Coplanar forces of magnitudes $17\text{N}$, $51\text{N}$ and $34\text{N}$ act at a point $O$ in the directions shown in the diagram, where $\tan \alpha = \frac{15}{8}$. (a) Find the magnitude and direction of the resultant of the three forces. (b) The force of magnitude $51\text{N}$ is replaced by a force of magnitude $P\text{N}$ acting in the same direction. The resultant of the three forces now acts in the positive $x$-direction. Find the value of $P$.

(a) 26.0 N at 15.6° above the positive x-axis · (b) 43.1

A particle $P$ moves in a straight line starting from a point $O$. At time $t\text{ s}$ after leaving $O$, the velocity, $v\text{ms}^{-1}$, of $P$ is given by $$ v = (15 - 2t)^2 $$ (a) Find the values of $t$ when the velocity of $P$ is $100\text{ms}^{-1}$. (b) Show that there is a particular value of $t$ for which the velocity and acceleration of the particle are both zero. (c) Find the displacement of $P$ from $O$ at the time that the velocity and acceleration of the…

(a) t = 2.5 or t = 12.5 · (b) t = 7.5 · (c) s = 562.5 m

9709/45 May/June 2025 — Mathematics Solved Past Paper

Q14MMedium-EasyEnergy, Work and Power

A box of mass $25\text{kg}$ is pulled $12\text{m}$ up a rough plane inclined at an angle of $8^\circ$ to the horizontal. The box moves up a line of greatest slope against a frictional force of $50\text{N}$ . The force pulling the box is parallel to the line of greatest slope. The box starts from rest and has speed $2.4\text{ms}^{-1}$ at the end of the $12\text{m}$ .

Find the work done by the pulling force.

DifficultyMedium-Easy

Worked solution

Approach

Use the work-energy principle. The work done by the pulling force is used to increase the box's gravitational potential energy, increase its kinetic energy, and overcome friction. The height gained is $12\sin 8^\circ$ .

Working

Gain in gravitational potential energy:

\Delta E_p = mg\Delta h = 25g \times 12\sin 8^\circ

Using $g = 10\,\text{ms}^{-2}$ :

\Delta E_p = 25 \times 10 \times 12\sin 8^\circ = 3000\sin 8^\circ = 417.519\ldots\ \text{J}

Gain in kinetic energy:

\Delta E_k = \frac{1}{2}mv^2 - \frac{1}{2}mu^2 = \frac{1}{2} \times 25 \times 2.4^2 - 0 = 72\ \text{J}

Work done against friction:

W_f = F_f d = 50 \times 12 = 600\ \text{J}

Therefore the work done by the pulling force is:

W = \Delta E_p + \Delta E_k + W_f = 417.519\ldots + 72 + 600 = 1089.519\ldots\ \text{J}

Answer

W = 1090\,\text{J} \quad (3\ \text{s.f.})

Final answer

1090 J

Detailed explanation

Walkthrough

The problem asks for the work done by the pulling force, not the size of the force itself. Since the box starts from rest and reaches a given speed after moving up a slope against friction, the most direct method is the work-energy principle.

Step 1: Identify the energy changes. As the box moves $12\ \text{m}$ up a plane inclined at $8^\circ$ , its vertical height increases by $12\sin 8^\circ$ . The gain in gravitational potential energy is $\Delta E_p = mg\Delta h$ . Using $g = 10\,\text{ms}^{-2}$ , this is $25 \times 10 \times 12\sin 8^\circ = 3000\sin 8^\circ \approx 417.5\ \text{J}$ .

Step 2: Find the gain in kinetic energy. It starts from rest, so initial kinetic energy is zero. Final kinetic energy is $\frac{1}{2}mv^2 = \frac{1}{2} \times 25 \times 2.4^2 = 72\ \text{J}$ .

Step 3: Account for friction. Friction opposes motion, so the pulling force must also do work equal to the frictional force times the distance moved along the plane: $50 \times 12 = 600\ \text{J}$ .

Step 4: Add the three contributions. The total work done by the pulling force equals the total energy transferred: $417.5 + 72 + 600 = 1089.5\ \text{J}$ , which rounds to $1090\ \text{J}$ .

Key Takeaways

The work-energy principle is useful when a force moves an object and energy is transferred between forms.
On a slope, the vertical height gained is $s\sin\theta$ , where $s$ is the distance along the slope and $\theta$ is the angle of inclination.
Work done against a constant frictional force is $F_f d$ .
Work done by a force can be found from energy changes without first finding the force itself.

Common Mistakes

Forgetting to include the work done against friction.
Using the distance along the slope as the vertical height in $\Delta E_p$ ; the height gain is $12\sin 8^\circ$ , not $12$ .
Using the final speed as if the box started with that speed; since it starts from rest, initial kinetic energy is zero.
Mixing up $\sin$ and $\cos$ when finding the height gain.
Stating the work done as $90.793\ \text{J}$ ; that value is the pulling force, not the work done.

Things to Be Careful About

Use the correct value of $g$ ; the mark scheme uses $g = 10\,\text{ms}^{-2}$ , giving $3000\sin 8^\circ$ .
The final answer should be rounded to a sensible degree of accuracy: $1090\ \text{J}$ (or $1089$ – $1090\ \text{J}$ ).
All energy terms here are positive because the box is gaining gravitational potential energy and kinetic energy while losing energy to friction; sign errors are common.
Work is force times distance, so do not confuse the pulling force with the work done by that force.

Techniques used

apply the work-energy principlecompute gravitational potential energy gaincompute kinetic energy gainaccount for work done against frictionsum energy transfers to find work done

Approach

Use the principle of conservation of linear momentum. The total momentum of the system (hammerhead and nail) immediately before the impact must equal the total momentum immediately after the impact, since there are no external horizontal forces and the impact is instantaneous.

Working

Let $v$ be the speed of the combined hammerhead and nail immediately after the impact.

Momentum before impact:

\begin{aligned} p_{\text{before}} &= m_{\text{hammer}} \times u_{\text{hammer}} + m_{\text{nail}} \times u_{\text{nail}} \\ &= 1.5 \times 32 + 0.005 \times 0 \\ &= 48 \text{ kg m s}^{-1} \end{aligned}

Momentum after impact:

\begin{aligned} p_{\text{after}} &= (m_{\text{hammer}} + m_{\text{nail}}) \times v \\ &= (1.5 + 0.005) v \\ &= 1.505 v \end{aligned}

Equating momentum before and after:

1.505 v = 48

Solving for $v$ :

v = \frac{48}{1.505} \approx 31.893687... \text{ ms}^{-1}

Rounding to 3 decimal places as required:

v = 31.894 \text{ ms}^{-1}

Answer

31.894 \text{ ms}^{-1}

Final answer

31.894

Detailed explanation

Walkthrough

The problem describes a direct impact where two objects (hammerhead and nail) stick together after collision. This is a perfectly inelastic collision, and the key principle to use is the conservation of linear momentum. Since the impact is instantaneous, we can ignore external forces like gravity during the collision itself.

