A car of mass $1400\text{ kg}$ travelling on a straight horizontal road accelerates uniformly from a speed of $15\text{ m s}^{-1}$ to $20\text{ m s}^{-1}$ over a distance of $175\text{ m}$. (a) Find the time taken to travel the $175\text{ m}$. (b) There is a constant resistance force of $800\text{ N}$ to the motion of the car. (i) Use an energy method throughout to find the average power of the car’s engine as the speed increases from $15\text{ m s}^{-1}$ to $20\text{ m…

(a) 10 s · (b)(i) 26250 W · (b)(ii) 32.8125 m/s

9709/42 Oct/Nov 2025 — Mathematics Solved Past Paper

Q13MMedium-EasyKinematics of Motion in a Straight Line

Fig. 1

The diagram shows the displacement-time graph for the motion of a particle. The particle starts from rest at a point $O$ and travels with constant acceleration $a\text{ m s}^{-2}$ , taking $4\text{ s}$ to move a distance of $x\text{ m}$ . The particle then returns to $O$ with constant speed of $40\text{ m s}^{-1}$ , over a period of $2\text{ s}$ .

Find the value of $x$ and the value of $a$ .

DifficultyMedium-Easy

Worked solution

Approach

Use the second phase of motion (constant speed return) to find the displacement $x$ at $t = 4$ s. Then use the first phase of motion (constant acceleration from rest) with the found value of $x$ to determine the acceleration $a$ .

Working

During the return phase, the particle travels from displacement $x$ back to $0$ at a constant speed of $40$ m s $^{-1}$ over $2$ s. The distance covered is:

\text{distance} = \text{speed} \times \text{time} = 40 \times 2 = 80 \text{ m}

Since the particle returns to the origin $O$ , the displacement at the end of the first phase must be $x = 80$ m.

For the first phase, the particle starts from rest ( $u = 0$ ), travels with constant acceleration $a$ , and takes $t = 4$ s to cover a displacement $s = x = 80$ m. Using the constant acceleration formula:

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

Substitute the known values:

80 = 0 \times 4 + \frac{1}{2} a (4)^2

80 = 8a

a = 10 \text{ m s}^{-2}

Answer

x = 80, \quad a = 10

Final answer

x = 80, a = 10

Detailed explanation

Walkthrough

First, analyze the second part of the journey where the particle returns to the origin. It moves at a constant speed of 40 m/s for 2 seconds. Multiplying speed by time gives the total distance covered, which is 80 m. Because it ends up back at the starting point $O$ , the maximum displacement reached at $t = 4$ s must be exactly 80 m. Thus, $x = 80$ .

Next, analyze the first part of the journey. The particle starts from rest, meaning initial velocity $u = 0$ . It accelerates uniformly at $a$ for 4 seconds to reach a displacement of 80 m. Substituting $s = 80$ , $u = 0$ , and $t = 4$ into the kinematic equation $s = ut + \frac{1}{2}at^2$ allows us to solve directly for $a$ , yielding $a = 10$ .

Key Takeaways

Distance equals speed multiplied by time when speed is constant.
The displacement-time graph's peak value corresponds to the total distance covered during the accelerating phase if the particle returns to the start.
The suvat equation $s = ut + \frac{1}{2}at^2$ is useful when time, displacement, and initial velocity are known.

Common Mistakes

Assuming the speed at the end of the first phase is needed before finding $x$ . The mark scheme allows condoning this assumption, but it is simpler to find $x$ directly from the return phase.
Forgetting that the particle returns to $O$ , meaning the displacement at $t=4$ is equal to the distance covered in the return phase.
Using $v = u + at$ without first finding $x$ , leading to two unknowns ( $v$ and $a$ ) and an unsolvable system without additional steps.

Things to Be Careful About

Ensure units are consistent (meters and seconds are already used).
The mark scheme notes that getting an equation relating $x$ and $a$ earns a method mark, but the final answers must be explicitly stated as $x = 80$ and $a = 10$ .
When using the return phase, remember it is constant speed, not constant acceleration, so simple $s = vt$ applies.

Techniques used

calculate distance from constant speed and timeapply suvat equation for constant accelerationsolve for displacement and acceleration

Similar questions

Q25MMediumForces and EquilibriumNewton's Laws of Motion

A railway locomotive of mass $240\,000\text{ kg}$ is towing a coach of mass $36\,000\text{ kg}$ down a hill inclined at an angle of $\sin^{-1} 0.04$ to the horizontal. The driving force produced by the locomotive is $450\,000\text{ N}$ and there are resistances to motion of $120\,000\text{ N}$ on the locomotive and $15\,000\text{ N}$ on the coach. The coupling between the locomotive and the coach is light, rigid and parallel to the hill.

Find the acceleration of the locomotive and the tension in the coupling.

DifficultyMedium

Worked solution

Approach

Take the positive direction down the hill. Resolve the weight of each vehicle along the hill using $\sin \theta = 0.04$ , so the downhill component of each weight is $mg \times 0.04$ . Apply Newton's second law separately to the locomotive and the coach, with the tension $T$ opposing the locomotive's motion and pulling the coach downhill. Eliminate $T$ by adding the equations to find $a$ , then substitute back to find $T$ .

Working

Let $a$ be the acceleration down the hill and $T$ be the tension in the coupling. Use $g = 10\text{ m s}^{-2}$ .

For the locomotive:

450000 + 240000(10)(0.04) - 120000 - T = 240000a

450000 + 96000 - 120000 - T = 240000a

426000 - T = 240000a

For the coach:

T + 36000(10)(0.04) - 15000 = 36000a

T + 14400 - 15000 = 36000a

T - 600 = 36000a

Add the two equations:

426000 - T + T - 600 = 240000a + 36000a

425400 = 276000a

a = \frac{425400}{276000} = \frac{709}{460} = 1.5413\ldots

So

a = 1.54\text{ m s}^{-2} \quad (3\text{ s.f.})

Substitute into the coach equation:

T = 36000a + 600 = 36000 \times \frac{709}{460} + 600 = 56086.956\ldots

So

T = 56100\text{ N} \quad (3\text{ s.f.})

Answer

Acceleration $= 1.54\text{ m s}^{-2}$ down the hill; tension in the coupling $= 56100\text{ N}$ .

