Mathematics 9709/43 — October/November 2025

Walkthrough

The problem asks for the deceleration at $t = 7$ s. On a velocity-time graph, the acceleration at any point is equal to the gradient of the graph at that point. We first identify which line segment contains $t = 7$ s. Looking at the graph, the segment from $t = 5$ to $t = 15$ covers this time. The velocity at $t = 5$ is $8$ ms$^{-1}$ and at $t = 15$ is $-8$ ms$^{-1}$. We calculate the gradient using the formula $\frac{v_2 - v_1}{t_2 - t_1}$. This gives $\frac{-8 - 8}{15 - 5} = -1.6$ ms$^{-2}$. Acceleration is $-1.6$ ms$^{-2}$, which means the velocity is decreasing at a rate of $1.6$ ms$^{-2}$. Deceleration is defined as the positive magnitude of a negative acceleration, so the deceleration is $1.6$ ms$^{-2}$.

Key Takeaways

• The gradient of a velocity-time graph represents acceleration.
• Deceleration is the magnitude of negative acceleration (or the rate at which speed is decreasing in the direction of motion, but here simply the positive value of negative acceleration).
• Identifying the correct time interval and corresponding coordinates on the graph is crucial.

Common Mistakes

• Forgetting that deceleration is a positive quantity and writing $-1.6$ ms$^{-2}$ (though the mark scheme condones this, it is technically the acceleration, not deceleration).
• Using the wrong coordinates, for example using $(0,0)$ and $(5,8)$ instead of the segment containing $t=7$.

Things to Be Careful About

• Ensure you read the coordinates correctly from the graph. The segment from $t=5$ to $t=15$ has endpoints $(5, 8)$ and $(15, -8)$.
• Deceleration is the magnitude of negative acceleration. If acceleration is $a$, deceleration is $-a$ (when $a < 0$).

Walkthrough

The problem asks for the total distance travelled by the athlete. On a velocity-time graph, the area under the curve represents displacement. However, total distance is the sum of the absolute values of the areas between the graph and the time axis, because distance is a scalar quantity and does not account for direction.

The graph crosses the time axis at $t = 10$ s and $t = 20$ s. This creates three main triangular regions:

• Region 1: $t = 0$ to $t = 10$. The velocity is positive. The shape is a triangle with base $10$ and height $8$. Area = $\frac{1}{2} \times 10 \times 8 = 40$ m.
• Region 2: $t = 10$ to $t = 20$. The velocity is negative. The shape is a triangle with base $10$ and height $8$ (taking the absolute value of velocity). Area = $\frac{1}{2} \times 10 \times 8 = 40$ m.
• Region 3: $t = 20$ to $t = 30$. The velocity is positive. The shape is a triangle with base $10$ and height $8$. Area = $\frac{1}{2} \times 10 \times 8 = 40$ m.

Total distance = $40 + 40 + 40 = 120$ m.

Alternatively, you can notice there are six identical smaller triangles, each with base $5$ s and height $8$ ms$^{-1}$. Area of one = $\frac{1}{2} \times 5 \times 8 = 20$ m. Total distance = $6 \times 20 = 120$ m.

Key Takeaways

• The area under a velocity-time graph gives displacement.
• Total distance is the sum of the absolute areas between the graph and the time axis.
• Areas below the time axis must be treated as positive when calculating total distance.

Common Mistakes

• Calculating displacement instead of distance by taking the area below the axis as negative (which would give $40 - 40 + 40 = 40$ m).
• Misreading the coordinates of the vertices of the triangles on the graph.

Things to Be Careful About

• Ensure you are calculating total distance, not displacement. The question states the athlete runs from A to B, back to A, and finishes at B, confirming the total path length is required.
• The graph consists of straight lines, so the areas are simple triangles. Use $\frac{1}{2} \times \text{base} \times \text{height}$.

Walkthrough

First convert the maximum power to watts: $105\text{ kW} = 105000\text{ W}$. The equation $P = Fv$ relates power, driving force and speed, so at $35\text{ m s}^{-1}$ the driving force is $3000\text{ N}$.

Next write the resultant force horizontally. The driving force acts forwards and the resistance $cv$, evaluated at $v = 35$, acts backwards. Newton's second law therefore gives:

Solving this linear equation gives $c = 80$.

