A particle $X$ moves in a straight line. The displacement of $X$ from $O$ at time $t\text{ s}$ after leaving $O$ is $s\text{ m}$, where $s = 0.3t^2 + 0.6t$ for $0 \le t \le 4$. (a) Find the velocity of $X$ at $t = 4$. (b) For $t > 4$, the acceleration of $X$ at time $t\text{ s}$ after leaving $O$ is $a\text{ ms}^{-2}$, where $a = 0.3t^{\frac{1}{2}}$. There is no change in the velocity of $X$ at $t = 4$. The velocity of $X$ at $t = T$ is $14.2\text{ ms}^{-1}$. (i) Find the…

(a) 3 m s^-1 · (b)(i) T = 16 · (b)(ii) 103.36 m

9709/43 May/June 2025 — Mathematics Solved Past Paper

Q1MediumMomentumEnergy, Work and Power

Two particles $P$ and $Q$ , of masses $0.1\text{ kg}$ and $0.3\text{ kg}$ respectively, are at rest on a smooth horizontal plane. $P$ is projected directly towards $Q$ with speed $4u\text{ ms}^{-1}$ . At the same instant, $Q$ is projected directly towards $P$ with speed $u\text{ ms}^{-1}$ . After $P$ and $Q$ collide, $P$ moves with speed $2\text{ ms}^{-1}$ and $Q$ moves with speed $4\text{ ms}^{-1}$ .

(a)

Find the two possible values of $u$ .

3M Similar

DifficultyMedium

Worked solution

Approach

Take the direction in which $P$ is initially projected as positive. Since both particles move on a smooth horizontal plane, momentum is conserved in the collision. The final speeds are known, but the directions after the collision are not, so test the possible sign choices for the final velocities and keep those giving a positive value of $u$ .

Working

Let positive be the direction of $P$ 's initial motion. Then initial velocities are:

v_P = 4u, \qquad v_Q = -u.

Conservation of linear momentum gives:

0.1(4u) + 0.3(-u) = 0.1 v_P' + 0.3 v_Q'.

After the collision, $|v_P'| = 2$ and $|v_Q'| = 4$ . Test the sign choices that give positive $u$ .

Case 1: $v_P' = 2$ , $v_Q' = 4$ .

0.1(4u) + 0.3(-u) = 0.1(2) + 0.3(4)

0.4u - 0.3u = 0.2 + 1.2

0.1u = 1.4

u = 14

Case 2: $v_P' = -2$ , $v_Q' = 4$ .

0.1(4u) + 0.3(-u) = 0.1(-2) + 0.3(4)

0.4u - 0.3u = -0.2 + 1.2

0.1u = 1.0

u = 10

The other two sign choices give $u = -14$ and $u = -10$ , which are not possible because $u$ is a speed.

Answer

u = 14 \quad \text{or} \quad u = 10

Final answer

u = 14 or u = 10

Detailed explanation

Walkthrough

This part uses conservation of momentum in a direct collision. The particles are on a smooth horizontal plane, so there is no external horizontal force and total momentum is conserved.

First choose a positive direction. It is convenient to take the direction in which $P$ is initially projected as positive. Since $Q$ is projected directly towards $P$ , $Q$ 's initial velocity is in the opposite direction, so it is $-u$ .

Write the conservation equation:

0.1(4u) + 0.3(-u) = 0.1 v_P' + 0.3 v_Q'.

The left-hand side simplifies to $0.1u$ . On the right-hand side, the final velocities are not known exactly because we are told speeds, not directions. Try the sign choices that can give positive $u$ .

If both final velocities are in the positive direction, $v_P' = 2$ and $v_Q' = 4$ , giving $u = 14$ . If $P$ rebounds but $Q$ continues in the positive direction, $v_P' = -2$ and $v_Q' = 4$ , giving $u = 10$ . The remaining combinations give negative $u$ , which cannot be a speed.

Thus the two possible values are $u = 14$ and $u = 10$ .

Key Takeaways

Momentum is a vector, so directions must be included in the conservation equation.
Speeds are magnitudes; the actual velocities after collision may be positive or negative.
In a direct collision on a smooth horizontal plane, total momentum is conserved.
Reject solutions that give a negative speed.

Common Mistakes

Forgetting that $Q$ 's initial velocity is $-u$ , not $+u$ .
Assuming both final velocities have the same sign.
Accepting negative values of $u$ .
Using only one of the two valid sign combinations and missing one value of $u$ .
The mark scheme requires a correct momentum equation with the correct number of terms; sign errors in the final velocities may be condoned only if the method is otherwise clear.

Things to Be Careful About

Be consistent with the sign convention throughout.
The masses are $0.1\text{ kg}$ and $0.3\text{ kg}$ ; include both masses on both sides.
The final speeds are $2$ and $4$ , but their signs are unknown.
Only positive values of $u$ are physically possible because $u$ is a speed.

Techniques used

apply conservation of linear momentumchoose a sign convention for directiontest possible post-collision velocity directionssolve linear equations for speed

(b)

Find the largest possible loss of kinetic energy in the collision.

2M Similar

DifficultyMedium-Easy

Worked solution

Approach

The loss of kinetic energy is the initial total kinetic energy minus the final total kinetic energy. Since the final kinetic energy is the same for both possible values of $u$ , the largest loss occurs with the larger value of $u$ found in part (a).

Working

For a value of $u$ , the initial kinetic energy is

\frac12(0.1)(4u)^2 + \frac12(0.3)(u)^2 = 0.8u^2 + 0.15u^2 = 0.95u^2.

The final kinetic energy is independent of $u$ :

\frac12(0.1)(2)^2 + \frac12(0.3)(4)^2 = 0.2 + 2.4 = 2.6\text{ J}.

Using the larger value $u = 14$ :

\text{Loss} = 0.95(14)^2 - 2.6 = 0.95(196) - 2.6 = 186.2 - 2.6 = 183.6\text{ J}.

For comparison, $u = 10$ gives $0.95(100) - 2.6 = 92.4\text{ J}$ , so the largest loss is $183.6\text{ J}$ .

Answer

183.6\text{ J}

Final answer

183.6 J

Detailed explanation

Walkthrough

Kinetic energy is a scalar, so direction does not matter; only the speeds matter. The total kinetic energy before the collision is

\frac12(0.1)(4u)^2 + \frac12(0.3)(u)^2.

