9709/42 Feb/Mar 2025 — Mathematics Solved Past Paper

Q14MMedium-EasyForces and Equilibrium

Fig. 1

Three coplanar forces of magnitudes $40\text{N}$ , $30\text{N}$ and $X\text{N}$ act at a point in the directions shown in the diagram.

Given that the forces are in equilibrium, find the values of $\theta$ and $X$ .

DifficultyMedium-Easy

Worked solution

Approach

Since the three forces are in equilibrium, the vector sum of all forces is zero. This means the sum of horizontal components is zero and the sum of vertical components is zero. We resolve the forces horizontally and vertically to form two equations.

Working

Resolve horizontally (taking right as positive):

X \cos \theta - 40 = 0

X \cos \theta = 40

Resolve vertically (taking upwards as positive):

30 - X \sin \theta = 0

X \sin \theta = 30

To find $X$ , square both equations and add them:

(X \cos \theta)^2 + (X \sin \theta)^2 = 40^2 + 30^2

X^2 (\cos^2 \theta + \sin^2 \theta) = 1600 + 900

X^2 = 2500

X = 50

To find $\theta$ , divide the vertical equation by the horizontal equation:

\frac{X \sin \theta}{X \cos \theta} = \frac{30}{40}

\tan \theta = 0.75

\theta = \tan^{-1}(0.75) \approx 36.869...^\circ

\theta = 36.9^\circ \text{ (to 1 d.p.)}

Answer

X = 50, \quad \theta = 36.9^\circ

Final answer

X = 50, theta = 36.9

Detailed explanation

Walkthrough

The problem states that three forces act at a point and are in equilibrium. Equilibrium means that the net force is zero, so the forces must balance out in every direction. We choose two perpendicular directions, horizontal and vertical, and resolve each force into components along these axes.

Horizontal resolution: The 40 N force acts to the left. The force $X$ has a horizontal component $X \cos \theta$ acting to the right. For equilibrium, these must be equal: $X \cos \theta = 40$ .
Vertical resolution: The 30 N force acts upwards. The force $X$ has a vertical component $X \sin \theta$ acting downwards. For equilibrium, these must be equal: $X \sin \theta = 30$ .
Finding X: We have a system of two equations. By squaring both and adding them, we use the identity $\cos^2 \theta + \sin^2 \theta = 1$ to eliminate $\theta$ and solve for $X$ . This gives $X = \sqrt{40^2 + 30^2} = 50$ .
Finding theta: By dividing the vertical equation by the horizontal equation, we eliminate $X$ and get $\tan \theta = 30/40 = 0.75$ . Taking the inverse tangent gives $\theta \approx 36.9^\circ$ .

Key Takeaways

When a system of forces is in equilibrium, the sum of components in any direction is zero.
Resolving forces into perpendicular components is a powerful technique for solving equilibrium problems.
Pythagoras' theorem and trigonometric ratios can be used to find the magnitude and direction of an unknown force from its components.

Common Mistakes

Forgetting to equate left and right forces correctly, leading to sign errors.
Mixing up sine and cosine when resolving the force $X$ , especially since $\theta$ is given with respect to the horizontal.
Not showing the method for finding $X$ or $\theta$ (e.g., just writing the answer without squaring and adding or dividing the equations).

Things to Be Careful About

Ensure that $\theta$ is measured correctly from the horizontal, as shown in the diagram. The horizontal component is $X \cos \theta$ and the vertical component is $X \sin \theta$ .
The mark scheme allows alternative methods such as using a triangle of forces (Pythagoras directly since 40N and 30N are perpendicular) or Lami's theorem, but resolving forces is the most systematic approach.
Give answers to an appropriate number of decimal places or significant figures as required (here, 1 d.p. or 3 s.f. is standard, $36.9^\circ$ is acceptable).

Techniques used

resolve forces horizontally and verticallyapply equilibrium conditionuse Pythagorean theorem to find resultant magnitudeuse inverse tangent to find angle

Approach

The deceleration phase is uniform, so use a constant-acceleration (suvat) equation that connects initial speed $V$ , final speed $0$ , acceleration $-2$ and distance $16$ . The equation $v^2 = u^2 + 2as$ involves all these quantities and avoids needing the time.

Working

For the deceleration phase:

u = V, \quad v = 0, \quad a = -2, \quad s = 16

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

0^2 = V^2 + 2(-2)(16)

0 = V^2 - 64

V^2 = 64

Since $V$ is a speed, $V = 8$ .

Answer

V = 8\ \text{ms}^{-1}

Final answer

V = 8 ms^-1

Detailed explanation

Walkthrough

The cyclist is slowing down uniformly at $2\text{ms}^{-2}$ and comes to rest after $16\text{m}$ . We know the final speed is $0$ , the acceleration is $-2$ (negative because it is a deceleration in the direction of motion), and the displacement is $16$ . The quantity we want is the initial speed $V$ . The best suvat equation is $v^2 = u^2 + 2as$ because it connects all four of these quantities and does not require the time.

Substitute $u = V$ , $v = 0$ , $a = -2$ and $s = 16$ :

0^2 = V^2 + 2(-2)(16)

This simplifies to $V^2 = 64$ , so $V = \pm 8$ . Since $V$ is a speed, it cannot be negative, so $V = 8$ . The mark scheme gives one mark for the correct equation in $V$ only and one mark for $V = 8$ only; writing $-8$ as well would lose the second mark.

Key Takeaways

In uniform acceleration problems, choose the suvat equation that contains the known and required quantities.
Deceleration is represented by a negative acceleration when the positive direction is the direction of motion.
Speed is a non-negative scalar, so reject negative roots.

Common Mistakes

Using $a = +2$ instead of $a = -2$ , which gives $V^2 + 64 = 0$ and no real solution.
Forgetting to square $V$ or $0$ when substituting into $v^2 = u^2 + 2as$ .
Giving $V = \pm 8$ ; the mark scheme requires $V = 8$ only.

