A lorry of mass $18000\text{ kg}$ is travelling along a straight road. (a) On a horizontal section of the road, the power of the lorry's engine is constant. There is a constant resistance to motion of $1600\text{ N}$. (i) The steady speed which the lorry can maintain with the engine working at power $P\text{ W}$ is $30\text{ m s}^{-1}$. Find the value of $P$. (ii) At an instant when the speed of the lorry is $16\text{ m s}^{-1}$, its engine is working at a power of $40\text{…

(a)(i) P = 48000 W · (a)(ii) a = 0.05 m s^-2 · (b) alpha = 1.40 degrees

9709/41 May/June 2025 — Mathematics Solved Past Paper

Q14MMediumForces and EquilibriumNewton's Laws of Motion

A block of mass $12\text{ kg}$ is being pulled by a rope up a rough plane. The plane is inclined at an angle of $20^\circ$ above the horizontal. The rope pulling the block is parallel to a line of greatest slope of the plane. The coefficient of friction between the block and the plane is $0.4$ . The acceleration of the block is $2\text{ m s}^{-2}$ .

Find the tension in the rope.

DifficultyMedium

Worked solution

Approach

Resolve forces perpendicular to the plane to find the normal reaction $R$ . Then use the friction model $F = \mu R$ to find the friction force, and apply Newton's second law parallel to the plane to solve for the tension $T$ .

Working

Step 1. Normal reaction. The block has no acceleration perpendicular to the plane, so the normal reaction balances the perpendicular component of the weight:

R = 12g \cos 20^\circ

With $g = 10\text{ m s}^{-2}$ :

R = 12 \times 10 \times \cos 20^\circ = 120 \cos 20^\circ = 112.763...\text{ N}

Step 2. Friction force. The friction opposes the motion up the plane, so it acts down the plane:

F = \mu R = 0.4 \times 120 \cos 20^\circ = 45.105...\text{ N}

Step 3. Newton's second law parallel to the plane. Take the positive direction as up the plane. The forces up the plane are the tension $T$ ; the forces down the plane are the friction $F$ and the weight component $12g \sin 20^\circ$ . The acceleration is $2\text{ m s}^{-2}$ up the plane:

T - F - 12g \sin 20^\circ = 12 \times 2

Substitute $F = 0.4 \times 12g \cos 20^\circ$ :

T - 0.4 \times 12g \cos 20^\circ - 12g \sin 20^\circ = 24

T - 45.105... - 41.042... = 24

T = 24 + 45.105... + 41.042... = 110.147...\text{ N}

Answer

T = 110\text{ N}

Final answer

110 N

Detailed explanation

Walkthrough

The block is pulled up a rough inclined plane with a constant acceleration of $2\text{ m s}^{-2}$ . Three forces act along the line of the plane: the tension $T$ pulling the block up the plane, the friction $F$ acting down the plane (friction always opposes the relative motion), and the component of the weight $12g \sin 20^\circ$ acting down the plane. Perpendicular to the plane, the normal reaction $R$ balances the component of the weight $12g \cos 20^\circ$ .

First, resolve perpendicular to the plane. Since the block does not accelerate in that direction, we have $R = 12g \cos 20^\circ = 112.763\text{ N}$ . This is the B1 mark.

Second, use the friction model $F = \mu R = 0.4 \times 112.763 = 45.105\text{ N}$ , acting down the plane.

Third, apply Newton's second law parallel to the plane. The net force up the plane equals mass times acceleration:

T - F - 12g \sin 20^\circ = 12 \times 2

Substitute $F = 0.4 \times 12g \cos 20^\circ$ to obtain an equation in $T$ only:

T - 0.4 \times 12g \cos 20^\circ - 12g \sin 20^\circ = 24

Evaluating the two known terms:

T - 45.105 - 41.042 = 24

T = 110.147\text{ N}

The tension is therefore $110$ N.

Key Takeaways

On an inclined plane, the weight must be resolved into components parallel and perpendicular to the plane.
The normal reaction is obtained from the perpendicular balance (no acceleration perpendicular to the plane).
Friction always opposes the motion and is given by $F = \mu R$ .
Newton's second law is applied along the direction of motion, with the net force equal to $ma$ .

Common Mistakes

Swapping $\sin 20^\circ$ and $\cos 20^\circ$ : the normal reaction uses $\cos 20^\circ$ , while the weight component parallel to the plane uses $\sin 20^\circ$ .
Forgetting to include the friction force or the weight component in the Newton's second law equation.
Using the full weight $12g$ instead of resolving it into components.
Using $g = 9.8$ instead of the syllabus value $g = 10$ .
Sign errors in the N2L equation (the mark scheme allows sign errors for the M1 mark, but the final answer requires correct signs).

Things to Be Careful About

The block is accelerating, so the right-hand side of Newton's second law is $12 \times 2 = 24$ , not zero.
Friction acts down the plane because the block is moving up the plane.
The mark scheme requires the equation to be shown with all four terms ( $T$ , $F$ , weight component, and $ma$ ) for the M1 mark.
The final answer must be within reason of $110$ N (AWRT 110) with correct working (CWO).

Techniques used

resolve weight into components parallel and perpendicular to the planeapply the friction model F = μRapply Newton's second law along the line of motionsolve the resulting linear equation for the tension

Approach

The resultant of the three forces acts perpendicular to $P$ . Since $P$ acts vertically upwards, the resultant $Q$ must act horizontally. This means the sum of the vertical components of all three forces is zero, and the sum of the horizontal components equals $Q$ .

Working

Resolving vertically (taking upwards as positive, in the direction of $P$ ):

The force $P$ acts upwards. The $5\text{ N}$ force has a vertical component of $5\sin 60°$ upwards. The $10\text{ N}$ force acts downwards and to the left, with a vertical component of $10\sin 30°$ downwards.