Final answer

a = 1.54 m s^-2, T = 56100 N

Detailed explanation

Walkthrough

The locomotive and coach move together down the hill, so they have the same acceleration. Choose downhill as positive. Since the angle is given by $\sin \theta = 0.04$ , the component of each weight along the hill is $mg \sin \theta = mg \times 0.04$ . The driving force acts downhill on the locomotive; the resistances oppose motion and act uphill. The tension in the coupling is an internal force: it pulls the coach downhill and pulls the locomotive uphill.

For the locomotive, the downhill forces are the driving force and the component of its weight; the uphill forces are its resistance and the tension. Newton's second law gives:

450000 + 240000(10)(0.04) - 120000 - T = 240000a

which simplifies to $426000 - T = 240000a$ .

For the coach, the downhill forces are the tension and the component of its weight; the uphill force is its resistance:

T + 36000(10)(0.04) - 15000 = 36000a

which simplifies to $T - 600 = 36000a$ .

Adding the two equations eliminates $T$ , because it appears with opposite signs. This gives $425400 = 276000a$ , so $a = 425400/276000 = 709/460 = 1.5413\ldots$ . Substituting this value into the coach equation gives $T = 36000a + 600 = 56086.956\ldots$ , so to 3 significant figures $T = 56100\text{ N}$ .

Key Takeaways

On an inclined plane, the component of weight along the plane is $mg \sin \theta$ .
Apply Newton's second law separately to each connected body; the internal tension cancels when the whole system is considered.
Use a consistent positive direction and treat forces opposing motion as negative.
When a quantity such as tension appears with opposite signs in two equations, adding the equations eliminates it.

Common Mistakes

Using $mg$ instead of the component $mg \sin \theta$ , or mixing up $\sin$ and $\cos$ .
Forgetting one of the resistances, especially the resistance on the coach.
Giving the tension the same sign in both equations, so it does not cancel when the equations are added.
Using only the total-mass equation and then failing to consider one body separately to find the tension.
Omitting the Newton's second law equation and jumping straight to an answer; the mark scheme requires a correct number of terms in the equation before solving.
Using a rounded value of $a$ when calculating $T$ can give slightly different but accepted values: using $a = 1.54$ gives $T = 56040\text{ N}$ from the coach equation or $56400\text{ N}$ from the locomotive equation; using $a = 1.541$ gives $56076\text{ N}$ or $56160\text{ N}$ . Using the exact fraction $a = 709/460$ avoids this ambiguity.

Things to Be Careful About

The mark scheme uses $g = 10\text{ m s}^{-2}$ ; state this assumption if it is not given.
The angle is $\sin^{-1} 0.04$ , so $\sin \theta = 0.04$ exactly; there is no need to find the angle itself.
The coupling is light and rigid, so the tension is the same throughout and both vehicles have the same acceleration.
Check the direction of each force: driving force and motion are downhill, resistances are uphill, and tension pulls the coach downhill while pulling the locomotive uphill.
The final answers are $a = 1.54\text{ m s}^{-2}$ and $T = 56100\text{ N}$ , to 3 significant figures.

Techniques used

resolve weight components along the hillapply Newton's second law to each bodyeliminate tension by adding equationssubstitute to find the tension

Similar questions

Q3MediumKinematics of Motion in a Straight LineEnergy, Work and PowerForces and Equilibrium

A car of mass $1400\text{ kg}$ travelling on a straight horizontal road accelerates uniformly from a speed of $15\text{ m s}^{-1}$ to $20\text{ m s}^{-1}$ over a distance of $175\text{ m}$ .

(a)

Find the time taken to travel the $175\text{ m}$ .

2M Similar

DifficultyEasy

Worked solution

Approach

Since the car accelerates uniformly, its average speed over the $175\text{ m}$ is the mean of the initial and final speeds. Use

\begin{aligned} s &= \frac{u+v}{2}\,t \end{aligned}

with $s = 175$ , $u = 15$ and $v = 20$ , then solve for $t$ .

Working

\begin{aligned} 175 &= \frac{15+20}{2}\,t \\ 175 &= 17.5\,t \\ t &= \frac{175}{17.5} \\ t &= 10\text{ s} \end{aligned}

Answer

t = 10\text{ s}

Final answer

10 s

Detailed explanation

Walkthrough

The car is accelerating uniformly, so the speed changes linearly from $15\text{ m s}^{-1}$ to $20\text{ m s}^{-1}$ . The average speed during this time is therefore

\frac{15+20}{2} = 17.5\text{ m s}^{-1}.

The car travels $175\text{ m}$ at this average speed, so $175 = 17.5t$ . Solving gives $t=10\text{ s}$ . This uses the formula $s=\frac{u+v}{2}t$ , which avoids needing to find the acceleration separately.

Key Takeaways

For uniform acceleration, average speed equals the average of the initial and final speeds.
If displacement and initial/final speeds are known, time can be found directly from $s=\frac{u+v}{2}t$ .

Common Mistakes

Dividing incorrectly: $\frac{175}{17.5}=10$ , not $0.1$ .
Using $v=u+at$ before finding $a$ ; this adds unnecessary work and can introduce errors.
Forgetting that the formula uses the average, so forgetting to divide $15+20$ by 2.

Things to Be Careful About

The formula is only valid when acceleration is constant; the question states this.
Ensure all speeds are in $\text{ m s}^{-1}$ and distances in metres before substituting.
The final answer must have units: $10\text{ s}$ , not just 10.

Techniques used

use the average speed of a uniformly accelerating bodyapply distance equals average speed times timesolve a linear equation for time

(b)

There is a constant resistance force of $800\text{ N}$ to the motion of the car.

6M

(i)

Use an energy method throughout to find the average power of the car’s engine as the speed increases from $15\text{ m s}^{-1}$ to $20\text{ m s}^{-1}$ .

4M Similar

DifficultyMedium

Worked solution

Approach

Use the work-energy principle. The work done by the engine is used to increase the kinetic energy of the car and to overcome the constant resistance. Average power is the total work done divided by the time taken from part (a).