Things to Be Careful About

Keep the units in mind: $V$ is in $\text{ms}^{-1}$ .
The equation must be in terms of $V$ only; the mark scheme awards B1 for that equation.
The sign of acceleration matters: deceleration at $2\text{ms}^{-2}$ means $a = -2$ if motion is taken as positive.

Techniques used

apply v^2 = u^2 + 2as to the deceleration phasetake the positive square root for speed

(b)

Find the total time for which she is in motion from the instant that she passes $O$ .

3M Similar

DifficultyMedium-Easy

Worked solution

Approach

The motion has three phases: acceleration, constant speed and deceleration. Find the time for each phase separately and add them. For the acceleration phase, use the average-speed form of the suvat equations because acceleration is uniform. For the constant-speed phase, use time = distance/speed. For the deceleration phase, use $v = u + at$ .

Working

Acceleration phase: $u = 4$ , $v = 8$ , $s = 42$ . Since the acceleration is uniform,

s = \frac{u+v}{2}t

42 = \frac{4+8}{2}t = 6t

t = 7\ \text{s}

Constant-speed phase: distance $50$ at speed $8$ :

t = \frac{50}{8} = 6.25\ \text{s}

Deceleration phase: $u = 8$ , $v = 0$ , $a = -2$ :

v = u + at

0 = 8 - 2t

t = 4\ \text{s}

Total time:

7 + 6.25 + 4 = 17.25 = \frac{69}{4}\ \text{s}

Answer

\frac{69}{4}\ \text{s} = 17.25\ \text{s}

Final answer

69/4 s = 17.25 s

Detailed explanation

Walkthrough

From part (a), $V = 8$ . The journey has three separate phases.

Acceleration phase: the cyclist goes from $4\text{ms}^{-1}$ to $8\text{ms}^{-1}$ over $42\text{m}$ . Because the acceleration is constant, the average speed is $\frac{4+8}{2} = 6\text{ms}^{-1}$ , so $t = \frac{42}{6} = 7\text{s}$ .
Constant-speed phase: she travels $50\text{m}$ at $8\text{ms}^{-1}$ , so $t = \frac{50}{8} = 6.25\text{s}$ .
Deceleration phase: she slows from $8\text{ms}^{-1}$ to rest at $2\text{ms}^{-2}$ . Using $v = u + at$ , $0 = 8 - 2t$ , so $t = 4\text{s}$ .

The total time is $7 + 6.25 + 4 = 17.25\text{s} = \frac{69}{4}\text{s}$ .

The mark scheme awards M1 for finding a time equation for at least one of the phases using their $V$ , a second M1 for finding time equations for the other two phases, and A1 for the final total. It also accepts answers rounding to 17.3 from correct working.

Key Takeaways

Split a multi-phase motion into intervals and handle each interval with the appropriate formula.
For uniform acceleration, $s = \frac{u+v}{2}t$ is often quicker than using the quadratic suvat equations.
At constant speed, time is simply distance divided by speed.
Total time is the sum of the times for all phases.

Common Mistakes

Using an acceleration formula for the constant-speed section.
Using $a = +2$ in the deceleration section instead of $a = -2$ .
Forgetting to include one of the three phases when adding the times.
If part (a) was wrong, using the wrong $V$ ; the mark scheme allows follow-through with their $V$ .
Giving a negative time; the mark scheme requires positive $t$ for each phase.

Things to Be Careful About

The acceleration phase uses the distance $42\text{m}$ , not $50\text{m}$ .
The constant-speed phase distance is $50\text{m}$ ; its time is $50/V$ .
The deceleration phase distance is $16\text{m}$ , but the time can be found directly from $v = u + at$ without using that distance.
The final answer is an exact fraction $\frac{69}{4}\text{s}$ ; the mark scheme also accepts 17.3 from correct work.

Techniques used

use the average-speed formula s = (u+v)/2 t for uniform accelerationapply v = u + at to the deceleration phasecalculate time at constant speed as distance divided by speed

Q3MediumEnergy, Work and Power

An aeroplane is flying at a constant speed.

(a)

The aeroplane is flying horizontally. The aeroplane’s engines are producing a constant power of $5500\text{kW}$ , and the aeroplane experiences a constant horizontal resistance force of $25\text{kN}$ .

Find the speed of the aeroplane.

2M Similar

DifficultyMedium-Easy

Worked solution

Approach

At constant speed the resultant force on the aeroplane is zero, so the driving force from the engines equals the resistance force. The engine power is $P = Fv$ , where $F$ is the driving force and $v$ is the speed. Convert the given quantities to consistent units and solve for $v$ .

Working

The power is $5500\text{ kW} = 5\,500\,000\text{ W}$ and the resistance is $25\text{ kN} = 25\,000\text{ N}$ .

Since the speed is constant, the driving force is $F = 25\,000\text{ N}$ .

Using $P = Fv$ :

5\,500\,000 = 25\,000 \, v

v = \frac{5\,500\,000}{25\,000} = 220

Answer

The speed of the aeroplane is $220\text{ m s}^{-1}$ .

Final answer

220 m s^-1

Detailed explanation

Walkthrough

The aeroplane is flying horizontally at constant speed. Constant speed means there is no acceleration, so the resultant horizontal force is zero. Therefore the forward driving force produced by the engines must exactly balance the resistance force of $25\text{ kN}$ .

The power of the engines tells us how quickly they do work. For a constant force $F$ moving at speed $v$ , the power is $P = Fv$ . Here the power is given in kilowatts and the force in kilonewtons. We can either work directly in kW and kN, or convert to W and N. Using $5500 = 25v$ gives $v = 220$ , so the speed is $220\text{ m s}^{-1}$ .

Key Takeaways

Constant speed in a straight line implies zero resultant force, so driving force equals resistance.
Power, force and speed are related by $P = Fv$ when the force is in the direction of motion.
Units such as kW and kN can be used together as long as they are consistent; here kW/kN gives m/s.

