9709/43 Oct/Nov 2024 — Mathematics Solved Past Paper

Q14MMediumEnergy, Work and Power

An athlete has mass $m\text{ kg}$ . The athlete runs along a horizontal road against a constant resistance force of magnitude $24\text{ N}$ . The total work done by the athlete in increasing his speed from $5\text{ m s}^{-1}$ to $6\text{ m s}^{-1}$ while running a distance of $50$ metres is $1541\text{ J}$ .

Find the value of $m$ .

DifficultyMedium

Worked solution

Approach

Use the work-energy principle. The work done by the athlete is used to increase the kinetic energy and to overcome the constant resistance.

Working

Initial kinetic energy:

KE_{\text{before}} = \frac{1}{2} m (5)^2 = 12.5m

Final kinetic energy:

KE_{\text{after}} = \frac{1}{2} m (6)^2 = 18m

Increase in kinetic energy:

18m - 12.5m = 5.5m

Work done against resistance:

W_{\text{resistance}} = 50 \times 24 = 1200

Work-energy equation:

5.5m + 1200 = 1541

Solve for $m$ :

5.5m = 341

m = \frac{341}{5.5} = 62

Answer

m = 62

Final answer

m = 62

Detailed explanation

Walkthrough

We are given the total work done by the athlete, which is 1541 J. This work is not all used to speed up: part of it is used to overcome the constant resistance of 24 N over 50 m. So the work-energy equation must include both the increase in kinetic energy and the work done against resistance.

First find the kinetic energy before and after. The kinetic energy formula is $KE = \frac{1}{2}mv^2$ . Before the speed increase, $v = 5$ , so $KE_{\text{before}} = \frac{1}{2}m(5^2) = 12.5m$ . After the speed increase, $v = 6$ , so $KE_{\text{after}} = \frac{1}{2}m(6^2) = 18m$ . The increase is therefore $18m - 12.5m = 5.5m$ .

Next, the work done against the constant resistance is force times distance: $24 \times 50 = 1200$ J.

Now apply the work-energy principle:

\text{total work done by athlete} = \text{increase in kinetic energy} + \text{work done against resistance}

So

1541 = 5.5m + 1200

Subtract 1200 from both sides:

5.5m = 341

Divide by 5.5:

m = \frac{341}{5.5} = 62

The athlete's mass is 62 kg.

An alternative method using constant acceleration would give the same result: from $6^2 = 5^2 + 2a(50)$ , $a = 0.11$ , and then the driving force is $24 + 0.11m$ , so the work done is $50(24 + 0.11m) = 1541$ , which again gives $m = 62$ .

Key Takeaways

Work done by a force can be split into useful energy change and energy lost to resistance.
Kinetic energy depends on $v^2$ , so the change in KE is $\frac{1}{2}m(v_2^2 - v_1^2)$ , not $\frac{1}{2}m(v_2 - v_1)^2$ .
Work done against a constant resistance is $Fd$ .
The work-energy equation is a powerful way to relate forces, distances and speeds without needing time.

Common Mistakes

Using $\frac{1}{2}m(6-5)^2$ for the change in kinetic energy. This is incorrect because kinetic energy is proportional to the square of speed, not to the square of the speed change.
Forgetting to include the 1200 J of work done against resistance, or subtracting it instead of adding it.
Writing the resistance work with an incorrect mass term, such as $(m-24) \times 50$ .
Not showing a dimensionally correct work-energy equation; the marks require all four relevant terms.
Giving an unsupported final answer; the final value should follow from a clearly formed equation.

Things to Be Careful About

Keep units consistent: mass in kg, speed in m/s, distance in m, work in J.
The total work done by the athlete is the input energy, so it equals the sum of the useful KE gain and the work against resistance.
The resistance force acts opposite to motion, so its work is positive in the energy-loss term when we write total work input = useful gain + loss.
If using the constant-acceleration alternative, make sure the acceleration is obtained from $v^2 = u^2 + 2as$ and that the driving force is $R + ma$ .

Techniques used

compute kinetic energy before and after a speed changecalculate work done against a constant resistanceapply the work-energy principlesolve the resulting linear equation

Approach

Set up a coordinate system with the positive $x$ -axis to the right and the positive $y$ -axis upward, matching the diagram. Resolve each of the four given forces into horizontal and vertical components. For equilibrium, the additional force must be equal in magnitude and opposite in direction to the resultant of the four forces.

Working

Resolving the four forces:

Force 1: $12\text{ N}$ at $30^\circ$ to the positive $y$ -axis (first quadrant)

\begin{aligned} F_{1x} &= 12\sin 30^\circ = 6 \text{ N} \\ F_{1y} &= 12\cos 30^\circ = 6\sqrt{3} \text{ N} \end{aligned}

Force 2: $24\text{ N}$ along the positive $x$ -axis

\begin{aligned} F_{2x} &= 24 \text{ N} \\ F_{2y} &= 0 \text{ N} \end{aligned}

Force 3: $8\text{ N}$ at $30^\circ$ below the positive $x$ -axis (fourth quadrant)

\begin{aligned} F_{3x} &= 8\cos 30^\circ = 4\sqrt{3} \text{ N} \\ F_{3y} &= -8\sin 30^\circ = -4 \text{ N} \end{aligned}

Force 4: $16\text{ N}$ at $30^\circ$ below the negative $x$ -axis (third quadrant)

\begin{aligned} F_{4x} &= -16\cos 30^\circ = -8\sqrt{3} \text{ N} \\ F_{4y} &= -16\sin 30^\circ = -8 \text{ N} \end{aligned}

Summing the components:

Total horizontal component:

\begin{aligned} \Sigma F_x &= 6 + 24 + 4\sqrt{3} - 8\sqrt{3} \\ &= 30 - 4\sqrt{3} \\ &\approx 23.07 \text{ N} \end{aligned}

Total vertical component:

\begin{aligned} \Sigma F_y &= 6\sqrt{3} + 0 - 4 - 8 \\ &= 6\sqrt{3} - 12 \\ &\approx -1.607 \text{ N} \end{aligned}

Equilibrium condition:

Let the additional force be $\mathbf{R} = (R_x, R_y)$ . For equilibrium:

\Sigma F_x + R_x = 0 \quad \Rightarrow \quad R_x = -(30 - 4\sqrt{3}) = 4\sqrt{3} - 30 \approx -23.07 \text{ N}