Key Takeaways

The power-speed relation $P = Fv$ gives the driving force produced by an engine at a given speed. The resistance here is speed-dependent, so it is $cv$, not a constant force. Newton's second law is applied by taking the resultant of all forces in the direction of motion.

Common Mistakes

• Forgetting to convert $105\text{ kW}$ to $105000\text{ W}$.
• Using $105000$ directly as the driving force instead of $\frac{105000}{35}$.
• Assuming the acceleration is zero in part (a). The mark scheme awards only the first mark if the equation is set up with $a = 0$.
• Omitting one of the three terms in Newton's second law, such as the resistance.

Things to Be Careful About

The speed at which the resistance is evaluated must be $35\text{ m s}^{-1}$ inside $cv$. Be careful with the signs: the driving force and resistance act in opposite directions, so the resultant is their difference. The final answer is exact, $c = 80$, and is clearly stated.

Walkthrough

The motorcyclist travels up the hill at steady speed, so the acceleration is zero. The driving force produced by the engine is still $\frac{105000}{v}$, because the power is maximum and power is $P = Fv$.

The resistance is now $80v$, using the value of $c$ found in part (a). The weight also has a component down the slope. Since the hill makes an angle $\theta$ with the horizontal, and $\sin\theta = \frac{3}{20}$, this component is:

Because the speed is steady, the resultant force along the slope is zero. Taking the direction up the hill as positive:

Multiply by $v$ and rearrange to obtain the quadratic:

Solving this quadratic gives one positive root and one negative root. Speed cannot be negative, so the negative root is rejected. The positive root is approximately $32.7\text{ m s}^{-1}$.

Key Takeaways

Steady speed means zero acceleration, so the resultant force is zero. The driving force from an engine at maximum power is $\frac{P}{v}$ and therefore varies with speed. On an incline, the weight contributes a component $mg\sin\theta$ along the slope, and it must be included in the force balance.

Common Mistakes

• Mixing up $\sin$ and $\cos$ when resolving the weight component. Here the component along the slope is $mg\sin\theta$.
• Forgetting the resistance term $80v$, or the weight component, or using the wrong sign for one of them.
• Not multiplying through by $v$ after forming $\frac{105000}{v} - 80v - 600 = 0$, leading to a non-quadratic equation.
• Stating the negative root without rejecting it. The mark scheme accepts only the positive speed, $32.7\text{ m s}^{-1}$.
• Using a value of $c$ other than $80$; part (b) must use $c = 80$.

Things to Be Careful About

The mark scheme allows the exact form $\frac{-15 + 5\sqrt{849}}{4}$ or the decimal $32.7\text{ m s}^{-1}$. If you write the negative root, it must be explicitly rejected. Remember that $\sin^{-1}\left(\frac{3}{20}\right)$ means $\sin\theta = \frac{3}{20}$; do not convert the angle unnecessarily. Finally, ensure all powers are in watts and all forces are in newtons.

Walkthrough

The block is pulled along a rough horizontal road by a force of 25 N acting at 36° above the horizontal. This force has a horizontal component $25\cos 36^\circ$ that pulls the block forward, and a vertical component $25\sin 36^\circ$ that lifts the block slightly. Because the block moves horizontally, it has no vertical acceleration, so the vertical forces must balance: the upward normal reaction $R$ plus the upward vertical component $25\sin 36^\circ$ equals the downward weight $4g$.

Using $g = 10$, we get $4g = 40$, so:

Note that $R$ is less than the weight (40 N) because the vertical component of the pulling force supports part of the block's weight.

The friction force is given by $F = \mu R = 0.4 \times 25.3054... = 10.1221...\text{ N}$. This friction opposes the motion.

Work done against friction is the friction force multiplied by the distance travelled:

This rounds to 1210 J to 3 significant figures, or 1215 J to 4 significant figures.

Key Takeaways

• When a force acts at an angle, resolve it into components: the vertical component affects the normal reaction, and the horizontal component does the useful work of pulling.
• The normal reaction is not always equal to the weight; it is reduced when a force has an upward vertical component.
• Work done against friction = friction force $\times$ distance, where friction force $= \mu R$.