Common Mistakes

Forgetting that constant speed means driving force equals resistance, not using the given resistance directly as the force in $P=Fv$ .
Mixing units incorrectly, e.g. using $5500$ with $25\,000$ without converting.
Writing $P = Fv$ with the wrong quantities or solving for $F$ instead of $v$ .

Things to Be Careful About

The mark scheme allows errors in the use of kN and kW for the method mark, but the final answer must be correct in m/s.
The force in $P=Fv$ must be the force in the direction of motion; here that is the horizontal driving force.
Since the motion is horizontal, gravitational potential energy is not involved in part (a).

Techniques used

equate driving force to resistance at constant speedapply P = Fvconvert units consistently

(b)

The aeroplane then ascends $300\text{m}$ in $50\text{s}$ , while maintaining the same speed. The resistance force is no longer constant, and the work done against the resistance force in ascending the $300\text{m}$ is $270\,000\text{kJ}$ . The mass of the aeroplane is $60\,000\text{kg}$ .

Find the average power of the aeroplane’s engines.

4M Similar

DifficultyMedium

Worked solution

Approach

The aeroplane maintains the same speed, so its kinetic energy does not change. All the work done by the engines goes into increasing gravitational potential energy and overcoming resistance. Use the work-energy principle:

\text{Work done by engines} = \text{gain in PE} + \text{work done against resistance}.

The work done by the engines over $50\text{ s}$ is $P_{\text{avg}} \times 50$ , so solve for $P_{\text{avg}}$ .

Working

Gain in gravitational potential energy:

\Delta PE = mgh = 60\,000 \times 10 \times 300 = 180\,000\,000\text{ J} = 180\,000\text{ kJ}.

Work done against resistance is $270\,000\text{ kJ} = 270\,000\,000\text{ J}$ .

Work-energy equation in kJ:

P_{\text{avg}} \times 50 = 180\,000 + 270\,000

P_{\text{avg}} \times 50 = 450\,000

P_{\text{avg}} = \frac{450\,000}{50} = 9000\text{ kW}.

Answer

The average power of the aeroplane’s engines is $9000\text{ kW}$ (or $9\,000\,000\text{ W}$ ).

Final answer

9000 kW

Detailed explanation

Walkthrough

The aeroplane ascends $300\text{ m}$ in $50\text{ s}$ while maintaining the same speed. Because the speed does not change, the kinetic energy of the aeroplane is the same at the start and end of the ascent. Therefore no work is used to change kinetic energy.

The engines must do work to increase the gravitational potential energy of the aeroplane and to overcome the resistance force. The gain in gravitational potential energy is:

\Delta PE = mgh = 60\,000 \times 10 \times 300 = 180\,000\,000\text{ J}.

The work done against the resistance is given as $270\,000\text{ kJ} = 270\,000\,000\text{ J}$ .

The work done by the engines over the $50\text{ s}$ is $P_{\text{avg}} \times 50$ , where $P_{\text{avg}}$ is the average power. Using the work-energy principle:

P_{\text{avg}} \times 50 = 180\,000\,000 + 270\,000\,000.

This gives $P_{\text{avg}} \times 50 = 450\,000\,000$ , so $P_{\text{avg}} = 9\,000\,000\text{ W} = 9000\text{ kW}$ .

Key Takeaways

When speed is constant, there is no change in kinetic energy, so work done by engines equals gain in PE plus work against resistance.
The work-energy principle can be written as work done by engines = change in PE + work done against resistance (plus change in KE if speed changes).
Average power is total work done divided by the time taken, or equivalently work done by engines $= P_{\text{avg}} \times t$ .

Common Mistakes

Omitting one of the three terms in the work-energy equation; the mark scheme requires a three-term equation.
Using $g = 9.8$ when the question expects $g = 10$ (the mark scheme uses $g = 10$ ).
Mixing kJ and J without converting; the mark scheme allows either consistent set of units, but not mixing them.
Forgetting to divide by $50\text{ s}$ to find average power.

Things to Be Careful About

The mark scheme awards the method mark for a dimensionally correct work-energy equation with three terms, and allows sign errors.
The resistance force is no longer constant in part (b), so you cannot use $P = Fv$ with the given resistance force directly; use the work-energy principle instead.
The work done against resistance is given as a total, so it should be added to the PE gain, not subtracted.
Use the same units throughout. In kJ: $60 \times 10 \times 300 = 180\,000$ kJ, and the resistance work is $270\,000$ kJ, giving total $450\,000$ kJ.
The mark scheme allows work done by engines instead of $\text{Power} \times 50$ in the equation.

Techniques used

calculate change in gravitational potential energyapply the work-energy principlerelate work done by engines to average power over time

Q4MediumNewton's Laws of MotionKinematics of Motion in a Straight Line

Fig. 2

Two particles $A$ and $B$ have masses $0.3\text{kg}$ and $0.1\text{kg}$ respectively. The particles are attached to the ends of a light inextensible string. The string passes over a fixed smooth pulley, and the particles hang vertically below the pulley. Both particles are initially at a height of $x\text{m}$ above horizontal ground (see diagram). The system is released from rest.

(a)

Find the tension in the string and the acceleration of the particles.

4M Similar

DifficultyMedium-Easy

Worked solution

Approach

Apply Newton's second law ( $F=ma$ ) to each particle separately. Particle A is heavier, so it accelerates downwards and Particle B accelerates upwards. The tension $T$ is the same in the string throughout. Solve the resulting simultaneous equations for acceleration $a$ and tension $T$ .

Working

Take $g = 10 \text{ m s}^{-2}$ .