Working

Change in kinetic energy:

\begin{aligned} \Delta KE &= \frac{1}{2} \times 1400 \times 20^2 - \frac{1}{2} \times 1400 \times 15^2 \\ &= 280000 - 157500 \\ &= 122500\text{ J} \end{aligned}

Work done against resistance:

800 \times 175 = 140000\text{ J}

Work done by the engine:

\begin{aligned} W &= 122500 + 140000 \\ &= 262500\text{ J} \end{aligned}

Average power:

\begin{aligned} \text{Average power} &= \frac{W}{t} \\ &= \frac{262500}{10} \\ &= 26250\text{ W} \end{aligned}

Answer

\text{Average power} = 26250\text{ W} = 26.25\text{ kW}

Final answer

26250 W

Detailed explanation

Walkthrough

Because the question requires an energy method, do not use Newton's second law. The engine does work to increase the car's kinetic energy and to overcome the constant resistance.

Change in kinetic energy:

\frac{1}{2} \times 1400 \times 20^2 - \frac{1}{2} \times 1400 \times 15^2 = 280000 - 157500 = 122500\text{ J}.

Work done against resistance:

800 \times 175 = 140000\text{ J}.

So the total work done by the engine is

122500 + 140000 = 262500\text{ J}.

The average power is total work divided by the time taken from part (a), $t=10\text{ s}$ :

\frac{262500}{10} = 26250\text{ W}.

This can also be written as $26.25\text{ kW}$ .

Key Takeaways

The work-energy principle for a car on level ground: work done by engine = increase in kinetic energy + work done against resistance.
Change in kinetic energy is the difference of two kinetic-energy terms, not the kinetic energy of the change in speed.
Average power over a time interval = total work done divided by total time.

Common Mistakes

Using $\frac{1}{2} \times 1400 \times (20-15)^2$ for the kinetic energy change. This is wrong; change in KE is $\frac{1}{2}mv^2 - \frac{1}{2}mu^2$ .
Omitting the work done against resistance, or adding it with the wrong sign.
Dividing by the distance instead of the time; average power is work per unit time.
Starting with Newton's second law; since the instruction is "use an energy method throughout", the N2L approach scores at most 2 marks.

Things to Be Careful About

Use the value of $t$ from part (a), $10\text{ s}$ , when converting work to average power. If $t$ was recalculated correctly, the mark scheme allows following that value.
All quantities must be in SI units: mass in kg, speeds in $\text{m s}^{-1}$ , force in N, distance in m, so energy is in J and power is in W.
The answer may be given as $26250\text{ W}$ or $26.25\text{ kW}$ ; both are acceptable with units.

Techniques used

compute the change in kinetic energycalculate work done against a constant resistanceapply the work-energy equationdivide total work by time to obtain average power

(ii)

Find the steady speed that the car could maintain on the horizontal road if the engine is working at the power found in (b)(i).

2M Similar

DifficultyMedium-Easy

Worked solution

Approach

At steady speed there is no acceleration, so the resultant horizontal force is zero. Therefore the engine's driving force equals the constant resistance, $800\text{ N}$ . With the engine working at the power from part (b)(i), use $P = Fv$ to find the speed.

Working

At constant speed, driving force $F$ satisfies

F = 800\text{ N}

Using $P = Fv$ with $P = 26250\text{ W}$ :

v = \frac{P}{F} = \frac{26250}{800} = 32.8125\text{ m s}^{-1}

As an exact fraction,

v = \frac{525}{16}\text{ m s}^{-1}

Answer

v = 32.8125\text{ m s}^{-1} = \frac{525}{16}\text{ m s}^{-1}

Final answer

32.8125 m/s

Detailed explanation

Walkthrough

Once the speed is steady, acceleration is zero. Newton's second law then says the net horizontal force on the car is zero. The two horizontal forces are the driving force forwards and the $800\text{ N}$ resistance backwards, so the driving force must be $800\text{ N}$ .

Power is the product of driving force and speed for motion in the direction of the force:

P = Fv.

Using $P = 26250\text{ W}$ and $F = 800\text{ N}$ gives

v = \frac{26250}{800} = 32.8125\text{ m s}^{-1}.

This is the steady speed the engine's power can sustain against the constant resistance.

Key Takeaways

Constant or steady speed means zero acceleration, hence zero resultant force.
For a vehicle moving in the direction of its driving force, $P = Fv$ links engine power, driving force and speed.
The resistance force fixes the driving force needed at steady speed.

Common Mistakes

Trying to use a nonzero acceleration with $F = ma$ ; at steady speed $a = 0$ , so there is no net force.
Using the mass of the car in $P = Fv$ ; power depends on force and speed, not mass.
Using the power as $26.25$ instead of $26250$ would give a speed 1000 times too small. Always use watts.

Things to Be Careful About

If the rounded value $26300\text{ W}$ is carried from part (b)(i), the speed becomes $32.875\text{ m s}^{-1}$ , which the mark scheme allows as an alternative from that rounded power.
The exact expected speed is $32.8125\text{ m s}^{-1}$ , or $\frac{525}{16}\text{ m s}^{-1}$ .
Include units in the final answer.

Techniques used

recognise that steady speed means zero accelerationequate driving force to resistanceapply P = Fvsolve for speed

Q4Medium-HardKinematics of Motion in a Straight LineMomentum

A particle $P$ of mass $0.1\text{ kg}$ is projected vertically upwards with speed $30\text{ m s}^{-1}$ from horizontal ground. At the same instant a particle $Q$ of mass $0.4\text{ kg}$ is projected vertically upwards with speed $10\text{ m s}^{-1}$ from a height of $15\text{ m}$ above the ground. $P$ and $Q$ move in the same vertical line.

(a)

Find the height above the ground at which $P$ and $Q$ collide.

4M Similar

DifficultyMedium

Worked solution

Approach

Write the height of each particle above the ground as a function of time using the suvat equation $s = ut + \frac{1}{2}at^2$ with $a = -g$ . Equate the two heights, solve for the collision time $t$ , then substitute back to find the height.

Working

Take $g = 10\text{ m s}^{-2}$ and upward as positive.