Since the resultant is horizontal, the vertical components sum to zero:

\begin{aligned} P + 5\sin 60° - 10\sin 30° &= 0 \\ P + 5 \times \frac{\sqrt{3}}{2} - 10 \times \frac{1}{2} &= 0 \\ P + \frac{5\sqrt{3}}{2} - 5 &= 0 \\ P &= 5 - \frac{5\sqrt{3}}{2} \\ P &\approx 0.670 \end{aligned}

Resolving horizontally (taking the direction of $Q$ as positive):

The force $P$ has no horizontal component. The $5\text{ N}$ force has a horizontal component of $5\cos 60°$ to the right. The $10\text{ N}$ force has a horizontal component of $10\cos 30°$ to the left. Since $10\cos 30° > 5\cos 60°$ , the resultant acts to the left:

\begin{aligned} Q &= 10\cos 30° - 5\cos 60° \\ &= 10 \times \frac{\sqrt{3}}{2} - 5 \times \frac{1}{2} \\ &= 5\sqrt{3} - \frac{5}{2} \\ &\approx 6.16 \end{aligned}

Answer

P = 5 - \frac{5\sqrt{3}}{2} \approx 0.670 \text{ N}, \quad Q = 5\sqrt{3} - \frac{5}{2} \approx 6.16 \text{ N}

Final answer

P = 0.670 N, Q = 6.16 N

Detailed explanation

Walkthrough

The question gives three forces acting at a point and tells us their resultant is perpendicular to the force $P$ . Since $P$ acts vertically upwards (as shown in the diagram), the resultant must act horizontally. This is the key insight: if the resultant has no vertical component, then the vertical components of all individual forces must cancel out.

Step 1: Resolve each force into vertical and horizontal components.

$P\text{ N}$ acts vertically upwards: vertical component $= +P$ , horizontal component $= 0$ .
$5\text{ N}$ acts at $60°$ above the horizontal to the right: vertical component $= +5\sin 60° = 5 \times \frac{\sqrt{3}}{2}$ , horizontal component $= +5\cos 60° = 5 \times \frac{1}{2}$ .
$10\text{ N}$ acts at $30°$ below the horizontal to the left: vertical component $= -10\sin 30° = -10 \times \frac{1}{2}$ , horizontal component $= -10\cos 30° = -10 \times \frac{\sqrt{3}}{2}$ .

Step 2: Apply the vertical equilibrium condition.

Since the resultant is horizontal, the sum of vertical components equals zero:

P + 5\sin 60° - 10\sin 30° = 0

Substituting the known values:

P + \frac{5\sqrt{3}}{2} - 5 = 0

Solving for $P$ :

P = 5 - \frac{5\sqrt{3}}{2} \approx 0.670

We check that $P > 0$ , which it is, so this is a valid physical answer.

Step 3: Compute the horizontal resultant $Q$ .

The horizontal components are $5\cos 60°$ to the right and $10\cos 30°$ to the left. Since $10\cos 30° = 5\sqrt{3} \approx 8.66$ is greater than $5\cos 60° = 2.5$ , the net horizontal force acts to the left with magnitude:

Q = 10\cos 30° - 5\cos 60° = 5\sqrt{3} - \frac{5}{2} \approx 6.16

Key Takeaways

When a resultant is stated to be perpendicular to one of the component forces, that provides a direct constraint: the components along the direction of that force must sum to zero.
Resolving forces into perpendicular components (typically horizontal and vertical) is the standard method for combining multiple coplanar forces.
Always check that computed magnitudes are positive; a negative result indicates an error in sign convention or setup.

Common Mistakes

Forgetting that the $10\text{ N}$ force has both a downward and a leftward component, leading to sign errors in the vertical resolution.
Using $\cos$ instead of $\sin$ (or vice versa) when resolving a force at an angle to the horizontal.
Assuming the resultant is zero; the problem states the resultant has magnitude $Q$ , not that the system is in equilibrium.
Not checking that $P$ is positive; the mark scheme explicitly requires this.

Things to Be Careful About

The angle $60°$ is measured from the horizontal to the $5\text{ N}$ force, so the vertical component uses $\sin 60°$ and the horizontal uses $\cos 60°$ .
The angle $30°$ is measured from the horizontal to the $10\text{ N}$ force (below the horizontal), so its vertical component is $-10\sin 30°$ (downwards) and horizontal is $-10\cos 30°$ (leftwards).
The mark scheme allows sign errors and sin/cos mix-ups in the method mark (M1), but the final answers for $P$ and $Q$ must be positive.
Exact answers ( $5 - \frac{5\sqrt{3}}{2}$ and $5\sqrt{3} - \frac{5}{2}$ ) should be given or rounded to 3 significant figures ( $0.670$ and $6.16$ ).

Techniques used

resolve forces into vertical and horizontal componentsapply condition that vertical resultant is zerocompute horizontal resultant as Q

Approach

First, determine the velocity of the cyclist at $t = 10$ s using the constant acceleration formulae. Then, use the area under the velocity-time graph during the deceleration phase (which represents distance) to find the duration of deceleration and thus the total time.

Working

Step 1: Find velocity at $t = 10$ s
The cyclist starts at $u = 3$ m s $^{-1}$ and accelerates at $a = 0.5$ m s $^{-2}$ for $t = 10$ s.

v = u + at

v = 3 + (0.5)(10) = 8 \text{ m s}^{-1}

The cyclist travels at this constant speed of $8$ m s $^{-1}$ from $t = 10$ s to $t = 40$ s.

Step 2: Analyze deceleration phase
The cyclist decelerates from $v = 8$ m s $^{-1}$ to rest ( $v = 0$ ). The distance covered during this phase is given as $80$ m.

The area under the velocity-time graph for this phase is a triangle with height $8$ and base $(T - 40)$ , where $T$ is the total time.

\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}

80 = \frac{1}{2} \times (T - 40) \times 8

80 = 4(T - 40)

20 = T - 40

T = 60

Answer

The total time taken for the journey is $60$ s.

Final answer

60 s

Detailed explanation

Walkthrough

The problem describes a journey in three stages: acceleration, constant speed, and deceleration. We are given a velocity-time graph description.

