9709/45 Oct/Nov 2025 — Mathematics Solved Past Paper

Q15MMedium-EasyForces and Equilibrium

Fig. 1

Coplanar forces of magnitudes $P\text{ N}$ , $Q\text{ N}$ , $32\text{ N}$ and $21\text{ N}$ act at a point in the directions shown in the diagram. The forces are in equilibrium.

Find the value of $P$ and the value of $Q$ .

DifficultyMedium-Easy

Worked solution

Approach

Resolve the forces into horizontal and vertical components. Since the system is in equilibrium, the sum of horizontal components is zero and the sum of vertical components is zero. This gives two simultaneous equations in $P$ and $Q$ .

Working

Resolve horizontally (taking right as positive):

P + 21\cos 47^\circ - 32\cos 35^\circ - Q\cos 65^\circ = 0 \quad \text{--- (1)}

Resolve vertically (taking upward as positive):

21\sin 47^\circ + 32\sin 35^\circ - Q\sin 65^\circ = 0 \quad \text{--- (2)}

From equation (2), solve for $Q$ :

Q\sin 65^\circ = 21\sin 47^\circ + 32\sin 35^\circ

Q\sin 65^\circ = 15.358 + 18.355 = 33.713

Q = \frac{33.713}{\sin 65^\circ} = 37.198

Substitute $Q$ into equation (1) to find $P$ :

P = 32\cos 35^\circ + Q\cos 65^\circ - 21\cos 47^\circ

P = 26.213 + 37.198 \times \cos 65^\circ - 14.322

P = 26.213 + 15.720 - 14.322 = 27.611

Rounding to three significant figures:

P = 27.6 \quad \text{and} \quad Q = 37.2

Answer

P = 27.6, \; Q = 37.2

Final answer

P = 27.6, Q = 37.2

Detailed explanation

Walkthrough

First, we establish a coordinate system. Taking the positive horizontal axis to the right and the positive vertical axis upwards, we resolve each force into its horizontal and vertical components using sine and cosine of the given angles.

For the horizontal direction, the forces $P$ and $21\text{ N}$ act to the right (positive), while $32\text{ N}$ and $Q\text{ N}$ act to the left (negative). Setting the sum of horizontal components to zero gives equation (1).

For the vertical direction, the forces $21\text{ N}$ and $32\text{ N}$ act upwards (positive), while $Q\text{ N}$ acts downwards (negative). Setting the sum of vertical components to zero gives equation (2).

Equation (2) contains only one unknown, $Q$ , so we can solve it directly. We calculate the numerical values of the sine terms, sum them, and divide by $\sin 65^\circ$ to find $Q$ .

Once $Q$ is known, we substitute its value into equation (1) and rearrange to solve for $P$ . We calculate the cosine terms, substitute $Q$ , and find the final value for $P$ .

Key Takeaways

Any system of coplanar forces in equilibrium can be analysed by resolving forces along two perpendicular axes.
The sum of components along each axis must independently equal zero.
Equations with only one unknown should be solved first to simplify the system.

Common Mistakes

Mixing up sine and cosine when resolving forces (e.g., using $\sin$ for the horizontal component of a force given at an angle to the horizontal).
Sign errors when setting up the equations (e.g., adding a leftward force instead of subtracting it).
Rounding intermediate values too early, which leads to incorrect final answers.

Things to Be Careful About

Always carry at least 4-5 decimal places in intermediate calculations to avoid rounding errors.
Ensure the angles are correctly matched to their respective sine or cosine components based on whether they are measured from the horizontal or vertical axis.
The mark scheme allows sign errors and inconsistent sin/cos mixes as long as they are consistent throughout the working, but final answers must be correct for the equations set up.

Techniques used

resolve forces horizontallyresolve forces verticallysolve simultaneous equations

Approach

The motion consists of three phases: constant acceleration from rest, constant velocity, and constant deceleration to rest. The velocity-time graph will be a trapezium starting and finishing on the time axis.

Working

Phase 1 (Acceleration): Accelerates from $0$ to $V$ at $0.5 \text{ m s}^{-2}$ . Time taken $= \frac{V}{0.5} = 2V$ . The graph is a straight line from $(0, 0)$ to $(2V, V)$ .
Phase 2 (Constant velocity): Maintains speed $V$ for $T$ seconds. The graph is a horizontal line from $(2V, V)$ to $(2V+T, V)$ .
Phase 3 (Deceleration): Decelerates from $V$ to $0$ at $0.2 \text{ m s}^{-2}$ . Time taken $= \frac{V}{0.2} = 5V$ . The graph is a straight line from $(2V+T, V)$ to $(2V+T+5V, 0) = (7V+T, 0)$ .
Total time: $7V + T = 246$ , so the graph ends at $(246, 0)$ .

The velocity-time graph is a trapezium with vertices at $(0, 0)$ , $(2V, V)$ , $(246-5V, V)$ , and $(246, 0)$ . The horizontal axis is $t \text{ (s)}$ and the vertical axis is $v \text{ (ms}^{-1}\text{)}$ . The values $V$ and $246$ are labelled on the axes.

Answer

A trapezium on $v$ - $t$ axes with vertices at $(0,0)$ , $(2V, V)$ , $(246-5V, V)$ , and $(246, 0)$ . The points $246$ on the $t$ -axis and $V$ on the $v$ -axis are labelled.

Final answer

Trapezium with vertices (0,0), (2V, V), (246-5V, V), and (246, 0)

Detailed explanation

Walkthrough

The athlete's motion has three distinct phases. First, they accelerate from rest at a constant rate, which produces a straight line with a positive gradient on a velocity-time graph. Second, they run at a constant speed, producing a horizontal line. Third, they decelerate at a constant rate to rest, producing a straight line with a negative gradient. Connecting these three phases gives a trapezium shape that starts and ends on the time axis (velocity = 0).