This simplifies to $0.95u^2$ . The final kinetic energy is

\frac12(0.1)(2)^2 + \frac12(0.3)(4)^2 = 2.6\text{ J}.

Because the final kinetic energy is fixed, the loss is larger when the initial kinetic energy is larger. Since initial kinetic energy is proportional to $u^2$ , use the larger value $u = 14$ from part (a). Then

0.95(14)^2 - 2.6 = 183.6\text{ J}.

Using $u = 10$ would give only $92.4\text{ J}$ , so the largest possible loss is $183.6\text{ J}$ .

Key Takeaways

Kinetic energy is $\frac12 mv^2$ and depends on speed squared.
Loss of kinetic energy is initial kinetic energy minus final kinetic energy.
When the final state is fixed, larger initial speed gives larger initial kinetic energy and therefore larger loss.

Common Mistakes

Using the smaller value of $u$ from part (a).
Forgetting to square the speeds when calculating kinetic energy.
Omitting the final kinetic energy and giving only the initial kinetic energy as the loss.
Using the mass of one particle only.
The mark scheme awards M1 for the correct difference expression and A1 for the final value; if only one value of $u$ was found in part (a), the method mark is not awarded.

Things to Be Careful About

The final answer should be in joules.
The loss is positive; if your expression gives a negative value, take the absolute value or reorder the subtraction.
The largest loss corresponds to the larger value of $u$ , not the smaller one.
Units: masses in kg and speeds in $\text{ ms}^{-1}$ give energy in joules.

Techniques used

calculate initial kinetic energycalculate final kinetic energysubtract to find energy lossselect the larger value of u

Q24MMedium-EasyNewton's Laws of Motion

A van of mass $4500\text{ kg}$ is towing a trailer of mass $350\text{ kg}$ along a straight horizontal road. The van and trailer are connected by a light rigid tow-bar which is parallel to the road. There are resistance forces of $X\text{ N}$ on the van and $120\text{ N}$ on the trailer. The driving force produced by the van's engine is $2500\text{ N}$ . The tension in the tow-bar is $T\text{ N}$ , and the acceleration of the van is $0.4\text{ ms}^{-2}$ .

Find the value of $X$ and the value of $T$ .

DifficultyMedium-Easy

Worked solution

Approach

Treat the van and trailer as separate objects. Choose the direction of motion as positive and apply Newton's second law, $F = ma$ , to each body.

Working

For the trailer, the tension $T$ pulls it forward and the resistance $120\text{ N}$ acts backward:

T - 120 = 350 \times 0.4

T - 120 = 140

T = 260\text{ N}

For the van, the driving force $2500\text{ N}$ acts forward, while the resistance $X$ and the tension $T$ both act backward:

2500 - X - T = 4500 \times 0.4

Substitute $T = 260$ :

2500 - X - 260 = 1800

2240 - X = 1800

X = 440\text{ N}

Answer

T = 260\text{ N}, \quad X = 440\text{ N}

Final answer

T = 260 N, X = 440 N

Detailed explanation

Walkthrough

The van and trailer move together, so both have the same acceleration $0.4\text{ ms}^{-2}$ . Choose the direction of motion as positive.

Start with the trailer. The only horizontal forces on the trailer are the tension $T$ in the tow-bar, pulling it forward, and the resistance $120\text{ N}$ , opposing motion. Newton's second law gives:

T - 120 = 350 \times 0.4

The right-hand side is the trailer's mass times its acceleration. This gives:

T - 120 = 140

so $T = 260\text{ N}$ .

Now consider the van. The driving force $2500\text{ N}$ acts forward. The resistance $X$ acts backward. The tension in the tow-bar also acts backward on the van, by Newton's third law, because the tow-bar pulls the trailer forward and therefore pulls the van backward. Thus:

2500 - X - T = 4500 \times 0.4

Substitute $T = 260$ :

2500 - X - 260 = 1800

2240 - X = 1800

X = 440\text{ N}

As a check, applying Newton's second law to the whole system gives:

2500 - X - 120 = (4500 + 350) \times 0.4 = 1940

so $2380 - X = 1940$ , hence $X = 440\text{ N}$ , which agrees.

Key Takeaways

This question tests setting up Newton's second law for each body in a connected system. The key steps are: identify all forces on each body, choose a positive direction, write $F_{\text{net}} = ma$ for each body, and solve the equations. The tension in a light tow-bar has the same magnitude at both ends but acts in opposite directions on the two bodies. Resistances always oppose motion.

Common Mistakes

Forgetting that the tension pulls backward on the van as well as forward on the trailer.
Writing $2500 - X + T = 4500 \times 0.4$ with the wrong sign for $T$ .
Using the trailer's mass in the van equation, or the van's mass in the trailer equation.
Adding the masses incorrectly when checking with the system equation.
Omitting the resistance on the trailer in the system equation.

Things to Be Careful About

The acceleration is the same for both bodies because the tow-bar is rigid and light.
All forces are horizontal, so only horizontal components are needed.
The mark scheme requires dimensionally correct equations with the correct number of terms; an unsupported final answer may not receive full marks.
Keep units consistent: masses in kg, forces in N, acceleration in $\text{ms}^{-2}$ .

Techniques used

apply Newton's second law to the trailerapply Newton's second law to the vanuse Newton's third law to determine the direction of the tension on the vansolve the resulting equations

Approach

Find the velocity at key times ( $t=20$ and $t=50$ ) using the given acceleration and graph features. Calculate the displacement up to $t=40$ by finding the area under the graph. Use the condition that the particle returns to $O$ (total displacement = 0) to set up an equation for $T$ using the area of the negative velocity region.

Working

Step 1: Find velocity at $t = 20$
The particle starts at $v = 5 ext{ ms}^{-1}$ and accelerates at $0.75 ext{ ms}^{-2}$ for $20 ext{ s}$ .

v = u + at = 5 + (0.75)(20) = 20 ext{ ms}^{-1}

Step 2: Find velocity at $t = 50$
From $t = 20$ to $t = 40$ , the velocity decreases from $20 ext{ ms}^{-1}$ to $0 ext{ ms}^{-1}$ over $20 ext{ s}$ . The acceleration (gradient) during this phase is:

a = \frac{0 - 20}{40 - 20} = -1 ext{ ms}^{-2}

The graph continues with this gradient to $t = 50$ :

v_{50} = 0 + (-1)(50 - 40) = -10 ext{ ms}^{-1}

Step 3: Calculate displacement at $t = 40$
The displacement is the area under the velocity-time graph from $t = 0$ to $t = 40$ .