Acceleration Phase ( $0 \le t \le 10$ ): The cyclist starts at $3$ m s $^{-1}$ and accelerates at $0.5$ m s $^{-2}$ . We use $v = u + at$ to find the speed at the end of this phase. $v = 3 + 0.5(10) = 8$ m s $^{-1}$ .
Constant Speed Phase ( $10 \le t \le 40$ ): The speed remains $8$ m s $^{-1}$ for $30$ seconds (from $t=10$ to $t=40$ ).
Deceleration Phase ( $40 \le t \le T$ ): The cyclist comes to rest. The distance covered is $80$ m. On a velocity-time graph, distance is the area under the curve. Since the graph is a straight line to the axis, the area is a triangle with height $8$ (the speed at $t=40$ ) and base $(T-40)$ .

Setting the area equal to $80$ : $\frac{1}{2} \times 8 \times (T-40) = 80$ . Solving this gives $T-40 = 20$ , so $T=60$ .

Key Takeaways

The area under a velocity-time graph represents displacement (or distance if motion is in one direction).
The gradient of a velocity-time graph represents acceleration.
Constant acceleration allows the use of $v = u + at$ .

Common Mistakes

Forgetting to use the correct initial velocity ( $3$ m s $^{-1}$ ) for the first phase.
Calculating the area of the deceleration triangle incorrectly (e.g., forgetting the $\frac{1}{2}$ ).
Assuming the total time is just the sum of the given times ( $10+30=40$ ) without calculating the deceleration time.

Things to Be Careful About

Ensure units are consistent (m s $^{-1}$ , s, m).
The base of the triangle for deceleration is $(T - 40)$ , not just $T$ .

Techniques used

apply constant acceleration formulaecalculate area under velocity-time graphsolve for total time

(b)

On the given axes, sketch a displacement-time graph for the cyclist's journey from $O$ to $P$ , showing on your graph the distances travelled after $10\text{ s}$ and $40\text{ s}$ .

Fig. 3.2

4M Similar

DifficultyMedium

Worked solution

Approach

Calculate the displacement at $t = 10$ s and $t = 40$ s by finding the area under the velocity-time graph up to those times. Then, sketch the displacement-time graph, noting that:

Acceleration ( $0-10$ s) produces a quadratic curve (parabola) with increasing gradient.
Constant speed ( $10-40$ s) produces a straight line with constant positive gradient.
Deceleration ( $40-60$ s) produces a quadratic curve (parabola) with decreasing gradient.

Working

Step 1: Distance after $10$ s
The area under the graph from $t=0$ to $t=10$ is a trapezium with parallel sides $3$ and $8$ , and height $10$ .

s_{10} = \frac{1}{2}(3 + 8)(10) = 55 \text{ m}

Step 2: Distance after $40$ s
The area from $t=10$ to $t=40$ is a rectangle with width $30$ and height $8$ .

\text{Area}_{10-40} = 8 \times 30 = 240 \text{ m}

Total distance at $t=40$ :

s_{40} = 55 + 240 = 295 \text{ m}

Step 3: Sketching the graph

$0 \le t \le 10$ : Curve starting at $(0,0)$ , concave up (gradient increasing from $3$ to $8$ ), ending at $(10, 55)$ .
$10 \le t \le 40$ : Straight line from $(10, 55)$ to $(40, 295)$ with gradient $8$ .
$40 \le t \le 60$ : Curve starting at $(40, 295)$ , concave down (gradient decreasing from $8$ to $0$ ), ending at $(60, 375)$ .

Answer

Distances: $55$ m at $10$ s, $295$ m at $40$ s. See sketch description below.

Final answer

Distances: 55 m and 295 m. Graph is a curve from (0,0) to (10,55), straight line to (40,295), then curve to (60,375).

Detailed explanation

Walkthrough

We need to construct a displacement-time ( $s-t$ ) graph from the velocity-time ( $v-t$ ) graph.

Calculate key points:
- At $t=10$ , displacement is the area of the trapezium under the first part of the $v-t$ graph: $\frac{1}{2}(3+8)(10) = 55$ m.
- At $t=40$ , we add the area of the rectangle (constant speed phase): $8 \times 30 = 240$ m. Total $s = 55 + 240 = 295$ m.
Determine graph shapes:
- $0-10$ s: Velocity is increasing linearly ( $v = 3 + 0.5t$ ). Displacement is the integral, $s = 3t + 0.25t^2$ . This is a quadratic curve opening upwards (convex). The gradient increases from $3$ to $8$ .
- $10-40$ s: Velocity is constant ( $v=8$ ). Displacement is linear, $s = 55 + 8(t-10)$ . This is a straight line with gradient $8$ .
- $40-60$ s: Velocity is decreasing linearly. Displacement is a quadratic curve opening downwards (concave). The gradient decreases from $8$ to $0$ .
Sketch: Plot points $(0,0)$ , $(10, 55)$ , $(40, 295)$ , and $(60, 375)$ (total distance $295+80=375$ ). Connect them with the appropriate curves and lines.

Key Takeaways

Displacement is the area under the velocity-time graph.
A linear velocity-time graph corresponds to a quadratic displacement-time graph.
A constant velocity-time graph corresponds to a linear displacement-time graph.
The gradient of the $s-t$ graph at any point equals the velocity at that time.

Common Mistakes

Calculating the area incorrectly (e.g., using triangle formula for the trapezium).
Sketching the acceleration phase as a straight line (it should be curved).
Not ensuring the graph is smooth/continuous at the junctions $t=10$ and $t=40$ .

Things to Be Careful About

The question asks to show distances after $10$ s and $40$ s on the graph. Ensure these values ( $55$ and $295$ ) are clearly marked or readable.
The scale on the axes might need to be chosen appropriately to fit $375$ m on the $s$ -axis and $60$ s on the $t$ -axis.

Techniques used

calculate displacement from area under velocity-time graphdetermine shape of displacement-time graph from velocitysketch graph with correct continuity and gradients

Q4MediumEnergy, Work and PowerNewton's Laws of MotionForces and Equilibrium

A lorry of mass $18000\text{ kg}$ is travelling along a straight road.