For $P$ , projected from the ground with $u = 30$ :

s_P = 30t - \frac{1}{2}gt^2

For $Q$ , projected from height $15\text{ m}$ with $u = 10$ , its height above the ground is:

s_Q = 15 + 10t - \frac{1}{2}gt^2

The particles collide when their heights above the ground are equal:

30t - \frac{1}{2}gt^2 = 15 + 10t - \frac{1}{2}gt^2

The $\frac{1}{2}gt^2$ terms cancel:

30t = 15 + 10t

20t = 15

t = 0.75\text{ s}

Substitute $t = 0.75$ into $s_P$ :

s_P = 30(0.75) - \frac{1}{2}(10)(0.75)^2 = 22.5 - 2.8125 = 19.6875\text{ m}

Answer

19.6875\text{ m} \quad \left(\text{or } \frac{315}{16}\text{ m, or } 19.7\text{ m to 3 s.f.}\right)

Final answer

19.6875 m (or 315/16 m, or 19.7 m to 3 s.f.)

Detailed explanation

Walkthrough

The key idea is to write down the height of each particle above the ground as a function of time, then set them equal because the collision happens when both are at the same height.

For particle $P$ , it starts at ground level with speed $30\text{ m s}^{-1}$ upwards. The suvat equation $s = ut + \frac{1}{2}at^2$ gives the displacement. Since the acceleration is $g$ downwards, we take $a = -g$ . So $s_P = 30t - \frac{1}{2}gt^2$ .

For particle $Q$ , it starts $15\text{ m}$ above the ground. Its displacement from its starting point is $10t - \frac{1}{2}gt^2$ , so its height above the ground is $15 + 10t - \frac{1}{2}gt^2$ .

When they collide, $s_P = s_Q$ . The $\frac{1}{2}gt^2$ terms are identical on both sides and cancel, leaving a simple linear equation $30t = 15 + 10t$ , giving $t = 0.75\text{ s}$ .

Finally, substitute $t = 0.75$ into either expression to get the height. Using $s_P$ : $30(0.75) - \frac{1}{2}(10)(0.75)^2 = 22.5 - 2.8125 = 19.6875\text{ m}$ .

Key Takeaways

The suvat equation $s = ut + \frac{1}{2}at^2$ describes displacement under constant acceleration.
When two particles are at the same position, their displacement expressions can be equated.
The initial height offset must be added to the displacement of the particle that starts above the ground.

Common Mistakes

Forgetting to add the $15\text{ m}$ initial height for $Q$ .
Using $a = +g$ instead of $-g$ (or inconsistent signs).
Thinking the collision time requires solving a quadratic — the $\frac{1}{2}gt^2$ terms cancel because both particles have the same acceleration.

Things to Be Careful About

Use a consistent sign convention (upward positive, $a = -g$ ).
Both particles accelerate at the same rate $g$ , so the quadratic terms cancel.
The mark scheme requires showing the suvat equation used (M1) and equating heights (DM1); unsupported answers are not credited.

Techniques used

write displacement equations using suvatequate heights of the two particlessolve linear equation for collision timesubstitute time to find collision height

(b)

When $P$ and $Q$ collide, they coalesce.

Find the speed of the combined particle at the instant that it reaches the ground.

5M Similar

DifficultyMedium-Hard

Worked solution

Approach

First find the velocities of $P$ and $Q$ just before the collision using $v = u + at$ . Then apply conservation of linear momentum for the coalescing collision to find the speed of the combined particle. Finally, use $v^2 = u^2 + 2as$ to find the speed of the combined particle when it reaches the ground.

Working

At $t = 0.75\text{ s}$ , using $v = u + at$ with $a = -g = -10$ :

v_P = 30 - 10(0.75) = 22.5\text{ m s}^{-1} \quad \text{(upwards)}

v_Q = 10 - 10(0.75) = 2.5\text{ m s}^{-1} \quad \text{(upwards)}

Both particles are moving upwards. Since they coalesce, momentum is conserved:

0.1(22.5) + 0.4(2.5) = (0.1 + 0.4)v

2.25 + 1.0 = 0.5v

v = 6.5\text{ m s}^{-1} \quad \text{(upwards)}

The combined particle is at height $19.6875\text{ m}$ with speed $6.5\text{ m s}^{-1}$ upwards. To find its speed when it reaches the ground, use $v^2 = u^2 + 2as$ with displacement $s = 19.6875\text{ m}$ downwards:

v^2 = 6.5^2 + 2(10)(19.6875) = 42.25 + 393.75 = 436

v = \sqrt{436} = 2\sqrt{109} \approx 20.9\text{ m s}^{-1}

Answer

20.9\text{ m s}^{-1} \quad \left(\text{or } 2\sqrt{109}\text{ m s}^{-1}\right)

Final answer

20.9 m/s (or 2√109 m/s)

Detailed explanation

Walkthrough

First, find how fast each particle is moving at the instant of collision. Use $v = u + at$ with $t = 0.75$ : $v_P = 30 - 10(0.75) = 22.5\text{ m s}^{-1}$ and $v_Q = 10 - 10(0.75) = 2.5\text{ m s}^{-1}$ , both upwards.

When the particles coalesce, it is a perfectly inelastic collision, so momentum is conserved: total momentum before equals total momentum after. $0.1 \times 22.5 + 0.4 \times 2.5 = (0.1 + 0.4)v$ , giving $v = 6.5\text{ m s}^{-1}$ upwards.

Now the combined particle (mass $0.5\text{ kg}$ ) is at height $19.6875\text{ m}$ moving upwards at $6.5\text{ m s}^{-1}$ . It will rise a little more, then fall back to the ground. To find its speed at the ground, use $v^2 = u^2 + 2as$ with the displacement being $19.6875\text{ m}$ downward. Taking upward as positive, $a = -g$ and $s = -19.6875$ , so $v^2 = 6.5^2 + 2(-10)(-19.6875) = 42.25 + 393.75 = 436$ . Therefore $v = \sqrt{436} \approx 20.9\text{ m s}^{-1}$ .

Alternatively, you could first find the extra height gained above $19.6875\text{ m}$ : $s = \frac{6.5^2}{2g} = 2.1125\text{ m}$ , so the maximum height is $21.8\text{ m}$ , then $v^2 = 2g \times 21.8 = 436$ , giving the same result.