For particle A (mass $0.3 \text{ kg}$ , moving downwards):

\begin{aligned} 0.3g - T &= 0.3a \\ 3 - T &= 0.3a \quad \text{--- (1)} \end{aligned}

For particle B (mass $0.1 \text{ kg}$ , moving upwards):

\begin{aligned} T - 0.1g &= 0.1a \\ T - 1 &= 0.1a \quad \text{--- (2)} \end{aligned}

Add equations (1) and (2) to eliminate $T$ :

3 - 1 = 0.3a + 0.1a

2 = 0.4a

a = 5 \text{ m s}^{-2}

Substitute $a = 5$ into equation (2) to find $T$ :

T - 1 = 0.1(5)

T = 1 + 0.5 = 1.5 \text{ N}

Answer

a = 5 \text{ m s}^{-2}, \quad T = 1.5 \text{ N}

Final answer

Acceleration = 5 m/s², Tension = 1.5 N

Detailed explanation

Walkthrough

We analyze the forces on each particle individually. Since the pulley is smooth and the string is light and inextensible, the tension $T$ is uniform throughout the string, and both particles have the same magnitude of acceleration $a$ .

Particle A ( $0.3 \text{ kg}$ ) is heavier, so the system accelerates with A moving down and B moving up. We apply Newton's second law ( $F_{\text{net}} = ma$ ) in the direction of motion for each particle.

For A, the forces are its weight ( $0.3g$ ) downwards and tension ( $T$ ) upwards. The net force is $0.3g - T$ , giving the equation $0.3g - T = 0.3a$ .

For B, the forces are tension ( $T$ ) upwards and its weight ( $0.1g$ ) downwards. The net force is $T - 0.1g$ , giving the equation $T - 0.1g = 0.1a$ .

Adding these two equations eliminates $T$ and allows us to solve for $a$ . Once $a$ is known, we substitute it back into either equation to find $T$ . Using $g = 10 \text{ m s}^{-2}$ , we get $a = 5 \text{ m s}^{-2}$ and $T = 1.5 \text{ N}$ .

Key Takeaways

In a pulley system with two hanging masses, write $F=ma$ for each mass separately, taking the direction of acceleration as positive for each. The tension is the same in both equations. Adding the equations eliminates tension and solves for acceleration directly.

Common Mistakes

Using the wrong sign for tension or weight (e.g., $T + 0.3g = 0.3a$ for particle A).
Forgetting that the acceleration magnitude is the same for both particles.
Using $g = 9.8$ when the mark scheme expects $g = 10$ (though $9.8$ is often accepted, the final answers here align with $g=10$ ).

Things to Be Careful About

Ensure the direction of positive acceleration is consistent: downwards for A and upwards for B. The tension must be the same variable in both equations.

Techniques used

apply Newton's second law to each particleform simultaneous equations for acceleration and tensionsolve equations to find numerical values

(b)

During the subsequent motion, $B$ does not reach the pulley. When $A$ reaches the ground, it comes to rest.

Given that the greatest height of $B$ above the ground is $1.2\text{m}$ , find the value of $x$ .

3M Similar

DifficultyMedium

Worked solution

Approach

The motion of particle B has two phases:

Accelerated phase: Both particles move with acceleration $a = 5 \text{ m s}^{-2}$ . Particle B rises a distance $x$ (same as A falls). Find the velocity $v$ of B at the end of this phase.
Free-fall phase: Particle A is at rest on the ground. Particle B continues moving upwards under gravity alone (acceleration $-g$ ) until its velocity becomes zero at its greatest height.

Use the constant acceleration formula $v^2 = u^2 + 2as$ for both phases and combine the heights.

Working

Phase 1: Accelerated motion
Initial velocity $u = 0$ , acceleration $a = 5 \text{ m s}^{-2}$ , displacement $s = x$ .

v^2 = 0 + 2(5)(x) = 10x

Phase 2: Free-fall motion
Initial velocity $u = v$ (where $v^2 = 10x$ ), final velocity $= 0$ , acceleration $= -g = -10 \text{ m s}^{-2}$ .
Let $s'$ be the additional height B rises after A hits the ground.

0 = v^2 + 2(-10)s'

20s' = v^2 = 10x

s' = 0.5x

Total height of B
B starts at height $x$ . During Phase 1, it rises $x$ (so it is at height $2x$ when A hits the ground). During Phase 2, it rises a further $s' = 0.5x$ .
Total greatest height $= x + x + s' = 2x + 0.5x = 2.5x$ .

Given that the greatest height is $1.2 \text{ m}$ :

2.5x = 1.2

x = \frac{1.2}{2.5} = 0.48

Answer

x = 0.48

Final answer

x = 0.48

Detailed explanation

Walkthrough

Once particle A hits the ground, the string goes slack. Particle B is no longer pulled by the string and continues to move upwards under the influence of gravity alone (free fall).

Phase 1 (Both particles moving):
Particle B starts from rest and accelerates upwards at $5 \text{ m s}^{-2}$ for a distance $x$ (since A falls distance $x$ to reach the ground). Using $v^2 = u^2 + 2as$ , the velocity squared at the moment A hits the ground is $v^2 = 0 + 2(5)(x) = 10x$ .

Phase 2 (B moving alone):
Particle B now has an initial upward velocity $v$ (where $v^2 = 10x$ ) and accelerates downwards at $g = 10 \text{ m s}^{-2}$ . It rises until its velocity is zero. Let the additional height be $s'$ . Using $v_f^2 = v_i^2 + 2as$ , we have $0 = 10x - 2(10)s'$ , which gives $s' = 0.5x$ .

Total height:
B starts at height $x$ . It rises $x$ during Phase 1 (reaching height $2x$ ) and rises a further $0.5x$ during Phase 2. Total height $= 2x + 0.5x = 2.5x$ . Setting this equal to $1.2$ gives $x = 0.48$ .

Key Takeaways

When a connected particle system is released and one mass hits a surface, the other mass continues with the velocity it had at that instant but now moves under gravity alone. Break the problem into distinct phases and link them using the velocity at the transition point.