\Sigma F_y + R_y = 0 \quad \Rightarrow \quad R_y = -(6\sqrt{3} - 12) = 12 - 6\sqrt{3} \approx 1.607 \text{ N}

Magnitude of $\mathbf{R}$ :

\begin{aligned} |\mathbf{R}| &= \sqrt{R_x^2 + R_y^2} \\ &= \sqrt{(30 - 4\sqrt{3})^2 + (6\sqrt{3} - 12)^2} \\ &= \sqrt{(23.07)^2 + (-1.607)^2} \\ &= \sqrt{532.2 + 2.58} \\ &= \sqrt{534.8} \\ &\approx 23.1 \text{ N} \end{aligned}

Direction of $\mathbf{R}$ :

Since $R_x < 0$ and $R_y > 0$ , the force lies in the second quadrant (above the negative $x$ -axis). The angle $\theta$ above the negative $x$ -axis is:

\begin{aligned} \theta &= \tan^{-1}\left(\frac{|R_y|}{|R_x|}\right) \\ &= \tan^{-1}\left(\frac{12 - 6\sqrt{3}}{30 - 4\sqrt{3}}\right) \\ &= \tan^{-1}\left(\frac{1.607}{23.07}\right) \\ &\approx 3.99^\circ \end{aligned}

Answer

The additional force has magnitude $23.1\text{ N}$ and acts at $3.99^\circ$ above the negative $x$ -axis (i.e., $176.0^\circ$ measured anticlockwise from the positive $x$ -axis).

Final answer

23.1 N at 3.99° above the negative x-axis

Detailed explanation

Walkthrough

Step 1: Set up a coordinate system.
We choose the positive $x$ -axis pointing right and the positive $y$ -axis pointing up, matching the dashed axes in the diagram. This gives us a consistent way to assign signs to each component.

Step 2: Resolve each force into components.
Each force is broken into horizontal ( $x$ ) and vertical ( $y$ ) parts using sine and cosine of the given angle. The key is to use the correct trigonometric ratio for each direction and to assign the correct sign based on which quadrant the force points into.

The $12\text{ N}$ force is $30^\circ$ from the vertical, so its horizontal component is $12\sin 30^\circ$ (positive, pointing right) and its vertical component is $12\cos 30^\circ$ (positive, pointing up).
The $24\text{ N}$ force is purely horizontal, so it contributes $24\text{ N}$ to $x$ and $0$ to $y$ .
The $8\text{ N}$ force is $30^\circ$ below the positive $x$ -axis, so its horizontal component is $8\cos 30^\circ$ (positive) and its vertical component is $-8\sin 30^\circ$ (negative, pointing down).
The $16\text{ N}$ force is $30^\circ$ below the negative $x$ -axis, so both components are negative: horizontal is $-16\cos 30^\circ$ and vertical is $-16\sin 30^\circ$ .

Step 3: Sum the components.
Add all horizontal components together to get the net horizontal force, and do the same for the vertical components. This gives the resultant of the four given forces.

Step 4: Apply the equilibrium condition.
For the system to be in equilibrium, the vector sum of all forces must be zero. The additional force must therefore be equal in magnitude and opposite in direction to the resultant of the four given forces. This means we negate both components of the resultant.

Step 5: Calculate the magnitude.
Use Pythagoras' theorem: $|\mathbf{R}| = \sqrt{R_x^2 + R_y^2}$ . Substitute the numerical values and compute.

Step 6: Determine the direction.
Use $\tan^{-1}(|R_y|/|R_x|)$ to find the acute angle the force makes with the horizontal. Then use the signs of $R_x$ and $R_y$ to place the force in the correct quadrant. Since $R_x < 0$ and $R_y > 0$ , the force is in the second quadrant, so the angle is measured above the negative $x$ -axis.

Key Takeaways

Resolving forces into perpendicular components is the standard method for handling multiple coplanar forces.
For equilibrium, the equilibrant force is the negative of the resultant: $\mathbf{R}_{\text{eq}} = -\sum \mathbf{F}_i$ .
The signs of the components determine the quadrant and must be used carefully when finding the direction angle.
Using $\tan^{-1}(|R_y|/|R_x|)$ gives the acute angle; the quadrant is determined separately from the component signs.

Common Mistakes

Using sine and cosine in the wrong places for a force given at an angle to the vertical. If the angle is to the vertical, the horizontal component uses sine and the vertical uses cosine.
Forgetting the signs of components. Forces in the third and fourth quadrants have negative components that must be included correctly.
Writing $\sin^{-1}(R_y/R_x)$ instead of $\tan^{-1}(R_y/R_x)$ to find the angle. The mark scheme explicitly warns that $\sin^{-1}((6\sqrt{3}-12)/(30-4\sqrt{3}))$ gives $4.0^\circ$ but scores M0A0 because it is the wrong method.
Not specifying the direction clearly. Simply stating $4^\circ$ without indicating the reference axis or quadrant loses the final mark.

Things to Be Careful About

The mark scheme allows a consistent sin/cos mix (e.g., swapping which force uses sin and which uses cos for a given angle) as long as it is applied consistently, but does not accept "forces to the left = forces to the right" without subsequent correction.
When using $\cos^{-1}$ or $\sin^{-1}$ to find $\theta$ , the magnitude $F$ must be given to several significant figures; otherwise, the angle will be wrong.
The angle $3.99^\circ$ must not be rounded to $4^\circ$ alone; the mark scheme allows $4.0^\circ$ but not simply $4^\circ$ .
Direction can be expressed in multiple equivalent ways: "above the negative $x$ -axis", "north of west", $176.0^\circ$ from the positive $x$ -axis, or a bearing of $274.0^\circ$ . Any clear, unambiguous description is acceptable.

Techniques used

resolve forces into horizontal and vertical componentssum components to find resultantapply equilibrium conditioncalculate magnitude using Pythagorasdetermine direction using inverse tangent

Approach

Since the car moves at constant speed, its acceleration is zero, so the resultant force along the slope is zero. Resolve the weight into a component down the slope, add the resistance, and equate the driving force to the total opposing force. Then use $P = Fv$ .

Working

Take $g = 10\text{ m s}^{-2}$ . The slope is inclined at angle $\theta$ where $\sin\theta = 0.08$ .

