Mathematics 9709/41 — October/November 2025

Walkthrough

Step 1: Interpret the diagram and determine angles.

The diagram shows four forces acting at a single point. We need to find the angle each force makes with the horizontal axis to resolve them into components.

• The $45\text{ N}$ force acts vertically upward, so it is purely in the $Y$-direction.
• The $28\text{ N}$ force is $35^\circ$ above the positive $x$-axis (first quadrant).
• The $72\text{ N}$ force is $50^\circ$ below the positive $x$-axis (fourth quadrant).
• The $35\text{ N}$ force is in the third quadrant. The diagram shows $70^\circ$ between the $35\text{ N}$ and $72\text{ N}$ forces. Since the $72\text{ N}$ force is $50^\circ$ below the positive $x$-axis, the angle from the negative $x$-axis to the $35\text{ N}$ force is $180^\circ - 50^\circ - 70^\circ = 60^\circ$ below the negative $x$-axis.

Step 2: Resolve into vertical components ($Y$-direction).

Using the convention that upward is positive:

• $45\text{ N}$: $+45$
• $28\text{ N}$: $+28\sin 35^\circ$ (upward component)
• $72\text{ N}$: $-72\sin 50^\circ$ (downward component)
• $35\text{ N}$: $-35\sin 60^\circ$ (downward component)

Sum: $Y = 45 + 28\sin 35^\circ - 72\sin 50^\circ - 35\sin 60^\circ = -24.4059\text{ N}$

Step 3: Resolve into horizontal components ($X$-direction).

Using the convention that rightward is positive:

• $45\text{ N}$: $0$ (no horizontal component)
• $28\text{ N}$: $+28\cos 35^\circ$ (rightward)
• $72\text{ N}$: $+72\cos 50^\circ$ (rightward)
• $35\text{ N}$: $-35\cos 60^\circ$ (leftward)

Sum: $X = 28\cos 35^\circ + 72\cos 50^\circ - 35\cos 60^\circ = 51.7169\text{ N}$

Step 4: Calculate the magnitude.

Step 5: Calculate the direction.

Since $X$ is positive (rightward) and $Y$ is negative (downward), the resultant points into the fourth quadrant: $25.3^\circ$ below the positive $x$-axis.

Key Takeaways

• When resolving forces at a point, always determine the angle each force makes with the horizontal (or vertical) axis from the diagram.
• Use a consistent sign convention (e.g., rightward and upward positive) and apply it uniformly to all components.
• The magnitude of the resultant is found using Pythagoras' theorem: $R = \sqrt{X^2 + Y^2}$.
• The direction is found using $\tan^{-1}\left(\frac{|Y|}{|X|}\right)$, and the quadrant must be determined from the signs of $X$ and $Y$.
• When angles are given between forces rather than directly with the axis, use geometry to find the required angles.

Common Mistakes

• Using the wrong angle for the $35\text{ N}$ force: the diagram shows $70^\circ$ between the $35\text{ N}$ and $72\text{ N}$ forces, not between the $35\text{ N}$ force and the horizontal. The correct angle with the negative horizontal is $60^\circ$.
• Sign errors when resolving components: forces pointing left or down must have negative signs in the respective component sums.
• Mixing up sine and cosine: the component along the axis the angle is measured from uses cosine, and the perpendicular component uses sine.
• Giving only an angle without specifying the reference direction (e.g., "below the positive $x$-direction" or a bearing).

Things to Be Careful About

• The $70^\circ$ angle in the diagram is between the $35\text{ N}$ and $72\text{ N}$ forces, not with the horizontal. Calculate: $180^\circ - 50^\circ - 70^\circ = 60^\circ$ below the negative $x$-axis.
• The direction must be stated clearly with a reference (e.g., "$25.3^\circ$ below the positive $x$-direction" or bearing $115^\circ$). Giving just an angle is insufficient.
• Ensure the final answer is given to an appropriate number of significant figures (3 s.f. is standard for this type of problem).

Walkthrough

The problem asks us to find the mass $m$ of particle $A$ using an energy method. The system is released from rest, and we are given the speed of $A$ just before it hits the ground. Since the string is light and inextensible, both particles move the same distance ($0.75 \text{ m}$) and have the same speed ($2 \text{ m s}^{-1}$) at any given time.

We apply the work-energy principle, which states that the total loss in potential energy equals the total gain in kinetic energy plus the work done against non-conservative forces (like resistance).

First, we calculate the loss in gravitational potential energy (PE) of particle $A$. As it falls $0.75 \text{ m}$ vertically, the loss is $m \times g \times 0.75$. Using $g = 10 \text{ m s}^{-2}$, this is $7.5m$.