Common Mistakes

• Using $R = 4g$ directly (ignoring the vertical component of the pulling force) — this scores no marks.
• Adding or subtracting the work done by the pulling force when only the work against friction is asked.
• Mixing up $\sin$ and $\cos$ when resolving the 25 N force.

Things to Be Careful About

• Use $g = 10\text{ m s}^{-2}$ (standard for this syllabus).
• The vertical component of the pulling force is $25\sin 36^\circ$ (opposite to the angle) and the horizontal component is $25\cos 36^\circ$ (adjacent to the angle).
• The work done against friction is a positive quantity.

Walkthrough

The work-energy principle states that the net work done on an object equals its change in kinetic energy. Here, the net work is the work done by the pulling force minus the work done against friction.

The kinetic energy at $A$ is $\frac{1}{2} \times 4 \times 7^2 = 98\text{ J}$. The kinetic energy at $B$ is $\frac{1}{2} \times 4 \times v_B^2 = 2v_B^2$. So the change in kinetic energy is $2v_B^2 - 98$.

The work done by the pulling force is its horizontal component times the distance:

The work done against friction was found in part (a) to be $1214.66...\text{ J}$.

Setting up the work-energy equation:

Solving:

Key Takeaways

• The work-energy principle links work done and kinetic energy change.
• The work done by the pulling force uses only its horizontal component ($25\cos 36^\circ \times$ distance).
• The work done against friction is subtracted because it removes energy from the block.

Common Mistakes

• Using $25 \times 120$ instead of $25\cos 36^\circ \times 120$ for the work done by the pulling force.
• Forgetting to subtract the work done against friction.
• Using the wrong sign convention (e.g., adding instead of subtracting the friction work).

Things to Be Careful About

• The work done against friction from part (a) must be used consistently.
• The kinetic energy change must be written with the correct sign: final minus initial.
• The final answer should be rounded to 3 significant figures: $25.6\text{ m s}^{-1}$.

Walkthrough

We are given the displacement of a particle as a function of time and asked for the value of $s$ when the velocity is a minimum, plus the speed at that instant.

The velocity is the rate of change of displacement, so we differentiate $s$ with respect to $t$:

The velocity is a quadratic in $t$. To find its minimum we need the derivative of $v$ with respect to $t$, which is the acceleration:

Setting this derivative to zero locates the stationary point of $v$; since $v$ is a quadratic with a positive coefficient of $t^2$, this point is a minimum. Solving $0.06t - 0.6 = 0$ gives $t = 10$.

With $t = 10$ we substitute back: into $s$ to get the displacement

and into $v$ to get the velocity $v = -5.07\text{ m s}^{-1}$. The speed is the magnitude of the velocity, so we drop the negative sign and report $5.07\text{ m s}^{-1}$. The negative velocity simply means the particle is moving in the negative direction, while the speed is a scalar, so it is always non-negative.

Key Takeaways

This question tests the kinematic link between displacement, velocity and acceleration through differentiation, and the important idea that the maximum or minimum of a function is found by setting its derivative to zero. It also reinforces the distinction between velocity (a vector, which can be negative) and speed (the magnitude of velocity, always non-negative).

Common Mistakes

• Forgetting to differentiate a second time and instead solving $v = 0$ to find the minimum velocity. The minimum of the velocity occurs where $a = 0$, not where $v = 0$.
• Treating $a = \frac{v}{t}$ as if it were the acceleration; this is not valid here because acceleration is the derivative of velocity, not velocity divided by time.
• Removing the negative sign from the velocity and saying the velocity is $5.07$; the velocity is $-5.07\text{ m s}^{-1}$ and only the speed is $5.07\text{ m s}^{-1}$.
• Reporting $s = 40.7$ instead of $s = -40.7$; the displacement is negative and must be kept negative.

Things to Be Careful About

• The mark scheme is strict: the negative displacement must be preserved, and the speed must be stated as the positive value $5.07$. Saying the speed is either $\pm 5.07$ is not acceptable.
• The time $t = 10$ must come from setting the correct two-term linear expression for $a$ equal to zero; it must follow from the correct three-term quadratic for $v$.
• The final marks in this part require the method marks to have already been earned, so keep all intermediate steps clearly shown.