Weight component down the slope:

1600g\sin\theta = 1600 \times 10 \times 0.08 = 1280\text{ N}

At constant speed, $a = 0$ , so:

DF - 240 - 1280 = 0

DF = 240 + 1280 = 1520\text{ N}

Power:

P = DF \times v = 1520 \times 32 = 48640\text{ W}

Answer

P = 48640\text{ W} = 48.64\text{ kW}

Final answer

48640 W

Detailed explanation

Walkthrough

Start by listing the forces acting on the car along the slope: the engine's driving force $DF$ acts up the slope, the resistance of $240\text{ N}$ acts down the slope, and the component of the car's weight down the slope is $mg\sin\theta$ . The normal reaction acts perpendicular to the slope and does not affect motion along the slope.

The slope angle is given by $\sin\theta = 0.08$ , so the weight component is

1600 \times 10 \times 0.08 = 1280\text{ N}.

Because the car travels at constant speed, its acceleration is zero. By Newton's first law, the resultant force along the slope must be zero, so the driving force must balance the resistance plus the weight component:

DF = 240 + 1280 = 1520\text{ N}.

Finally, power is the rate at which the driving force does work, so

P = Fv = 1520 \times 32 = 48640\text{ W}.

Key Takeaways

This part combines two ideas: equilibrium when acceleration is zero, and the power formula $P = Fv$ . It also shows the importance of resolving the weight into a component along an inclined plane.

Common Mistakes

Forgetting the weight component $mg\sin\theta$ and using only the $240\text{ N}$ resistance.
Using $\sin 0.08$ instead of using $\sin\theta = 0.08$ ; the angle is defined by its sine, so the component is $mg \times 0.08$ .
Omitting $g$ when calculating the weight component.
Leaving the answer in watts when the mark scheme also allows kilowatts, but without stating units if using kW.

Things to Be Careful About

Use $g = 10\text{ m s}^{-2}$ , as is standard in this syllabus. The expression $\sin^{-1}0.08$ gives the angle, but you do not need to find the angle itself; you only need $\sin\theta = 0.08$ . Keep the calculation exact where possible; if you round the power too early, part (b) may be affected.

Techniques used

resolve forces parallel to the slopeapply the equilibrium condition for constant speedcompute power using P = Fv

(b)

Find the acceleration of the car when its speed is $24\text{ m s}^{-1}$ and the engine is working at $95\%$ of the power found in (a).

3M Similar

DifficultyMedium

Worked solution

Approach

The engine now works at 95% of the power found in part (a). Use $P = Fv$ to find the new driving force at $24\text{ m s}^{-1}$ , then apply Newton's second law along the slope to find the acceleration.

Working

Engine power:

P = 0.95 \times 48640 = 46208\text{ W}

Driving force at $v = 24\text{ m s}^{-1}$ :

DF = \frac{P}{v} = \frac{46208}{24} = \frac{5776}{3}\text{ N} \approx 1925.33\text{ N}

Newton's second law up the slope:

DF - 240 - 1600g\sin\theta = 1600a

\frac{5776}{3} - 240 - 1280 = 1600a

Simplify the left-hand side:

\frac{5776}{3} - 1520 = \frac{5776 - 4560}{3} = \frac{1216}{3}\text{ N}

Therefore:

1600a = \frac{1216}{3}

a = \frac{1216}{4800} = \frac{19}{75}\text{ m s}^{-2} \approx 0.253\text{ m s}^{-2}

Answer

a = 0.253\text{ m s}^{-2}\quad \left(\text{or } \frac{19}{75}\text{ m s}^{-2}\right)

Final answer

0.253 m/s^2

Detailed explanation

Walkthrough

The engine now supplies only 95% of the power from part (a), so the new power is

0.95 \times 48640 = 46208\text{ W}.

At a speed of $24\text{ m s}^{-1}$ , the driving force is found from $P = Fv$ :

DF = \frac{46208}{24} = \frac{5776}{3}\text{ N} \approx 1925.33\text{ N}.

Now the car is accelerating, so the forces are not balanced. Apply Newton's second law along the slope, taking the direction up the slope as positive:

DF - 240 - 1600g\sin\theta = 1600a.

Substitute the values:

\frac{5776}{3} - 240 - 1280 = 1600a.

The left-hand side simplifies to $\frac{1216}{3}\text{ N}$ , so

a = \frac{1216}{4800} = \frac{19}{75}\text{ m s}^{-2} \approx 0.253\text{ m s}^{-2}.

Key Takeaways

This part shows how power, driving force and acceleration are linked. Once the driving force is known from $P = Fv$ , Newton's second law gives the acceleration. It also reinforces that on an incline the weight component $mg\sin\theta$ always acts down the slope.

Common Mistakes

Using the full power from part (a) instead of 95% of it.
Forgetting the weight component or the resistance when applying Newton's second law.
Using the wrong speed: the driving force must be calculated at $24\text{ m s}^{-1}$ , not at $32\text{ m s}^{-1}$ .
Rounding the power to $48600\text{ W}$ and then using it in part (b) without following through correctly; the mark scheme allows this only if the subsequent acceleration is consistent.

Things to Be Careful About

The mark scheme allows follow-through from part (a), so use your part (a) power unless told otherwise. Make sure the equation is dimensionally correct: every term must be a force in newtons. The final answer is $0.253\text{ m s}^{-2}$ ; an answer of $0.25$ is not accepted because it is rounded too much. Keep the exact fraction $\frac{19}{75}$ if you want an exact answer.

Techniques used

find driving force from power and speedapply Newton's second law along the slopesolve for acceleration

Q46MMediumMomentumEnergy, Work and Power

Two particles, $A$ and $B$ , of masses $3\text{ kg}$ and $6\text{ kg}$ respectively, lie on a smooth horizontal plane. Initially, $B$ is at rest and $A$ is moving towards $B$ with speed $8\text{ m s}^{-1}$ . After $A$ and $B$ collide, $A$ moves with speed $2\text{ m s}^{-1}$ .

Find the greater of the two possible total losses of kinetic energy due to the collision.

DifficultyMedium

Worked solution

Approach

Use conservation of linear momentum in the direction of $A$ 's initial motion. Since the speed of $A$ after the collision is $2\ \text{m s}^{-1}$ , $A$ may either continue in the same direction or rebound, so the two possible velocities of $A$ are $v_A = +2$ and $v_A = -2$ . For each case, use conservation of momentum to find $B$ 's velocity, then calculate the kinetic energy before and after the collision and find the loss. Finally choose the greater loss.