Common Mistakes

Assuming B stops when A hits the ground. B has upward velocity and will continue to rise.
Forgetting that B is already at height $2x$ (initial $x$ + rise $x$ ) when A hits the ground, and only adding the extra rise $s'$ to get the total height.
Using the wrong acceleration for the second phase (must be $-g$ , not $-a$ ).

Things to Be Careful About

Ensure you account for the initial height $x$ and the height gained during the first phase ( $x$ ) when calculating the total maximum height. The displacement in the second phase is only the additional height gained.

Techniques used

apply suvat equations for the accelerating phaseapply suvat equations for the free-fall phaselink the two phases using final velocity of the first phasesum displacement components to find total height

Q5Medium-HardMomentumKinematics of Motion in a Straight Line

Fig. 3

Three particles $P$ , $Q$ and $R$ , of masses $0.6\text{kg}$ , $0.4\text{kg}$ and $0.8\text{kg}$ respectively, are at rest in a straight line on a smooth horizontal plane. The distance from $P$ to $Q$ is $3\text{m}$ , and the distance from $Q$ to $R$ is also $3\text{m}$ (see diagram). $P$ is projected directly towards $Q$ with speed $3\text{ms}^{-1}$ . After $P$ and $Q$ collide, $P$ continues to move in the same direction with speed $1.5\text{ms}^{-1}$ .

(a)

Find the speed of $Q$ after the collision.

2M Similar

DifficultyMedium-Easy

Worked solution

Approach

Apply the principle of conservation of linear momentum to the collision between P and Q. Since the plane is smooth, there are no external horizontal forces, so total momentum is conserved.

Working

Let $v$ be the speed of $Q$ after the collision. All motion is in the same direction, so we can treat speeds as positive scalars in the momentum equation.

Initial momentum before collision:

\begin{aligned} p_{\text{initial}} &= m_P u_P + m_Q u_Q \\ &= (0.6)(3) + (0.4)(0) \\ &= 1.8 \text{ kg m s}^{-1} \end{aligned}

Final momentum after collision:

\begin{aligned} p_{\text{final}} &= m_P v_P + m_Q v \\ &= (0.6)(1.5) + (0.4)v \\ &= 0.9 + 0.4v \end{aligned}

Equating initial and final momentum:

1.8 = 0.9 + 0.4v

0.9 = 0.4v

v = \frac{0.9}{0.4} = 2.25 \text{ m s}^{-1}

Answer

2.25 \text{ m s}^{-1}

Final answer

2.25 m s^-1

Detailed explanation

Walkthrough

The problem states that $P$ and $Q$ collide on a smooth horizontal plane. "Smooth" means there is no friction, so the only horizontal forces during the brief collision are the internal contact forces between $P$ and $Q$ . By Newton's third law, these are equal and opposite, meaning the total momentum of the system $P+Q$ is conserved.

We set up the equation: (mass of $P$ ) $\times$ (initial speed of $P$ ) + (mass of $Q$ ) $\times$ (initial speed of $Q$ ) = (mass of $P$ ) $\times$ (final speed of $P$ ) + (mass of $Q$ ) $\times$ (final speed of $Q$ ).

Substituting the given values: $0.6 \times 3 + 0.4 \times 0 = 0.6 \times 1.5 + 0.4 \times v$ . Solving this linear equation gives $v = 2.25$ .

Key Takeaways

Conservation of momentum applies to collisions where no external horizontal forces act.
Momentum is a vector quantity; here all motion is in one direction, so we use positive signs for velocities in the direction of initial motion.

Common Mistakes

Forgetting that $Q$ is initially at rest (using $u_Q = 3$ or similar).
Sign errors if assuming directions incorrectly (though here all final velocities are in the same direction as initial motion for $P$ ).
Arithmetic errors in solving $1.8 = 0.9 + 0.4v$ .

Things to Be Careful About

Ensure units are consistent (kg and m/s give momentum in kg m/s).
The question asks for speed, which is a positive scalar. The result $2.25$ is positive, confirming $Q$ moves in the same direction as $P$ was initially travelling.

Techniques used

apply conservation of momentumsolve linear equation for unknown velocity

(b)

In the subsequent collision between $Q$ and $R$ , these particles coalesce.

Find the speed of the combined particle after this collision.

1M Similar

DifficultyMedium-Easy

Worked solution

Approach

When $Q$ collides with $R$ , they coalesce (stick together). This is a perfectly inelastic collision. Momentum is still conserved. The final mass is the sum of the masses of $Q$ and $R$ .

Working

Let $w$ be the speed of the combined particle ( $Q+R$ ) after collision.

Before collision:

$Q$ is moving at $2.25 \text{ m s}^{-1}$ (from part a).
$R$ is at rest ( $0 \text{ m s}^{-1}$ ).

p_{\text{initial}} = m_Q v_Q + m_R u_R = (0.4)(2.25) + (0.8)(0) = 0.9 \text{ kg m s}^{-1}

After collision:

Combined mass $= 0.4 + 0.8 = 1.2 \text{ kg}$ .
Speed $= w$ .

p_{\text{final}} = (m_Q + m_R) w = 1.2 w

Equating momentum:

0.9 = 1.2 w

w = \frac{0.9}{1.2} = 0.75 \text{ m s}^{-1}

Answer

0.75 \text{ m s}^{-1}

Final answer

0.75 m s^-1

Detailed explanation

Walkthrough

$Q$ continues moving and hits $R$ . Since they coalesce, they move together as a single object of mass $0.4 + 0.8 = 1.2$ kg. We apply conservation of momentum again. The initial momentum is just the momentum of $Q$ (since $R$ is stationary). The final momentum is the combined mass times the new common speed.

$0.4 \times 2.25 = (0.4 + 0.8) \times w \Rightarrow 0.9 = 1.2w \Rightarrow w = 0.75$ .

Key Takeaways

In a coalescing collision, the final mass is the sum of the individual masses.
Momentum is conserved even when kinetic energy is not (this is an inelastic collision).