Walkthrough

The direction of motion of a particle changes exactly when the velocity passes through zero. So to find this instant we set the velocity equal to zero:

This is a quadratic equation. To make the arithmetic cleaner, multiply by 100 and divide by 3:

Factorising gives

so $t = 23$ or $t = -3$. The value $t = -3$ corresponds to a time before the particle passes through $O$, and since time is taken as positive after passing $O$, we reject it and use $t = 23$.

Now substitute $t = 23$ into the acceleration expression $a = 0.06t - 0.6$:

The acceleration at the instant the direction of motion changes is therefore $0.78\text{ m s}^{-2}$.

Key Takeaways

This part tests the kinematic fact that a change of direction occurs when the velocity is zero, and then requires solving a quadratic and substituting into the derivative of the velocity. It emphasises that the acceleration at such an instant need not be zero — here the particle is momentarily at rest but still accelerating.

Common Mistakes

• Setting the acceleration equal to zero instead of the velocity. The direction of motion changes when $v = 0$, not when $a = 0$.
• Including the negative root $t = -3$ without rejecting it. Time must be positive here, so $t = -3$ is invalid.
• Forgetting that this part relies on the correct three-term quadratic for $v$ from part (a); using a truncated form of the velocity will lose the method mark.
• Substituting into the velocity instead of the acceleration when finding the required value.

Things to Be Careful About

• The mark scheme awards the method mark in this part only if it is dependent on the correct velocity expression; the value $t = 23$ must be stated in this part even if it was previously found.
• The final answer must be positive: only $a = 0.78\text{ m s}^{-2}$ is accepted. If the negative value $-0.78$ is obtained, it must be explicitly rejected.
• Make sure every step is shown, since the method is implied by a correct answer but is otherwise only recoverable from visible working.

Walkthrough

First, we determine the direction of motion. Since particle $B$ ($6\text{ kg}$) is heavier than particle $A$ ($2\text{ kg}$), $B$ will accelerate downwards and $A$ will accelerate upwards. We define the positive direction for each particle as its direction of acceleration.

For $A$, the upward forces are tension $T$ and downward weight $2g$. Newton's second law gives $T - 2g = 2a$.

For $B$, the downward forces are weight $6g$ and upward tension $T$. Newton's second law gives $6g - T = 6a$.

Adding these two equations eliminates $T$ and allows us to solve for $a$: $4g = 8a$, so $a = 0.5g = 5\text{ m s}^{-2}$.

Substituting $a = 5$ back into the first equation gives $T = 2(10) + 2(5) = 30\text{ N}$.

Key Takeaways

In a connected particles system with a smooth pulley, the tension is the same throughout the string and the magnitude of acceleration is the same for both particles. Always define a consistent positive direction for each particle relative to its expected motion.

Common Mistakes

• Assuming the heavier particle moves upwards.
• Using the wrong sign for acceleration in the equations of motion.
• Forgetting that $g$ is typically taken as $10\text{ m s}^{-2}$ in these problems unless specified otherwise.

Things to Be Careful About

Ensure that the tension $T$ is positive in the equation where it acts in the positive direction, and negative (or on the opposite side) where it acts in the negative direction. The acceleration $a$ must be taken as a positive magnitude.

Walkthrough

Particle $A$ is projected downwards at $1.2\text{ m s}^{-1}$, so particle $B$ is initially projected upwards at $1.2\text{ m s}^{-1}$. The acceleration of $B$ is $5\text{ m s}^{-2}$ downwards (from part a).

We set upwards as positive. Thus $u = 1.2$, $a = -5$, and $s = -3.2$ (since $B$ ends up $3.2\text{ m}$ below its starting point).

Substituting into $s = ut + \frac{1}{2}at^2$ gives $-3.2 = 1.2t - 2.5t^2$. Rearranging gives $25t^2 - 12t - 32 = 0$.

Solving this quadratic yields $t = \frac{6 + 2\sqrt{209}}{25} \approx 1.4\text{ s}$.

Key Takeaways

When connected particles move, their initial velocities are equal in magnitude but opposite in direction. Careful sign conventions are essential in suvat equations.

Common Mistakes

• Taking the initial velocity of $B$ as downwards instead of upwards.
• Using $a = -g$ instead of the calculated acceleration $a = -5\text{ m s}^{-2}$.
• Discarding the positive root of the quadratic equation.