Working

Take the positive direction to be $A$ 's initial direction of motion.

Total momentum before the collision:

(3)(8)+(6)(0)=24\ \text{kg m s}^{-1}

Case 1: $A$ continues in the same direction, $v_A=+2$

24=3(2)+6v_B

24=6+6v_B \quad\Rightarrow\quad v_B=3\ \text{m s}^{-1}

Case 2: $A$ rebounds, $v_A=-2$

24=3(-2)+6v_B

24=-6+6v_B \quad\Rightarrow\quad v_B=5\ \text{m s}^{-1}

So the two possible outcomes are $(v_A,v_B)=(2,3)$ and $(-2,5)$ .

The initial kinetic energy is

KE_{\text{initial}}=\frac{1}{2}(3)(8^2)=96\ \text{J}

For $(v_A,v_B)=(2,3)$ :

KE_{\text{final}}=\frac{1}{2}(3)(2^2)+\frac{1}{2}(6)(3^2)=6+27=33\ \text{J}

\text{Loss}=96-33=63\ \text{J}

For $(v_A,v_B)=(-2,5)$ :

KE_{\text{final}}=\frac{1}{2}(3)(2^2)+\frac{1}{2}(6)(5^2)=6+75=81\ \text{J}

\text{Loss}=96-81=15\ \text{J}

The greater of the two possible losses is

63\ \text{J}

Answer

63\ \text{J}

Final answer

63 J

Detailed explanation

Walkthrough

The plane is smooth, so no external horizontal force acts during the collision. We can therefore use conservation of linear momentum. However, momentum is a vector, so the direction of $A$ after impact matters. We only know $A$ 's speed after the collision is $2\ \text{m s}^{-1}$ . If we choose the positive direction to be the direction in which $A$ was originally moving, then the possible velocities of $A$ are $+2$ and $-2$ .

Start with the total momentum before the collision. $B$ is at rest, so the initial momentum is entirely due to $A$ :

(3)(8)+(6)(0)=24\ \text{kg m s}^{-1}

For the first possibility, $A$ continues forwards with $v_A=+2$ . Substituting into momentum conservation gives $v_B=3$ ; for the second possibility, $A$ rebounds with $v_A=-2$ , giving $v_B=5$ .

Next, kinetic energy. Initially only $A$ is moving:

KE_{\text{initial}}=\frac{1}{2}(3)(8^2)=96\ \text{J}

After the collision, both particles may be moving, so add their kinetic energies. Use the squares of their speeds. For the $(2,3)$ outcome the final KE is $33\ \text{J}$ , so the loss is $63\ \text{J}$ . For the $(-2,5)$ outcome the final KE is $81\ \text{J}$ , so the loss is $15\ \text{J}$ . The greater loss is therefore $63\ \text{J}$ .

Key Takeaways

Momentum is conserved during a collision as long as no external horizontal force acts.
Momentum is a vector: use signed velocities, not merely speeds, in the momentum equation.
Kinetic energy is a scalar and uses speed squared, so the direction does not affect the energy formula, but the possible speed values differ.
When a problem says a particle's speed after collision, the direction may be unknown; consider both directions.

Common Mistakes

Using $+2$ for $A$ 's velocity and never considering $v_A=-2$ . If both outcomes are found, the mark scheme requires both speeds to be correct and the greater loss to be identified.
Writing the momentum equation with speeds instead of signed velocities, especially in the rebound case.
Forgetting that $B$ is initially at rest, so its initial momentum is zero.
Computing the kinetic energy loss as only one particle's change instead of subtracting the total final kinetic energy from the total initial kinetic energy.
Giving the smaller loss, $15\ \text{J}$ , as the answer instead of $63\ \text{J}$ .

Things to Be Careful About

Clearly state a positive direction before applying momentum conservation.
In the rebound case, $v_A=-2$ ; omitting the minus sign gives $v_B=3$ instead of $5$ and misses the other possible outcome.
Kinetic energy after the collision must include both particles: $\frac{1}{2}m_Av_A^2+\frac{1}{2}m_Bv_B^2$ .
All speeds are in $\text{m s}^{-1}$ , masses in $\text{kg}$ , so energy is in joules ( $\text{J}$ ).
The marking scheme allows a solution that finds only one possible loss, but if both possible losses are found, both must be correct and the greater loss must be selected.

Techniques used

Apply conservation of linear momentum in one dimensionConsider both possible post-collision directions of ACompute total kinetic energy before and after the collisionCompare the kinetic-energy losses and select the greater

Approach

The particle is on the point of moving, so it is in limiting equilibrium. We resolve forces vertically and horizontally, then apply the condition $F = \mu R$ where $\mu = 0.5$ .

Working

Identify the forces acting on the particle:

Weight: $W = 12g = 12 \times 10 = 120\text{ N}$ acting vertically downwards
Normal reaction: $R$ acting vertically upwards
Tension: $T$ acting at $30^\circ$ above the horizontal
Friction: $F$ acting horizontally opposite to the direction of motion

Resolve vertically (upwards positive):

\begin{aligned} R + T\sin 30^\circ &= 12g \\ R + 0.5T &= 120 \\ R &= 120 - 0.5T \end{aligned}

Resolve horizontally (in the direction of impending motion):

F = T\cos 30^\circ = \frac{\sqrt{3}}{2}T

Apply the limiting friction condition $F = \mu R$ :

\frac{\sqrt{3}}{2}T = 0.5(120 - 0.5T)

\frac{\sqrt{3}}{2}T = 60 - 0.25T

T\left(\frac{\sqrt{3}}{2} + 0.25\right) = 60

T = \frac{60}{\frac{\sqrt{3}}{2} + 0.25}

T = \frac{60}{0.86603 + 0.25} = \frac{60}{1.11603}

T = 53.7622\ldots

T \approx 53.8\text{ N}

Answer

T = 53.8\text{ N}

Final answer

T = 53.8 N

Detailed explanation

Walkthrough

The particle is on the point of moving, which means it is in limiting equilibrium. At this point, the friction force is at its maximum value and equals $F = \mu R$ .

Step 1: Identify all forces. The particle experiences four forces: its weight $W = 12g$ acting downward, the normal reaction $R$ from the plane acting upward, the tension $T$ in the string at $30^\circ$ above the horizontal, and the friction force $F$ acting horizontally opposing the impending motion. The diagram shows the tension pulling at $30^\circ$ above the horizontal plane.