Common Mistakes

Using the wrong final mass (e.g., forgetting to add $0.8$ ).
Carrying forward the speed of $P$ instead of $Q$ (though $P$ is not involved in this collision).

Things to Be Careful About

This is a "follow through" mark from part (a). If the student got part (a) wrong, they can still get marks here using their incorrect value from (a), as long as the method is correct.

Techniques used

apply conservation of momentum for inelastic collisionhandle coalescing particles

(c)

Find the time that it takes from when $P$ is initially projected until the instant at which $P$ collides with the combined particle.

4M Similar

DifficultyMedium-Hard

Worked solution

Approach

We need to find the total time from $t=0$ until $P$ collides with the combined particle ( $QR$ ). This involves breaking the motion into phases:

Time for $P$ to reach $Q$ .
Time for $Q$ to reach $R$ (and coalesce).
Time for $P$ to catch up with the combined particle $QR$ after the second collision.

Working

Phase 1: $P$ travels to $Q$
Distance $= 3$ m, speed of $P = 3$ m/s.

t_1 = \frac{3}{3} = 1 \text{ s}

At $t = 1$ s, $P$ and $Q$ collide. $P$ is now at $x = 3$ m (taking $P$ 's start as origin), moving at $1.5$ m/s. $Q$ is at $x = 3$ m, moving at $2.25$ m/s. $R$ is at $x = 6$ m, at rest.

Phase 2: $Q$ travels to $R$
Distance $Q$ to $R = 3$ m. Speed of $Q = 2.25$ m/s.

t_2 = \frac{3}{2.25} = \frac{3}{9/4} = \frac{12}{9} = \frac{4}{3} \text{ s}

Total time elapsed when $Q$ collides with $R$ : $T_1 = 1 + \frac{4}{3} = \frac{7}{3}$ s.

At this instant ( $t = 7/3$ s):

The combined particle $QR$ is at $x = 6$ m (position of $R$ ), moving at $0.75$ m/s.
$P$ has been travelling from $x=3$ at $1.5$ m/s for duration $t_2 = 4/3$ s.
Distance $P$ travelled in this phase $= 1.5 \times \frac{4}{3} = 2$ m.
Position of $P = 3 + 2 = 5$ m.

Phase 3: $P$ catches up with $QR$
At $t = 7/3$ s:

Separation distance between $P$ and $QR = 6 - 5 = 1$ m.
Speed of $P = 1.5$ m/s.
Speed of $QR = 0.75$ m/s.
Relative speed of approach $= 1.5 - 0.75 = 0.75$ m/s.

Time for $P$ to catch $QR$ :

t_3 = \frac{\text{separation}}{\text{relative speed}} = \frac{1}{0.75} = \frac{1}{3/4} = \frac{4}{3} \text{ s}

Total time:

T_{\text{total}} = t_1 + t_2 + t_3 = 1 + \frac{4}{3} + \frac{4}{3} = 1 + \frac{8}{3} = \frac{11}{3} \text{ s}

T_{\text{total}} \approx 3.67 \text{ s}

Answer

\frac{11}{3} \text{ s} \quad (\text{or } 3.67 \text{ s})

Final answer

11/3 s

Detailed explanation

Walkthrough

This is a multi-stage problem. We track the timeline of events.

Start to first collision ( $P$ hits $Q$ ): $P$ covers 3m at 3 m/s. Time = 1s. At this point, $P$ is at 3m, $Q$ is at 3m. $P$ slows to 1.5 m/s, $Q$ speeds up to 2.25 m/s.
First collision to second collision ( $Q$ hits $R$ ): $Q$ must cover the 3m gap to $R$ . Time = $3 / 2.25 = 4/3$ s. During this same time, $P$ is moving at 1.5 m/s, covering $1.5 \times (4/3) = 2$ m. So $P$ is now at $3+2=5$ m. $Q$ hits $R$ at position 6m. They stick together and move at 0.75 m/s.
Catch-up phase: Now $P$ is at 5m (speed 1.5 m/s) and the combined mass is at 6m (speed 0.75 m/s). The gap is 1m. $P$ is faster, so it catches up. Relative speed is $1.5 - 0.75 = 0.75$ m/s. Time to close gap = $1 / 0.75 = 4/3$ s.

Total time = $1 + 4/3 + 4/3 = 11/3$ s.

Key Takeaways

Break complex motion into distinct time intervals based on events (collisions).
Use constant velocity equations ( $s = vt$ ) for each interval.
Relative speed is key for catch-up problems: $v_{\text{relative}} = v_1 - v_2$ .

Common Mistakes

Forgetting to add the initial time $t_1 = 1$ s to the final catch-up time.
Calculating the position of $P$ incorrectly during the second phase.
Using the wrong relative speed (e.g., adding speeds instead of subtracting).

Things to Be Careful About

The question asks for time from when $P$ is initially projected, so we must include the time for the first collision.
Ensure all time intervals are added correctly: $1 + 4/3 + 4/3 = 11/3$ .
Mark scheme allows alternative methods (e.g., setting up equations of motion for position vs time from $t=0$ or from the second collision), but the phase-by-phase approach is most transparent.

Techniques used

calculate time for constant velocity motiondetermine relative distance and speed for catch-up problemsum time intervals

Q6MediumForces and EquilibriumNewton's Laws of Motion

Fig. 4

A block of mass $12\text{kg}$ is placed on a rough plane inclined at an angle of $\alpha$ to the horizontal, where $\tan \alpha = 0.5$ . A force of $X\text{N}$ is applied to the block, directly up the plane (see diagram). The coefficient of friction between the block and the plane is $\mu$ .

(a)

It is given that $\mu = 0.15$ and $X = 20$ .

Find the time that it takes for the block to move $2\text{m}$ down the plane from rest.