\text{Area}_1 = \frac{1}{2}(5 + 20)(20) + \frac{1}{2}(20)(20) = 250 + 200 = 450 ext{ m}

Step 4: Use the return condition to find $T$
The particle returns to $O$ at time $T$ , so the total displacement is $0$ . The negative displacement from $t = 40$ to $t = T$ must cancel the positive displacement of $450 ext{ m}$ .

The region from $t = 40$ to $t = T$ is a triangle with base $(T - 40)$ and height $-10$ (at $t = 50$ ):

\text{Area}_2 = \frac{1}{2}(T - 40)(-10) = -5(T - 40)

Set total displacement to zero:

450 - 5(T - 40) = 0

450 = 5(T - 40)

90 = T - 40

T = 130

Answer

T = 130

Final answer

T = 130

Detailed explanation

Walkthrough

First, determine the velocity at the peak of the graph ( $t = 20$ ) using the initial velocity and acceleration. The particle starts at $5 ext{ ms}^{-1}$ and accelerates at $0.75 ext{ ms}^{-2}$ , reaching $20 ext{ ms}^{-1}$ at $t = 20$ .

Next, analyze the deceleration phase. From $t = 20$ to $t = 40$ , the velocity drops from $20 ext{ ms}^{-1}$ to $0 ext{ ms}^{-1}$ , giving a gradient (acceleration) of $-1 ext{ ms}^{-2}$ . This gradient continues until $t = 50$ , where the velocity becomes $-10 ext{ ms}^{-1}$ .

Calculate the total positive displacement up to $t = 40$ by finding the area under the graph. This is a trapezium from $t = 0$ to $20$ and a triangle from $t = 20$ to $40$ , totaling $450 ext{ m}$ .

Since the particle returns to the starting point $O$ at time $T$ , the total displacement must be zero. The area below the time axis from $t = 40$ to $t = T$ represents negative displacement. This forms a triangle with base $(T - 40)$ and height $-10$ . Setting the sum of areas to zero allows solving for $T$ .

Key Takeaways

The area under a velocity-time graph represents displacement.
Areas below the time axis represent negative displacement (motion in the opposite direction).
Returning to the starting point means the total signed area under the graph is zero.

Common Mistakes

Forgetting that displacement is the signed area; using only positive areas.
Calculating the area of the negative region incorrectly (e.g., using the wrong base or height).
Not recognizing that the gradient from $t = 20$ to $t = 40$ continues to $t = 50$ .

Things to Be Careful About

Ensure the velocity at $t = 50$ is correctly calculated as $-10 ext{ ms}^{-1}$ (speed is $10 ext{ ms}^{-1}$ , but velocity is negative for the area calculation).
The base of the negative triangle is $(T - 40)$ , not $(T - 50)$ , if using the full triangle from $t = 40$ to $t = T$ .
The final answer must be a positive time value greater than $50$ .

Techniques used

read velocity from graph at specific timescalculate area under velocity-time graph for displacementapply displacement condition to find unknown time

(b)

Find the acceleration of the particle from $t = 50$ to $t = T$ .

2M Similar

DifficultyMedium-Easy

Worked solution

Approach

Acceleration is the gradient of the velocity-time graph. Identify the velocity at the start ( $t = 50$ ) and end ( $t = T = 130$ ) of the interval and calculate the gradient.

Working

From part (a), we know:

At $t = 50$ , $v = -10 ext{ ms}^{-1}$
At $t = T = 130$ , $v = 0 ext{ ms}^{-1}$

The acceleration is the change in velocity divided by the time taken:

a = \frac{v_f - v_i}{t_f - t_i} = \frac{0 - (-10)}{130 - 50}

a = \frac{10}{80} = \frac{1}{8} = 0.125 ext{ ms}^{-2}

Answer

a = 0.125 ext{ ms}^{-2}

Final answer

0.125 ms^-2

Detailed explanation

Walkthrough

Acceleration is defined as the rate of change of velocity, which corresponds to the gradient of the velocity-time graph. For the interval from $t = 50$ to $t = T$ , the graph is a straight line, so the acceleration is constant.

Using the values from part (a):

Initial velocity at $t = 50$ is $-10 ext{ ms}^{-1}$ .
Final velocity at $t = 130$ is $0 ext{ ms}^{-1}$ .

Calculate the gradient: $\frac{0 - (-10)}{130 - 50} = \frac{10}{80} = 0.125 ext{ ms}^{-2}$ . The positive value indicates the particle is accelerating in the positive direction (slowing down its negative velocity).

Key Takeaways

Acceleration is the gradient of a velocity-time graph.
A positive gradient on a graph where velocity is negative indicates the object is decelerating (slowing down) in the negative direction.

Common Mistakes

Using speed instead of velocity (getting $-0.125 ext{ ms}^{-2}$ instead of positive).
Calculating the time interval incorrectly (e.g., using $130 - 50 = 80$ is correct, but sometimes students use $130 - 40$ ).

Things to Be Careful About

Pay attention to signs: velocity is $-10$ , not $10$ . The change in velocity is $0 - (-10) = +10$ .
The question asks for acceleration, which is a vector; the sign matters.

Techniques used

calculate acceleration from gradientuse velocity-time graph endpoints

Q46MMediumForces and Equilibrium

Fig. 4.1

Three blocks $P$ , $Q$ and $R$ , of masses $25\text{ kg}$ , $20\text{ kg}$ and $m\text{ kg}$ respectively, are held in equilibrium by three light inextensible strings $OP$ , $OQ$ and $OR$ . The strings $OP$ and $OR$ both pass over small fixed smooth pulleys $A$ and $B$ respectively, with $P$ and $R$ hanging vertically below the pulleys. The block $Q$ hangs vertically below the point $O$ . The angle between $OA$ and the vertical is $30^\circ$ and the angle $BOQ = \alpha^\circ$ (see diagram).

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

DifficultyMedium

Worked solution

Approach

Identify the tensions in the three strings from the weights of the hanging masses P, Q, and R. Apply the equilibrium condition at the junction point O by resolving the forces horizontally and vertically. This produces two equations in the two unknowns $m$ and $\alpha$ , which can be solved simultaneously.