Step 2: Resolve vertically. Taking upward as positive, the upward forces are $R$ and the vertical component of tension $T\sin 30^\circ$ . The downward force is the weight $12g = 120\text{ N}$ (using $g = 10\text{ m s}^{-2}$ ). Since there is no vertical acceleration:

R + T\sin 30^\circ = 120

This gives $R = 120 - 0.5T$ . Note that the vertical component of tension reduces the normal reaction because it lifts the particle slightly off the plane.

Step 3: Resolve horizontally. The horizontal component of tension is $T\cos 30^\circ = \frac{\sqrt{3}}{2}T$ , and this is balanced by friction $F$ since the particle is in equilibrium:

F = T\cos 30^\circ = \frac{\sqrt{3}}{2}T

Step 4: Apply the limiting friction condition. At the point of moving, $F = \mu R = 0.5R$ . Substituting the expressions from Steps 2 and 3:

\frac{\sqrt{3}}{2}T = 0.5(120 - 0.5T)

Step 5: Solve for $T$ . Expand and collect terms:

\frac{\sqrt{3}}{2}T = 60 - 0.25T

T\left(\frac{\sqrt{3}}{2} + 0.25\right) = 60

T = \frac{60}{\frac{\sqrt{3}}{2} + 0.25} \approx 53.8\text{ N}

Key Takeaways

When a particle is on the point of moving, friction is at its maximum: $F = \mu R$ .
Resolving forces in two perpendicular directions gives two equations that can be solved simultaneously.
A tension at an angle to the horizontal has both a horizontal component (driving motion) and a vertical component (reducing the normal reaction and hence friction).

Common Mistakes

Forgetting that the vertical component of tension reduces the normal reaction: writing $R = 12g$ instead of $R = 12g - T\sin 30^\circ$ .
Using $g = 9.8$ when the mark scheme expects $g = 10$ (common in Cambridge papers).
Resolving incorrectly by mixing up $\sin$ and $\cos$ for the $30^\circ$ angle.
Not showing enough working to eliminate $T$ or $R$ before stating the final answer.

Things to Be Careful About

The particle is on the point of moving, not moving yet. This means acceleration is zero and $F = \mu R$ (limiting friction), not $F < \mu R$ .
Always use $g = 10\text{ m s}^{-2}$ unless otherwise stated in Cambridge A-Level questions.
The normal reaction is not simply $mg$ when there is a vertical component of an applied force. Here, $R = 12g - T\sin 30^\circ$ .
Ensure the final answer is given to an appropriate number of significant figures (3 sf here gives $53.8\text{ N}$ ).

Techniques used

resolve forces verticallyresolve forces horizontallyapply limiting friction condition F = μRsolve simultaneous equations for tension

(b)

Given instead that the particle is accelerating at $0.2\text{ m s}^{-2}$ , find the value of $T$ .

3M Similar

DifficultyMedium

Worked solution

Approach

The particle is now accelerating at $0.2\text{ m s}^{-2}$ . We apply Newton's second law horizontally, while still resolving vertically to find $R$ . The friction is still at its limiting value $F = \mu R$ since the particle is moving.

Working

Resolve vertically (same as part (a)):

\begin{aligned} R + T\sin 30^\circ &= 12g \\ R &= 120 - 0.5T \end{aligned}

Apply Newton's second law horizontally:

The resultant horizontal force equals mass times acceleration:

T\cos 30^\circ - F = ma

T\cos 30^\circ - F = 12 \times 0.2 = 2.4

Substitute $F = \mu R = 0.5R$ and $R = 120 - 0.5T$ :

T\cos 30^\circ - 0.5(120 - 0.5T) = 2.4

\frac{\sqrt{3}}{2}T - 60 + 0.25T = 2.4

T\left(\frac{\sqrt{3}}{2} + 0.25\right) = 62.4

T = \frac{62.4}{\frac{\sqrt{3}}{2} + 0.25}

T = \frac{62.4}{1.11603}

T = 55.9127\ldots

T \approx 55.9\text{ N}

Answer

T = 55.9\text{ N}

Final answer

T = 55.9 N

Detailed explanation

Walkthrough

In part (b), the particle is no longer in equilibrium — it is accelerating at $0.2\text{ m s}^{-2}$ . This means we must use Newton's second law ( $F_{\text{net}} = ma$ ) rather than the equilibrium condition.

Step 1: Resolve vertically. The vertical situation is unchanged from part (a). There is still no vertical acceleration, so:

R + T\sin 30^\circ = 120

R = 120 - 0.5T

Step 2: Apply Newton's second law horizontally. The horizontal forces are the driving component $T\cos 30^\circ$ (forward) and friction $F$ (backward). The resultant equals $ma$ :

T\cos 30^\circ - F = 12 \times 0.2 = 2.4

Step 3: Substitute the friction expression. Since the particle is moving, friction is at its limiting value $F = \mu R = 0.5R$ . Substituting $R = 120 - 0.5T$ :

F = 0.5(120 - 0.5T) = 60 - 0.25T

Step 4: Form and solve the equation. Substitute into the N2L equation:

\frac{\sqrt{3}}{2}T - (60 - 0.25T) = 2.4

\frac{\sqrt{3}}{2}T - 60 + 0.25T = 2.4

T\left(\frac{\sqrt{3}}{2} + 0.25\right) = 62.4

T = \frac{62.4}{1.11603} \approx 55.9\text{ N}

Key Takeaways

When a particle is accelerating, use $F_{\text{net}} = ma$ instead of equilibrium equations.
The vertical resolution (finding $R$ ) is often the same as in the equilibrium case, even when horizontal motion is accelerating.
The friction force is still $F = \mu R$ when the particle is moving (kinetic friction at limiting value).
The tension must be larger than in part (a) because it must both overcome friction and provide the net force for acceleration.

Common Mistakes

Using the equilibrium condition $T\cos 30^\circ = F$ instead of $T\cos 30^\circ - F = ma$ .
Using the value of $T$ from part (a) in part (b) — each part must be solved independently.
Forgetting that the vertical component of tension still reduces the normal reaction.
Sign errors when subtracting friction from the driving force.