6M Similar

DifficultyMedium

Worked solution

Approach

The block moves down the plane, so friction acts up the plane. We resolve forces perpendicular to the plane to find the normal reaction $R$ , then use $F = \mu R$ to find the frictional force. Applying Newton's second law along the plane gives the acceleration, which we use with the kinematic equation $s = ut + \frac{1}{2}at^2$ to find the time.

Working

Since $\tan \alpha = 0.5 = \frac{1}{2}$ , we can construct a right triangle with opposite $= 1$ , adjacent $= 2$ , and hypotenuse $= \sqrt{5}$ . Therefore:

\sin \alpha = \frac{1}{\sqrt{5}}, \quad \cos \alpha = \frac{2}{\sqrt{5}}

Step 1: Resolve perpendicular to the plane to find $R$ .

The forces perpendicular to the plane are the normal reaction $R$ (upward, away from the plane) and the component of weight $12g \cos \alpha$ (downward, into the plane). Since there is no acceleration perpendicular to the plane:

R = 12g \cos \alpha = 12g \times \frac{2}{\sqrt{5}} = \frac{24g}{\sqrt{5}}

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

R = \frac{240}{\sqrt{5}} = 48\sqrt{5} \text{ N}

Step 2: Calculate the frictional force $F$ .

Since the block is moving, friction is at its limiting value:

F = \mu R = 0.15 \times 48\sqrt{5} = 7.2\sqrt{5} \text{ N}

Step 3: Apply Newton's second law along the plane (downward positive).

The forces acting along the plane are:

Weight component down the plane: $12g \sin \alpha = 12 \times 10 \times \frac{1}{\sqrt{5}} = \frac{120}{\sqrt{5}} = 24\sqrt{5} \text{ N}$
Applied force $X = 20 \text{ N}$ up the plane
Friction $F = 7.2\sqrt{5} \text{ N}$ up the plane (opposing motion)

12g \sin \alpha - X - F = 12a

24\sqrt{5} - 20 - 7.2\sqrt{5} = 12a

16.8\sqrt{5} - 20 = 12a

a = \frac{16.8\sqrt{5} - 20}{12} = \frac{16.8\sqrt{5}}{12} - \frac{20}{12} = 1.4\sqrt{5} - \frac{5}{3}

Converting to a single fraction:

a = \frac{1.4\sqrt{5} \times 3 - 5}{3} = \frac{4.2\sqrt{5} - 5}{3} = \frac{21\sqrt{5} - 25}{15}

Numerically:

a = \frac{21 \times 2.2361 - 25}{15} = \frac{46.958 - 25}{15} = \frac{21.958}{15} \approx 1.464 \text{ m/s}^2

Step 4: Use kinematics to find $t$ .

The block moves $s = 2 \text{ m}$ from rest ( $u = 0$ ) with constant acceleration $a$ :

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

2 = 0 + \frac{1}{2} \times 1.464 \times t^2

t^2 = \frac{4}{1.464} \approx 2.732

t = \sqrt{2.732} \approx 1.65 \text{ s}

Answer

t = 1.65 \text{ s}

Final answer

1.65 s

Detailed explanation

Walkthrough

Step 1: Determine trigonometric values from the given angle.

We are told $\tan \alpha = 0.5 = \frac{1}{2}$ . This means we can imagine a right-angled triangle where the side opposite to $\alpha$ is $1$ and the side adjacent is $2$ . The hypotenuse is then $\sqrt{1^2 + 2^2} = \sqrt{5}$ . From this triangle, we read off:

\sin \alpha = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{1}{\sqrt{5}}, \quad \cos \alpha = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{2}{\sqrt{5}}

These values will be used throughout the solution.

Step 2: Resolve forces perpendicular to the plane.

The block is not accelerating perpendicular to the plane, so the net force in that direction is zero. The only two forces with components perpendicular to the plane are the normal reaction $R$ (pushing away from the surface) and the component of the block's weight pressing into the surface, which is $12g \cos \alpha$ . Setting these equal:

R = 12g \cos \alpha = 12 \times 10 \times \frac{2}{\sqrt{5}} = \frac{240}{\sqrt{5}} = 48\sqrt{5} \text{ N}

Step 3: Calculate the frictional force.

Since the block is moving down the plane, friction opposes the motion and acts up the plane. For a moving block on a rough surface, the friction is at its limiting value $F = \mu R$ :

F = 0.15 \times 48\sqrt{5} = 7.2\sqrt{5} \text{ N}

Step 4: Apply Newton's second law along the incline.

Taking the downward direction along the plane as positive, the forces are:

Weight component down the plane: $12g \sin \alpha = 120 \times \frac{1}{\sqrt{5}} = 24\sqrt{5} \text{ N}$
Applied force $X = 20 \text{ N}$ up the plane (negative)
Friction $F = 7.2\sqrt{5} \text{ N}$ up the plane (negative)

Newton's second law gives:

12g \sin \alpha - X - F = ma

24\sqrt{5} - 20 - 7.2\sqrt{5} = 12a

16.8\sqrt{5} - 20 = 12a

Solving for $a$ :

a = \frac{16.8\sqrt{5} - 20}{12} = \frac{21\sqrt{5} - 25}{15} \approx 1.464 \text{ m/s}^2

Step 5: Use the constant acceleration formula to find time.

With initial velocity $u = 0$ , displacement $s = 2 \text{ m}$ , and acceleration $a \approx 1.464 \text{ m/s}^2$ :

s = ut + \frac{1}{2}at^2 \implies 2 = \frac{1}{2} \times 1.464 \times t^2

t^2 = \frac{4}{1.464} \approx 2.732 \implies t \approx 1.65 \text{ s}

Key Takeaways

When $\tan \alpha$ is given as a simple fraction, construct a right triangle to find $\sin \alpha$ and $\cos \alpha$ exactly.
On an inclined plane, always resolve forces perpendicular to the plane first to find the normal reaction $R$ .
Friction always opposes the direction of motion (or intended motion), so its direction must be determined from the physical context.
Newton's second law along the incline combines weight components, applied forces, and friction to yield acceleration.
The kinematic equation $s = ut + \frac{1}{2}at^2$ is the natural choice when time is the unknown and acceleration is constant.

