g = (2.5 + 0.5)a

Substitute $F = 0.5g\cos 30°$ :

2.5g\sin 30° - 0.5g\cos 30° - 0.5g = 3a

Factor out $g$ on the left:

g\bigl(2.5 \cdot 0.5 - 0.5 \cdot \tfrac{\sqrt{3}}{2} - 0.5\bigr) = 3a

g\bigl(1.25 - 0.25\sqrt{3} - 0.5\bigr) = 3a

g\bigl(0.75 - 0.25\sqrt{3}\bigr) = 3a

\frac{g(3 - \sqrt{3})}{12} = a

With $g = 9.8$ :

a = \frac{9.8(3 - 1.7321)}{12} = \frac{9.8 \times 1.2679}{12} = \frac{12.425}{12} = 1.0355

Answer

a \approx 1.04\text{ m s}^{-2}

(Using $g = 10$ , the answer is $\frac{15-5\sqrt{3}}{6} \approx 1.06$ m s $^{-2}$ , which matches the mark scheme.)

Final answer

a ≈ 1.04 m s⁻²

Detailed explanation

Walkthrough

We have two particles connected by an inextensible string passing over a pulley at the top of a rough inclined plane. Because the string is inextensible, both particles share the same magnitude of acceleration $a$ . When P slides down the plane, Q rises vertically.

Step 1 — Normal reaction on P. Resolving the weight of P perpendicular to the plane and using the fact that P has no acceleration perpendicular to the surface gives $R = 2.5g\cos 30°$ .

Step 2 — Friction on P. The plane is rough with $\mu = 0.2$ , so the friction force is $F = \mu R = 0.2 \times 2.5g\cos 30° = 0.5g\cos 30°$ . Because P is moving down the plane, friction acts up the plane to oppose this motion.

Step 3 — Newton's second law along the plane for P. Taking the down-the-plane direction as positive, the three forces contributing along the plane are: the weight component $2.5g\sin 30°$ (positive), the tension $T$ (negative, up the plane), and the friction $F$ (negative, up the plane). This yields equation (i): $2.5g\sin 30° - T - F = 2.5a$ .

Step 4 — Newton's second law for Q. For the hanging mass Q, taking upward as positive: $T - 0.5g = 0.5a$ .

Step 5 — Eliminate the tension. Adding (i) and (ii) cancels $T$ and gives the system equation $2.5g\sin 30° - F - 0.5g = 3a$ .

Step 6 — Substitute and solve. Replacing $F$ with $0.5g\cos 30°$ and using $g = 9.8$ , the right-hand side becomes $3a = 9.8(1.25 - 0.25\sqrt{3} - 0.5) = 12.425 \times 0.25 \approx 3.107$ , giving $a \approx 1.04$ m s $^{-2}$ .

Key Takeaways

Connected particles on either side of a pulley share the same magnitude of acceleration.
On a rough plane, friction always opposes the direction of motion, so it acts up the plane when P slides down.
The cleanest approach is to write Newton's second law for each particle and then add the equations to eliminate the unknown tension.

Common Mistakes

Forgetting to take the friction direction as opposite to motion (i.e., placing it on the wrong side of the equation).
Using $0.5g\cos 30°$ (the friction on a 0.5 kg particle) instead of $0.2 \times 2.5g\cos 30°$ (the friction on the 2.5 kg particle actually on the plane). The mark scheme explicitly warns about this: " $F = 0.5g\cos 30°$ is wrong; it should use the mass of P."
Mixing up the sine and cosine: the component along the plane is $mg\sin\theta$ , not $mg\cos\theta$ .
Treating the two particles as if they have different accelerations.

Things to Be Careful About

The friction model $F = \mu R$ only applies once the contact is rough; with $\mu = 0.2$ and the system accelerating down the plane, this is the relevant case.
The mark scheme accepts $a \approx 1.06$ (using $g = 10$ ); with $g = 9.8$ the equivalent is $1.04$ , which is a slightly different numerical value but uses identical working. Both are "correct work"; the choice of $g$ is convention.
The mass of P (2.5 kg) — not the mass of Q (0.5 kg) — must be used in the friction calculation.

Techniques used

resolve forces on inclined plane into componentscalculate normal reaction and friction forceapply Newton's second law to each particleeliminate tension to solve for common acceleration

(b)

Use an energy method to find the speed of the particles at the instant when they are at the same vertical height.

5M Similar

DifficultyMedium

Worked solution

Approach

Let $x$ be the distance P slides down the plane; the string being inextensible means Q rises the same distance $x$ . Use the geometric condition that P and Q are at the same vertical height to find $x$ . Then apply the work–energy principle to the whole system: total KE gained equals the net loss of gravitational PE minus the work done against friction.

Working

Step 1 — Geometry to find $x$ .

When P moves a distance $x$ down the plane, its vertical drop is $x\sin 30° = x/2$ . Q rises a vertical distance $x$ . The initial vertical separation is $2$ m (Q is $2$ m below P), so the new vertical separation is $2 - x/2 - x = 2 - 3x/2$ . Setting this to zero:

2 - \frac{3x}{2} = 0 \quad\Longrightarrow\quad x = \frac{4}{3}\text{ m}

So P's vertical drop is $\frac{4}{3} \cdot \frac{1}{2} = \frac{2}{3}$ m, and Q's vertical rise is $\frac{4}{3}$ m.

Step 2 — Work–energy equation.

Both particles have the same speed $v$ at the instant in question. Take the system as a whole:

\frac{1}{2}(2.5)v^2 + \frac{1}{2}(0.5)v^2 = 2.5g\left(\frac{2}{3}\right) - 0.5g\left(\frac{4}{3}\right) - F\left(\frac{4}{3}\right)

where $F = 0.5g\cos 30° = 0.5g\cdot\frac{\sqrt{3}}{2} = \frac{\sqrt{3}\,g}{4}$ from part (a).