Things to Be Careful About

The particle is accelerating, so $F_{\text{net}} = ma \neq 0$ . This is the key difference from part (a).
Do not use $T$ from part (a); the tension is different because the particle is now accelerating.
The friction is still $F = \mu R$ because the particle is moving across the rough surface.
The answer $T = 55.9\text{ N}$ is larger than in part (a) ( $53.8\text{ N}$ ), which makes physical sense: more tension is needed to accelerate the particle.

Techniques used

apply Newton's second law horizontallyresolve forces vertically for normal reactionsubstitute limiting friction expressionsolve equation for tension

Q6MediumKinematics of Motion in a Straight Line

A particle moves in a straight line. It starts from rest, at time $t = 0$ , and accelerates at $0.6t\text{ m s}^{-2}$ for $4\text{ s}$ , reaching a speed of $V\text{ m s}^{-1}$ . The particle then travels at $V\text{ m s}^{-1}$ for $11\text{ s}$ , and finally slows down, with constant deceleration, stopping after a further $5\text{ s}$ .

(a)

Show that $V = 4.8$ .

1M Similar

(b)

Sketch a velocity-time graph for the motion.

3M Similar

DifficultyMedium-Easy

Worked solution

Approach

The motion consists of three distinct phases. We sketch the velocity-time graph by identifying the shape and key coordinates for each phase.

Working

Phase 1 ( $0 \le t \le 4$ ):
The velocity is $v = 0.3t^2$ . This is a quadratic curve starting at $(0, 0)$ and ending at $(4, 4.8)$ , with increasing gradient.

Phase 2 ( $4 < t \le 15$ ):
The particle travels at constant speed $V = 4.8$ . This is a horizontal line from $(4, 4.8)$ to $(15, 4.8)$ .

Phase 3 ( $15 < t \le 20$ ):
The particle decelerates uniformly to rest. This is a straight line from $(15, 4.8)$ to $(20, 0)$ .

Answer

A velocity-time graph with a quadratic curve from $(0,0)$ to $(4, 4.8)$ , a horizontal line from $(4, 4.8)$ to $(15, 4.8)$ , and a straight line from $(15, 4.8)$ to $(20, 0)$ .

Final answer

Graph with quadratic from (0,0) to (4, 4.8), horizontal line to (15, 4.8), straight line to (20, 0)

Detailed explanation

Walkthrough

The motion is divided into three time intervals. From $t=0$ to $t=4$ , velocity is $v=0.3t^2$ , which is a parabola opening upwards starting at the origin. From $t=4$ to $t=15$ , velocity is constant at $4.8$ , giving a horizontal line. From $t=15$ to $t=20$ , the particle stops with constant deceleration, giving a straight line with negative slope from $(15, 4.8)$ to $(20, 0)$ .

Key Takeaways

A velocity-time graph visually represents motion phases. Quadratic velocity gives a curved graph; constant velocity gives a horizontal line; constant acceleration gives a straight line.

Common Mistakes

Drawing the first phase as a straight line instead of a curve. Forgetting to label the key coordinates $(4, 4.8)$ , $(15, 4.8)$ , and $(20, 0)$ .

Things to Be Careful About

The graph does not need to be to scale, but the shape must be correct. The first section must be a curve with increasing gradient (convex), not a straight line.

Techniques used

sketch velocity-time graphidentify phases of motionrecognise curvature from integration

(c)

Find an expression, in terms of $t$ , for the velocity of the particle for $15 \le t \le 20$ .

2M Similar

DifficultyMedium-Easy

Worked solution

Approach

For $15 \le t \le 20$ , the particle undergoes constant deceleration. Find the acceleration using the gradient of the velocity-time graph, then use $v = u + at$ to find the velocity expression.

Working

The velocity changes from $4.8 \text{ m s}^{-1}$ at $t = 15$ to $0 \text{ m s}^{-1}$ at $t = 20$ .

The acceleration $a$ is:

a = \frac{0 - 4.8}{20 - 15} = \frac{-4.8}{5} = -0.96 \text{ m s}^{-2}

Using $v = u + a(t - t_0)$ where $u = 4.8$ at $t_0 = 15$ :

\begin{aligned} v &= 4.8 - 0.96(t - 15) \\ &= 4.8 - 0.96t + 14.4 \\ &= 19.2 - 0.96t \end{aligned}

Answer

v = 19.2 - 0.96t

Final answer

v = 19.2 - 0.96t

Detailed explanation

Walkthrough

In the third phase ( $15 \le t \le 20$ ), the velocity-time graph is a straight line, meaning acceleration is constant. We calculate this acceleration as the gradient of the line: $(0 - 4.8) / (20 - 15) = -0.96$ . Then we use the equation $v = u + at$ . Since the phase starts at $t=15$ with $u=4.8$ , we write $v = 4.8 - 0.96(t - 15)$ . Expanding this gives $v = 19.2 - 0.96t$ .

Key Takeaways

The gradient of a velocity-time graph gives acceleration. The suvat equation $v = u + at$ can be applied from any starting time $t_0$ by using $(t - t_0)$ .

Common Mistakes

Using $v = 4.8 - 0.96t$ directly without adjusting for the start time $t=15$ . Forgetting the negative sign for deceleration.

Things to Be Careful About

The expression is valid only for $15 \le t \le 20$ . If you use $v = u + at$ , remember that $t$ must be the time elapsed since $t=15$ , so use $(t - 15)$ .

Techniques used

calculate constant decelerationapply suvat equation v = u + at

(d)

Find the total distance travelled by the particle.

4M Similar

DifficultyMedium

Worked solution

Approach

The total distance is the area under the velocity-time graph. We calculate the area for each of the three phases separately and sum them.