Things to Be Careful About

Ensure the displacement $s$ is negative if the particle ends up below its starting position and upwards is defined as positive. The time must be positive.

Walkthrough

We break the motion of $A$ into three phases based on the forces acting on it.

In Phase 1, $A$ is projected downwards at $1.2\text{ m s}^{-1}$ but accelerates upwards at $5\text{ m s}^{-2}$. Using $v^2 = u^2 + 2as$, we find it travels $0.144\text{ m}$ downwards before momentarily stopping. This is the lowest point.

In Phase 2, $A$ accelerates upwards at $5\text{ m s}^{-2}$ for a distance of $3.2\text{ m}$ (the distance $B$ falls). Using $v^2 = u^2 + 2as$, we find its speed when $B$ hits the ground is $\sqrt{33.44} \approx 5.78\text{ m s}^{-1}$. At this point, $A$ is $6.4\text{ m}$ above the ground.

In Phase 3, $B$ hits the ground and the string goes slack. $A$ is now in free fall, moving upwards at $5.78\text{ m s}^{-1}$ with acceleration $-10\text{ m s}^{-2}$. It travels a further $1.672\text{ m}$ upwards before stopping. This is the highest point.

The total distance between the lowest and highest points is the sum of the downward distance ($0.144\text{ m}$), the distance from initial to $B$-hit ($3.2\text{ m}$), and the upward free-fall distance ($1.672\text{ m}$), giving $5.016\text{ m}$.

Key Takeaways

Connected particle problems often involve multiple phases of motion with different accelerations. Carefully track the position and velocity at each transition point.

Common Mistakes

• Forgetting that $A$ continues to move upwards after $B$ hits the ground.
• Using $a = -5\text{ m s}^{-2}$ in Phase 3 instead of $a = -10\text{ m s}^{-2}$ (free fall).
• Miscalculating the height of $A$ when $B$ hits the ground (it is $6.4\text{ m}$, not $3.2\text{ m}$).

Things to Be Careful About

The string only provides tension while it is taut. Once $B$ hits the ground, $A$ is in free fall with acceleration $g$. Ensure you use the correct acceleration for each phase.

Walkthrough

We start by choosing a positive direction. It is convenient to take P's initial direction of motion as positive. Since Q is moving directly towards P, Q's velocity is negative: $-3u$.

Before the collision, the total momentum is the sum of each particle's momentum: $m \times u$ for P and $2 \times (-3u)$ for Q.

After the collision, both particles reverse direction. P now moves with velocity $-\frac{u}{3}$. Q's new velocity is unknown, call it $v$. Because Q's direction is reversed, $v$ must be positive.

Conservation of momentum says the total momentum before equals the total momentum after. This gives an equation we solve for $v$:

Rearranging gives $v = \frac{2mu}{3} - 3u$.

Since Q's direction was reversed, $v$ must be positive. Setting $\frac{2mu}{3} - 3u > 0$ and dividing by $u$ gives $\frac{2m}{3} > 3$, so $m > 4.5$.

Key Takeaways

• Conservation of linear momentum applies to direct collisions on a smooth horizontal plane.
• Choosing and consistently applying a sign convention is essential — velocities in opposite directions get opposite signs.
• The phrase "direction of motion is reversed" translates directly into a sign condition on the velocity.

Common Mistakes

• Giving Q's initial velocity as $+3u$ instead of $-3u$ — this is a sign error. The mark scheme allows this sign error for the method mark (M1) but not for the accuracy marks.
• Forgetting to include the factor of 2 (Q's mass) in the momentum terms.
• Using trial and improvement with particular values of $m$ instead of solving the inequality — the mark scheme awards no accuracy mark for this unless the boundary $m = 4.5$ is also shown.
• Not explicitly stating $v > 0$ before forming the inequality.

Things to Be Careful About

• The mark scheme requires the inequality to be derived correctly: $v > 0 \Rightarrow m > 4.5$. Simply stating $m > 4.5$ without showing $v > 0$ may lose the final mark.
• If $g$ is introduced with either mass, the method mark is allowed but no accuracy marks follow.
• The final accuracy mark depends on the previous accuracy mark (it is a "correct working only" mark).