Key Takeaways

A velocity-time graph for constant acceleration, constant velocity, and constant deceleration forms a trapezium. The area under this graph represents the total distance travelled, and the gradients of the sloping lines represent the acceleration and deceleration.

Common Mistakes

Drawing a triangle instead of a trapezium (forgetting the constant velocity phase).
Failing to label the key values $V$ and $246$ on the axes.
Drawing the sloping lines with incorrect relative gradients (the deceleration gradient should be less steep than the acceleration gradient since $0.2 < 0.5$ ).

Things to Be Careful About

The question only asks for a sketch, so exact scale is not required. However, the shape must be a trapezium, and the labels $V$ and $246$ must be present. Ignore gradients for marking purposes, but ensure the deceleration line is less steep than the acceleration line to reflect the given values.

Techniques used

sketch velocity-time graphidentify trapezium shape

(b)

Find $T$ in terms of $V$ , and hence show that $7V^2 - 492V + 2700 = 0$ .

4M Similar

DifficultyMedium

Worked solution

Approach

Use the kinematic equation $v = u + at$ to find the time spent accelerating and decelerating in terms of $V$ . Then use the total time to find $T$ in terms of $V$ . Finally, use the area under the velocity-time graph (which is a trapezium) to set up an equation equal to the total distance of $1350 \text{ m}$ , and simplify to get the required quadratic.

Working

Time for each phase:
Time accelerating ( $t_1$ ):

v = u + at \implies V = 0 + 0.5 t_1 \implies t_1 = \frac{V}{0.5} = 2V

Time decelerating ( $t_2$ ):

0 = V - 0.2 t_2 \implies t_2 = \frac{V}{0.2} = 5V

Find $T$ in terms of $V$ :
Total time is $246 \text{ s}$ :

t_1 + T + t_2 = 246 \implies 2V + T + 5V = 246 \implies T = 246 - 7V

Equation for total distance:
The total distance is the area under the velocity-time graph. The graph is a trapezium with parallel sides of length $246$ and $T$ , and height $V$ :

\text{Area} = \frac{1}{2}(a + b)h

\text{Total distance} = \frac{1}{2}(246 + T)V = 1350

Substitute $T = 246 - 7V$ :

\frac{1}{2}(246 + 246 - 7V)V = 1350

\frac{1}{2}(492 - 7V)V = 1350

Multiply both sides by 2:

(492 - 7V)V = 2700

Expand:

492V - 7V^2 = 2700

Rearrange into standard quadratic form:

7V^2 - 492V + 2700 = 0

Answer

7V^2 - 492V + 2700 = 0

Final answer

7V^2 - 492V + 2700 = 0

Detailed explanation

Walkthrough

First, we find the time spent in each phase using $v = u + at$ . For acceleration, $V = 0 + 0.5t_1$ , so $t_1 = 2V$ . For deceleration, $0 = V - 0.2t_2$ , so $t_2 = 5V$ . The total time is the sum of the three phases: $2V + T + 5V = 246$ , which gives $T = 246 - 7V$ .

Next, we use the fact that the total distance is the area under the velocity-time graph. The graph is a trapezium with parallel sides $246$ (the total time) and $T$ (the time at constant speed), and height $V$ (the maximum speed). The area formula gives $\frac{1}{2}(246 + T)V = 1350$ . Substituting $T = 246 - 7V$ yields $\frac{1}{2}(492 - 7V)V = 1350$ . Multiplying by 2 and expanding gives $492V - 7V^2 = 2700$ , which rearranges to $7V^2 - 492V + 2700 = 0$ .

Key Takeaways

The area under a velocity-time graph equals the displacement. For constant acceleration phases, you can use $v = u + at$ to find times, and the trapezium area formula to find total distance efficiently.

Common Mistakes

Forgetting to include the constant velocity phase when calculating total time or distance.
Using the wrong formula for the area of a trapezium (e.g., using only the rectangle area $VT$ and forgetting the triangles).
Algebraic errors when expanding and rearranging the equation into standard quadratic form.

Things to Be Careful About

Ensure all terms are present when setting up the equation. If you substitute $T$ back into the trapezium area equation and get $1350 = 1350$ , you have not formed a valid equation for $V$ . You must use an independent equation (like total time) to eliminate $T$ before substituting into the distance equation.

Techniques used

calculate time for each phase using v=u+atcalculate total distance using area under graphform and simplify quadratic equation

(c)

Find the value of $V$ , giving a justification for your answer.

2M Similar

DifficultyMedium-Easy

Worked solution

Approach

Solve the quadratic equation $7V^2 - 492V + 2700 = 0$ using the quadratic formula. Then use physical constraints (e.g., $T > 0$ or total time for acceleration and deceleration must be less than $246 \text{ s}$ ) to reject the larger root.

Working

Solve $7V^2 - 492V + 2700 = 0$ :

V = \frac{-(-492) \pm \sqrt{(-492)^2 - 4(7)(2700)}}{2(7)}

V = \frac{492 \pm \sqrt{242064 - 75600}}{14}

V = \frac{492 \pm \sqrt{166464}}{14}

V = \frac{492 \pm 408}{14}

This gives two possible values for $V$ :

V = \frac{492 + 408}{14} = \frac{900}{14} = \frac{450}{7} \approx 64.3 \text{ m s}^{-1}

V = \frac{492 - 408}{14} = \frac{84}{14} = 6 \text{ m s}^{-1}

Justification for rejecting $V = \frac{450}{7}$ :
If $V = \frac{450}{7}$ , the time spent at constant speed would be:

T = 246 - 7V = 246 - 7\left(\frac{450}{7}\right) = 246 - 450 = -204 \text{ s}

Since time cannot be negative, $V = \frac{450}{7}$ is not physically feasible. Alternatively, the time to decelerate alone would be $\frac{450/7}{0.2} = \frac{2250}{7} \approx 321.4 \text{ s}$ , which exceeds the total time of $246 \text{ s}$ .