(a)

On a horizontal section of the road, the power of the lorry's engine is constant. There is a constant resistance to motion of $1600\text{ N}$ .

3M

(i)

The steady speed which the lorry can maintain with the engine working at power $P\text{ W}$ is $30\text{ m s}^{-1}$ .

Find the value of $P$ .

1M Similar

(ii)

At an instant when the speed of the lorry is $16\text{ m s}^{-1}$ , its engine is working at a power of $40\text{ kW}$ .

Find the acceleration of the lorry at this instant.

2M Similar

DifficultyMedium-Easy

Worked solution

Approach

Use $P = Fv$ to find the driving force at the given speed. Then apply Newton's second law horizontally: the resultant force is the driving force minus the resistance, and this equals $ma$ .

Working

Driving force at $v = 16\text{ m s}^{-1}$ :

F = \frac{P}{v} = \frac{40000}{16} = 2500\text{ N}

Resultant force along the road:

F - 1600 = 2500 - 1600 = 900\text{ N}

Newton's second law:

900 = 18000a

Therefore:

a = \frac{900}{18000} = 0.05\text{ m s}^{-2}

Answer

a = 0.05\text{ m s}^{-2}

Final answer

a = 0.05 m s^-2

Detailed explanation

Walkthrough

The engine power is given in kilowatts, so first convert $40\text{ kW}$ to $40000\text{ W}$ . The driving force is obtained from $P = Fv$ : at a lower speed the same power produces a larger driving force. Here $F = 40000/16 = 2500\text{ N}$ . The resistance is $1600\text{ N}$ , so the net forward force is $2500 - 1600 = 900\text{ N}$ . Newton's second law $F_{\text{net}} = ma$ then gives $900 = 18000a$ , so $a = 0.05\text{ m s}^{-2}$ .

Key Takeaways

Power, driving force and speed are linked by $P = Fv$ . To find acceleration, combine the driving force from the engine with all resistances and apply Newton's second law. Units must be consistent: convert kilowatts to watts before calculating.

Common Mistakes

Using $40$ instead of $40000$ when calculating the driving force. Forgetting to subtract the resistance. Dividing the resultant force by the wrong quantity instead of the mass. Sign errors in the equation of motion; the mark scheme allows sign errors as long as the method has three dimensionally correct terms.

Things to Be Careful About

The driving force and resistance act in opposite directions, so one must be subtracted. The acceleration is positive because the driving force is greater than the resistance. Keep all units as newtons, watts, metres per second and kilograms.

Techniques used

convert power to wattsfind driving force from power and speedapply Newton's second law to find acceleration

(b)

When the lorry has reached a speed of $20\text{ m s}^{-1}$ , it begins to ascend a section of road inclined at an angle $\alpha^\circ$ to the horizontal. The engine now works at a power of $120\text{ kW}$ . There is no change in the lorry's speed as it ascends the hill. The constant resistance to motion remains $1600\text{ N}$ .

Find the value of $\alpha$ .

3M Similar

DifficultyMedium

Worked solution

Approach

At constant speed up the incline the acceleration is zero, so the resultant force parallel to the slope is zero. Find the driving force from $P = Fv$ , resolve the weight of the lorry down the slope as $mg\sin\alpha$ , and balance the forces along the slope.

Working

Driving force:

F = \frac{P}{v} = \frac{120000}{20} = 6000\text{ N}

Taking $g = 10\text{ m s}^{-2}$ , the component of the weight down the slope is:

18000g\sin\alpha = 180000\sin\alpha

Since the speed is constant, resolving parallel to the slope gives:

6000 - 1600 - 180000\sin\alpha = 0

4400 = 180000\sin\alpha

\sin\alpha = \frac{4400}{180000} = 0.024444\ldots

\alpha = \sin^{-1}(0.024444\ldots) = 1.4007\ldots^\circ

Answer

\alpha = 1.40^\circ

Final answer

alpha = 1.40 degrees

Detailed explanation

Walkthrough

The lorry ascends the hill without changing speed, so its acceleration is zero. Therefore the forces parallel to the slope must balance. The engine provides a driving force up the slope. From $P = Fv$ , with $P = 120000\text{ W}$ and $v = 20\text{ m s}^{-1}$ , the driving force is $6000\text{ N}$ . Down the slope there are two contributions: the constant resistance of $1600\text{ N}$ and the component of the lorry's weight, $mg\sin\alpha = 180000\sin\alpha$ using $g = 10\text{ m s}^{-2}$ . Balancing these gives $6000 = 1600 + 180000\sin\alpha$ , so $\sin\alpha = 4400/180000$ . Taking inverse sine gives $\alpha = 1.40^\circ$ .

Key Takeaways

On an incline, the weight component along the slope is $mg\sin\alpha$ . Constant speed means equilibrium along the direction of motion, so the resultant force is zero. The power-speed relation $P = Fv$ gives the driving force needed to set up the force balance.

Common Mistakes

Forgetting to include the resistance when balancing forces. Using $\cos\alpha$ instead of $\sin\alpha$ for the component of weight down the slope. Failing to convert $120\text{ kW}$ to $120000\text{ W}$ . Omitting the mass from the weight component. The mark scheme allows sign errors and a missing $g$ in the method mark, but the final equation must be correct.

Things to Be Careful About

The angle is in degrees, so the final answer should be given as $1.40^\circ$ . Use $g = 10\text{ m s}^{-2}$ as in the mark scheme. Ensure the driving force is greater than the resistance plus the weight component; otherwise the lorry could not maintain its speed up the hill.

Techniques used

find driving force from power and speedresolve the weight component along the inclineapply the equilibrium condition for constant speedsolve for the angle using inverse sine

Q5MediumMomentumEnergy, Work and Power

When a particle $P$ of mass $m\text{ kg}$ has speed $u\text{ m s}^{-1}$ , its momentum is $4\text{ N s}$ and its kinetic energy is $16\text{ J}$ .