Calculate initial momentum: The hammerhead is moving downwards at $32 \text{ ms}^{-1}$ with mass $1.5 \text{ kg}$ . The nail is stationary ( $0 \text{ ms}^{-1}$ ) with mass $0.005 \text{ kg}$ . Total initial momentum is simply $1.5 \times 32 = 48 \text{ kg m s}^{-1}$ .
Calculate final momentum: After impact, the two masses move together as a single combined mass of $1.5 + 0.005 = 1.505 \text{ kg}$ . Let their common velocity be $v$ . The final momentum is $1.505v$ .
Equate and solve: By conservation of momentum, $1.505v = 48$ . Solving this gives $v \approx 31.894 \text{ ms}^{-1}$ .

Key Takeaways

In direct impact problems where objects stick together, total linear momentum is conserved.
The combined mass after impact is the sum of the individual masses.
Always ensure units are consistent and round the final answer to the specified number of decimal places.

Common Mistakes

Forgetting to add the mass of the nail to the hammerhead mass in the final momentum term (using $1.5v$ instead of $1.505v$ ).
Including the weight ( $mg$ ) in the momentum equation during the impact phase. Momentum conservation applies to the instantaneous collision, where impulsive forces dominate and external forces like gravity are negligible.
Rounding too early in the calculation, which can lead to errors in subsequent parts.

Things to Be Careful About

The question asks for the answer correct to 3 decimal places. Ensure you keep full precision during calculation and only round at the final step.
The nail is initially stationary, so its initial momentum is zero.
Direction matters in momentum, but since all motion is vertically downwards, we can treat downward as positive and ignore vector notation for this 1D problem.

Techniques used

apply conservation of linear momentumsolve for final velocity

(b)

There is a constant force resisting the motion of magnitude $25\,000\text{N}$ .

Calculate the distance the nail is driven into the wood.

3M Similar

DifficultyMedium

Worked solution

Approach

After the impact, the combined hammerhead and nail move downwards into the wood. There are two forces acting on the combined mass: its weight acting downwards and the constant resistive force from the wood acting upwards. Use Newton's second law to find the deceleration, then use the constant acceleration formulae (suvat) to find the distance driven into the wood.

Working

Step 1: Find the acceleration using Newton's second law

Let the downward direction be positive. The combined mass is $M = 1.5 + 0.005 = 1.505 \text{ kg}$ .

Forces acting on the system:

Weight downwards: $W = Mg = 1.505 \times 10 = 15.05 \text{ N}$ (using $g = 10 \text{ m s}^{-2}$ )
Resistive force upwards: $R = 25000 \text{ N}$

Resultant force $F = W - R = 15.05 - 25000 = -24984.95 \text{ N}$

Using $F = Ma$ :

\begin{aligned} -24984.95 &= 1.505 \times a \\ a &= \frac{-24984.95}{1.505} \\ a &\approx -16601.29568 \text{ m s}^{-2} \end{aligned}

The negative sign indicates the acceleration is upwards (deceleration), which is expected as the nail is slowing down.

Step 2: Use suvat equations to find the distance $s$

We know:

Initial velocity $u = 31.894 \text{ ms}^{-1}$ (from part a)
Final velocity $v = 0 \text{ ms}^{-1}$ (the nail comes to rest)
Acceleration $a = -16601.29568 \text{ m s}^{-2}$
Distance $s = ?$

Using the equation $v^2 = u^2 + 2as$ :

\begin{aligned} 0 &= (31.894)^2 + 2 \times (-16601.29568) \times s \\ 0 &= 1017.227236 - 33202.59136 s \\ 33202.59136 s &= 1017.227236 \\ s &= \frac{1017.227236}{33202.59136} \\ s &\approx 0.030636983 \text{ m} \end{aligned}

Rounding to 3 significant figures (or appropriate precision):

s = 0.0306 \text{ m}

Answer

0.0306 \text{ m}

Final answer

0.0306

Detailed explanation

Walkthrough

Once the hammer and nail are moving together, they act as a single body of mass $1.505 \text{ kg}$ penetrating the wood. We need to find how far they travel before stopping.

Identify forces: The body is subject to gravity pulling it down ( $W = Mg$ ) and the wood's resistance pushing up ( $R = 25000 \text{ N}$ ). Since the resistance is much larger than the weight, the net force is upwards, causing deceleration.
Calculate acceleration: Apply Newton's second law ( $F = ma$ ). Taking downward as positive, the resultant force is $Mg - R$ . With $g = 10 \text{ m s}^{-2}$ , $W = 15.05 \text{ N}$ . Net force $= 15.05 - 25000 = -24984.95 \text{ N}$ . Acceleration $a = F/M = -24984.95 / 1.505 \approx -16601.3 \text{ m s}^{-2}$ .
Apply kinematics: Use $v^2 = u^2 + 2as$ . The initial speed $u$ is the speed calculated in part (a), $31.894 \text{ ms}^{-1}$ . The final speed $v$ is $0$ (it stops). Solve for $s$ .
Calculate distance: $0 = (31.894)^2 + 2(-16601.29568)s$ . Solving gives $s \approx 0.0306 \text{ m}$ .

An alternative method is to use the work-energy principle: the change in kinetic energy equals the work done by the net force. $\frac{1}{2}Mu^2 = (R - Mg)s$ , which yields the same result.

Key Takeaways

Newton's second law can be used to find acceleration when forces are known.
The suvat equations link velocity, acceleration, and displacement.
Sign conventions are crucial: if downward is positive, upward forces and deceleration are negative.
The weight of the object is often negligible compared to large resistive forces, but it must still be included for correctness.

Common Mistakes

Omitting the weight term ( $Mg$ ) in the force calculation. The mark scheme notes that omitting weight gives $a = -16611.29...$ and results in zero marks for the method.
Using the wrong mass (e.g., using only $1.5 \text{ kg}$ instead of the combined $1.505 \text{ kg}$ ).
Sign errors in the acceleration or in the suvat equation (e.g., $v^2 = u^2 - 2as$ without adjusting signs correctly).
Using the pre-impact speed ( $32 \text{ ms}^{-1}$ ) instead of the post-impact speed ( $31.894 \text{ ms}^{-1}$ ) for the deceleration phase.