Thus, the only valid solution is $V = 6$ .

Answer

V = 6

Final answer

V = 6

Detailed explanation

Walkthrough

First, apply the quadratic formula to $7V^2 - 492V + 2700 = 0$ . The discriminant is $(-492)^2 - 4(7)(2700) = 242064 - 75600 = 166464$ . The square root of $166464$ is $408$ . This gives $V = \frac{492 \pm 408}{14}$ , yielding $V = \frac{450}{7}$ or $V = 6$ .

Next, we must check which value is physically meaningful. We know $T = 246 - 7V$ . If $V = \frac{450}{7}$ , then $T = 246 - 450 = -204$ , which is impossible since time cannot be negative. Therefore, we reject $V = \frac{450}{7}$ and accept $V = 6$ .

Key Takeaways

When solving physics problems that lead to quadratic equations, always check the solutions against physical constraints (e.g., time > 0, speed < reasonable human limit). Extraneous mathematical solutions often arise and must be discarded.

Common Mistakes

Stating both values of $V$ without rejecting the invalid one.
Failing to show the calculation that proves the rejected root is invalid (e.g., just saying "it's too fast" without calculating $T < 0$ ).
Arithmetic errors when evaluating the quadratic formula or the discriminant.

Things to Be Careful About

The question asks for a justification. You must explicitly show why the larger root is rejected, for example by calculating $T$ and showing it is negative, or by showing the deceleration time exceeds the total time. Simply stating $T > 0$ without reference to the other root may not earn full marks.

Techniques used

solve quadratic equationapply physical constraints to reject extraneous root

Q3MediumEnergy, Work and PowerForces and EquilibriumNewton's Laws of Motion

A car of mass $800\text{ kg}$ is moving on a straight road.

When the car is moving at a constant speed of $20\text{ m s}^{-1}$ on a horizontal section of the road, the engine of the car is working at $P\text{ W}$ .

When the car is moving at a constant speed of $12\text{ m s}^{-1}$ up a section of the road inclined at $\sin^{-1} 0.15$ to the horizontal, the engine of the car is also working at $P\text{ W}$ .

On both sections of the road there is a constant force of magnitude $R\text{ N}$ resisting the motion of the car.

(a)

Find the value of $R$ and the value of $P$ .

5M Similar

DifficultyMedium

Worked solution

Approach

Let $F_1$ be the driving force on the horizontal section and $F_2$ that on the inclined section. Since the car moves at constant speed, its acceleration is zero. Use $P = Fv$ to express the driving forces, then apply Newton's second law parallel to the road on each section.

Working

On the horizontal section, the driving force is

F_1 = \frac{P}{20}

and, since the acceleration is zero,

\frac{P}{20} - R = 0 \quad \Longrightarrow \quad P = 20R.

On the inclined section, let $\alpha = \sin^{-1} 0.15$ , so $\sin \alpha = 0.15$ . The component of the weight down the slope is $800g \sin \alpha = 800g \times 0.15$ . The driving force is

F_2 = \frac{P}{12}.

Applying Newton's second law parallel to the slope at constant speed,

\frac{P}{12} - R - 800g \times 0.15 = 0.

Using $P = 20R$ , we have $P/12 = 5R/3$ . Taking $g = 10\text{ m s}^{-2}$ , the weight component is $1200\text{ N}$ , so

\frac{5R}{3} - R - 1200 = 0.

Hence

\frac{2R}{3} = 1200 \quad \Longrightarrow \quad R = 1800.

Then

P = 20R = 20(1800) = 36000.

Answer

R = 1800\text{ N}, \qquad P = 36000\text{ W}.

Final answer

R = 1800 N, P = 36000 W

Detailed explanation

Walkthrough

The important observation is that constant speed means zero acceleration, so the resultant force on the car is zero. On each road section, resolve parallel to the road. Use $P = Fv$ to write the engine's driving force as $P/v$ ; this gives $P/20$ on the flat road and $P/12$ on the slope. On the flat road the only horizontal forces are the driving force and the resistance, so $P/20 - R = 0$ . On the slope, the weight has a component $800g\sin\alpha = 800g \times 0.15$ down the slope, which must be included as an extra resisting term. This gives the equation $P/12 - R - 800g \times 0.15 = 0$ . The two equations are then solved simultaneously. Replacing $P/12$ by $5R/3$ uses $P = 20R$ and makes the algebra straightforward.

Key Takeaways

A constant speed implies $a = 0$ , so the resultant force along the direction of motion is zero.
The relation $P = Fv$ connects engine power to the driving force at a given speed.
On an incline, the component of the weight parallel to the plane is $mg\sin\alpha$ .
Two equations in two unknowns can be solved by substitution.

Common Mistakes

Giving only the final values without showing the equations. The mark scheme awards marks for the equations, so unsupported answers score no marks.
Using $\cos\alpha$ instead of $\sin\alpha$ for the component of weight down the slope.
Forgetting the resistance on the slope, or using two different resistance forces in the two equations.
Rounding $\sin^{-1}0.15$ to $8.6^\circ$ and then using $\sin 8.6^\circ$ ; this can prevent the final answers from being correct to 3 significant figures.