(a)

Find the value of $m$ and the value of $u$ .

3M Similar

DifficultyMedium-Easy

Worked solution

Approach

Use the definitions of momentum and kinetic energy to form two equations in $m$ and $u$ , then solve them simultaneously.

Working

Momentum is given by $p = mu$ , so

mu = 4

Kinetic energy is given by $E_k = \frac{1}{2}mu^2$ , so

\frac{1}{2}mu^2 = 16

From $mu = 4$ , substitute $u = \frac{4}{m}$ into the kinetic energy equation:

\frac{1}{2}m\left(\frac{4}{m}\right)^2 = 16

\frac{1}{2}m \cdot \frac{16}{m^2} = 16

\frac{8}{m} = 16

m = 0.5

Then

u = \frac{4}{0.5} = 8

Answer

m = 0.5\text{ kg}, \quad u = 8\text{ m s}^{-1}

Final answer

m = 0.5 kg, u = 8 m/s

Detailed explanation

Walkthrough

We are told that the momentum of the particle is $4\text{ N s}$ and its kinetic energy is $16\text{ J}$ . Momentum is mass times velocity, so $mu = 4$ . Kinetic energy is $\frac{1}{2}mu^2$ , so $\frac{1}{2}mu^2 = 16$ . These are two equations in the two unknowns $m$ and $u$ . From the momentum equation, $u = \frac{4}{m}$ . Substituting into the kinetic energy equation gives $\frac{1}{2}m\left(\frac{4}{m}\right)^2 = 16$ , which simplifies to $\frac{8}{m} = 16$ , so $m = 0.5$ . Then $u = \frac{4}{0.5} = 8$ . The units are kg for mass and m/s for speed.

Key Takeaways

Momentum and kinetic energy are both expressed in terms of mass and speed.
Two independent equations are needed to find two unknowns.
A quick check: $\frac{1}{2}(0.5)(8^2) = 16\text{ J}$ and $0.5(8) = 4\text{ N s}$ .

Common Mistakes

Forgetting the factor of $\frac{1}{2}$ in kinetic energy.
Mixing up momentum $mu$ and kinetic energy $\frac{1}{2}mu^2$ .
Substituting incorrectly, for example using $u = 4m$ instead of $u = \frac{4}{m}$ .

Things to Be Careful About

Momentum is a vector, but in this one-dimensional context we use its magnitude.
Units: momentum in N s is equivalent to kg m/s; kinetic energy in J.
The mark scheme awards a mark for each correct equation and one for the final values, so show both equations clearly.

Techniques used

write the momentum equationwrite the kinetic energy equationsolve simultaneous equations

(b)

$P$ is now projected on a smooth horizontal surface with speed $v\text{ m s}^{-1}$ directly towards a particle $Q$ of mass $1.25\text{ kg}$ which is stationary. After $P$ and $Q$ collide, the velocity of $P$ is $w\text{ m s}^{-1}$ and the velocity of $Q$ is $2w\text{ m s}^{-1}$ . The loss of kinetic energy in the collision is $25\text{ J}$ .

Find the value of $v$ and the value of $w$ .

4M Similar

DifficultyMedium

Worked solution

Approach

Use conservation of linear momentum during the collision to relate $v$ and $w$ . Then use the given kinetic energy loss to form a second equation and solve for $w$ and $v$ .

Working

From part (a), $m = 0.5\text{ kg}$ .

Before the collision, only $P$ is moving, so the total momentum is

0.5v

After the collision:

0.5w + 1.25(2w) = 0.5w + 2.5w = 3w

Conservation of momentum gives:

0.5v = 3w

v = 6w

Total kinetic energy before the collision:

\frac{1}{2}(0.5)v^2 = 0.25v^2

Total kinetic energy after the collision:

\frac{1}{2}(0.5)w^2 + \frac{1}{2}(1.25)(2w)^2 = 0.25w^2 + 2.5w^2 = 2.75w^2

The loss of kinetic energy is $25\text{ J}$ , so

0.25v^2 - 2.75w^2 = 25

Substitute $v = 6w$ :

0.25(6w)^2 - 2.75w^2 = 25

9w^2 - 2.75w^2 = 25

6.25w^2 = 25

w^2 = 4

Since $w$ is a speed, $w = 2$ . Then

v = 6(2) = 12

Answer

v = 12\text{ m s}^{-1}, \quad w = 2\text{ m s}^{-1}

Final answer

v = 12 m/s, w = 2 m/s

Detailed explanation

Walkthrough

From part (a), the mass of $P$ is $0.5\text{ kg}$ . Before the collision, $P$ moves at speed $v$ and $Q$ is stationary, so the total momentum is $0.5v$ . After the collision, $P$ moves at $w$ and $Q$ moves at $2w$ , so the total momentum is $0.5w + 1.25(2w) = 3w$ . Since no external horizontal force acts during the collision, momentum is conserved, giving $0.5v = 3w$ , so $v = 6w$ .

Next, write down the total kinetic energy before and after the collision. Before: $\frac{1}{2}(0.5)v^2 = 0.25v^2$ . After: $\frac{1}{2}(0.5)w^2 + \frac{1}{2}(1.25)(2w)^2 = 0.25w^2 + 2.5w^2 = 2.75w^2$ . The loss of kinetic energy is the initial value minus the final value, so $0.25v^2 - 2.75w^2 = 25$ . Substitute $v = 6w$ to get $9w^2 - 2.75w^2 = 25$ , so $6.25w^2 = 25$ , giving $w^2 = 4$ . Since $w$ is a speed, $w = 2$ , and then $v = 6(2) = 12$ .

Key Takeaways

Conservation of linear momentum applies in direct impact problems when no external impulse acts.
Kinetic energy is not necessarily conserved in a collision; here the loss is given.
Reduce the two equations to one variable before solving.

Common Mistakes

Using the wrong mass for $P$ ; it must be $0.5\text{ kg}$ from part (a).
Forgetting to include $Q$ 's kinetic energy after the collision.
Writing the loss as final minus initial instead of initial minus final. The mark scheme allows $\pm 25$ , but the equation must be consistent.
Using $mg$ instead of $m$ in momentum or kinetic energy; the mark scheme gives no credit for this.