Things to Be Careful About

Use the value of $g$ consistent with your exam board (here $g=10 \text{ m s}^{-2}$ is implied by the mark scheme values).
Carry forward the unrounded value from part (a) or use at least 3 significant figures to avoid compounding errors. The mark scheme allows $31.9$ but penalizes using only 3sf if it leads to incorrect intermediate steps in some methods, though here $31.9$ gives $0.030649...$ which is accepted.
Ensure units are consistent: mass in kg, force in N, distance in m.

Techniques used

apply Newton's second law to find accelerationuse suvat equations to find distance

Q35MMediumKinematics of Motion in a Straight Line

A train travels $4.8\text{km}$ between two stations, $A$ and $B$ . The train starts from rest at $A$ and accelerates at a constant rate until it reaches a speed of $30\text{ms}^{-1}$ . It then travels at this constant speed for $T$ seconds, before decelerating at a constant rate, coming to rest at $B$ . The total time for the journey is $180\text{s}$ .

Find the value of $T$ and hence find the distance moved by the train while travelling at the constant speed of $30\text{ms}^{-1}$ .

DifficultyMedium

Worked solution

Approach

Convert the total distance to metres and split the journey into three phases: acceleration, constant speed and deceleration. Use $s = \frac{1}{2}(u+v)t$ for the phases where the speed changes, and $s = vt$ for the constant-speed phase. Then add the three displacements and use the total time to eliminate the unknown acceleration and deceleration times.

Working

Let $t_1$ be the accelerating time, $T$ the constant-speed time and $t_3$ the decelerating time. The total time is

t_1 + T + t_3 = 180

For the accelerating phase, from rest to $30\text{ ms}^{-1}$ :

s_1 = \frac{1}{2}(0+30)t_1 = 15t_1

For the constant-speed phase:

s_2 = 30T

For the decelerating phase, from $30\text{ ms}^{-1}$ to rest:

s_3 = \frac{1}{2}(30+0)t_3 = 15t_3

The total distance is $4.8\text{ km} = 4800\text{ m}$ , so

4800 = s_1 + s_2 + s_3 = 15t_1 + 30T + 15t_3

Factor the first and last terms:

4800 = 15(t_1 + t_3) + 30T

Since $t_1 + t_3 = 180 - T$ :

4800 = 15(180 - T) + 30T

Simplify:

4800 = 2700 - 15T + 30T = 2700 + 15T

15T = 2100

T = 140

Distance travelled at constant speed:

30 \times 140 = 4200\text{ m}

Answer

$T = 140\text{ s}$ and the distance travelled at constant speed is $4200\text{ m}$ (or $4.2\text{ km}$ ).

Final answer

T = 140 s; distance at constant speed = 4200 m (4.2 km)

Detailed explanation

Walkthrough

The journey has three distinct phases: speeding up from rest to $30\text{ ms}^{-1}$ , travelling at a steady $30\text{ ms}^{-1}$ for $T$ seconds, and slowing down to rest. We do not know the acceleration, the deceleration, or the times of the first and last phases, but we do know the total time and the total distance.

For a phase in which speed changes uniformly, the displacement is the average speed multiplied by time. Since the train starts at $0$ and reaches $30$ , the average speed during acceleration is $15\text{ ms}^{-1}$ , so $s_1 = 15t_1$ . Similarly, during deceleration from $30$ to $0$ , the average speed is again $15\text{ ms}^{-1}$ , so $s_3 = 15t_3$ . During the middle phase the speed is constant at $30$ , so $s_2 = 30T$ .

Adding these gives the total distance in metres. The total time tells us that $t_1 + t_3 = 180 - T$ . Substituting this into the distance equation removes the two unknown times and leaves a simple equation in $T$ . Solving it gives $T = 140$ . Finally, multiplying $30$ by $140$ gives the distance covered at constant speed.

Key Takeaways

The key idea is that the total distance is the sum of the displacements of the separate phases, and the total time is the sum of the times of the separate phases. Using $s = \frac{1}{2}(u+v)t$ for uniformly accelerated motion avoids needing the actual acceleration values. Alternatively, this problem can be viewed as the area under a speed-time graph: the graph is a trapezium with height $30$ , bottom base $180$ and top base $T$ , so its area is $\frac{1}{2} \times 30 \times (180 + T)$ .

Common Mistakes

Using the same variable for all three times. The mark scheme specifically says the same variable must not be used for all three phases.
Forgetting to convert $4.8\text{ km}$ to $4800\text{ m}$ before forming the equation.
Writing the middle-phase distance as $30(180 - T)$ ; that would be the distance if the whole remaining time were at constant speed, but the first and last phases are not at constant speed.
Assuming the acceleration and deceleration are equal; this is not given and is not needed.

Things to Be Careful About

Keep all distances in the same units. Use metres throughout.
Remember that $t_1 + T + t_3 = 180$ , so $t_1 + t_3 = 180 - T$ , not $180 + T$ .
The distance at constant speed is $30 \times T$ , not the total distance.
If using the trapezium method, the bases are $180$ and $T$ , and the height is $30$ ; equating the area to $4800$ gives the same equation.
After finding $T$ , the mark scheme allows follow-through: distance $= 30 \times \text{their } T$ .

Techniques used

apply suvat displacement formulae to each phaseexpress total distance as sum of segment displacementseliminate unknown times using total timesolve linear equation for T

Approach

Resolve each of the three forces into horizontal ( $x$ ) and vertical ( $y$ ) components. Use the given $\tan \alpha = \frac{15}{8}$ to find $\sin \alpha$ and $\cos \alpha$ . Sum the components to find the resultant vector, then calculate its magnitude and direction.

Working

From $\tan \alpha = \frac{15}{8}$ , we can construct a right-angled triangle with opposite side $15$ and adjacent side $8$ . The hypotenuse is $\sqrt{15^2 + 8^2} = \sqrt{225 + 64} = \sqrt{289} = 17$ .

Thus, $\sin \alpha = \frac{15}{17}$ and $\cos \alpha = \frac{8}{17}$ .