Things to Be Careful About

Choose a positive direction (up the slope for the inclined section) and be consistent with signs.
The force from the engine is not $P$ ; it is $P/v$ in newtons.
Include the $mg\sin\alpha$ term exactly, with $\sin\alpha = 0.15$ , not an approximated angle.
With $g = 10\text{ m s}^{-2}$ , the weight component is $1200\text{ N}$ , leading to $R = 1800\text{ N}$ and $P = 36000\text{ W}$ .

Techniques used

apply the power relationship P = Fvresolve forces parallel to the roaduse zero-acceleration equilibrium equationssolve a pair of simultaneous equations

(b)

Find the acceleration of the car when it is moving at $10\text{ m s}^{-1}$ up the inclined section of the road with the engine working at $32\text{ kW}$ .

3M Similar

DifficultyMedium-Easy

Worked solution

Approach

The engine now works at $32\text{ kW}$ , which is $32000\text{ W}$ . At speed $10\text{ m s}^{-1}$ , use $P = Fv$ to find the driving force. Then apply Newton's second law parallel to the incline, using $R = 1800\text{ N}$ from part (a) and the weight component $800g\sin\alpha = 1200\text{ N}$ .

Working

The driving force is

F = \frac{32000}{10} = 3200\text{ N}.

Take the direction up the slope as positive. Newton's second law gives

F - R - 800g \times 0.15 = 800a.

Substitute $F = 3200$ , $R = 1800$ and $g = 10$ :

3200 - 1800 - 1200 = 800a.

200 = 800a.

Therefore

a = \frac{200}{800} = 0.25\text{ m s}^{-2}.

Answer

a = 0.25\text{ m s}^{-2}.

Final answer

a = 0.25 m s^-2

Detailed explanation

Walkthrough

Part (b) uses the same ideas but now there is acceleration. First convert the power to watts: $32\text{ kW} = 32000\text{ W}$ . The driving force is obtained from $P = Fv$ : $F = 32000/10 = 3200\text{ N}$ . Resolve parallel to the slope, taking up the slope as positive. The driving force acts up the slope, while the resistance $R$ and the component $800g\sin\alpha$ act down the slope. Newton's second law gives $F - R - 800g\sin\alpha = 800a$ . Substitute $R = 1800\text{ N}$ from part (a) and the weight component $1200\text{ N}$ , then solve $3200 - 1800 - 1200 = 800a$ to obtain $200 = 800a$ , so $a = 0.25\text{ m s}^{-2}$ .

Key Takeaways

Convert kW to W before using $P = Fv$ .
The driving force from the engine is $P/v$ , not the power itself.
On an incline, Newton's second law must include the component of weight parallel to the plane.
A previous part may supply a needed value, such as $R$ , for later calculations.

Common Mistakes

Forgetting to convert $32\text{ kW}$ to $32000\text{ W}$ .
Omitting either the resistance or the weight component when writing the Newton's second law equation.
Using the value of $P$ from part (a) instead of the new power $32\text{ kW}$ .
Using $ma$ on the wrong side or with a sign error; since motion is up the slope, use $F - R - mg\sin\alpha = ma$ .

Things to Be Careful About

Use the $R$ from part (a) consistently; it is the same resistance force on both sections.
The weight component is $mg\sin\alpha$ with $\sin\alpha = 0.15$ , so it is $1200\text{ N}$ when $g = 10\text{ m s}^{-2}$ .
If an approximate angle such as $8.63^\circ$ is used, the answer may be $0.249\text{ m s}^{-2}$ , which the mark scheme condones; using $8.6^\circ$ would not be correct to 3 significant figures.

Techniques used

compute driving force from power and speedapply Newton's second law parallel to the inclinesubstitute the resistance and the component of weightsolve for the acceleration

Q4MediumKinematics of Motion in a Straight Line

A particle $P$ moves in a straight line. At time $t\text{ s}$ after leaving a point $O$ on the line, the acceleration $a\text{ m s}^{-2}$ of $P$ is given by $a = kt - 3$ , where $k$ is a positive constant. At time $t = 0$ , the velocity of $P$ is $1\text{ m s}^{-1}$ .

(a)

Given that $P$ is never at instantaneous rest, show that $k > 4.5$ .

4M Similar

DifficultyMedium

Worked solution

Approach

Integrate the acceleration to obtain the velocity, use the initial condition to fix the constant, then require the quadratic for $v=0$ to have no real roots by setting its discriminant negative.

Working

Acceleration is

a = kt - 3

Integrating with respect to $t$ :

v = \frac{1}{2}kt^2 - 3t + c

At $t=0$ , $v=1$ , so $c=1$ :

v = \frac{1}{2}kt^2 - 3t + 1

For $P$ never to be at instantaneous rest, $v=0$ must have no real solutions. The quadratic is

\frac{1}{2}kt^2 - 3t + 1 = 0

Its discriminant is

(-3)^2 - 4\left(\frac{1}{2}k\right)(1) = 9 - 2k

For no real roots:

9 - 2k < 0 \implies k > 4.5

Answer

k > 4.5

Final answer

k > 4.5

Detailed explanation

Walkthrough

We are told that the acceleration is a function of time, so the velocity is found by integrating $a$ with respect to $t$ . The initial condition $v=1$ at $t=0$ determines the constant of integration.

The phrase 'never at instantaneous rest' means that the velocity is never zero. Since the velocity is a quadratic function of $t$ , the equation $v=0$ must have no real solutions. A quadratic has no real roots exactly when its discriminant is negative. Setting $9-2k<0$ gives the required inequality.

Key Takeaways

This question combines calculus with the algebraic condition for a quadratic to have no real roots. It shows that 'instantaneous rest' corresponds to $v=0$ , and that a quadratic velocity function stays away from zero only when its discriminant is negative.