Next, we calculate the gain in PE of particle $B$. As $B$ moves $0.75 \text{ m}$ up the plane, its vertical height increases by $0.75 \times \sin(\theta)$. Since $\sin(\theta) = 0.6$, the vertical rise is $0.45 \text{ m}$. The gain in PE is $3 \times 10 \times 0.45 = 13.5 \text{ J}$.

Then, we find the gain in kinetic energy (KE) of the entire system. Both masses reach a speed of $2 \text{ m s}^{-1}$ from rest. The KE gained by $B$ is $\frac{1}{2} \times 3 \times 2^2 = 6 \text{ J}$, and the KE gained by $A$ is $\frac{1}{2} \times m \times 2^2 = 2m \text{ J}$. Total KE gain is $6 + 2m$.

We also account for the work done against the constant resistance force. The resistance is $\frac{10}{3} \text{ N}$ acting over a distance of $0.75 \text{ m}$, giving work done $= \frac{10}{3} \times 0.75 = 2.5 \text{ J}$.

Equating the energy loss to the energy gains and work done:

Solving this linear equation yields $5.5m = 22$, so $m = 4$.

Key Takeaways

• The work-energy principle is a powerful alternative to Newton's second law for systems with multiple moving parts and resistive forces.
• When a particle moves along an inclined plane, its change in gravitational potential energy depends only on its vertical height change, which is $d \sin \theta$ where $d$ is the distance moved along the slope.
• Connected particles moving on an inextensible string share the same speed and distance moved.

Common Mistakes

• Forgetting to include the work done against resistance in the energy balance equation.
• Calculating the PE gain of $B$ using the distance moved along the slope ($0.75 \text{ m}$) instead of the vertical height change ($0.75 \times 0.6 = 0.45 \text{ m}$).
• Using $g = 9.8 \text{ m s}^{-2}$ instead of $g = 10 \text{ m s}^{-2}$, which is standard for this syllabus unless specified otherwise (and leads to an incorrect non-integer answer).
• Forgetting that both particles gain kinetic energy, not just the one that is falling.

Things to Be Careful About

• Ensure all energy terms are in consistent units (joules for energy and work, kg for mass, m/s for speed, m/s² for $g$).
• The resistance force acts opposite to the motion of $B$, so it represents energy lost from the mechanical system (work done against resistance must be added to the energy gains side of the equation, or subtracted from the energy loss side).
• Always verify that the final value of $m$ is physically reasonable (positive mass).

Walkthrough

Step 1: Identify all forces. The particle $P$ is subject to four forces: its weight $mg$ acting vertically downwards, the applied force of $8mg$ N directed at angle $\theta^\circ$ to the line of greatest slope, the normal reaction $R$ perpendicular to the plane acting away from it, and the friction force $F$ acting up the plane since the particle is on the point of slipping downwards.

Step 2: Resolve perpendicular to the plane. The weight $mg$ can be split into two components: $mg\sin\theta$ parallel to the plane (downwards) and $mg\cos\theta$ perpendicular into the plane. The applied force $8mg$ makes angle $\theta^\circ$ with the line of greatest slope, so its component perpendicular to the plane (into the plane) is $8mg\sin\theta$. Since there is no motion perpendicular to the plane, the normal reaction balances these: $R = mg\cos\theta + 8mg\sin\theta$.

Step 3: Resolve parallel to the plane. The weight component down the plane is $mg\sin\theta$. The applied force component down the plane is $8mg\cos\theta$. Friction $F$ acts up the plane to oppose the impending downward slip. Setting the net force parallel to the plane to zero (equilibrium): $F = mg\sin\theta + 8mg\cos\theta$.

Step 4: Apply limiting friction. Since $P$ is on the point of slipping, friction is at its maximum value: $F = \mu R = 0.5R$. Substituting the expressions for $F$ and $R$:

Step 5: Solve the trigonometric equation. Dividing through by $mg$ (since $m \neq 0$ and $g \neq 0$) gives $\sin\theta + 8\cos\theta = 0.5\cos\theta + 4\sin\theta$. Rearranging yields $7.5\cos\theta = 3\sin\theta$, so $\tan\theta = 2.5$. Therefore $\theta = \arctan(2.5) \approx 68.2^\circ$.