Resolve forces horizontally (positive to the right):

\begin{aligned} R_x &= 51 \cos \alpha + 34 \cos \alpha - 17 \sin \alpha \\ &= 51 \left(\frac{8}{17}\right) + 34 \left(\frac{8}{17}\right) - 17 \left(\frac{15}{17}\right) \\ &= 24 + 16 - 15 \\ &= 25 \text{ N} \end{aligned}

Resolve forces vertically (positive upwards):

\begin{aligned} R_y &= 51 \sin \alpha - 34 \sin \alpha - 17 \cos \alpha \\ &= 51 \left(\frac{15}{17}\right) - 34 \left(\frac{15}{17}\right) - 17 \left(\frac{8}{17}\right) \\ &= 45 - 30 - 8 \\ &= 7 \text{ N} \end{aligned}

Calculate the magnitude of the resultant $R$ :

R = \sqrt{R_x^2 + R_y^2} = \sqrt{25^2 + 7^2} = \sqrt{625 + 49} = \sqrt{674} \approx 26.0 \text{ N}

Calculate the direction $\beta$ of the resultant with respect to the positive $x$ -axis:

\tan \beta = \frac{R_y}{R_x} = \frac{7}{25}

\beta = \tan^{-1}\left(\frac{7}{25}\right) \approx 15.6^\circ

Answer

The magnitude of the resultant is $26.0 \text{ N}$ (or $\sqrt{674} \text{ N}$ ), and its direction is $15.6^\circ$ above the positive $x$ -axis.

Final answer

26.0 N at 15.6° above the positive x-axis

Detailed explanation

Walkthrough

First, we determine the exact values of $\sin \alpha$ and $\cos \alpha$ from the given $\tan \alpha = \frac{15}{8}$ . By drawing a right triangle with opposite side 15 and adjacent side 8, the hypotenuse is 17, giving $\sin \alpha = \frac{15}{17}$ and $\cos \alpha = \frac{8}{17}$ .

Next, we resolve each force into horizontal ( $x$ ) and vertical ( $y$ ) components. The 51 N force has a positive $x$ -component ( $51 \cos \alpha$ ) and a positive $y$ -component ( $51 \sin \alpha$ ). The 34 N force has a positive $x$ -component ( $34 \cos \alpha$ ) and a negative $y$ -component ( $-34 \sin \alpha$ ). The 17 N force is perpendicular to the 34 N force and points into the third quadrant; its $x$ -component is negative ( $-17 \sin \alpha$ ) and its $y$ -component is negative ( $-17 \cos \alpha$ ).

Summing the $x$ -components gives $R_x = 25$ N, and summing the $y$ -components gives $R_y = 7$ N. The magnitude of the resultant is found using Pythagoras' theorem: $R = \sqrt{25^2 + 7^2} = \sqrt{674} \approx 26.0$ N. The direction is found using $\beta = \tan^{-1}(7/25) \approx 15.6^\circ$ above the positive $x$ -axis.

Key Takeaways

When given $\tan \theta$ , always find $\sin \theta$ and $\cos \theta$ using a right triangle or identities before resolving forces.
Careful attention to the signs of components based on the quadrant each force lies in is essential.
The resultant magnitude and direction are found using $R = \sqrt{R_x^2 + R_y^2}$ and $\tan \beta = R_y / R_x$ .

Common Mistakes

Forgetting to find $\sin \alpha$ and $\cos \alpha$ and trying to use $\tan \alpha$ directly in component equations.
Sign errors when resolving components, especially for the 17 N force which is in the third quadrant.
Mixing up sine and cosine for the 17 N force; remember its angle with the horizontal is $90^\circ - \alpha$ , so its horizontal component uses $\sin \alpha$ and vertical uses $\cos \alpha$ .

Things to Be Careful About

Ensure all forces are resolved into the same two perpendicular directions (usually horizontal and vertical).
The direction of the resultant must be specified clearly, e.g., ' $15.6^\circ$ above the positive $x$ -axis', to avoid ambiguity.

Techniques used

resolve forces into horizontal and vertical componentscalculate resultant magnitude using Pythagoras theoremcalculate resultant direction using inverse tangent

(b)

The force of magnitude $51\text{N}$ is replaced by a force of magnitude $P\text{N}$ acting in the same direction. The resultant of the three forces now acts in the positive $x$ -direction.

Find the value of $P$ .

2M Similar

DifficultyMedium-Easy

Worked solution

Approach

The $51 \text{ N}$ force is replaced by a force of magnitude $P \text{ N}$ in the same direction. For the resultant to act in the positive $x$ -direction, the vertical ( $y$ ) component of the resultant must be zero. Set the sum of the vertical components to zero and solve for $P$ .

Working

The vertical components of the three forces are:

$P \text{ N}$ force: $P \sin \alpha = P \left(\frac{15}{17}\right)$ (upwards)
$34 \text{ N}$ force: $-34 \sin \alpha = -34 \left(\frac{15}{17}\right)$ (downwards)
$17 \text{ N}$ force: $-17 \cos \alpha = -17 \left(\frac{8}{17}\right)$ (downwards)

Set the sum of vertical components to zero:

P \left(\frac{15}{17}\right) - 34 \left(\frac{15}{17}\right) - 17 \left(\frac{8}{17}\right) = 0

Simplify the equation:

\frac{15P}{17} - 30 - 8 = 0

\frac{15P}{17} = 38

15P = 38 \times 17 = 646

P = \frac{646}{15} \approx 43.1

Answer

P = 43.1

Final answer

43.1

Detailed explanation

Walkthrough

For the resultant to act purely in the positive $x$ -direction, there must be no net force in the vertical ( $y$ ) direction. This means the sum of all vertical components must equal zero.

The vertical component of the new force $P$ is $P \sin \alpha = P(15/17)$ upwards. The vertical component of the 34 N force is $-34 \sin \alpha = -34(15/17) = -30$ downwards. The vertical component of the 17 N force is $-17 \cos \alpha = -17(8/17) = -8$ downwards.

Setting the sum to zero: $P(15/17) - 30 - 8 = 0$ , which gives $15P/17 = 38$ . Solving for $P$ yields $P = 646/15 \approx 43.1$ .

Key Takeaways

If a resultant is specified to act in a particular direction, the components perpendicular to that direction must sum to zero.
This reduces a vector problem to a simple scalar equation in one unknown.

Common Mistakes

Including the horizontal components in the equation when only the vertical condition is needed.
Sign errors when setting up the vertical equilibrium equation.

Things to Be Careful About

Ensure that only the components perpendicular to the required resultant direction are set to zero. The horizontal components will sum to the magnitude of the new resultant, but that is not required here.