Common Mistakes

Using $v=at$ instead of integrating $a$ .
Forgetting the constant of integration and not using $v=1$ at $t=0$ .
Using $\le 0$ instead of $<0$ ; a zero discriminant would give one instant of rest.
Making an algebraic error when simplifying $9-2k<0$ .

Things to Be Careful About

The inequality is strict: $k > 4.5$ , not $k \ge 4.5$ . The coefficient of $t^2$ is $\frac{1}{2}k$ , so the discriminant is $9-2k$ . Since $k$ is given as positive, the result is consistent with the domain of the problem.

Techniques used

integrate acceleration to obtain velocityapply initial condition to find constant of integrationset discriminant of quadratic to be negativededuce inequality for parameter

(b)

Given instead that $k = 2.5$ , find the total distance travelled by $P$ in the interval $1 \le t \le 3$ .

6M Similar

DifficultyMedium

Worked solution

Approach

Substitute $k=2.5$ into the velocity expression, find the times in the interval when $v=0$ , split the interval at that time, integrate $v$ to get displacement on each sub-interval, and add the magnitudes.

Working

With $k=2.5 = \frac{5}{2}$ :

v = \frac{1}{2}\left(\frac{5}{2}\right)t^2 - 3t + 1 = \frac{5}{4}t^2 - 3t + 1

Set $v=0$ :

\frac{5}{4}t^2 - 3t + 1 = 0

Multiply by 4:

5t^2 - 12t + 4 = 0

Solve:

t = \frac{12 \pm \sqrt{144 - 80}}{10} = \frac{12 \pm 8}{10}

So

t = 2 \quad \text{or} \quad t = 0.4

Only $t=2$ lies in $1 \le t \le 3$ , so the particle changes direction at $t=2$ . Integrate $v$ :

s = \int \left(\frac{5}{4}t^2 - 3t + 1\right)dt = \frac{5}{12}t^3 - \frac{3}{2}t^2 + t + c

At the limits:

s(1) = \frac{5}{12} - \frac{3}{2} + 1 = -\frac{1}{12}

s(2) = \frac{5}{12}(8) - \frac{3}{2}(4) + 2 = -\frac{2}{3}

s(3) = \frac{5}{12}(27) - \frac{3}{2}(9) + 3 = \frac{3}{4}

Distance from $t=1$ to $t=2$ :

\left|s(2)-s(1)\right| = \left|-\frac{2}{3} + \frac{1}{12}\right| = \frac{7}{12}

Distance from $t=2$ to $t=3$ :

\left|s(3)-s(2)\right| = \left|\frac{3}{4} + \frac{2}{3}\right| = \frac{17}{12}

Total distance:

\frac{7}{12} + \frac{17}{12} = 2

Answer

2\text{ m}

Final answer

2 m

Detailed explanation

Walkthrough

From part (a), the velocity is $v = \frac{1}{2}kt^2 - 3t + 1$ . With $k=2.5$ , this becomes $v = \frac{5}{4}t^2 - 3t + 1$ . To find where the particle changes direction, solve $v=0$ . The roots are $t=2$ and $t=0.4$ ; only $t=2$ is inside $1 \le t \le 3$ .

Because the velocity changes sign at $t=2$ , the displacement from $1$ to $3$ is not the total distance. We integrate $v$ separately on $1$ to $2$ and $2$ to $3$ , take absolute values, and add them.

Key Takeaways

Total distance is the sum of the magnitudes of displacements over intervals where the direction of motion does not change. To find where direction changes, solve $v=0$ and check which roots lie in the given interval.

Common Mistakes

Integrating from $1$ to $3$ directly and giving the net displacement $\frac{5}{6}$ .
Forgetting to solve $v=0$ before integrating.
Including $t=0.4$ as a split point even though it is outside the interval.
Sign errors when evaluating the definite integrals.

Things to Be Careful About

Use absolute values for each segment. The total distance is positive. The roots of $5t^2-12t+4=0$ are $t=2$ and $t=0.4$ ; only $t=2$ matters here. Keep the units as metres.

Techniques used

substitute value of k into velocity expressionsolve quadratic equation for times when velocity is zerointegrate velocity to obtain displacementsplit interval at turning point and sum absolute displacements

Q5MediumNewton's Laws of MotionForces and EquilibriumKinematics of Motion in a Straight Line

Fig. 2

The diagram shows a particle $P$ of mass $6\text{ kg}$ on a rough plane inclined at an angle of $30^\circ$ to the horizontal. Two light inextensible strings are attached to $P$ . The strings pass over small smooth pulleys, which are fixed at the ends of the plane. The non-vertical parts of the string are parallel to a line of greatest slope of the plane. Particles $Q$ and $R$ , of masses $5\text{ kg}$ and $2\text{ kg}$ respectively, hang vertically at the ends of the strings.

Both strings are taut, and the system is released from rest.

It is given that the tension in the string attached to $Q$ is twice the tension in the string attached to $R$ .

(a)

Find, in terms of $g$ , the tension in each of the strings and the magnitude of the acceleration of the particles.

5M Similar

DifficultyMedium

Worked solution

Approach

Let $T_1$ be the tension in the string connecting $P$ to $Q$ (bottom pulley), and $T_2$ be the tension in the string connecting $P$ to $R$ (top pulley). We are given $T_1 = 2T_2$ .

Since $Q$ (5 kg) is heavier than $R$ (2 kg), $Q$ will accelerate downward, $R$ will accelerate upward, and $P$ will accelerate down the plane toward the bottom pulley. All three particles share the same magnitude of acceleration $a$ .