Working

The tensions in the strings are equal to the weights of the hanging masses (since the pulleys are smooth):

T_P = 25g, \quad T_Q = 20g, \quad T_R = mg

At point O, the three forces $T_P$ , $T_Q$ , and $T_R$ are in equilibrium. Resolve forces horizontally (taking left as positive):

T_P \sin 30^\circ = T_R \sin \alpha

Substitute the tensions:

25g \sin 30^\circ = mg \sin \alpha \quad \text{--- (1)}

Resolve forces vertically (taking upwards as positive):

T_P \cos 30^\circ = T_Q + T_R \cos \alpha

Substitute the tensions and rearrange:

25g \cos 30^\circ = 20g + mg \cos \alpha

mg \cos \alpha = 25g \cos 30^\circ - 20g \quad \text{--- (2)}

Divide equation (1) by equation (2) to eliminate $m$ and $g$ :

\frac{mg \sin \alpha}{mg \cos \alpha} = \frac{25g \sin 30^\circ}{25g \cos 30^\circ - 20g}

\tan \alpha = \frac{25 \sin 30^\circ}{25 \cos 30^\circ - 20}

Calculate the value:

\tan \alpha = \frac{25 \times 0.5}{25 \times \frac{\sqrt{3}}{2} - 20} = \frac{12.5}{21.6506\dots - 20} = \frac{12.5}{1.6506\dots} \approx 7.5739

\alpha = \tan^{-1}(7.5739\dots) \approx 82.4775^\circ

Rounding to 3 significant figures:

\alpha = 82.5

To find $m$ , square and add equations (1) and (2) after removing $g$ :

(mg \sin \alpha)^2 + (mg \cos \alpha)^2 = (25g \sin 30^\circ)^2 + (25g \cos 30^\circ - 20g)^2

(mg)^2 = g^2 \left[ (25 \sin 30^\circ)^2 + (25 \cos 30^\circ - 20)^2 \right]

m^2 = (12.5)^2 + (21.6506\dots - 20)^2

m^2 = 156.25 + (1.6506\dots)^2 = 156.25 + 2.7245\dots = 158.9745\dots

m = \sqrt{158.9745\dots} \approx 12.6085

Rounding to 3 significant figures:

m = 12.6

Answer

\alpha = 82.5, \quad m = 12.6

Final answer

α = 82.5, m = 12.6

Detailed explanation

Walkthrough

First, we recognize that the tension in each string is simply the weight of the mass it supports, because the pulleys are smooth and the system is in equilibrium. Thus, $T_P = 25g$ , $T_Q = 20g$ , and $T_R = mg$ . At the junction point O, three forces meet: $T_P$ pulling up and to the left at 30° to the vertical, $T_Q$ pulling straight down, and $T_R$ pulling down and to the right at an angle $\alpha$ to the vertical.

To apply the equilibrium condition, we resolve these forces into horizontal and vertical components. Horizontally, the leftward pull of $T_P$ must balance the rightward pull of $T_R$ , giving $T_P \sin 30^\circ = T_R \sin \alpha$ . Vertically, the upward pull of $T_P$ must balance the downward pulls of both $T_Q$ and the vertical component of $T_R$ , giving $T_P \cos 30^\circ = T_Q + T_R \cos \alpha$ .

Substituting the known tensions into these equations yields two equations with two unknowns ( $m$ and $\alpha$ ). By isolating $mg \cos \alpha$ in the vertical equation and dividing the horizontal equation by it, we eliminate $m$ and $g$ , leaving a single equation for $\alpha$ . Solving this gives $\alpha \approx 82.5^\circ$ .

Finally, to find $m$ , we use the Pythagorean identity $\sin^2 \alpha + \cos^2 \alpha = 1$ by squaring and adding the two original equations (with $g$ factored out), which directly gives $m^2$ . Taking the square root yields $m \approx 12.6$ .

Key Takeaways

In equilibrium problems with smooth pulleys, the tension in a string supporting a hanging mass is equal to the weight of that mass ( $T = mg$ ).
When multiple forces act at a point in equilibrium, resolving them in two perpendicular directions (horizontal and vertical) produces a system of independent equations.
Simultaneous equations involving $m \sin \alpha$ and $m \cos \alpha$ can be solved by dividing to find $\tan \alpha$ , and by squaring and adding to find $m$ using the identity $\sin^2 \theta + \cos^2 \theta = 1$ .

Common Mistakes

Forgetting to include $g$ in the tension expressions or dropping it during calculation, which leads to incorrect final values.
Resolving incorrectly: mixing up sine and cosine components, or getting the directions of the vertical components wrong (e.g., thinking $T_R$ has an upward vertical component when it actually pulls downwards).
Attempting to find $m$ and $\alpha$ separately without eliminating one first, leading to complex algebraic manipulations.

Things to Be Careful About

Always check the signs of the components based on the direction of the forces. Here, $T_P$ pulls upwards and leftwards, while $T_Q$ and $T_R$ pull downwards and rightwards (horizontally).
Ensure the angle $\alpha$ is measured correctly from the vertical, as shown in the diagram, so its horizontal component uses $\sin \alpha$ and vertical uses $\cos \alpha$ .
Round final answers to 3 significant figures as is standard in Cambridge A-Level Mechanics, unless otherwise specified.

Techniques used

determine string tensions from hanging massesresolve forces horizontally and vertically at a pointapply equilibrium condition for concurrent forcessolve simultaneous trigonometric equations

Approach

At constant power, the driving force at speed $v$ is $F_D = P/v$ . On level ground at steady speed the acceleration is zero, so the driving force equals the resistance. This determines $k$ . Then use Newton's second law at the other speed.

Working

At steady speed $50\text{ ms}^{-1}$ , the driving force is

F_D = \frac{62500}{50} = 1250\text{ N}

The resistance is $k(50)^2 = 2500k$ . Since the van is in equilibrium along the road,

1250 = 2500k

k = 0.5

At speed $25\text{ ms}^{-1}$ ,

F_D = \frac{62500}{25} = 2500\text{ N}

and the resistance is

0.5(25)^2 = 312.5\text{ N}

Newton's second law gives

2500 - 312.5 = 2500a

2187.5 = 2500a

a = 0.875\text{ m s}^{-2}

Answer

k = 0.5, \qquad a = 0.875\text{ m s}^{-2}

Final answer

k = 0.5; acceleration = 0.875 m s^-2

Detailed explanation

Walkthrough

At steady speed the acceleration is zero, so the resultant force along the horizontal road is zero. The engine provides a driving force $F_D$ . The power equation $P = F_D v$ lets you find $F_D$ at the steady speed, and this must equal the resistance $kv^2$ ; substituting $v = 50$ gives $k = 0.5$ . Then at $25\text{ ms}^{-1}$ , repeat $P = F_D v$ to find the new driving force, calculate the resistance, and use $F_{\text{net}} = ma$ to get the acceleration.