Techniques used

apply condition for resultant to act in a specific directionsolve linear equation for unknown force magnitude

Q5MediumEnergy, Work and PowerNewton's Laws of Motion

A car of mass $1500\text{ kg}$ is travelling along a straight horizontal road.

(a)

It is given that there is a constant resistance to motion. The engine of the car is working at $24\text{kW}$ while the car is travelling at a constant speed of $32\text{ms}^{-1}$ . The power is now increased to $28\text{kW}$ .

Find the acceleration of the car at the instant it is travelling at a speed of $36\text{ms}^{-1}$ .

4M Similar

DifficultyMedium

Worked solution

Approach

At constant speed the resultant force is zero, so the driving force equals the resistance. Use $P = Fv$ to find the resistance from the initial power and speed. Then, after the power is increased, find the new driving force at $36\text{ m s}^{-1}$ and apply Newton's second law to find the acceleration.

Working

Initially the car travels at constant speed, so the driving force $D$ balances the resistance $R$ .

P = Dv

D = \frac{24000}{32} = 750\text{ N}

Therefore

R = 750\text{ N}

After the power is increased to $28\text{ kW} = 28000\text{ W}$ , at speed $v = 36\text{ m s}^{-1}$ the driving force is

D = \frac{28000}{36} = \frac{7000}{9}\text{ N} \approx 777.78\text{ N}

Apply Newton's second law:

D - R = ma

\frac{28000}{36} - 750 = 1500a

\frac{7000}{9} - \frac{6750}{9} = 1500a

\frac{250}{9} = 1500a

a = \frac{250}{9 \times 1500} = \frac{250}{13500} = \frac{1}{54}\text{ m s}^{-2}

Answer

The acceleration of the car is

a = \frac{1}{54}\text{ m s}^{-2} \approx 0.0185\text{ m s}^{-2}

Final answer

a = 1/54 m s^-2 (0.0185 m s^-2)

Detailed explanation

Walkthrough

The car is initially travelling at constant speed, so its acceleration is zero and the driving force must exactly balance the resistance. The engine power and speed are given, so the driving force can be found from $P = Fv$ : $F = P/v$ . Substituting $P = 24000\text{ W}$ and $v = 32\text{ m s}^{-1}$ gives the resistance as $750\text{ N}$ .

When the power is increased to $28000\text{ W}$ , the driving force at the instant the speed is $36\text{ m s}^{-1}$ is $28000/36$ , which is larger than the resistance. The resultant force is therefore $D - R$ . Newton's second law, $F_{\text{net}} = ma$ , gives $D - R = 1500a$ . Substituting the values and solving gives $a = 1/54\text{ m s}^{-2}$ .

Key Takeaways

The relationship $P = Fv$ connects engine power, driving force and speed.
Constant speed means zero resultant force, so driving force equals resistance.
Newton's second law is used to relate the net force to acceleration.
All quantities must be in SI units: watts, newtons, kilograms and metres per second.

Common Mistakes

Forgetting to convert kilowatts to watts, e.g. using $24$ instead of $24000$ .
Using the initial speed $32$ when calculating the new driving force after the power increase.
Making sign errors in Newton's second law. The mark scheme allows sign errors for the method mark, but the final acceleration must have the correct sign.
Omitting the mass $1500$ from the equation.

Things to Be Careful About

The resistance remains constant in this part because it is stated to be constant.
The driving force is evaluated at the instant the speed is $36\text{ m s}^{-1}$ , not at the original speed.
The final answer can be given as $\frac{1}{54}\text{ m s}^{-2}$ or $0.0185\text{ m s}^{-2}$ ; both are accepted.
Use the exact fraction to avoid rounding errors in later working.

Techniques used

convert power to wattsuse P = Fv to find driving forceequate driving force and resistance at constant speedapply Newton's second lawsolve for acceleration

(b)

It is given instead that the resistance to the motion of the car is $(340 + 4v)\text{N}$ when the speed of the car is $v\text{ms}^{-1}$ .

When the engine is working at $20\text{kW}$ , the car is travelling at constant speed.

Find this constant speed.

3M Similar

DifficultyMedium-Easy

Worked solution

Approach

At constant speed the resultant force is zero, so the driving force equals the resistance. Use $P = Fv$ with $F = 340 + 4v$ to form an equation in $v$ . Solve the resulting quadratic and choose the positive speed.

Working

Let the constant speed be $v\text{ m s}^{-1}$ . At constant speed, driving force equals resistance:

F = 340 + 4v

The engine power is $20\text{ kW} = 20000\text{ W}$ , so using $P = Fv$ :

20000 = (340 + 4v)v

20000 = 340v + 4v^2

Rearrange into a quadratic equation:

4v^2 + 340v - 20000 = 0

Divide by 4:

v^2 + 85v - 5000 = 0

Factorise:

(v + 125)(v - 40) = 0

Therefore

v = -125 \quad \text{or} \quad v = 40

Since speed cannot be negative,

v = 40\text{ m s}^{-1}

Answer

The constant speed is $40\text{ m s}^{-1}$ .

Final answer

40 m s^-1

Detailed explanation

Walkthrough

Because the car travels at constant speed, its acceleration is zero and the driving force must equal the resistance. The resistance is given as a function of speed, $340 + 4v$ . The power equation $P = Fv$ then becomes $20000 = (340 + 4v)v$ . Expanding gives a quadratic equation. Rearranging and dividing by $4$ gives $v^2 + 85v - 5000 = 0$ , which factorises as $(v + 125)(v - 40) = 0$ . The two solutions are $v = -125$ and $v = 40$ . A speed cannot be negative, so the constant speed is $40\text{ m s}^{-1}$ .

Key Takeaways

At constant speed, driving force equals resistance.
Power, force and speed are related by $P = Fv$ .
When resistance depends on speed, substituting into $P = Fv$ produces a quadratic equation.
Physical constraints, such as speed being non-negative, are used to select the correct root.

Common Mistakes

Not converting $20\text{ kW}$ to $20000\text{ W}$ . The mark scheme allows using $20$ for the method marks, but the correct final answer requires the correct units.
Making a sign error when rearranging the quadratic.
Forgetting to reject the negative root $v = -125$ .
Using $P = F/v$ or $P = F + v$ instead of $P = Fv$ .

Things to Be Careful About

The quadratic has two roots; only the positive one is physically possible.
The final answer must be exactly $40\text{ m s}^{-1}$ ; no other speed is accepted.
If using the quadratic formula, ensure the discriminant is computed correctly.
The resistance is in newtons when $v$ is in metres per second.