Things to Be Careful About

Momentum is a vector, but all motion is in one direction, so signs can be taken positive.
The mark scheme requires using the value of $m$ from part (a).
The final answer requires both $v$ and $w$ ; finding only one may lose the final mark.
Since speeds are positive, take the positive square root when solving $w^2 = 4$ .

Techniques used

apply conservation of linear momentumexpress kinetic energy before and after collisionform the kinetic energy loss equationsolve simultaneous equations

Q6MediumNewton's Laws of MotionForces and EquilibriumEnergy, Work and Power

Fig. 6.1

Two particles, $P$ and $Q$ , of masses $0.3\text{ kg}$ and $0.6\text{ kg}$ respectively, are attached to the ends of a light inextensible string. The string passes over a smooth pulley fixed at a point $B$ where the inclined planes $AB$ and $BC$ meet. $P$ lies on the smooth plane $AB$ which is inclined at an angle $\theta^\circ$ to the horizontal where $\sin \theta^\circ = 0.4$ . $Q$ lies on the plane $BC$ which is inclined at $30^\circ$ to the horizontal. The string is taut and the particles can move on lines of greatest slope of the two planes (see diagram). The particles are released from rest.

(a)

It is given that the plane $BC$ is smooth.

Find the tension in the string and the acceleration of $Q$ .

5M Similar

DifficultyMedium

Worked solution

Approach

Apply Newton's second law to each particle separately. Since Q (0.6 kg) is on the steeper plane (30°) and has greater mass than P (0.3 kg on a plane with sin θ = 0.4, so θ ≈ 24°), Q accelerates down plane BC and P accelerates up plane AB. Write equations of motion for both particles and solve simultaneously.

Working

For particle Q (moving down plane BC at 30°), taking the direction of motion as positive:

0.6g \sin 30^\circ - T = 0.6a

For particle P (moving up plane AB where $\sin \theta^\circ = 0.4$ ), taking the direction of motion as positive:

T - 0.3g \times 0.4 = 0.3a

Adding the two equations to eliminate T:

0.6g \sin 30^\circ - 0.3g \times 0.4 = (0.3 + 0.6)a

Using $g = 10 \text{ m s}^{-2}$ :

0.6 \times 10 \times 0.5 - 0.3 \times 10 \times 0.4 = 0.9a

3.0 - 1.2 = 0.9a

1.8 = 0.9a

a = 2 \text{ m s}^{-2}

Substituting $a = 2$ into the equation for P to find T:

T = 0.3 \times 10 \times 0.4 + 0.3 \times 2

T = 1.2 + 0.6 = 1.8 \text{ N}

Answer

T = 1.8 \text{ N}, \quad a = 2 \text{ m s}^{-2}

Final answer

T = 1.8 N, a = 2 m s^-2

Detailed explanation

Walkthrough

First, we identify the direction of motion. Particle Q has mass 0.6 kg on a 30° plane, so the component of its weight down the plane is $0.6g \sin 30^\circ = 3.0$ N (using $g = 10$ ). Particle P has mass 0.3 kg on a plane with $\sin \theta^\circ = 0.4$ , so the component of its weight down the plane is $0.3g \times 0.4 = 1.2$ N. Since 3.0 > 1.2, Q moves down and P moves up.

We apply Newton's second law ( $F_{net} = ma$ ) to each particle along the direction of the string. For Q, the forces along the plane are the component of weight down the plane ( $0.6g \sin 30^\circ$ ) and the tension pulling up ( $T$ ), giving $0.6g \sin 30^\circ - T = 0.6a$ . For P, the tension pulls up the plane and the weight component pulls down, giving $T - 0.3g \times 0.4 = 0.3a$ .

Adding these equations eliminates T because the tension is the same throughout the light inextensible string. This gives a single equation in $a$ , which we solve to get $a = 2$ m s $^{-2}$ . Substituting back gives $T = 1.8$ N.

Key Takeaways

For connected particles over a smooth pulley, apply Newton's second law to each particle separately along the direction of motion.
The tension is the same throughout a light inextensible string passing over a smooth pulley.
Adding the two equations eliminates the tension when the particles have the same acceleration magnitude.

Common Mistakes

Using the wrong component of weight (using cos instead of sin for the inclined plane).
Forgetting that both particles have the same acceleration magnitude $a$ .
Sign errors when setting up equations of motion (not defining a consistent positive direction for each particle).
Missing $g$ in the weight components.

Things to Be Careful About

Always use $g = 10$ m s $^{-2}$ unless otherwise stated in Cambridge exams.
The angle $\theta^\circ$ is given via $\sin \theta^\circ = 0.4$ , not $\theta^\circ = 24^\circ$ exactly; use the given sine value directly.
Both particles accelerate at the same rate because the string is inextensible.

Techniques used

apply Newton's second law to each particleresolve weight components along inclined planessolve simultaneous equations for acceleration and tension

(b)

It is given instead that the plane $BC$ is rough. The work done against the frictional force when $Q$ moves $2\text{ m}$ down the plane is $1.8\text{ J}$ . You should assume that $P$ does not reach the pulley and that $Q$ does not reach $C$ .

Use an energy method to find the speed of $Q$ when it has moved $2\text{ m}$ down the plane.

4M Similar

DifficultyMedium

Worked solution

Approach

Use the work-energy principle: the net loss in gravitational potential energy equals the gain in kinetic energy plus the work done against frictional forces. Calculate the vertical displacement of each particle when Q moves 2 m down the plane.