Key Takeaways

• When resolving forces on an inclined plane, always resolve parallel and perpendicular to the plane, not horizontally and vertically.
• The weight $mg$ always splits into $mg\sin\theta$ (parallel, down the plane) and $mg\cos\theta$ (perpendicular, into the plane).
• An applied force at an angle to the line of greatest slope must itself be resolved into components parallel and perpendicular to the plane.
• "On the point of slipping" means friction is at its limiting value $F = \mu R$, and the direction of friction opposes the impending motion.

Common Mistakes

• Resolving forces horizontally and vertically instead of parallel and perpendicular to the plane, which makes the equations unnecessarily complicated.
• Getting the direction of friction wrong: since $P$ is on the point of slipping down, friction acts up the plane.
• Misidentifying which component of the $8mg$ force is parallel vs perpendicular to the plane. The component along the plane is $8mg\cos\theta$ and the component into the plane is $8mg\sin\theta$.
• Forgetting to divide by $mg$ when simplifying the equation, leading to an unsolvable expression.
• Sign errors when rearranging terms to form $\tan\theta = k$.

Things to Be Careful About

• The angle $\theta$ appears both as the inclination of the plane and as the angle the applied force makes with the line of greatest slope. These are the same variable, which is what makes the problem solvable.
• Ensure that the component of the $8mg$ force into the plane is $8mg\sin\theta$ (not $8mg\cos\theta$), since $\theta$ is measured from the line of greatest slope.
• The coefficient of friction is $0.5$, so $\mu R = 0.5R$, not $2R$.
• The final answer should be given to 3 significant figures as $68.2^\circ$.

Walkthrough

This is a multi-stage mechanics problem. There are three distinct phases, and each uses a different principle.

Phase 1: Motion of $A$ before the collision. Since the plane is smooth, there is no friction. The only force acting on $A$ along the plane is the component of its weight, $mg\sin\theta$. By Newton's second law, $F = ma$, so $mg\sin\theta = ma$, giving $a = g\sin\theta$. With $\sin\theta = 0.2$ and $g = 10$, this gives $a = 2\ \text{m s}^{-2}$. This is the acceleration down the plane.

Phase 2: Finding $A$'s speed just before the collision. $A$ is released from rest ($u = 0$) and slides for 2 seconds. Using the suvat equation $v = u + at$ gives $v = 0 + 2 \times 2 = 4\ \text{m s}^{-1}$ down the plane. This is the speed needed for the momentum calculation.

Phase 3: The collision. During the collision, the only significant forces are internal to the two-particle system, so linear momentum is conserved. We choose a positive direction (up the plane) and write the total momentum before the collision equal to the total momentum after. Before: $B$ moves up at $3.5$ (positive) and $A$ moves down at $4$ (negative). After: $B$ moves down at $0.5$ (negative) and $A$ moves up at unknown speed $u$ (positive). Solving the momentum equation gives $u = 0.8\ \text{m s}^{-1}$.

Phase 4: Distance $A$ travels up the plane after the collision. After the collision, $A$ moves up the plane at $0.8\ \text{m s}^{-1}$. The acceleration is still $2\ \text{m s}^{-2}$ down the plane, which is a deceleration of $-2\ \text{m s}^{-2}$ up the plane. $A$ comes to rest momentarily at its highest point, so $v = 0$. Using $v^2 = u^2 + 2as$ gives $s = 0.16\ \text{m}$.

Key Takeaways

• On a smooth inclined plane, the acceleration along the plane is $g\sin\theta$, independent of the mass.
• Momentum is conserved during a collision, provided we treat the two particles as an isolated system and use masses (not weights) in the equation.
• A consistent sign convention is essential: choose a positive direction and apply it to every velocity.
• The suvat equations apply to each phase of motion separately; the initial velocity of each phase comes from the final velocity of the previous phase.

Common Mistakes

• Using weight ($mg$) instead of mass ($m$) in the momentum equation. The mark scheme explicitly states "M1A0 only if using weight rather than mass."
• Sign errors in the momentum equation. The mark scheme allows sign errors for the method mark (DM1) but the final answer (A1) requires correct signs.
• Forgetting that after the collision, $A$ is still subject to the same acceleration down the plane, so it decelerates as it moves up.
• Using $g = 9.8$ instead of $g = 10$. The mark scheme uses $g \times 0.2 = 2$, i.e. $g = 10$.

Things to Be Careful About

• Choose the positive direction up the plane and be consistent throughout the momentum equation.
• The acceleration of $A$ is down the plane in both phases — before the collision it speeds $A$ up, after the collision it slows $A$ down.
• The collision is instantaneous, so $A$'s position doesn't change during it; only its velocity changes.
• At the highest point of $A$'s motion, its velocity is momentarily zero — this is why we set $v = 0$.