Key Takeaways

Conservation of linear momentum applies to collisions; for coalescing particles, the combined mass moves with a common velocity.
The velocities used in momentum must be the velocities at the instant of collision, not the initial velocities.
suvat can be applied to the combined particle after the collision to find its final speed.

Common Mistakes

Using the initial speeds ( $30$ and $10$ ) instead of the speeds at collision ( $22.5$ and $2.5$ ) in the momentum equation.
Sign errors in the momentum equation (all velocities are upward here, so signs are consistent, but it is easy to slip).
Using weight $mg$ instead of mass $m$ in the momentum equation — the mark scheme explicitly disallows this.
Forgetting that the combined particle rises before falling, or mishandling the sign of the displacement.

Things to Be Careful About

The momentum equation must use masses ( $0.1$ and $0.4$ ), not weights.
The mark scheme awards the B1 for the velocity expressions only if the M marks in part (a) were awarded.
The final speed is a magnitude; the direction is downward but the question asks for speed.
Use $g = 10\text{ m s}^{-2}$ consistently.

Techniques used

find velocities at collision using suvatapply conservation of momentum for coalescing particlesuse suvat to find final speed at ground

Q56MMediumNewton's Laws of MotionKinematics of Motion in a Straight Line

Fig. 2

A block of mass $5\text{ kg}$ is being pulled straight down a line of greatest slope of a rough plane by a force of magnitude $20\text{ N}$ . The plane is inclined at an angle of $10^\circ$ to the horizontal and the $20\text{ N}$ force acts at an angle of $35^\circ$ above the line of greatest slope of the plane (see diagram). The coefficient of friction between the block and the plane is $0.4$ . The speed of the block when it passes a point $O$ is $2\text{ m s}^{-1}$ .

Find the speed of the block when it has moved $3\text{ m}$ down the plane from $O$ .

DifficultyMedium

Worked solution

Approach

Resolve the forces acting on the block perpendicular to the plane to find the normal reaction force $R$ . Then resolve the forces parallel to the plane (down the line of greatest slope) to find the acceleration $a$ using Newton's second law. Finally, use the constant acceleration formula $v^2 = u^2 + 2as$ to find the final speed.

Working

Let $R$ be the normal reaction force, $F$ be the frictional force, and $a$ be the acceleration down the plane. The block moves down the plane, so friction acts up the plane.

Resolving perpendicular to the plane:

The forces perpendicular to the plane are the normal reaction $R$ (upwards), the component of weight $mg \cos 10^\circ$ (downwards into the plane), and the perpendicular component of the pulling force $20 \sin 35^\circ$ (upwards away from the plane).

\begin{aligned} R + 20 \sin 35^\circ &= 5g \cos 10^\circ \\ R &= 5(9.8) \cos 10^\circ - 20 \sin 35^\circ \\ R &= 49 \cos 10^\circ - 20 \sin 35^\circ \\ R &\approx 48.2556 - 11.4715 \\ R &\approx 36.7841 \text{ N} \end{aligned}

Finding the frictional force:

Since the block is moving, we use the limiting friction model $F = \mu R$ :

F = 0.4 \times 36.7841 \approx 14.7136 \text{ N}

Resolving parallel to the plane (down the slope):

The forces acting down the slope are the component of weight $mg \sin 10^\circ$ and the parallel component of the pulling force $20 \cos 35^\circ$ . The frictional force $F$ acts up the slope.

\begin{aligned} 20 \cos 35^\circ + 5g \sin 10^\circ - F &= 5a \\ 20 \cos 35^\circ + 5(9.8) \sin 10^\circ - 14.7136 &= 5a \\ 16.7836 + 8.5882 - 14.7136 &= 5a \\ 25.0654 - 14.7136 &= 5a \\ 10.3518 &= 5a \\ a &\approx 1.9904 \text{ m s}^{-2} \end{aligned}

Finding the final speed:

Using the constant acceleration formula $v^2 = u^2 + 2as$ with $u = 2 \text{ m s}^{-1}$ , $a \approx 1.9904 \text{ m s}^{-2}$ , and $s = 3 \text{ m}$ :

\begin{aligned} v^2 &= 2^2 + 2(1.9904)(3) \\ v^2 &= 4 + 11.9424 \\ v^2 &= 15.9424 \\ v &= \sqrt{15.9424} \\ v &\approx 3.99 \text{ m s}^{-1} \end{aligned}

Answer

The speed of the block when it has moved 3 m down the plane is approximately $3.99 \text{ m s}^{-1}$ .

Final answer

3.99 m s^-1

Detailed explanation

Walkthrough

First, we identify all the forces acting on the 5 kg block. The weight $mg$ acts vertically downwards. The pulling force of 20 N acts at an angle of 35° above the line of greatest slope. The plane is inclined at 10° to the horizontal. Friction opposes the motion, so it acts up the slope.

We resolve the forces perpendicular to the inclined plane to find the normal reaction force $R$ . The component of weight perpendicular to the plane is $5g \cos 10^\circ$ acting into the plane. The pulling force has a component $20 \sin 35^\circ$ acting away from the plane. Setting the sum of forces perpendicular to the plane to zero gives $R = 5g \cos 10^\circ - 20 \sin 35^\circ$ . Substituting $g = 9.8$ gives $R \approx 36.7841$ N.

Next, we calculate the frictional force using $F = \mu R = 0.4 \times 36.7841 \approx 14.7136$ N.

Then, we resolve the forces parallel to the plane down the line of greatest slope. The component of weight down the slope is $5g \sin 10^\circ$ . The component of the pulling force down the slope is $20 \cos 35^\circ$ . The frictional force $F$ acts up the slope. Applying Newton's second law, $20 \cos 35^\circ + 5g \sin 10^\circ - F = 5a$ , we solve for the acceleration $a \approx 1.9904$ m s $^{-2}$ .

Finally, we use the kinematic equation $v^2 = u^2 + 2as$ with initial speed $u = 2$ m s $^{-1}$ , acceleration $a \approx 1.9904$ m s $^{-2}$ , and distance $s = 3$ m to find the final speed $v \approx 3.99$ m s $^{-1}$ .