Working

When Q moves 2 m down plane BC (inclined at 30°), the vertical drop is:

h_Q = 2 \sin 30^\circ = 2 \times 0.5 = 1 \text{ m}

When Q moves 2 m down, P moves 2 m up plane AB (where $\sin \theta^\circ = 0.4$ ), so the vertical rise of P is:

h_P = 2 \times 0.4 = 0.8 \text{ m}

Loss in PE for Q:

\Delta PE_Q = 0.6g \times h_Q = 0.6 \times 10 \times 1 = 6 \text{ J}

Gain in PE for P:

\Delta PE_P = 0.3g \times h_P = 0.3 \times 10 \times 0.8 = 2.4 \text{ J}

Net loss in PE:

\Delta PE_{net} = 6 - 2.4 = 3.6 \text{ J}

Gain in KE (both particles move at speed $v$ ):

\Delta KE = \frac{1}{2}(0.3 + 0.6)v^2 = 0.45v^2

Work done against friction on Q: $1.8 \text{ J}$

Applying the work-energy principle (net loss in PE = gain in KE + work done against friction):

0.45v^2 = 3.6 - 1.8

0.45v^2 = 1.8

v^2 = \frac{1.8}{0.45} = 4

v = 2 \text{ m s}^{-1}

Answer

v = 2 \text{ m s}^{-1}

Final answer

v = 2 m s^-1

Detailed explanation

Walkthrough

The work-energy principle states that the net work done on the system equals the change in kinetic energy. Here, the forces doing work are gravity (conservative) and friction (non-conservative).

When Q moves 2 m down the 30° plane, it drops vertically by $2 \sin 30^\circ = 1$ m, losing $0.6 \times 10 \times 1 = 6$ J of PE. Simultaneously, P moves 2 m up the plane with $\sin \theta^\circ = 0.4$ , rising vertically by $2 \times 0.4 = 0.8$ m, gaining $0.3 \times 10 \times 0.8 = 2.4$ J of PE. The net loss in PE is $6 - 2.4 = 3.6$ J.

This net PE loss goes into two places: the kinetic energy of both particles (which move at the same speed $v$ since they're connected by an inextensible string) and the work done against friction (1.8 J). Setting up the equation $0.45v^2 = 3.6 - 1.8$ gives $v^2 = 4$ , so $v = 2$ m s $^{-1}$ .

Key Takeaways

The work-energy principle is a powerful alternative to Newton's second law for connected particle problems.
When using energy methods, calculate the net change in PE for the entire system, not just individual particles.
Both particles share the same speed at any instant because they're connected by an inextensible string.
Work done against friction is always positive and represents energy removed from the mechanical system.

Common Mistakes

Forgetting to include the PE change of particle P (only calculating Q's PE change).
Using the wrong vertical displacement (e.g., using 2 m directly instead of $2 \sin 30^\circ$ ).
Sign errors in the work-energy equation (adding friction work instead of subtracting it).
Missing $g$ in the PE calculations.

Things to Be Careful About

The question asks for an energy method; using Newton's second law with kinematics (as shown in the special case of the mark scheme) would not earn full marks for this part.
The work done against friction is given as 1.8 J directly, so you don't need to calculate the friction force or coefficient.
Both particles start from rest, so the initial KE is zero.

Techniques used

calculate changes in gravitational potential energy for both particlesapply the work-energy principleaccount for work done against friction

Q7MediumKinematics of Motion in a Straight Line

A particle $X$ moves along a straight track, starting from a point $O$ at time $t = 0$ . The displacement of $X$ from $O$ at time $t\text{ s}$ is $s\text{ m}$ , where $s = 3t^{\frac{3}{2}} - 6t$ .

(a)

Find the time at which $X$ is instantaneously at rest, and hence find the total distance travelled by $X$ between $t = 0$ and $t = 16$ .

6M Similar

DifficultyMedium

Worked solution

Approach

Differentiate the displacement $s = 3t^{3/2} - 6t$ to obtain the velocity. Set $v = 0$ to find the instant when $X$ is instantaneously at rest. Since the particle reverses direction at that instant, split the interval at $t = \frac{16}{9}$ and add the magnitudes of the two displacement changes.

Working

Differentiate:

v = \frac{ds}{dt} = \frac{9}{2}t^{1/2} - 6

Set $v = 0$ :

\frac{9}{2}t^{1/2} - 6 = 0

t^{1/2} = \frac{4}{3}

t = \frac{16}{9}

So $X$ is instantaneously at rest at $t = \frac{16}{9}$ s.

Let $F(t) = 3t^{3/2} - 6t$ . Then:

F(0) = 0

F\left(\frac{16}{9}\right) = 3\left(\frac{16}{9}\right)^{3/2} - 6\left(\frac{16}{9}\right) = \frac{64}{9} - \frac{96}{9} = -\frac{32}{9}

F(16) = 3(16)^{3/2} - 6(16) = 192 - 96 = 96

The total distance is the sum of the absolute changes in displacement:

\text{distance} = \left|F\left(\frac{16}{9}\right) - F(0)\right| + \left|F(16) - F\left(\frac{16}{9}\right)\right|

= \frac{32}{9} + \left(96 + \frac{32}{9}\right) = \frac{32}{9} + \frac{896}{9} = \frac{928}{9}

Answer

t = \frac{16}{9}\text{ s}, \qquad \text{total distance} = \frac{928}{9}\text{ m}

Final answer

t = 16/9 s; total distance = 928/9 m

Detailed explanation

Walkthrough

The displacement of $X$ is $s = 3t^{3/2} - 6t$ . To find when the particle is at rest, differentiate $s$ with respect to $t$ to get velocity:

v = \frac{ds}{dt} = \frac{9}{2}t^{1/2} - 6

Setting $v = 0$ gives $t^{1/2} = \frac{4}{3}$ , so $t = \frac{16}{9}$ . At this instant the particle changes direction, so the total distance is not simply $F(16) - F(0)$ .

Evaluate the displacement at the three key times. At $t = 0$ , $F(0)=0$ . At $t = \frac{16}{9}$ , the particle has moved backwards to $F\left(\frac{16}{9}\right) = -\frac{32}{9}$ . At $t = 16$ , it has moved forwards to $F(16)=96$ . The distance travelled is therefore the distance backwards from $0$ to $-\frac{32}{9}$ , plus the distance forwards from $-\frac{32}{9}$ to $96$ :

\frac{32}{9} + \left(96 + \frac{32}{9}\right) = \frac{928}{9}

Key Takeaways

This question tests the difference between displacement and distance. Displacement is a signed quantity; distance is the total length of path travelled. Whenever a particle changes direction, the interval must be split at the instant when $v=0$ and the absolute changes in displacement added.