Techniques used

convert power to wattsuse P = Fv with speed-dependent resistanceform and solve quadratic equationreject negative speed

Q6MediumKinematics of Motion in a Straight Line

A particle $P$ moves in a straight line starting from a point $O$ . At time $t\text{ s}$ after leaving $O$ , the velocity, $v\text{ms}^{-1}$ , of $P$ is given by

v = (15 - 2t)^2

(a)

Find the values of $t$ when the velocity of $P$ is $100\text{ms}^{-1}$ .

2M Similar

DifficultyMedium-Easy

Worked solution

Approach

Since the velocity is given as a function of time, set $v = 100$ and solve for $t$ . Taking square roots will produce two linear equations.

Working

Set the velocity equal to 100:

(15 - 2t)^2 = 100

Take square roots:

15 - 2t = \pm 10

If $15 - 2t = 10$ :

-2t = -5

t = 2.5

If $15 - 2t = -10$ :

-2t = -25

t = 12.5

Answer

t = 2.5 \quad \text{or} \quad t = 12.5

Final answer

t = 2.5 or t = 12.5

Detailed explanation

Walkthrough

We need to find the times at which the velocity is $100\text{ m s}^{-1}$ . The velocity function is $v = (15 - 2t)^2$ , so we replace $v$ by $100$ and solve

(15 - 2t)^2 = 100.

The left-hand side is a square, so $15 - 2t$ can be either $10$ or $-10$ . Solving each of these linear equations gives a different time. The two answers correspond to the particle having velocity $100\text{ m s}^{-1}$ on two separate occasions.

Key Takeaways

A squared velocity expression can be equated to a number and solved for time. When a squared quantity equals a positive number, both the positive and negative square roots must be considered. This produces two possible times.

Common Mistakes

Taking only the positive square root and missing $t = 12.5$ .
Expanding $(15-2t)^2$ and making an algebra error before solving.
Forgetting to include units or mixing up seconds with other units.

Things to Be Careful About

When $(15 - 2t)^2 = 100$ , the quantity $15 - 2t$ may be $\pm 10$ . Both equations must be solved. The accepted answers are $t = 2.5$ or $t = 12.5$ , equivalently $t = \frac52$ or $t = \frac{25}{2}$ .

Techniques used

set the velocity expression equal to a given valuetake square roots of both sidessolve the resulting linear equations for time

(b)

Show that there is a particular value of $t$ for which the velocity and acceleration of the particle are both zero.

3M Similar

DifficultyMedium

Worked solution

Approach

Acceleration is the derivative of velocity with respect to time. Differentiate $v=(15-2t)^2$ , then solve the equations $a=0$ and $v=0$ and show that they give the same time.

Working

Differentiate $v$ with respect to $t$ :

a = \frac{dv}{dt} = 2(15 - 2t)(-2) = -4(15 - 2t)

This is often written as

a = 8t - 60

Set $a = 0$ :

8t - 60 = 0

8t = 60

t = 7.5

Set $v = 0$ :

(15 - 2t)^2 = 0

15 - 2t = 0

t = 7.5

Therefore the same value $t = 7.5$ makes both the velocity and the acceleration zero.

Answer

t = 7.5 \text{ s}

Final answer

t = 7.5

Detailed explanation

Walkthrough

Acceleration is obtained by differentiating velocity with respect to time. Using the chain rule on $v=(15-2t)^2$ ,

a = 2(15-2t)(-2) = 8t - 60.

Setting $a=0$ gives $8t-60=0$ , so $t=7.5$ . To prove the velocity is also zero at this time, we solve $v=0$ :

(15-2t)^2 = 0 \quad \Longrightarrow \quad 15-2t = 0 \quad \Longrightarrow \quad t = 7.5.

Because both equations give the same value of $t$ , this is the required particular value.

Key Takeaways

The acceleration is the derivative of the velocity with respect to time. A square can only be zero when its base is zero. Showing that two equations share the same solution establishes that both velocity and acceleration vanish at the same time.

Common Mistakes

Using $a = \frac{v}{t}$ : this is not the correct way to find acceleration from a variable velocity and scores no method mark.
Finding only $t$ from $a=0$ and not verifying that $v=0$ at that time.
Expanding the velocity incorrectly before differentiating.

Things to Be Careful About

The acceleration is $a = 8t - 60$ after simplification, but any equivalent expression such as $-4(15-2t)$ is acceptable. The final value must be $t = 7.5$ exactly. When solving $v=0$ , only $15-2t=0$ is needed because the square is zero only when its base is zero.

Techniques used

differentiate velocity with respect to time to find accelerationset the acceleration equal to zero and solve for timeverify that the same time makes the velocity zero

(c)

Find the displacement of $P$ from $O$ at the time that the velocity and acceleration of the particle are both zero.

4M Similar

DifficultyMedium

Worked solution

Approach

Displacement is obtained by integrating velocity with respect to time. Since the particle starts from $O$ , $s = 0$ when $t = 0$ , so the constant of integration is zero when the polynomial form is used. Then evaluate the displacement at $t = 7.5$ from part (b).

Working

Expand the velocity:

v = (15 - 2t)^2 = 225 - 60t + 4t^2

Integrate to obtain displacement:

s = \int (4t^2 - 60t + 225)\,dt = \frac{4}{3}t^3 - 30t^2 + 225t + c

Because $s=0$ at $t=0$ , $c=0$ . At $t=7.5$ :

s(7.5) = \frac{4}{3}(7.5)^3 - 30(7.5)^2 + 225(7.5)

Evaluate the terms:

\frac{4}{3}(7.5)^3 = 562.5,\quad 30(7.5)^2 = 1687.5,\quad 225(7.5) = 1687.5

So

s(7.5) = 562.5 - 1687.5 + 1687.5 = 562.5

Therefore the displacement from $O$ is $562.5$ m.

Answer

s = 562.5 \text{ m} = \frac{1125}{2} \text{ m}

Final answer

s = 562.5 m

Detailed explanation

Walkthrough

Displacement is the integral of velocity with respect to time. We first expand the velocity:

v = (15-2t)^2 = 225 - 60t + 4t^2.

Integrating term by term gives

s = \frac43t^3 - 30t^2 + 225t + c.

Because the particle starts from $O$ , the displacement is zero when $t=0$ , so $c=0$ . We then substitute $t=7.5$ , the time found in part (b), into this expression. The arithmetic gives $s(7.5)=562.5$ m.