Working

Phase 1 ( $0 \le t \le 4$ ):
Distance is the integral of velocity:

\begin{aligned} s_1 &= \int_0^4 0.3t^2 \, dt \\ &= \left[ \frac{0.3}{3}t^3 \right]_0^4 \\ &= \left[ 0.1t^3 \right]_0^4 \\ &= 0.1(4)^3 - 0.1(0)^3 \\ &= 6.4 \text{ m} \end{aligned}

Phase 2 ( $4 < t \le 15$ ):
Distance is the area of a rectangle:

s_2 = 4.8 \times (15 - 4) = 4.8 \times 11 = 52.8 \text{ m}

Phase 3 ( $15 < t \le 20$ ):
Distance is the area of a triangle:

s_3 = \frac{1}{2} \times 4.8 \times (20 - 15) = \frac{1}{2} \times 4.8 \times 5 = 12 \text{ m}

Total distance:

s_{\text{total}} = 6.4 + 52.8 + 12 = 71.2 \text{ m}

Answer

71.2 \text{ m}

Final answer

71.2

Detailed explanation

Walkthrough

Distance travelled is the area under the velocity-time graph. We split this into three parts corresponding to the three phases of motion. For the first phase, we integrate the velocity function $v = 0.3t^2$ from $t=0$ to $t=4$ , giving $0.1t^3$ evaluated from 0 to 4, which is $6.4$ m. For the second phase, the graph is a rectangle with height $4.8$ and width $11$ , giving $52.8$ m. For the third phase, the graph is a triangle with base $5$ and height $4.8$ , giving $12$ m. Adding these gives $71.2$ m.

Key Takeaways

Distance is the area under a velocity-time graph. For curved sections, integration is required. For straight-line sections, geometric area formulas (rectangle, triangle, trapezium) can be used.

Common Mistakes

Using $s = vt$ for the first phase instead of integrating. Forgetting to add the distances from all three phases. Miscalculating the width of the rectangle or triangle.

Things to Be Careful About

Ensure limits of integration are correct. The mark scheme notes that using $s = vt$ for the first phase scores 0 marks. You can also calculate the total area using a trapezium for the last two phases: $\frac{(11 + 16) \times 4.8}{2} = 64.8$ , then add $6.4$ to get $71.2$ .

Techniques used

integrate velocity to find distancecalculate area under velocity-time graphsum distances from multiple phases

Q7MediumNewton's Laws of MotionKinematics of Motion in a Straight Line

Fig. 3

Two particles, $A$ and $B$ , of masses $3\text{ kg}$ and $5\text{ kg}$ respectively, are connected by a light inextensible string that passes over a fixed smooth pulley. The particles are held with the string taut and its straight parts vertical. Particle $A$ is $1\text{ m}$ above a horizontal plane, and particle $B$ is $2\text{ m}$ above the plane (see diagram). The particles are released from rest. In the subsequent motion, $A$ does not reach the pulley, and after $B$ reaches the plane it remains in contact with the plane.

(a)

Find the tension in the string and the time taken for $B$ to reach the plane.

6M Similar

DifficultyMedium

Worked solution

Approach

Since particle B (5 kg) is heavier than particle A (3 kg), B accelerates downward and A accelerates upward with the same magnitude of acceleration $a$ . Apply Newton's second law to each particle to find $a$ and the tension $T$ . Then use suvat equations to find the time for B to travel 2 m from rest.

Working

Newton's second law for each particle:

For A (moving upward, mass 3 kg):

T - 3g = 3a

For B (moving downward, mass 5 kg):

5g - T = 5a

Finding acceleration:

Adding the two equations to eliminate $T$ :

5g - 3g = 3a + 5a

2g = 8a

a = \frac{g}{4} = \frac{10}{4} = 2.5 \text{ m/s}^2

Finding tension:

Substituting $a = 2.5$ into the equation for A:

T = 3g + 3a = 3(10) + 3(2.5) = 30 + 7.5 = 37.5 \text{ N}

Finding time for B to reach the plane:

Particle B starts from rest ( $u = 0$ ), travels $s = 2$ m with acceleration $a = 2.5$ m/s². Using $s = ut + \frac{1}{2}at^2$ :

2 = 0 + \frac{1}{2}(2.5)t^2

t^2 = \frac{4}{2.5} = 1.6

t = \sqrt{1.6} = \frac{\sqrt{40}}{5} = \frac{2\sqrt{10}}{5} \approx 1.26 \text{ s}

Answer

Tension $T = 37.5$ N, time $t = \frac{2\sqrt{10}}{5} \approx 1.26$ s

Final answer

T = 37.5 N, t = 2√10/5 ≈ 1.26 s

Detailed explanation

Walkthrough

Step 1: Determine the direction of motion.
Particle B has mass 5 kg and particle A has mass 3 kg. Since B is heavier, when released from rest, B will accelerate downward and A will accelerate upward. The string is inextensible, so both particles have the same magnitude of acceleration $a$ .

Step 2: Apply Newton's second law to each particle.
For particle A moving upward, the forces are tension $T$ upward and weight $3g$ downward. Taking upward as positive:

T - 3g = 3a

For particle B moving downward, the forces are weight $5g$ downward and tension $T$ upward. Taking downward as positive:

5g - T = 5a

Step 3: Solve for acceleration.
Adding the two equations eliminates $T$ :

(5g - T) + (T - 3g) = 5a + 3a

2g = 8a \implies a = \frac{g}{4} = 2.5 \text{ m/s}^2

Step 4: Solve for tension.
Substituting $a = 2.5$ into $T = 3g + 3a$ :

T = 3(10) + 3(2.5) = 30 + 7.5 = 37.5 \text{ N}

Step 5: Find the time for B to reach the plane.
B starts from rest and must travel 2 m. Using $s = ut + \frac{1}{2}at^2$ with $u = 0$ , $s = 2$ , $a = 2.5$ :

2 = \frac{1}{2}(2.5)t^2 \implies t^2 = 1.6 \implies t = \sqrt{1.6} = \frac{2\sqrt{10}}{5} \approx 1.26 \text{ s}

Key Takeaways

For connected particles over a smooth pulley, apply Newton's second law to each particle separately, choosing the positive direction as the direction of motion for each.
The tension is the same throughout a light inextensible string over a smooth pulley.
Once acceleration is found, suvat equations can be used to find time, velocity, or displacement.

Common Mistakes

Taking the wrong sign for acceleration in the equations of motion (must be consistent: positive direction is the direction of motion for each particle).
Forgetting that the string is inextensible, so both particles have the same magnitude of acceleration.
Using the wrong displacement for B (it is 2 m, not 1 m).
Not showing enough working to earn method marks — the mark scheme requires seeing the equations and the attempt to solve for $T$ and $t$ .

Things to Be Careful About

Use $g = 10$ m/s² as is standard in Cambridge A-Level unless otherwise stated.
The mark scheme allows an alternative energy method: using $5g \times 2 - 3g \times 2 = \frac{1}{2}(5+3)v^2$ to find $v = \sqrt{10}$ , then $2 = \frac{0+v}{2}t$ to find $t$ . Both methods are acceptable.
If a candidate finds $t = \sqrt{1.6}$ without simplifying, they only get the final A mark if they attempt to evaluate it; $t = \sqrt{1.6}$ alone scores A0.