Working

Apply Newton's second law to $Q$ (taking downward as positive):

5g - T_1 = 5a

Apply Newton's second law to $R$ (taking upward as positive):

T_2 - 2g = 2a

Substitute $T_1 = 2T_2$ into the first equation:

5g - 2T_2 = 5a

From the second equation, express $T_2$ in terms of $a$ :

T_2 = 2a + 2g

Substitute this into the modified first equation:

5g - 2(2a + 2g) = 5a

5g - 4a - 4g = 5a

g = 9a

a = \frac{1}{9}g

Now find $T_2$ :

T_2 = 2\left(\frac{1}{9}g\right) + 2g = \frac{2g}{9} + \frac{18g}{9} = \frac{20g}{9}

And $T_1$ :

T_1 = 2T_2 = 2 \times \frac{20g}{9} = \frac{40g}{9}

Answer

a = \frac{1}{9}g, \quad T_{PQ} = \frac{40g}{9}, \quad T_{PR} = \frac{20g}{9}

Final answer

a = g/9, T_PQ = 40g/9, T_PR = 20g/9

Detailed explanation

Walkthrough

Step 1: Identify the direction of motion.
Particle $Q$ (5 kg) is heavier than particle $R$ (2 kg), so when the system is released, $Q$ accelerates downward, $R$ accelerates upward, and $P$ accelerates down the inclined plane toward the bottom pulley. All three particles have the same acceleration magnitude $a$ because the strings are inextensible.

Step 2: Set up Newton's second law for each hanging particle.
For $Q$ , taking downward as positive: the weight $5g$ acts downward and tension $T_1$ acts upward, so $5g - T_1 = 5a$ .

For $R$ , taking upward as positive: the tension $T_2$ acts upward and weight $2g$ acts downward, so $T_2 - 2g = 2a$ .

Step 3: Use the given tension relationship.
We are told $T_1 = 2T_2$ . Substitute this into the equation for $Q$ to get $5g - 2T_2 = 5a$ .

Step 4: Solve the simultaneous equations.
From the equation for $R$ , express $T_2 = 2a + 2g$ . Substitute into the modified equation for $Q$ : $5g - 2(2a + 2g) = 5a$ , which simplifies to $g = 9a$ , giving $a = g/9$ .

Step 5: Find the tensions.
Substitute $a = g/9$ back to get $T_2 = 20g/9$ and $T_1 = 40g/9$ .

Key Takeaways

When connected particles share an inextensible string, they all have the same acceleration magnitude.
Newton's second law can be applied to each particle independently, choosing the direction of motion as positive for each.
A given relationship between tensions (here $T_1 = 2T_2$ ) provides the extra equation needed to solve the system.

Common Mistakes

Using the same tension symbol for both strings — the problem explicitly states the tensions are different.
Getting the signs wrong in Newton's second law equations (e.g., writing $T_1 - 5g = 5a$ for $Q$ when $Q$ accelerates downward).
Not substituting the tension relationship correctly, leading to an equation with two unknowns.

Things to Be Careful About

The acceleration $a$ must be the same for all three particles due to the inextensible strings.
Tensions must be expressed in terms of $g$ as the final answer, not as a numerical value (unless $g$ is substituted throughout).
The direction of acceleration for each particle must be consistent with the chosen positive direction in the equation.

Techniques used

apply Newton's second law to hanging particlesuse given tension relationshipsolve simultaneous equations

(b)

Find the coefficient of friction between $P$ and the plane.

5M Similar

DifficultyMedium

Worked solution

Approach

Particle $P$ (mass 6 kg) is on a rough plane inclined at $30^\circ$ . The forces acting on $P$ are:

Weight $6g$ acting vertically downward
Normal reaction $R_P$ perpendicular to the plane
Friction $F$ opposing motion (acting up the plane since $P$ accelerates down)
Tension $T_1 = \frac{40g}{9}$ pulling down the plane (toward $Q$ )
Tension $T_2 = \frac{20g}{9}$ pulling up the plane (toward $R$ )

Working

Resolve perpendicular to the plane:

R_P = 6g \cos 30^\circ = 6g \times \frac{\sqrt{3}}{2} = 3\sqrt{3}\,g

Apply Newton's second law along the plane (downward positive):

6g \sin 30^\circ + T_1 - T_2 - F = 6a

Substitute known values ( $a = \frac{g}{9}$ , $T_1 = \frac{40g}{9}$ , $T_2 = \frac{20g}{9}$ ):

6g \sin 30^\circ + \frac{40g}{9} - \frac{20g}{9} - F = 6 \times \frac{g}{9}

3g + \frac{20g}{9} - F = \frac{6g}{9}

\frac{27g}{9} + \frac{20g}{9} - \frac{6g}{9} = F

F = \frac{41g}{9}

Use the friction model $F = \mu R_P$ :

\frac{41g}{9} = \mu \times 3\sqrt{3}\,g

\mu = \frac{41}{27\sqrt{3}} = \frac{41\sqrt{3}}{81}

\mu \approx 0.877

Answer

\mu = \frac{41\sqrt{3}}{81} \approx 0.877

Final answer

μ = 41√3/81 ≈ 0.877

Detailed explanation

Walkthrough

Step 1: Resolve forces perpendicular to the plane.
The component of $P$ 's weight perpendicular to the plane is $6g \cos 30^\circ$ . Since there is no acceleration perpendicular to the plane, the normal reaction balances this: $R_P = 3\sqrt{3}\,g$ .