Equivalently, integrating directly using the reverse chain rule gives

s = -\frac16(15-2t)^3 + 562.5,

and at $t=7.5$ the first term is zero, so the same answer is obtained.

Key Takeaways

Integrating velocity gives displacement. A definite integral from the starting time to the required time gives the displacement from the starting point. The constant of integration is determined by the initial condition $s=0$ when $t=0$ .

Common Mistakes

Using $s=vt$ for a non-constant velocity; this is incorrect and scores no method mark.
Leaving a non-zero constant of integration in the final answer; the mark scheme does not allow the final A1 if $+c$ is still present.
Making sign errors when using the reverse chain rule form $-\frac16(15-2t)^3$ .
Forgetting to use the value of $t$ from part (b), or using the wrong limit.

Things to Be Careful About

The displacement is measured from $O$ , so the lower limit is $t=0$ . In the expanded form the constant is zero; in the directly integrated form the constant is $562.5$ . Both approaches give the same final displacement. The units are metres, so the final answer should be quoted as $562.5$ m or $\frac{1125}{2}$ m.

Techniques used

expand the squared velocity expressionintegrate the velocity to obtain displacementapply the initial condition at the starting pointevaluate the definite integral between the starting time and the required time

Q7MediumForces and EquilibriumEnergy, Work and Power

Fig. 7

Two particles $A$ and $B$ of masses $6.5\text{ kg}$ and $m\text{kg}$ respectively are connected by a light inextensible string that passes over a smooth pulley. The pulley is fixed at the top of a rough slope which is at an angle of $\alpha$ to the horizontal ground, where $\tan \alpha = \frac{5}{12}$ . $A$ is on the rough slope and $B$ hangs below the pulley (see diagram). The coefficient of friction between the slope and $A$ is $0.4$ .

(a)

Given that the system is in equilibrium, find the set of possible values of $m$ .

7M Similar

DifficultyMedium

Worked solution

Approach

From $\tan\alpha = \frac{5}{12}$ , we identify $\sin\alpha = \frac{5}{13}$ and $\cos\alpha = \frac{12}{13}$ using a 5-12-13 right triangle.

Set up equilibrium equations for both particles. Since friction can act in either direction along the rough slope, there are two limiting cases: A about to slide down (friction acts up the slope) and A about to slide up (friction acts down the slope). The system is in equilibrium for all values of $m$ between these two limits.

Working

Trigonometric values:

\sin\alpha = \frac{5}{13}, \quad \cos\alpha = \frac{12}{13}

Forces on particle B (hanging, equilibrium):

T = mg

Forces on particle A (on the slope):

Component of weight down the slope:

6.5g\sin\alpha = 6.5g \times \frac{5}{13} = \frac{32.5g}{13} = 2.5g = 25\text{ N}

Normal reaction (perpendicular to slope):

R = 6.5g\cos\alpha = 6.5g \times \frac{12}{13} = \frac{78g}{13} = 6g = 60\text{ N}

Limiting friction:

F = \mu R = 0.4 \times 60 = 24\text{ N}

Case 1: A is about to slide down the slope (friction acts up the slope)

Resolving along the slope (up positive):

T + F - 25 = 0

mg + 24 = 25

mg = 1 \implies m = 0.1

Case 2: A is about to slide up the slope (friction acts down the slope)

Resolving along the slope (up positive):

T - 25 - F = 0

mg = 25 + 24 = 49

m = 4.9

For the system to remain in equilibrium, $m$ must lie between these two limiting values:

0.1 \leq m \leq 4.9

Answer

0.1 \leq m \leq 4.9

Final answer

0.1 ≤ m ≤ 4.9

Detailed explanation

Walkthrough

Step 1: Determine trigonometric values.

We are given $\tan\alpha = \frac{5}{12}$ . This corresponds to a right triangle with opposite side 5, adjacent side 12, and hypotenuse $\sqrt{5^2 + 12^2} = \sqrt{169} = 13$ . Therefore:

\sin\alpha = \frac{5}{13}, \quad \cos\alpha = \frac{12}{13}

Step 2: Analyse forces on particle B.

Particle B hangs vertically and is in equilibrium. The only forces acting on it are its weight $mg$ downward and the tension $T$ upward. Equilibrium gives:

T = mg

Step 3: Analyse forces on particle A on the slope.

Particle A has mass 6.5 kg on a rough slope at angle $\alpha$ . The forces are:

Weight $6.5g$ $6.5 g$ acting vertically downward, resolved into:
- Component down the slope: $6.5g\sin\alpha = 6.5g \times \frac{5}{13} = 25$ N
- Component perpendicular to the slope: $6.5g\cos\alpha = 6.5g \times \frac{12}{13} = 60$ N
Normal reaction $R = 60$ N perpendicular to the slope (outward)
Tension $T$ up the slope
Friction $F$ along the slope (direction depends on impending motion)

Step 4: Calculate limiting friction.

F_{\max} = \mu R = 0.4 \times 60 = 24 \text{ N}

Step 5: Consider the two limiting equilibrium cases.

Since the system is in equilibrium but friction can act in either direction, we need both cases:

Case 1 (m small): B is light, so A tends to slide down. Friction acts up the slope to oppose this. Resolving along the slope:
$T + F = 25 \implies mg + 24 = 25 \implies m = 0.1$
Case 2 (m large): B is heavy, so A tends to slide up. Friction acts down the slope to oppose this. Resolving along the slope:
$T = 25 + F \implies mg = 25 + 24 = 49 \implies m = 4.9$

Step 6: Combine into a range.

For equilibrium, $m$ must not be so small that A slides down, nor so large that A slides up. Therefore:

0.1 \leq m \leq 4.9

Key Takeaways

When a system is in equilibrium on a rough surface, friction can act in either direction, giving two limiting cases.
Always resolve weight into components parallel and perpendicular to the inclined plane.
The normal reaction equals the perpendicular component of weight (when no other perpendicular forces act).
Limiting friction is $F = \mu R$ where $R$ is the normal reaction.
The range of values for equilibrium is found by considering both limiting friction directions.

Common Mistakes

Forgetting that friction can act in both directions, leading to only one value of $m$ instead of a range.
Using $\sin\alpha$ or $\cos\alpha$ without substituting the actual numerical values from $\tan\alpha = \frac{5}{12}$ .
Incorrectly resolving the weight component: the component down the slope is $mg\sin\alpha$ , not $mg\cos\alpha$ .
Forgetting to multiply by $g = 10$ when converting mass to weight (though some mark scheme entries allow leaving $g$ in, the final numerical values require it).
Not expressing the final answer as an inequality or interval.