Techniques used

apply Newton's second law to each particlesolve simultaneous equations for acceleration and tensionuse suvat to find time from rest

(b)

Find the time for which $A$ is at least $3.25\text{ m}$ above the plane.

4M Similar

DifficultyMedium

Worked solution

Approach

When B reaches the plane, A has risen 2 m (to height 3 m) and the string becomes slack. After this, A moves upward under gravity alone with $a = -g$ . Find the velocity of A when the string goes slack using suvat, then use suvat again to find the two times when A is at height 3.25 m. The time at least 3.25 m is the difference between these times.

Working

Velocity of A when the string goes slack:

A has risen $s = 2$ m from rest with $a = 2.5$ m/s². Using $v^2 = u^2 + 2as$ :

v^2 = 0 + 2(2.5)(2) = 10

v = \sqrt{10} \text{ m/s}

At this moment, A is at height $1 + 2 = 3$ m above the plane.

Motion of A after the string goes slack:

A continues upward under gravity alone with $a = -g = -10$ m/s². We need A to be at least 3.25 m above the plane, i.e., rise $s = 0.25$ m above its current height of 3 m.

Using $s = ut + \frac{1}{2}at^2$ with $s = 0.25$ , $u = \sqrt{10}$ , $a = -10$ :

0.25 = \sqrt{10}\,t + \frac{1}{2}(-10)t^2

0.25 = \sqrt{10}\,t - 5t^2

5t^2 - \sqrt{10}\,t + 0.25 = 0

Solving the quadratic:

Using the quadratic formula:

t = \frac{\sqrt{10} \pm \sqrt{10 - 4(5)(0.25)}}{2(5)}

t = \frac{\sqrt{10} \pm \sqrt{10 - 5}}{10}

t = \frac{\sqrt{10} \pm \sqrt{5}}{10}

The two solutions are:

t_1 = \frac{\sqrt{10} - \sqrt{5}}{10} \approx 0.0926 \text{ s}

t_2 = \frac{\sqrt{10} + \sqrt{5}}{10} \approx 0.5398 \text{ s}

Time at least 3.25 m above the plane:

\Delta t = t_2 - t_1 = \frac{2\sqrt{5}}{10} = \frac{\sqrt{5}}{5} = \frac{1}{\sqrt{5}} \approx 0.447 \text{ s}

Answer

Time = $\frac{\sqrt{5}}{5}$ s $\approx 0.447$ s

Final answer

√5/5 ≈ 0.447 s

Detailed explanation

Walkthrough

Step 1: Understand what happens when B reaches the plane.
When B hits the plane, it stops. Since the string is inextensible, the string goes slack — there is no longer any tension. Particle A, which was rising, continues upward but now only under the influence of gravity (no tension pulling it up).

At the moment B hits the plane, A has risen 2 m (same distance B fell), so A is at height $1 + 2 = 3$ m above the plane.

Step 2: Find the velocity of A when the string goes slack.
During the first phase of motion, A accelerated at $a = 2.5$ m/s² from rest over $s = 2$ m. Using $v^2 = u^2 + 2as$ :

v^2 = 0 + 2(2.5)(2) = 10 \implies v = \sqrt{10} \text{ m/s}

Step 3: Set up the equation for A's motion after the string goes slack.
After the string goes slack, A moves under gravity alone: $a = -g = -10$ m/s² (upward is positive). We want to find when A is at height 3.25 m, i.e., when it has risen $s = 0.25$ m from its current height of 3 m.

Using $s = ut + \frac{1}{2}at^2$ :

0.25 = \sqrt{10}\,t - 5t^2

Rearranging:

5t^2 - \sqrt{10}\,t + 0.25 = 0

Step 4: Solve the quadratic.

t = \frac{\sqrt{10} \pm \sqrt{10 - 5}}{10} = \frac{\sqrt{10} \pm \sqrt{5}}{10}

$t_1 = \frac{\sqrt{10} - \sqrt{5}}{10} \approx 0.0926$ s (A reaches 3.25 m on the way up)
$t_2 = \frac{\sqrt{10} + \sqrt{5}}{10} \approx 0.5398$ s (A reaches 3.25 m on the way down)

Step 5: Find the time interval.

\Delta t = t_2 - t_1 = \frac{2\sqrt{5}}{10} = \frac{\sqrt{5}}{5} \approx 0.447 \text{ s}

Key Takeaways

When a connected particle system has one particle hit a surface, the string may go slack and the other particle continues under gravity alone.
Always identify the change in physical situation and adjust the equations of motion accordingly.
A quadratic from suvat gives two times: one on the way up and one on the way down. The time at or above a height is the difference.

Common Mistakes

Forgetting that the string goes slack when B hits the plane and continuing to use $a = 2.5$ m/s² throughout.
Using the wrong displacement: A needs to rise only 0.25 m above 3 m, not 3.25 m from the ground.
Making sign errors with $a = -10$ m/s².
Not solving the quadratic correctly or not finding both roots.

Things to Be Careful About

The mark scheme provides alternative methods: finding the maximum height (3.5 m) and computing the time from 0.25 m below the top, or using energy methods. All are acceptable.
When using energy: PE gain = KE loss gives $3g(0.25) = \frac{1}{2}(3)(10) - \frac{1}{2}(3)w^2$ , yielding $w = \sqrt{5}$ at height 3.25 m, then $0 = \sqrt{5} - 10t$ gives $t = \frac{\sqrt{5}}{10}$ , doubled for the round trip.
Do not allow sign errors in energy equations.
The final answer $\frac{\sqrt{5}}{5}$ can also be written as $\frac{1}{\sqrt{5}}$ or $0.447$ s.

Techniques used

recognise string goes slack when B reaches planefind velocity of A at moment string goes slack using suvatapply suvat under gravity alone to find times at given heightsolve quadratic equation for time

Mathematics 9709/43 — October/November 2024

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Case 1: AAA continues in the same direction, vA=+2v_A=+2vA​=+2

Case 2: AAA rebounds, vA=−2v_A=-2vA​=−2

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Case 1: $A$ continues in the same direction, $v_A=+2$

Case 2: $A$ rebounds, $v_A=-2$