Key Takeaways

Resolving forces perpendicular to an inclined plane is essential to find the normal reaction force when there are applied forces at an angle to the plane.
The limiting friction model $F = \mu R$ must be used when an object is moving on a rough surface.
Newton's second law is applied parallel to the direction of motion to find acceleration.
Constant acceleration kinematic equations can then be used to find final velocities over a given distance.

Common Mistakes

Forgetting to include the perpendicular component of the applied 20 N force when resolving perpendicular to the plane, leading to an incorrect normal reaction $R$ .
Using the wrong component of the 20 N force (using sine instead of cosine for the parallel component).
Forgetting that friction acts up the slope when the block moves down the slope.
Using $g = 9.81$ or $g = 10$ without checking if the mark scheme expects a specific value (here $g = 9.8$ is standard unless specified).

Things to Be Careful About

Ensure all force components are correctly identified as acting into or away from the plane.
The angle 35° is measured from the line of greatest slope, so the parallel component uses $\cos 35^\circ$ and the perpendicular component uses $\sin 35^\circ$ .
When using $v^2 = u^2 + 2as$ , ensure the signs of $a$ and $s$ are consistent with the direction of motion. Here, both are positive down the slope.
The mark scheme allows for a work-energy approach as an alternative, which involves calculating the work done by each force and equating it to the change in kinetic energy.

Techniques used

resolve forces perpendicular to the inclined planeresolve forces parallel to the inclined planeapply limiting friction formulause constant acceleration kinematic equation

Similar questions

Q68MMedium-HardKinematics of Motion in a Straight Line

A particle starts from rest at a point $O$ . The acceleration of the particle at time $t\text{ s}$ after leaving $O$ is $a\text{ m s}^{-2}$ , where

a = 2(t + 1)^{-\frac{1}{2}} - 1

for $t \ge 0$ .

Find the distance that the particle travels from $O$ until the time at which its acceleration is zero.

DifficultyMedium-Hard

Worked solution

Approach

The acceleration is variable, so use integration. Since $a = \frac{dv}{dt}$ , integrate $a$ with respect to $t$ to find $v$ , using the initial condition that the particle starts from rest ( $v=0$ when $t=0$ ). Then integrate $v$ with respect to $t$ to find displacement $s$ , using $s=0$ when $t=0$ (or a definite integral from $0$ to the required time). First find the time when $a=0$ .

Working

Set $a=0$ :

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

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

t+1 = 4

t = 3

Integrate to find velocity:

v = \int \left(2(t+1)^{-\frac{1}{2}} - 1\right) dt

v = 4(t+1)^{\frac{1}{2}} - t + c

Using $v=0$ at $t=0$ :

0 = 4(1)^{\frac{1}{2}} - 0 + c \Rightarrow c = -4

So

v = 4(t+1)^{\frac{1}{2}} - t - 4

Integrate to find displacement:

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

s = \frac{8}{3}(t+1)^{\frac{3}{2}} - \frac{1}{2}t^2 - 4t + k

Using $s=0$ at $t=0$ :

0 = \frac{8}{3}(1)^{\frac{3}{2}} - 0 - 0 + k \Rightarrow k = -\frac{8}{3}

So

s(t) = \frac{8}{3}(t+1)^{\frac{3}{2}} - \frac{1}{2}t^2 - 4t - \frac{8}{3}

Distance from $O$ until $t=3$ :

s(3) = \frac{8}{3}(4)^{\frac{3}{2}} - \frac{1}{2}(3)^2 - 4(3) - \frac{8}{3}

s(3) = \frac{8}{3}(8) - \frac{9}{2} - 12 - \frac{8}{3}

s(3) = \frac{64}{3} - \frac{9}{2} - 12 - \frac{8}{3}

s(3) = \frac{56}{3} - \frac{9}{2} - 12 = \frac{13}{6}

Since $v>0$ for $0 \le t \le 3$ , distance travelled equals displacement.

Answer

\frac{13}{6}\text{ m} \quad \text{or} \quad 2.17\text{ m}

Final answer

13/6 m or 2.17 m

Detailed explanation

Walkthrough

We are told the acceleration is $a = 2(t+1)^{-\frac{1}{2}} - 1$ and the particle starts from rest at $O$ . The question asks for the distance travelled up to the moment when acceleration is zero.

First find the time when acceleration is zero. Set $2(t+1)^{-\frac{1}{2}} - 1 = 0$ . Rearranging: $2(t+1)^{-\frac{1}{2}} = 1$ , so $(t+1)^{\frac{1}{2}} = 2$ . Squaring gives $t+1=4$ , so $t=3$ . This is the upper limit of the motion we need.

Because the acceleration is not constant, the constant-acceleration (suvat) equations cannot be used. Instead, use $a = \frac{dv}{dt}$ , so $v = \int a\,dt$ . Integrate: $v = \int \left(2(t+1)^{-\frac{1}{2}} - 1\right)dt = 4(t+1)^{\frac{1}{2}} - t + c$ . The particle starts from rest, so $v=0$ when $t=0$ . Substitute: $0 = 4(1)^{\frac{1}{2}} - 0 + c$ , so $c=-4$ . Hence $v = 4(t+1)^{\frac{1}{2}} - t - 4$ .

Next, displacement is $s = \int v\,dt$ . Integrating: $s = \int \left(4(t+1)^{\frac{1}{2}} - t - 4\right)dt = \frac{8}{3}(t+1)^{\frac{3}{2}} - \frac{1}{2}t^2 - 4t + k$ . Since the particle starts at $O$ , $s=0$ when $t=0$ . Substitute: $0 = \frac{8}{3}(1)^{\frac{3}{2}} - 0 - 0 + k$ , so $k = -\frac{8}{3}$ .

Now evaluate at $t=3$ : $s(3) = \frac{8}{3}(4)^{\frac{3}{2}} - \frac{1}{2}(3)^2 - 4(3) - \frac{8}{3} = \frac{64}{3} - \frac{9}{2} - 12 - \frac{8}{3} = \frac{13}{6}$ .