Step 2: Apply Newton's second law along the plane.
Taking the direction of motion (down the plane) as positive, the forces along the plane are:

Component of weight down the plane: $6g \sin 30^\circ = 3g$
Tension $T_1$ pulling down the plane: $\frac{40g}{9}$
Tension $T_2$ pulling up the plane: $\frac{20g}{9}$
Friction $F$ opposing motion (up the plane)

The equation is: $3g + \frac{40g}{9} - \frac{20g}{9} - F = 6 \times \frac{g}{9}$ .

Step 3: Solve for friction $F$ .
Simplifying gives $F = \frac{41g}{9}$ .

Step 4: Apply $F = \mu R_P$ .
Substitute $F = \frac{41g}{9}$ and $R_P = 3\sqrt{3}\,g$ to find $\mu = \frac{41}{27\sqrt{3}} = \frac{41\sqrt{3}}{81} \approx 0.877$ .

Key Takeaways

On an inclined plane, resolve weight into components parallel ( $mg \sin \theta$ ) and perpendicular ( $mg \cos \theta$ ) to the plane.
Friction opposes the direction of motion (or impending motion) and is given by $F = \mu R$ at limiting equilibrium.
The normal reaction on an inclined plane is $mg \cos \theta$ , not $mg$ .

Common Mistakes

Forgetting to include both tensions in the equation for $P$ .
Using $R_P = 6g$ instead of $R_P = 6g \cos 30^\circ$ .
Getting the direction of friction wrong — it must oppose the actual motion of $P$ .
Not substituting the values from part (a) correctly.

Things to Be Careful About

The friction force $F$ acts up the plane because $P$ accelerates down the plane.
The coefficient of friction is dimensionless — check that $g$ cancels out in the final calculation.
Rationalise the denominator: $\frac{41}{27\sqrt{3}} = \frac{41\sqrt{3}}{81}$ .

Techniques used

resolve forces perpendicular to the planeapply Newton's second law along the planeuse F = μR to find coefficient of friction

(c)

It is given that when the system is released from rest, $P$ is at the midpoint of the plane. In the subsequent motion, $R$ does not reach the pulley at the top of the plane, and $P$ takes $1.5\text{ s}$ to reach the pulley at the bottom of the plane.

Find the total length of the plane.

2M Similar

DifficultyMedium-Easy

Worked solution

Approach

Particle $P$ starts at the midpoint of the plane and accelerates down the plane with constant acceleration $a = \frac{g}{9}$ . It takes $t = 1.5$ s to reach the bottom pulley. We need to find the distance $d$ from the midpoint to the bottom, then double it to get the total length.

Working

Use the suvat equation $s = ut + \frac{1}{2}at^2$ with $u = 0$ , $t = 1.5$ , and $a = \frac{g}{9}$ :

d = 0 \times 1.5 + \frac{1}{2} \times \frac{g}{9} \times (1.5)^2

d = \frac{1}{2} \times \frac{g}{9} \times 2.25

Using $g = 10$ m/s²:

d = \frac{1}{2} \times \frac{10}{9} \times 2.25 = \frac{1}{2} \times \frac{10}{9} \times \frac{9}{4} = \frac{10}{8} = 1.25 \text{ m}

The total length of the plane is twice this distance (since $P$ started at the midpoint):

L = 2d = 2 \times 1.25 = 2.5 \text{ m}

Answer

L = 2.5 \text{ m}

Final answer

2.5 m

Detailed explanation

Walkthrough

Step 1: Identify the known quantities.
Particle $P$ starts from rest ( $u = 0$ ) at the midpoint of the plane. It accelerates down the plane with $a = \frac{g}{9} = \frac{10}{9}$ m/s² (using $g = 10$ m/s²). The time taken to reach the bottom pulley is $t = 1.5$ s.

Step 2: Calculate the distance from midpoint to bottom.
Using $s = ut + \frac{1}{2}at^2$ : $d = 0 + \frac{1}{2} \times \frac{10}{9} \times (1.5)^2 = \frac{1}{2} \times \frac{10}{9} \times 2.25 = 1.25$ m.

Step 3: Find the total length.
Since $P$ started at the midpoint, the total length is $L = 2d = 2 \times 1.25 = 2.5$ m.

Key Takeaways

The suvat equation $s = ut + \frac{1}{2}at^2$ is the most direct tool when initial velocity, acceleration, and time are known.
Always check what the question is asking — here it asks for the total length, not just the distance travelled.

Common Mistakes

Forgetting that $P$ starts at the midpoint, so the total length is twice the distance travelled.
Using $g = 9.8$ or $g = 9.81$ when the mark scheme expects $g = 10$ (though the mark scheme does condone other values).
Using $v = u + at$ first and then $s = \frac{1}{2}(u+v)t$ , which is valid but more work.

Things to Be Careful About

The acceleration from part (a) is in terms of $g$ . The mark scheme accepts $g = 10$ , $g = 9.8$ , or $g = 9.81$ , but the final answer of 2.5 m follows from $g = 10$ .
Ensure units are consistent: acceleration in m/s², time in s, distance in m.

Techniques used

identify initial conditions and accelerationapply suvat equation s = ut + ½at²relate displacement to total length

Q68MMedium-HardMomentumEnergy, Work and Power

Two particles $A$ and $B$ of masses $km$ and $m$ respectively, where $k$ and $m$ are constants, are free to move in a straight line on a smooth horizontal plane. Particle $A$ is projected towards $B$ with speed $2u$ and at the same instant $B$ is projected towards $A$ with speed $u$ .

The particles collide. After the collision the speed of $A$ is $u$ and both particles move in the same direction as $A$ ’s original motion.

It is given that $35\%$ of the total kinetic energy is lost in the collision. Find, in terms of $u$ , the speed of $B$ after the collision.