Key Takeaways

This question tests the constant-power relation $P = F_D v$ , the dependence of resistance on speed squared, and the use of Newton's second law. To find acceleration you must use the instantaneous driving force at that speed, not a fixed value.

Common Mistakes

Using the driving force found at $50\text{ ms}^{-1}$ again at $25\text{ ms}^{-1}$ ; this gives the wrong acceleration.
Forgetting to convert $62.5\text{ kW}$ into $62500\text{ W}$ .
Confusing the resistance with the driving force or omitting one of the three terms in Newton's second law.
Not showing an equation linking $k$ to the steady state; the question says "show that", so a clear equation is required.

Things to Be Careful About

Keep all quantities in SI units: power in watts, speed in $\text{ m s}^{-1}$ , force in newtons. At steady speed on a horizontal road, the acceleration is zero so the net force is zero. The resistance is $kv^2$ , so it increases with the square of the speed.

Techniques used

use P = Fv to find driving force at a given speedequate driving force and resistance at steady speedapply Newton's second law to find accelerationevaluate resistance using kv squared

(b)

The van begins to ascend a hill inclined at an angle $\theta^\circ$ to the horizontal. The van travels along a line of greatest slope of the hill. The speed of the van at the start of the hill is $20\text{ ms}^{-1}$ , and its acceleration is $5a\text{ ms}^{-2}$ . Later, on the same hill, the speed of the van is $30\text{ ms}^{-1}$ , and its acceleration is $a\text{ ms}^{-2}$ . The power of the van's engine remains at $62.5\text{ kW}$ , and the resistance force remains at $0.5v^2\text{ N}$ .

Find the value of $a$ and the value of $\theta$ .

5M Similar

DifficultyMedium-Hard

Worked solution

Approach

Resolve along the line of greatest slope. The driving force at speed $v$ is $F_D = P/v$ , the resistance is $0.5v^2$ , and the component of weight down the slope is $mg\sin\theta$ . Write Newton's second law for both speeds and solve the two equations simultaneously.

Working

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

2500(10)\sin\theta = 25000\sin\theta

At speed $30\text{ ms}^{-1}$ with acceleration $a$ :

\frac{62500}{30} - 0.5(30)^2 - 25000\sin\theta = 2500a

2083.33\ldots - 450 - 25000\sin\theta = 2500a

1633.33\ldots - 25000\sin\theta = 2500a \qquad (1)

At speed $20\text{ ms}^{-1}$ with acceleration $5a$ :

\frac{62500}{20} - 0.5(20)^2 - 25000\sin\theta = 2500(5a)

3125 - 200 - 25000\sin\theta = 12500a

2925 - 25000\sin\theta = 12500a \qquad (2)

Subtract equation (1) from equation (2):

(2925 - 1633.33\ldots) = (12500 - 2500)a

1291.66\ldots = 10000a

a = \frac{31}{240} = 0.12916\ldots\text{ m s}^{-2}

Substitute into equation (2):

2925 - 25000\sin\theta = 12500\times\frac{31}{240}

2925 - 25000\sin\theta = 1614.58\ldots

25000\sin\theta = 1310.41\ldots

\sin\theta = 0.05241\ldots

\theta = \arcsin(0.05241\ldots) = 3.00^\circ

Answer

a = 0.129\text{ m s}^{-2}, \qquad \theta = 3.0^\circ

Final answer

a = 0.129 m s^-2 (31/240); theta = 3.0 degrees

Detailed explanation

Walkthrough

At each instant on the hill the engine still supplies constant power, so the driving force is not fixed; it is $P/v$ . The van's weight $mg$ has a component $mg\sin\theta$ down the slope, opposing the motion. Therefore Newton's second law up the slope is

\frac{P}{v} - 0.5v^2 - mg\sin\theta = ma.

Write this equation with the two given pairs $(v,a) = (30,a)$ and $(v,a) = (20,5a)$ . Both equations contain the same unknown $\sin\theta$ , so subtracting them eliminates $\sin\theta$ and gives $a$ . Substitute $a$ back to find $\sin\theta$ , then take the inverse sine to get $\theta$ in degrees.

Key Takeaways

This is a constant-power motion problem with variable driving force. It combines $P = F_D v$ , the quadratic resistance law, the weight component on an incline, and simultaneous equations from Newton's second law. The key idea is that the same unknowns appear in both equations, so subtracting them eliminates the angle.

Common Mistakes

Using a fixed driving force value instead of $P/v$ at each speed.
Putting the weight component on the wrong side or using $\cos\theta$ instead of $\sin\theta$ .
Swapping the accelerations: $30\text{ m s}^{-1}$ corresponds to $a$ , and $20\text{ m s}^{-1}$ to $5a$ .
Forgetting that the van travels up the slope, so the weight component and resistance oppose the driving force.
Using $g = 9.8$ when the mark scheme uses $g = 10$ ; the line $25000\sin\theta$ shows the intended value.

Things to Be Careful About

Use SI units throughout: $62.5\text{ kW} = 62500\text{ W}$ . The acceleration is instantaneous, so every speed needs its own driving force $P/v$ . When subtracting the equations, check the sign of every term. Give $\theta$ in degrees and retain sufficient accuracy; $a = 0.129\text{ m s}^{-2}$ and $\theta = 3.0^\circ$ match the mark scheme.

Techniques used

use P = Fv to find driving force at each speedresolve weight component on an inclined planeform two Newton's second law equationssolve simultaneous equations by eliminationinterpret the acceleration and angle results

Q6MediumNewton's Laws of MotionKinematics of Motion in a Straight LineEnergy, Work and Power

Fig. 6.1

The diagram shows the vertical cross-section $ABC$ of a rough waterslide. The section $AB$ is a straight line of length $5\text{ m}$ inclined at an angle of $\theta$ to the horizontal, where $\sin\theta = 0.8$ . The point $B$ is $2.5\text{ m}$ above the level of $C$ . A man of mass $80\text{ kg}$ , modelled as a particle, slides down the waterslide, starting from rest at $A$ . The coefficient of friction between the man and the straight section of the waterslide is $0.1$ .