We should check whether the particle reverses direction before $t=3$ . Since $a>0$ for $0<t<3$ (because $2(t+1)^{-\frac{1}{2}}>1$ ), the velocity is increasing from $0$ , so $v>0$ throughout. Therefore the distance travelled equals the displacement, $\frac{13}{6}$ m.

Key Takeaways

When acceleration depends on time, integrate to get velocity and integrate again to get displacement.
Initial conditions (rest at O) are used to find constants of integration.
The time when acceleration is zero is found by solving $a=0$ , not by guessing.
Distance equals displacement only if the particle does not change direction; here velocity stays positive on the interval.

Common Mistakes

Mark scheme: treating $2(t+1)^{-\frac{1}{2}} - 1$ as $(2t+2)^{-\frac{1}{2}} - 1$ or dropping the $-1$ scores no marks for the first step.
Using $v = at$ or $s = vt$ : these are only valid for constant acceleration, so they score M0.
Forgetting the constant of integration, or setting $c=0$ without using $v=0$ at $t=0$ .
Forgetting that $(t+1)^{\frac{3}{2}} = 1$ when $t=0$ , not $0$ , when evaluating the lower limit.
Using the wrong time; the upper limit must be the value of $t$ found from $a=0$ .

Things to Be Careful About

The mark scheme says the acceleration-zero step is CWO (correct working only): you must show the rearrangement to $(t+1)^{\frac{1}{2}}=2$ and $t=3$ .
For the integration method marks, the scheme requires an attempt of the form $A(t+1)^{\frac{1}{2}} - t$ with $A \neq 2$ for velocity, and a similar form for displacement.
If no integration of $v$ is seen, only a special-case mark for 2.17 is available, so full working must be shown.
Check units: the final distance is in metres.

Techniques used

set acceleration to zero and solve for tintegrate acceleration with respect to timeapply initial condition v=0 at t=0integrate velocity with respect to timeevaluate displacement at the required time

Similar questions

Q7MediumForces and EquilibriumKinematics of Motion in a Straight LineNewton's Laws of Motion

Fig. 7.1

Fig. 7.2

Four coplanar forces of magnitudes $20\text{ N}$ , $30\text{ N}$ , $15\text{ N}$ and $25\text{ N}$ act at a point $P$ in the directions shown in Fig. 7.1. The forces act in a vertical plane. The resultant of these forces has magnitude $S\text{ N}$ and acts at an angle $\alpha^\circ$ below the horizontal as shown in Fig. 7.2.

(a)

Find the value of $S$ and the value of $\alpha$ .

6M Similar

DifficultyMedium-Easy

Worked solution

Approach

Resolve each of the four forces into horizontal and vertical components, sum them separately, then use Pythagoras' theorem to find $S$ and the inverse tangent to find $\alpha$ .

Working

Take the positive $x$ -direction to the right and the positive $y$ -direction upward.

Horizontal component $X$

X = 30\cos 30° + 15\cos 45° - 20\cos 20°

X = 25.9808 + 10.6066 - 18.7939 = 17.7935\text{ N}

Vertical component $Y$

Y = 30\sin 30° - 15\sin 45° + 20\sin 20° - 25

Y = 15 - 10.6066 + 6.8404 - 25 = -13.7662\text{ N}

The negative sign means the vertical component is $13.7662\text{ N}$ downward.

Magnitude of the resultant

S = \sqrt{X^2 + Y^2} = \sqrt{17.7935^2 + 13.7662^2}

S = \sqrt{316.61 + 189.51} = \sqrt{506.12}

S = 22.5\text{ N}

Angle below the horizontal

\alpha = \tan^{-1}\left(\frac{|Y|}{X}\right) = \tan^{-1}\left(\frac{13.7662}{17.7935}\right) = \tan^{-1}(0.7737)

\alpha = 37.7°

Answer

S = 22.5\text{ N}, \quad \alpha = 37.7°

Final answer

S = 22.5 N, α = 37.7°

Detailed explanation

Walkthrough

The four coplanar forces act at a single point $P$ , so the resultant is found by vector addition. The standard approach is to resolve each force into its horizontal and vertical components, add all horizontal components together, add all vertical components together, and then combine these two perpendicular components to get the resultant.

When resolving a force of magnitude $F$ at an angle $\theta$ above the horizontal, the horizontal component is $F\cos\theta$ and the vertical component is $F\sin\theta$ . If the force is below the horizontal, the vertical component becomes $-F\sin\theta$ . If the force points to the left, the horizontal component becomes $-F\cos\theta$ .

For the $30\text{ N}$ force at $30°$ above horizontal pointing to the right: horizontal = $+30\cos 30°$ , vertical = $+30\sin 30°$ .
For the $15\text{ N}$ force at $45°$ below horizontal pointing to the right: horizontal = $+15\cos 45°$ , vertical = $-15\sin 45°$ .
For the $20\text{ N}$ force at $20°$ above horizontal pointing to the LEFT: horizontal = $-20\cos 20°$ , vertical = $+20\sin 20°$ .
For the $25\text{ N}$ force pointing straight down: horizontal = $0$ , vertical = $-25$ .

Summing horizontally: $X = 17.7935\text{ N}$ to the right.
Summing vertically: $Y = -13.7662\text{ N}$ (i.e., $13.7662\text{ N}$ downward).

The magnitude of the resultant is found using Pythagoras' theorem:

S = \sqrt{X^2 + Y^2} = \sqrt{17.7935^2 + 13.7662^2} = 22.5\text{ N}

The angle $\alpha$ below the horizontal is found using the inverse tangent of the ratio of the (absolute) vertical to horizontal components:

\alpha = \tan^{-1}\left(\frac{13.7662}{17.7935}\right) = 37.7°

Key Takeaways

To find the resultant of multiple coplanar forces, resolve each into horizontal ( $F\cos\theta$ ) and vertical ( $F\sin\theta$ ) components.
The sign of each component depends on the direction of the force (right/left for horizontal, up/down for vertical).
The magnitude of the resultant is found using Pythagoras' theorem, and the angle is found using inverse trigonometric ratios.
The angle is found by $\tan^{-1}(|Y|/X)$ because the resultant is below the horizontal (so the vertical component is negative).