DifficultyMedium-Hard

Worked solution

Approach

Use conservation of linear momentum to relate the unknown speed $v_B$ to $k$ , then write the kinetic energy before and after the collision. Since 35% of the kinetic energy is lost, 65% remains, so set $0.65 \times KE_{\text{before}} = KE_{\text{after}}$ . Solve the resulting quadratic and reject the root that is physically impossible.

Working

Take the positive direction to be A's original motion. Particle $B$ is initially moving towards $A$ , so its initial velocity is $-u$ . After the collision both particles move in A's original direction, so A's velocity is $u$ and B's speed is $v_B > 0$ .

Conservation of linear momentum:

km(2u) + m(-u) = km(u) + m v_B

Simplify:

2ku - u = ku + v_B

v_B = (k - 1)u

Kinetic energy before the collision:

KE_{\text{before}} = \frac{1}{2}km(2u)^2 + \frac{1}{2}mu^2 = 2kmu^2 + \frac{1}{2}mu^2

Kinetic energy after the collision:

KE_{\text{after}} = \frac{1}{2}km u^2 + \frac{1}{2}m v_B^2

Since 35% is lost, 65% remains:

0.65\left(2kmu^2 + \frac{1}{2}mu^2\right) = \frac{1}{2}km u^2 + \frac{1}{2}m v_B^2

Substitute $v_B = (k - 1)u$ :

0.65\left(2kmu^2 + \frac{1}{2}mu^2\right) = \frac{1}{2}km u^2 + \frac{1}{2}m(k - 1)^2 u^2

Divide by $mu^2$ and multiply by 2:

2.6k + 0.65 = k + (k - 1)^2 = k^2 - k + 1

So

k^2 - 3.6k + 0.35 = 0

Solve:

k = \frac{3.6 \pm \sqrt{3.6^2 - 4(0.35)}}{2} = \frac{3.6 \pm \sqrt{11.56}}{2} = \frac{3.6 \pm 3.4}{2}

Thus $k = 3.5$ or $k = 0.1$ . Since $v_B = (k - 1)u > 0$ , we must have $k > 1$ , so reject $k = 0.1$ .

Therefore

v_B = (3.5 - 1)u = 2.5u

Answer

v_B = 2.5u

Final answer

v_B = 2.5u

Detailed explanation

Walkthrough

Start by choosing a positive direction. Since $A$ is projected towards $B$ with speed $2u$ and $B$ towards $A$ with speed $u$ , if A's original direction is taken as positive then B's initial velocity is $-u$ . After the collision both particles move in A's original direction, so A's velocity is $u$ and B's unknown speed $v_B$ is positive.

Apply conservation of linear momentum: the total momentum before equals the total momentum after. This gives $km(2u) + m(-u) = km(u) + m v_B$ , which simplifies to $v_B = (k - 1)u$ . This relation is essential because it removes one unknown and links $v_B$ to $k$ .

Next write the kinetic energy before the collision: $\frac{1}{2}km(2u)^2 + \frac{1}{2}mu^2 = 2kmu^2 + \frac{1}{2}mu^2$ . After the collision, $KE_{\text{after}} = \frac{1}{2}km u^2 + \frac{1}{2}m v_B^2$ .

The statement that 35% of the kinetic energy is lost means 65% remains, so the correct equation is $0.65 \times KE_{\text{before}} = KE_{\text{after}}$ , not $0.35 \times KE_{\text{after}}$ . Substituting $v_B = (k - 1)u$ into this equation gives an equation in $k$ only.

Divide by $mu^2$ and multiply by 2 to obtain $2.6k + 0.65 = k^2 - k + 1$ , then rearrange to $k^2 - 3.6k + 0.35 = 0$ . Solving this quadratic gives $k = 3.5$ or $k = 0.1$ . The value $k = 0.1$ must be rejected because it would give $v_B = (0.1 - 1)u = -0.9u$ , which contradicts the fact that both particles move in A's original direction. Hence $k = 3.5$ and $v_B = 2.5u$ .

Key Takeaways

In a direct collision, linear momentum is conserved, but kinetic energy is not necessarily conserved.
A direction convention must be chosen and used consistently; velocities, not speeds, appear in the momentum equation.
A fractional energy loss means the retained fraction is $1 - \text{loss}$ , so a 35% loss leaves 65% of the original kinetic energy.
Physical constraints, such as the direction of motion after impact, can be used to reject an otherwise valid algebraic root.

Common Mistakes

Taking B's initial momentum as positive instead of negative, because B is moving towards A.
Writing $KE_{\text{after}} = 0.35 \times KE_{\text{before}}$ instead of $0.65 \times KE_{\text{before}} = KE_{\text{after}}$ .
Leaving both $k$ and $v_B$ in the energy equation instead of substituting $v_B = (k - 1)u$ .
Accepting $k = 0.1$ without checking that it gives $v_B = -0.9u$ , which is impossible here.
Including $g$ with the masses in the momentum equation; the mark scheme penalises this.

Things to Be Careful About

Use B's initial velocity as $-u$ , not $u$ , in the momentum equation.
The speed $v_B$ after the collision must be positive, so $k > 1$ ; this is the reason $k = 0.1$ is rejected.
When cancelling, remember that $m$ and $u$ are non-zero constants, so dividing by $mu^2$ is valid.
The energy-loss equation must be set up the correct way round: 65% of the kinetic energy before equals the kinetic energy after.
If the mark scheme is followed strictly, showing the quadratic and the rejection of the invalid root is necessary for full marks.

Techniques used

apply conservation of linear momentum to a direct impactwrite kinetic energy before and after the collisionuse the given fractional energy loss to form an equationsolve a quadratic and reject the invalid root

Mathematics 9709/45 — October/November 2025

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