(a)

Find the speed of the man at $B$ .

5M Similar

DifficultyMedium

Worked solution

Approach

Resolve the forces acting on the man perpendicular and parallel to the slope $AB$ . Use Newton's second law to find the acceleration down the slope, then apply the suvat equation $v^2 = u^2 + 2as$ to find the speed at $B$ .

Working

Since $\sin\theta = 0.8$ , we have $\cos\theta = \sqrt{1 - 0.8^2} = 0.6$ .

Resolve perpendicular to the slope. The normal reaction force $R$ balances the component of weight perpendicular to the slope:

R = mg\cos\theta = 80g \times 0.6

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

R = 80 \times 10 \times 0.6 = 480\,\text{N}

The friction force opposes motion down the slope:

F = \mu R = 0.1 \times 480 = 48\,\text{N}

Resolve parallel to the slope (downwards positive). The component of weight down the slope is $mg\sin\theta$ :

mg\sin\theta - F = ma

80 \times 10 \times 0.8 - 48 = 80a

640 - 48 = 80a

592 = 80a

a = 7.4\,\text{m s}^{-2}

The man starts from rest at $A$ , so $u = 0$ . The distance $AB = 5\,\text{m}$ . Using $v^2 = u^2 + 2as$ :

v^2 = 0 + 2 \times 7.4 \times 5 = 74

v = \sqrt{74} \approx 8.60\,\text{m s}^{-1}

Answer

v = 8.60\,\text{m s}^{-1}

Final answer

8.60 m s⁻¹

Detailed explanation

Walkthrough

Step 1: Find $\cos\theta$
We are given $\sin\theta = 0.8$ . Using the Pythagorean identity $\sin^2\theta + \cos^2\theta = 1$ , we get $\cos\theta = \sqrt{1 - 0.64} = 0.6$ . This is needed to resolve the weight into components perpendicular and parallel to the slope.

Step 2: Resolve forces perpendicular to the slope
There is no acceleration perpendicular to the slope, so the normal reaction $R$ balances the perpendicular component of weight. The weight $mg$ acts vertically downwards, and its component perpendicular to the slope is $mg\cos\theta$ . Thus $R = mg\cos\theta = 80 \times 10 \times 0.6 = 480\,\text{N}$ .

Step 3: Calculate the friction force
Friction is $F = \mu R = 0.1 \times 480 = 48\,\text{N}$ . This acts up the slope, opposing the motion.

Step 4: Resolve forces parallel to the slope and apply Newton's second law
The component of weight down the slope is $mg\sin\theta = 80 \times 10 \times 0.8 = 640\,\text{N}$ . The net force down the slope is $640 - 48 = 592\,\text{N}$ . By Newton's second law, $F_{\text{net}} = ma$ , so $a = 592/80 = 7.4\,\text{m s}^{-2}$ .

Step 5: Use suvat to find velocity at B
The man starts from rest ( $u = 0$ ) and travels $s = 5\,\text{m}$ with constant acceleration $a = 7.4\,\text{m s}^{-2}$ . Using $v^2 = u^2 + 2as$ , we get $v^2 = 2 \times 7.4 \times 5 = 74$ , so $v = \sqrt{74} \approx 8.60\,\text{m s}^{-1}$ .

An alternative energy method: the loss in gravitational PE is $mg \times 5\sin\theta = 3200\,\text{J}$ . The work done against friction is $F \times 5 = 240\,\text{J}$ . By the work-energy principle, $3200 - 240 = \frac{1}{2}mv^2$ , giving $v^2 = 74$ and $v = 8.60\,\text{m s}^{-1}$ .

Key Takeaways

On an inclined plane, resolve weight into components: $mg\sin\theta$ parallel to the slope (downwards) and $mg\cos\theta$ perpendicular to the slope.
The normal reaction equals the perpendicular component of weight when there is no acceleration perpendicular to the surface.
Friction is $F = \mu R$ and acts opposite to the direction of motion (or intended motion).
Newton's second law $F_{\text{net}} = ma$ can be applied along any direction; for constant acceleration problems on a straight line, suvat equations are the natural follow-up.
The work-energy principle provides an alternative approach that can avoid finding acceleration explicitly.

Common Mistakes

Using $\sin\theta$ instead of $\cos\theta$ for the normal reaction, or vice versa. Remember: perpendicular to the slope uses $\cos\theta$ and parallel uses $\sin\theta$ .
Forgetting that friction is $\mu R$ and not $\mu mg$ . The normal reaction on an inclined plane is $mg\cos\theta$ , not $mg$ .
Using the wrong suvat equation or forgetting that $u = 0$ .
In the energy method, forgetting to multiply the friction force by the distance $5\,\text{m}$ when computing work done against friction.

Things to Be Careful About

The mark scheme accepts $g = 10\,\text{m s}^{-2}$ (as evidenced by the expected answer $8.60\,\text{m s}^{-1}$ ). Using $g = 9.8$ gives $v \approx 8.52\,\text{m s}^{-1}$ , which would not match the expected answer.
The answer must be given to at least 3 significant figures: $8.60\,\text{m s}^{-1}$ , not $8.6\,\text{m s}^{-1}$ .
The acceleration is constant because all forces (weight, normal reaction, friction) are constant on the straight section $AB$ .
In the energy method, the distance in the work-done-against-friction term must be $5\,\text{m}$ (the length of $AB$ ), not the vertical height.

Techniques used

resolve forces on inclined planeapply Newton's second lawuse suvat equation to find velocity

(b)

It is given that there is no change in the speed of the man when passing through $B$ and that his speed at $C$ is $11\text{ ms}^{-1}$ .

Find the work done against the resistance force as the man moves from $B$ to $C$ .

4M Similar

DifficultyMedium

Worked solution

Approach

Apply the work-energy principle between points $B$ and $C$ . The change in kinetic energy equals the work done by gravity minus the work done against resistance. Use the speed found in part (a) for the speed at $B$ .

Working

From part (a), the speed at $B$ is $v_B = \sqrt{74}\,\text{m s}^{-1}$ , so $v_B^2 = 74\,\text{m}^2\text{s}^{-2}$ .