Things to Be Careful About

The coefficient of friction $\mu = 0.4$ applies at limiting equilibrium. For values of $m$ strictly between 0.1 and 4.9, the actual friction is less than $\mu R$ and the system is in stable equilibrium.
Always verify that $R > 0$ (particle A remains in contact with the slope), which is satisfied here since $R = 60$ N.
The answer must be expressed as $0.1 \leq m \leq 4.9$ or equivalent interval notation $[0.1, 4.9]$ . Writing just $m = 0.1$ or $m = 4.9$ alone is incomplete.

Techniques used

resolve weight into components parallel and perpendicular to slopecalculate normal reaction and limiting frictionform equilibrium equations for both particlesconsider both limiting friction directions for equilibrium range

(b)

It is given instead that $m = 12$ and the particles are released from rest with the string taut.

Use an energy method to find the speed of the particles when each particle has moved $0.6\text{m}$ . You may assume that this occurs before $A$ reaches the pulley or $B$ reaches the ground.

5M Similar

DifficultyMedium

Worked solution

Approach

Since $m = 12 > 4.9$ (from part a), particle B is heavier than the maximum mass for equilibrium. When released, B will descend and A will move up the slope. Use the work-energy principle for the system: the net loss in gravitational potential energy equals the gain in kinetic energy plus the work done against friction.

Working

Direction of motion:

Since $m = 12 > 4.9$ , particle B descends and particle A moves up the slope by $0.6$ m.

Change in gravitational potential energy:

Particle B descends $0.6$ m:

\Delta PE_B = -12g \times 0.6 = -12 \times 10 \times 0.6 = -72 \text{ J}

Particle A moves up the slope by $0.6$ m, rising vertically by $0.6\sin\alpha = 0.6 \times \frac{5}{13}$ m:

\Delta PE_A = +6.5g \times 0.6 \times \frac{5}{13} = 6.5 \times 10 \times 0.6 \times \frac{5}{13} = 39 \times \frac{5}{13} = 15 \text{ J}

Net change in PE of the system:

\Delta PE = 15 - 72 = -57 \text{ J}

Work done against friction:

Friction acts down the slope (opposing A's upward motion):

W_{\text{friction}} = F \times d = 24 \times 0.6 = 14.4 \text{ J}

Change in kinetic energy:

Both particles start from rest and reach speed $v$ :

\Delta KE = \frac{1}{2}(12 + 6.5)v^2 = \frac{1}{2} \times 18.5 \times v^2 = 9.25v^2

Work-energy equation:

Loss in PE = Gain in KE + Work done against friction:

72 - 15 - 14.4 = 9.25v^2

42.6 = 9.25v^2

v^2 = \frac{42.6}{9.25} = \frac{852}{185}

v = \sqrt{\frac{42.6}{9.25}} \approx 2.15 \text{ m/s}

Answer

v = 2.15 \text{ m/s}

Final answer

2.15 m/s

Detailed explanation

Walkthrough

Step 1: Determine direction of motion.

From part (a), the system is in equilibrium for $0.1 \leq m \leq 4.9$ . Since $m = 12 > 4.9$ , particle B is too heavy and will descend, pulling particle A up the slope. Both particles move with the same speed (inextensible string) and the same acceleration.

Step 2: Calculate the change in gravitational potential energy.

Particle B (mass 12 kg) descends $0.6$ m:

\Delta PE_B = -m_B g h = -12 \times 10 \times 0.6 = -72 \text{ J}

Particle A (mass 6.5 kg) moves up the slope by $0.6$ m. The vertical height gained is:

h_A = 0.6 \sin\alpha = 0.6 \times \frac{5}{13} \text{ m}

\Delta PE_A = +m_A g h_A = 6.5 \times 10 \times 0.6 \times \frac{5}{13} = 39 \times \frac{5}{13} = 15 \text{ J}

Step 3: Calculate work done against friction.

Friction acts down the slope (opposing A's upward motion). The friction force is:

F = \mu R = 0.4 \times 60 = 24 \text{ N}

Work done against friction over distance $0.6$ m:

W_{\text{friction}} = F \times d = 24 \times 0.6 = 14.4 \text{ J}

Step 4: Apply the work-energy principle.

For the system, the net loss in gravitational potential energy equals the gain in kinetic energy plus the work done against friction:

|\Delta PE_B| - \Delta PE_A - W_{\text{friction}} = \Delta KE

72 - 15 - 14.4 = \frac{1}{2}(12 + 6.5)v^2

42.6 = 9.25v^2

Step 5: Solve for v.

v^2 = \frac{42.6}{9.25} = \frac{852}{185} \approx 4.6054

v = \sqrt{4.6054} \approx 2.146 \approx 2.15 \text{ m/s}

Key Takeaways

When using the work-energy principle for a system of connected particles, treat the entire system together rather than individual particles (unless tension is needed).
The change in PE for a particle on an inclined plane depends on the vertical height change, which is $d\sin\alpha$ where $d$ is the distance moved along the slope.
Work done against friction is always positive (energy is lost from the mechanical system).
The total KE of the system is $\frac{1}{2}(m_A + m_B)v^2$ since both particles have the same speed.

Common Mistakes

Getting the direction of motion wrong: if A moves up the slope, friction acts down the slope, adding to the energy loss.
Forgetting that particle A also gains PE when moving up the slope (only accounting for B's PE loss).
Using the wrong component: the vertical rise is $0.6\sin\alpha$ , not $0.6\cos\alpha$ .
Including tension in the energy equation: tension does equal and opposite work on the two particles, so it cancels out for the system.
Sign errors in the energy equation: be consistent about what is a gain and what is a loss.

Things to Be Careful About

The question specifically asks for an energy method. Using Newton's second law to find acceleration first, then kinematics, would not earn full marks for this method requirement (though alternative methods are accepted in the mark scheme, the primary method expected is direct energy).
The work done against friction must use the correct friction value ( $24$ N), not a variable expression.
The distance $0.6$ m is the distance each particle moves along its path (along the slope for A, vertically for B), not a vertical distance for A.
The answer should be given to 3 significant figures: $2.15$ m/s.

Techniques used

identify direction of motion from equilibrium analysiscalculate change in gravitational potential energy for both particlescalculate work done against frictionapply work-energy principle to the systemsolve for final speed

Mathematics 9709/45 — May/June 2025

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