Common Mistakes

Mixing up sine and cosine when resolving forces at an angle to the horizontal.
Forgetting to consider the direction of each force (e.g., treating the $20\text{ N}$ force as if it points to the right).
Using the wrong sign convention, leading to an incorrect magnitude or angle.
Computing the angle as $\alpha = \tan^{-1}(X/|Y|)$ instead of $\alpha = \tan^{-1}(|Y|/X)$ , which would give a different (complementary) angle.

Things to Be Careful About

Be careful with the direction of each force when setting up the sign convention.
The $20\text{ N}$ force is to the LEFT, so its horizontal component is negative.
The $15\text{ N}$ force is BELOW the horizontal, so its vertical component is negative.
The $25\text{ N}$ force is DOWNWARD, so its vertical component is negative.
The angle $\alpha$ is measured from the positive $x$ -axis (horizontal) downward, so it is in the fourth quadrant.
The horizontal and vertical components are perpendicular, so Pythagoras' theorem applies.

Techniques used

resolve forces into horizontal and vertical componentscompute resultant magnitude using Pythagorascompute angle using inverse tangent

(b)

A small ring of mass $0.6\text{ kg}$ is threaded on a rough straight horizontal wire. The four forces shown in Fig. 7.1 act on the ring and are in the same vertical plane as the wire. The ring starts from rest and takes $3\text{ s}$ to travel a distance of $2\text{ m}$ along the wire.

Find the coefficient of friction between the ring and the wire.

5M Similar

DifficultyMedium

Worked solution

Approach

Use the constant acceleration (suvat) formula with the given $s$ , $u$ , and $t$ to find the acceleration. Then apply Newton's second law along the wire to find the friction force, use vertical equilibrium to find the normal reaction, and finally use $F = \mu R$ to find $\mu$ .

Working

Step 1: Find the acceleration

Using $s = ut + \frac{1}{2}at^2$ with $u = 0$ , $s = 2$ , and $t = 3$ :

2 = 0 + \frac{1}{2}a(3)^2 = 4.5a

a = \frac{4}{9}\text{ m/s}^2

Step 2: Apply Newton's second law along the wire

The horizontal component of the applied forces (from part (a)) is $X = 17.7935\text{ N}$ in the direction of motion. Friction $F$ acts opposite to the motion (backwards).

X - F = ma

17.7935 - F = 0.6 \times \frac{4}{9}

F = 17.7935 - 0.2667 = 17.5268\text{ N}

Step 3: Find the normal reaction using vertical equilibrium

The ring is on a horizontal wire, so the normal reaction $R$ is vertical. The vertical forces on the ring are:

The vertical component of the applied forces: $13.7662\text{ N}$ downward
The weight of the ring: $0.6g$ downward
The normal reaction: $R$ upward

For vertical equilibrium:

R = 13.7662 + 0.6g

Using $g = 9.8\text{ m/s}^2$ :

R = 13.7662 + 0.6 \times 9.8 = 13.7662 + 5.88 = 19.6462\text{ N}

Step 4: Apply the friction model

Using $F = \mu R$ :

17.5268 = \mu \times 19.6462

\mu = \frac{17.5268}{19.6462} = 0.892

Answer

\mu = 0.892

Final answer

μ = 0.892 (using g = 9.8 m/s²)

Detailed explanation

Walkthrough

The ring is on a horizontal wire, so we treat the motion as one-dimensional along the wire. The first step is to find the acceleration using the given information that the ring starts from rest ( $u = 0$ ) and travels $2\text{ m}$ in $3\text{ s}$ . Using $s = ut + \frac{1}{2}at^2$ , we get $a = 4/9\text{ m/s}^2$ .

Next, we apply Newton's second law along the wire. The horizontal component of the four applied forces (from part (a)) is $X = 17.7935\text{ N}$ in the direction of motion. Friction $F$ acts opposite to the motion. So $X - F = ma$ , giving $F = 17.5268\text{ N}$ .

Then, we apply vertical equilibrium. The wire is horizontal, so the normal reaction $R$ is vertical. The vertical forces on the ring are: the vertical component of the applied forces ( $13.7662\text{ N}$ downward), the weight of the ring ( $0.6g$ downward), and the normal reaction $R$ upward. For vertical equilibrium:

R = 13.7662 + 0.6g

With $g = 9.8$ , $R = 19.6462\text{ N}$ .

Finally, we use the friction model $F = \mu R$ to find:

\mu = \frac{F}{R} = \frac{17.5268}{19.6462} = 0.892

(Note: If $g = 10$ is used instead, $R = 19.7662\text{ N}$ and $\mu = 0.887$ .)

Key Takeaways

This problem combines kinematics (suvat), Newton's second law, vertical equilibrium, and the friction model $F = \mu R$ .
The horizontal component of the applied forces is the same as the $X$ value from part (a).
The normal reaction is found by balancing the vertical forces (including the weight of the ring).
The friction force opposes the motion and is the only unknown horizontal force besides $X$ .
The acceleration comes from the kinematic data, not from the force analysis.

Common Mistakes

Forgetting to include the weight of the ring in the vertical equilibrium.
Treating the vertical component of the applied forces as upward (it is downward).
Using the wrong sign for friction in Newton's second law (it opposes motion, so it should be subtracted).
Forgetting to find the acceleration first using the suvat equation.
Using the wrong value of $g$ (the Cambridge standard is $9.8$ or $9.81$ ).

Things to Be Careful About

The vertical component of the applied forces is $13.7662\text{ N}$ downward, not upward.
The weight of the ring ( $0.6g$ ) adds to the normal reaction, not subtracts from it.
The acceleration is positive (in the direction of motion), so $X > F$ (the applied force is greater than friction).
The value of $g$ used (typically $9.8$ or $9.81$ ) affects the final answer; using $g = 10$ gives $\mu \approx 0.887$ .
The mark scheme uses the student's value of $X$ and $|Y|$ from part (a) with follow-through, so a sign error in part (a) can still earn marks in part (b) if the method is correct.

Techniques used

apply constant acceleration formula to find accelerationapply Newton's second law horizontally to find frictionapply vertical equilibrium to find normal reactionuse friction model F = μR to find coefficient

Mathematics 9709/42 — October/November 2025

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