The left-hand side simplifies:

1.5\,v^2 = \frac{5g}{3} - \frac{2g}{3} - \frac{\sqrt{3}\,g}{4} \cdot \frac{4}{3}

1.5\,v^2 = \frac{5g}{3} - \frac{2g}{3} - \frac{\sqrt{3}\,g}{3}

1.5\,v^2 = \frac{g(5 - 2 - \sqrt{3})}{3} = \frac{g(3 - \sqrt{3})}{3}

v^2 = \frac{2g(3 - \sqrt{3})}{9}

With $g = 9.8$ :

v^2 = \frac{2 \times 9.8 \times 1.2679}{9} = \frac{24.85}{9} = 2.761

v = 1.66\text{ m s}^{-1}

Answer

v \approx 1.66\text{ m s}^{-1}

(Using $g = 10$ , the answer is $\sqrt{2(30-10\sqrt{3})/9} \approx 1.68$ m s $^{-1}$ , which matches the mark scheme.)

Final answer

v ≈ 1.66 m s⁻¹

Detailed explanation

Walkthrough

The question asks for the speed "at the instant when they are at the same vertical height", so the first step is to figure out how far each particle has moved at that instant. The energy equation cannot be set up until we know that distance.

Step 1 — Geometric condition. Let $x$ be the distance P travels down the slope. By the inextensibility of the string, Q rises the same distance $x$ vertically. P's vertical drop is $x\sin 30° = x/2$ . Setting the new vertical separation to zero:
$2 - x/2 - x = 0 \Rightarrow x = 4/3\text{ m}.$
P drops $2/3$ m; Q rises $4/3$ m. The mark scheme also lists an alternative: a time-based method using $s = \tfrac{1}{2}at^2$ , which gives the same $t$ and hence the same distance — this is the "Special Case" route and only earns 2 marks.

Step 2 — Kinetic-energy change. Both particles share the same speed (same string, same inextensibility), so the total KE gained is $\tfrac{1}{2}(2.5 + 0.5)v^2 = 1.5\,v^2$ .

Step 3 — Change in gravitational PE. P loses PE equal to $2.5g \times (2/3)$ ; Q gains PE equal to $0.5g \times (4/3)$ . The net PE change is $5g/3 - 2g/3 = g$ (with $g$ in the appropriate units). The mark scheme accepts either calculating each term separately or using the combined factor $\tfrac{3}{4}g \cdot x$ (since $2.5 \times \tfrac{1}{2} - 0.5 = 0.75 = 3/4$ ).

Step 4 — Work done against friction. From part (a), $F = 0.2 \times 2.5g\cos 30° = 0.5g\cos 30°$ . Over the distance $4/3$ m that P slides, the work done against friction is $F \cdot (4/3) = \sqrt{3}\,g/3$ .

Step 5 — Assemble the work–energy equation. KE gained = net PE lost − work done against friction:
$1.5\,v^2 = g - \frac{\sqrt{3}\,g}{3} = \frac{g(3 - \sqrt{3})}{3}.$
With $g = 9.8$ , $v^2 \approx 2.76$ , so $v \approx 1.66$ m s $^{-1}$ .

Key Takeaways

The work–energy principle for a system reads: total KE gained = total PE lost − energy dissipated (e.g., against friction).
When two particles are connected by an inextensible string, both share the same speed (and the same magnitude of acceleration) at every instant, so the two KE terms share a single $v$ .
The geometric condition "P and Q at the same height" converts a vertical-distance problem into a single equation in the unknown distance $x$ — this step is required before any energy balance can be set up.
Friction always dissipates energy: the work done against friction is $\mu R \cdot d$ , where $d$ is the distance the contact surface moves (here, the distance P slides down the plane).

Common Mistakes

Forgetting to subtract the PE gained by Q (a frequent error, since the system has two particles moving in opposite directions vertically).
Using the wrong distance in the friction work — e.g., using $2/3$ (P's vertical drop) instead of $4/3$ (P's distance along the plane). Friction is a contact force along the plane, so the relevant distance is along the plane, not vertical.
Mixing up the friction formula: $F = 0.2 \times 2.5g\cos 30°$ , not $0.2 \times 0.5g\cos 30°$ . Only P is on the rough plane.
Failing to find $x$ first and just plugging in $x = 2$ m (the initial separation) or $x = 1$ m.
Using a "time-based constant acceleration" approach: the mark scheme caps this at 2 marks because it skips the energy reasoning the question explicitly asks for ("use an energy method").

Things to Be Careful About

The string is over a pulley, so the distance moved by P along the plane equals the distance moved by Q vertically.
The mark scheme accepts the alternative of working with $y$ (the vertical distance of Q below P's starting point); the relation is $(2 - y)\sin 30° = y$ , giving $y = 2/3$ m and $2 - y = 4/3$ m. Either parameterisation works.
The mark scheme gives $v \approx 1.68$ m s $^{-1}$ using $g = 10$ ; with $g = 9.8$ the equivalent is $1.66$ m s $^{-1}$ . Both come from the same equation $v^2 = 2g(3 - \sqrt{3})/9$ .
The "use an energy method" instruction is mandatory for full credit on this part: students who find $v$ purely from $v^2 = 2as$ cap at 2 marks.

Techniques used

determine distance from geometric condition on the pulley systemcalculate change in gravitational potential energy for each particlecalculate work done against frictionapply work-energy principle to solve for the final speed

Mathematics 9709/42 — May/June 2024

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