The speed at $C$ is $v_C = 11\,\text{m s}^{-1}$ .

Change in kinetic energy from $B$ to $C$ :

\Delta KE = \frac{1}{2}mv_C^2 - \frac{1}{2}mv_B^2 = \frac{1}{2} \times 80 \times 11^2 - \frac{1}{2} \times 80 \times 74

\Delta KE = 40 \times 121 - 40 \times 74 = 4840 - 2960 = 1880\,\text{J}

Change in gravitational potential energy from $B$ to $C$ :

Point $B$ is $2.5\,\text{m}$ above $C$ , so the man descends $2.5\,\text{m}$ . The loss in PE is:

\Delta PE = mg \times 2.5 = 80 \times 10 \times 2.5 = 2000\,\text{J}

Work-energy principle:

The work done by gravity (loss in PE) minus the work done against resistance $W$ equals the gain in KE:

\Delta PE - W = \Delta KE

2000 - W = 1880

W = 2000 - 1880 = 120\,\text{J}

Answer

W = 120\,\text{J}

Final answer

120 J

Detailed explanation

Walkthrough

Step 1: Identify the speeds at B and C
From part (a), the speed at $B$ is $v_B = \sqrt{74}\,\text{m s}^{-1}$ , so $v_B^2 = 74$ . The speed at $C$ is given as $v_C = 11\,\text{m s}^{-1}$ . The question states there is no change in speed when passing through $B$ , so we use $v_B^2 = 74$ directly.

Step 2: Calculate the change in kinetic energy from B to C
$\Delta KE = \frac{1}{2}mv_C^2 - \frac{1}{2}mv_B^2 = \frac{1}{2}(80)(11^2) - \frac{1}{2}(80)(74) = 4840 - 2960 = 1880\,\text{J}$ . The KE increases by $1880\,\text{J}$ .

Step 3: Calculate the change in gravitational potential energy from B to C
Point $B$ is $2.5\,\text{m}$ above $C$ . The man descends $2.5\,\text{m}$ , so the loss in PE is $mg \times 2.5 = 80 \times 10 \times 2.5 = 2000\,\text{J}$ .

Step 4: Apply the work-energy principle
The work-energy principle states that the net work done on the man equals his change in KE. The forces doing work are gravity (positive, since he descends) and the resistance force (negative, since it opposes motion). Let $W$ be the work done against resistance (a positive quantity). Then:

$\text{Work done by gravity} - \text{Work done against resistance} = \Delta KE$

$2000 - W = 1880$

$W = 120\,\text{J}$

Key Takeaways

The work-energy principle can be applied between any two points, not just from the start.
When using the work-energy principle, be clear about what $W$ represents: work done by the resistance force is negative, while work done against resistance is positive.
The change in PE depends only on the vertical height difference, not the path taken.
Always use the correct speed at each point; do not include the PE change from $A$ to $C$ when only considering the motion from $B$ to $C$ .

Common Mistakes

Including the PE change from $A$ to $C$ instead of just from $B$ to $C$ . The question asks for work done from $B$ to $C$ , so only the $2.5\,\text{m}$ height difference matters.
Getting the sign wrong: writing $W - 2000 = 1880$ instead of $2000 - W = 1880$ , which gives $W = 120\,\text{J}$ vs $W = -120\,\text{J}$ . The work done against resistance must be positive.
Using $v_B = 8.6$ (rounded) instead of $v_B^2 = 74$ (exact), which gives $W = 118.4\,\text{J}$ instead of $120\,\text{J}$ . The mark scheme allows $118.4$ from using $8.60$ from part (a), but the exact answer is $120\,\text{J}$ .
Forgetting that the man descends from $B$ to $C$ , so PE is lost (positive work done by gravity).

Things to Be Careful About

The mark scheme explicitly states that using the change in PE from $A$ to $C$ gives M0 (no method mark). Only the $2.5\,\text{m}$ height difference from $B$ to $C$ should be used.
The answer must be positive: work done against resistance is $120\,\text{J}$ , not $-120\,\text{J}$ .
Use $v_B^2 = 74$ (exact value from part a) rather than the rounded $v_B = 8.60$ to get the exact answer $120\,\text{J}$ . The mark scheme allows $118.4$ if $8.60$ is used from part (a).
The work-energy equation must be dimensionally correct: all terms must be in joules (or equivalently, $\text{kg m}^2\text{s}^{-2}$ ).

Techniques used

calculate change in kinetic energycalculate change in gravitational potential energyapply work-energy principle to find work done against resistance

Q7MediumKinematics of Motion in a Straight Line

A particle $X$ moves in a straight line. The displacement of $X$ from $O$ at time $t\text{ s}$ after leaving $O$ is $s\text{ m}$ , where $s = 0.3t^2 + 0.6t$ for $0 \le t \le 4$ .

(a)

Find the velocity of $X$ at $t = 4$ .

2M Similar

(b)

For $t > 4$ , the acceleration of $X$ at time $t\text{ s}$ after leaving $O$ is $a\text{ ms}^{-2}$ , where $a = 0.3t^{\frac{1}{2}}$ . There is no change in the velocity of $X$ at $t = 4$ . The velocity of $X$ at $t = T$ is $14.2\text{ ms}^{-1}$ .

8M

(i)

Find the value of $T$ .

4M Similar

DifficultyMedium

Worked solution

Approach

For $t > 4$ , integrate the acceleration to obtain the velocity. Use the fact that the velocity does not change at $t = 4$ to determine the constant of integration, then set the velocity equal to $14.2$ and solve for $T$ .

Working

From part (a), the velocity at $t = 4$ is $3 \text{ m s}^{-1}$ .