Common Mistakes

Using $F(16)-F(0)=96$ as the total distance, which ignores the reversal of direction.
Forgetting to find the time at which $v=0$ before calculating distance.
Losing a sign when evaluating $F\left(\frac{16}{9}\right)$ .
Not using absolute values for the two parts of the distance.

Things to Be Careful About

The mark scheme requires a clear differentiation step: the power of $t$ must decrease by 1 in at least one term with a change of coefficient in the same term. It also requires both displacement changes to be combined correctly. Since $F\left(\frac{16}{9}\right)$ is negative, the distance from $t = \frac{16}{9}$ to $t = 16$ is $F(16) - F\left(\frac{16}{9}\right) = 96 + \frac{32}{9}$ , not $96 - \frac{32}{9}$ .

Techniques used

differentiate displacement to obtain velocitysolve velocity equal to zerosplit the interval at the turning pointsum absolute displacement changes to find distance

(b)

A second particle $Y$ moves along another straight track, starting from a point $P$ at time $t = 0$ . The acceleration of $Y$ at time $t\text{ s}$ is $a\text{ m s}^{-2}$ , where $a = 0.8 - 0.6t$ . The velocity of $Y$ when it leaves $P$ is $7.5\text{ m s}^{-1}$ .

When the velocity of $Y$ is $-9.6\text{ m s}^{-1}$ , show that the displacement of $X$ from $O$ is equal to the displacement of $Y$ from $P$ .

7M Similar

DifficultyMedium

Worked solution

Approach

Integrate the acceleration to obtain velocity, using the initial velocity to determine the constant. Set $v = -9.6$ and solve for $t$ . Then integrate the velocity to obtain displacement, and evaluate both $s_X$ and $s_Y$ at the resulting time.

Working

Integrate $a = 0.8 - 0.6t$ :

v = \int (0.8 - 0.6t)\,dt = 0.8t - 0.3t^2 + c

When $t=0$ , $v=7.5$ , so $c=7.5$ :

v = 0.8t - 0.3t^2 + 7.5

Set $v = -9.6$ :

0.8t - 0.3t^2 + 7.5 = -9.6

0.3t^2 - 0.8t - 17.1 = 0

Multiplying by 10:

3t^2 - 8t - 171 = 0

(3t + 19)(t - 9) = 0

So $t = 9$ or $t = -\frac{19}{3}$ . Since time is positive, $t = 9$ .

Integrate the velocity of $Y$ :

s_Y = \int (0.8t - 0.3t^2 + 7.5)\,dt = 0.4t^2 - 0.1t^3 + 7.5t + c

Since $Y$ starts at $P$ when $t=0$ , $s_Y(0)=0$ , so $c=0$ :

s_Y = 0.4t^2 - 0.1t^3 + 7.5t

At $t=9$ :

s_Y = 0.4(9)^2 - 0.1(9)^3 + 7.5(9) = 32.4 - 72.9 + 67.5 = 27

For $X$ :

s_X = 3(9)^{3/2} - 6(9) = 3(27) - 54 = 81 - 54 = 27

Therefore $s_X = s_Y = 27$ at $t=9$ .

Answer

s_X = s_Y = 27\text{ m at } t = 9\text{ s}

Final answer

At t = 9 s, s_X = s_Y = 27 m

Detailed explanation

Walkthrough

For $Y$ , the acceleration is $a = 0.8 - 0.6t$ . Integrate with respect to $t$ to find velocity:

v = \int (0.8 - 0.6t)\,dt = 0.8t - 0.3t^2 + c

The phrase “velocity when it leaves $P$ ” gives the initial condition $v(0)=7.5$ , so $c=7.5$ . Setting $v=-9.6$ gives a quadratic equation:

0.8t - 0.3t^2 + 7.5 = -9.6

Rearrange to $0.3t^2 - 0.8t - 17.1 = 0$ , or $3t^2 - 8t - 171 = 0$ . Factorising gives $(3t+19)(t-9)=0$ , so $t=9$ or $t=-\frac{19}{3}$ . The negative root is rejected because time starts at $t=0$ .

Now integrate the velocity of $Y$ to find displacement:

s_Y = \int (0.8t - 0.3t^2 + 7.5)\,dt = 0.4t^2 - 0.1t^3 + 7.5t + c

Since $Y$ starts at $P$ , $s_Y(0)=0$ , so the constant is zero. At $t=9$ :

s_Y = 0.4(9)^2 - 0.1(9)^3 + 7.5(9) = 27

For $X$ , substitute $t=9$ into $s_X = 3t^{3/2} - 6t$ :

s_X = 3(9)^{3/2} - 6(9) = 27

The two displacements are equal.

Key Takeaways

This question combines two integrations: acceleration to velocity and velocity to displacement. It also uses an initial condition to determine the constant of integration. Solving a quadratic is needed to find the time when a given velocity occurs.

Common Mistakes

Forgetting to include the constant of integration when integrating acceleration.
Using the wrong initial condition: the velocity when $Y$ leaves $P$ is $v(0)=7.5$ , not $v(0)=0$ .
Rejecting the negative root without checking that time must be positive.
Forgetting to integrate velocity again to get displacement.

Things to Be Careful About

The mark scheme allows follow-through: the displacement integral can be obtained from the candidate's velocity expression. The final comparison must be made at the positive time $t=9$ . When evaluating $s_Y$ , be careful with the signs of $-0.1(9)^3$ ; the calculation gives $32.4 - 72.9 + 67.5 = 27$ .

Techniques used

integrate acceleration to obtain velocitysolve a quadratic for the required timeintegrate velocity to obtain displacementevaluate displacements at the common time

Mathematics 9709/41 — May/June 2025

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