For $t > 4$ , the acceleration is

a = 0.3t^{\frac{1}{2}}

Integrate with respect to $t$ :

v = \int 0.3t^{\frac{1}{2}} \, dt = 0.3 \cdot \frac{t^{\frac{3}{2}}}{\frac{3}{2}} + c = 0.2t^{\frac{3}{2}} + c

Since there is no change in velocity at $t = 4$ , $v = 3$ when $t = 4$ :

3 = 0.2(4)^{\frac{3}{2}} + c

Now

4^{\frac{3}{2}} = 8 \quad \Rightarrow \quad 0.2(4)^{\frac{3}{2}} = 1.6

So

3 = 1.6 + c \quad \Rightarrow \quad c = 1.4

Hence

v = 0.2t^{\frac{3}{2}} + 1.4

At $t = T$ , $v = 14.2$ :

14.2 = 0.2T^{\frac{3}{2}} + 1.4

12.8 = 0.2T^{\frac{3}{2}}

T^{\frac{3}{2}} = 64

T = 64^{\frac{2}{3}} = 16

Answer

T = 16

Final answer

T = 16

Detailed explanation

Walkthrough

For $t > 4$ , acceleration is given as a function of time, so velocity is obtained by integration: $v = \int a \, dt$ . The integration of $0.3t^{1/2}$ gives $0.2t^{3/2} + c$ . The constant $c$ is unknown, but the velocity is continuous at $t = 4$ : the velocity just before $t = 4$ equals the velocity just after $t = 4$ . From part (a), that common velocity is $3 \text{ m s}^{-1}$ . Substituting $t = 4$ and $v = 3$ gives $c = 1.4$ . Then set the resulting expression equal to $14.2$ and solve $0.2T^{3/2} = 12.8$ , so $T^{3/2} = 64$ . Raising both sides to the power $2/3$ gives $T = 64^{2/3} = 16$ .

Key Takeaways

Acceleration is the derivative of velocity, so velocity is the integral of acceleration.
A continuity condition at a switch time provides the initial condition needed to find the constant of integration.
Fractional powers can be solved by raising both sides to the reciprocal power.

Common Mistakes

Using $v = at$ instead of integrating; this is not valid when acceleration is not constant.
Forgetting the constant of integration, or not using the continuity condition to find it.
Writing the integrated form as $k(t-4)^{3/2}$ rather than $kt^{3/2}$ ; the acceleration is a function of $t$ , not of time since $t = 4$ .
Arithmetic errors with $4^{3/2} = 8$ or $64^{2/3} = 16$ .

Things to Be Careful About

The mark scheme requires an attempt at integrating $0.3t^{1/2}$ to a form $kt^{3/2}$ . The constant must be found using the velocity at $t = 4$ from part (a). The final equation must be set equal to $14.2$ . Also remember $T > 4$ ; the solution $T = 16$ satisfies this.

Techniques used

integrate acceleration with respect to timeapply continuity of velocity at t = 4 to determine the constant of integrationsolve a fractional-power equation for T

(ii)

Find the total distance travelled by $X$ between $t = 0$ and $t = T$ .

4M Similar

DifficultyMedium

Worked solution

Approach

The velocity is positive on both intervals, so distance travelled equals displacement. First use the given displacement formula to find the distance in the first 4 seconds. Then integrate the velocity found in part (b)(i) to obtain the displacement from $t = 4$ to $t = T$ , and add the two distances.

Working

For $0 \le t \le 4$ , the displacement is

s = 0.3t^2 + 0.6t

At $t = 4$ :

s(4) = 0.3(4)^2 + 0.6(4) = 4.8 + 2.4 = 7.2

So the distance travelled in the first 4 seconds is $7.2 \text{ m}$ .

For $t > 4$ , from part (b)(i),

v = 0.2t^{\frac{3}{2}} + 1.4

Integrate to find displacement:

s = \int \left(0.2t^{\frac{3}{2}} + 1.4\right) dt = 0.08t^{\frac{5}{2}} + 1.4t + C

Let $F(t) = 0.08t^{\frac{5}{2}} + 1.4t$ . The displacement from $t = 4$ to $t = 16$ is

F(16) - F(4)

Compute:

F(16) = 0.08(16)^{\frac{5}{2}} + 1.4(16)

16^{\frac{5}{2}} = 1024 \quad \Rightarrow \quad 0.08(16)^{\frac{5}{2}} = 81.92

F(16) = 81.92 + 22.4 = 104.32

F(4) = 0.08(4)^{\frac{5}{2}} + 1.4(4)

4^{\frac{5}{2}} = 32 \quad \Rightarrow \quad 0.08(4)^{\frac{5}{2}} = 2.56

F(4) = 2.56 + 5.6 = 8.16

So the displacement from $t = 4$ to $t = 16$ is

104.32 - 8.16 = 96.16

Total distance:

7.2 + 96.16 = 103.36

Answer

\text{Total distance} = 103.36 \text{ m}

Final answer

103.36 m

Detailed explanation

Walkthrough

Since the velocity is positive for both parts of the motion, the particle never reverses direction, so distance travelled is the same as displacement. For the first 4 seconds, substitute $t = 4$ into the given displacement formula to get $7.2 \text{ m}$ . For $t > 4$ , integrate the velocity $v = 0.2t^{3/2} + 1.4$ to get $s = 0.08t^{5/2} + 1.4t + C$ . To find the displacement between $t = 4$ and $t = 16$ , use the definite integral $F(16) - F(4)$ , where $F$ is the antiderivative with $C = 0$ . This gives $104.32 - 8.16 = 96.16 \text{ m}$ . Finally add the first 4 seconds' distance: $7.2 + 96.16 = 103.36 \text{ m}$ .

Key Takeaways

Integrating velocity gives displacement; a definite integral between two times gives the change in displacement over that interval.
When motion is in one direction only, distance equals displacement.
For piecewise motion, split the problem at the change in the motion rule and add the contributions.

Common Mistakes

Forgetting to add the $7.2 \text{ m}$ travelled in the first 4 seconds.
Using the wrong limits, for example evaluating $F(16) - F(0)$ with the second-interval formula, which would use an incorrect constant.
Arithmetic errors with $16^{5/2} = 1024$ and $4^{5/2} = 32$ .
If a constant of integration is included, using an incorrect value; to make displacement continuous at $t = 4$ , the constant must be $-0.96$ .

Things to Be Careful About

The mark scheme requires the first 4 seconds distance to be found separately. The integration of the velocity from part (b)(i) must be of the form $\alpha t^{5/2} + \beta t$ . When using limits, they must be $4$ and $T = 16$ . The final answer should be $103.36 \text{ m}$ (or $\frac{2584}{25} \text{ m}$ ); do not stop at $96.16 \text{ m}$ , which is only the distance from $t = 4$ to $t = 16$ .

Techniques used

calculate distance in the first interval using the given displacementintegrate velocity to obtain a displacement functionevaluate a definite integral between the required limitsadd the distances from the two intervals

Mathematics 9709/43 — May/June 2025

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