Common Mistakes

Forgetting that friction acts up the plane when the block moves down (or vice versa). The direction of friction must be consistent with the direction of motion.
Using $\sin \alpha$ and $\cos \alpha$ in the wrong places. The component of weight parallel to the plane is $mg \sin \alpha$ and perpendicular is $mg \cos \alpha$ .
Not using the exact values of $\sin \alpha$ and $\cos \alpha$ derived from $\tan \alpha = 0.5$ , leading to rounding errors.
Sign errors in Newton's second law: ensuring all forces are assigned the correct sign based on the chosen positive direction.
Using $g = 9.8$ or $g = 9.81$ when the mark scheme expects $g = 10$ . Always check which value of $g$ is standard for the paper.

Things to Be Careful About

The direction of friction is critical. Since the block moves down the plane, friction acts up the plane. If the block were moving up, friction would act down.
The value of $g$ must be consistent. This mark scheme uses $g = 10 \text{ m/s}^2$ , which gives the clean exact answer $a = \frac{21\sqrt{5} - 25}{15}$ .
When using $s = ut + \frac{1}{2}at^2$ , ensure $u = 0$ since the block starts from rest.
The acceleration is positive (down the plane), confirming the block indeed moves down as stated in the question.
Round the final answer to 3 significant figures: $t = 1.65 \text{ s}$ .

Techniques used

resolve forces perpendicular to plane to find normal reactioncalculate limiting friction using F = μRapply Newton's second law along the inclineuse constant acceleration formula to find time

(b)

It is given instead that $\mu \neq 0.15$ and that when $X = 10$ , the block is on the point of moving down the plane.

Find the value of $\mu$ and the value of $X$ for which the block is on the point of moving up the plane.

4M Similar

DifficultyMedium

Worked solution

Approach

The block is on the point of moving, so it is in limiting equilibrium and friction is at its maximum value $F = \mu R$ . The direction of friction depends on the direction of impending motion. We set up two equations:

When $X = 10$ and the block is on the point of moving down, friction acts up the plane.
When the block is on the point of moving up, friction acts down the plane, and we solve for the new $X$ .

Working

From part (a), we have:

\sin \alpha = \frac{1}{\sqrt{5}}, \quad \cos \alpha = \frac{2}{\sqrt{5}}, \quad R = 48\sqrt{5} \text{ N}

Case 1: Block on the point of moving down the plane ( $X = 10$ ).

Friction acts up the plane (opposing impending downward motion). Resolving forces parallel to the plane (taking up the plane as positive):

X + F - 12g \sin \alpha = 0

Substituting known values:

10 + F - 12 \times 10 \times \frac{1}{\sqrt{5}} = 0

10 + F - \frac{120}{\sqrt{5}} = 0

10 + F - 24\sqrt{5} = 0

Solving for $F$ :

F = 24\sqrt{5} - 10 = 43.665 \text{ N}

Since the block is in limiting equilibrium, $F = \mu R$ :

\mu = \frac{F}{R} = \frac{24\sqrt{5} - 10}{48\sqrt{5}}

Simplifying:

\mu = \frac{24\sqrt{5}}{48\sqrt{5}} - \frac{10}{48\sqrt{5}} = \frac{1}{2} - \frac{10}{48\sqrt{5}} = \frac{1}{2} - \frac{5}{24\sqrt{5}}

Rationalizing:

\mu = \frac{1}{2} - \frac{5\sqrt{5}}{24 \times 5} = \frac{1}{2} - \frac{\sqrt{5}}{24} = \frac{12 - \sqrt{5}}{24}

Numerically:

\mu = \frac{12 - 2.2361}{24} = \frac{9.7639}{24} \approx 0.407

Case 2: Block on the point of moving up the plane.

Now friction acts down the plane (opposing impending upward motion). Resolving forces parallel to the plane (taking up the plane as positive):

X - F - 12g \sin \alpha = 0

The friction is still at its limiting value $F = \mu R = 43.665 \text{ N}$ (since $\mu$ and $R$ are unchanged):

X - 43.665 - 24\sqrt{5} = 0

X = 43.665 + 24\sqrt{5} = 43.665 + 53.666 = 97.331 \text{ N}

Exact form:

X = (24\sqrt{5} - 10) + 24\sqrt{5} = 48\sqrt{5} - 10

Numerically:

X = 48 \times 2.2361 - 10 = 107.331 - 10 = 97.331 \approx 97.3 \text{ N}

Answer

\mu = 0.407, \quad X = 97.3 \text{ N}

Final answer

μ = 0.407, X = 97.3 N

Detailed explanation

Walkthrough

Understanding the two limiting cases.

When a block is on a rough inclined plane and a force is applied, there are two critical scenarios:

Impending motion down the plane: The applied force is too small to hold the block, so it is about to slide down. Friction opposes this by acting up the plane.
Impending motion up the plane: The applied force is large enough to push the block up, so it is about to slide up. Friction opposes this by acting down the plane.

In both cases, the block is in limiting equilibrium: acceleration is zero, and friction is at its maximum value $F = \mu R$ .

Case 1: Finding $\mu$ when $X = 10$ and motion is impending down.

Since the block is about to move down, friction acts up the plane. Taking forces parallel to the plane with the upward direction as positive:

Forces up the plane: applied force $X = 10 \text{ N}$ and friction $F$
Forces down the plane: weight component $12g \sin \alpha = 24\sqrt{5} \text{ N}$

Equilibrium gives:

X + F = 12g \sin \alpha

10 + F = 24\sqrt{5}

F = 24\sqrt{5} - 10 \approx 43.665 \text{ N}

Since $F = \mu R$ and $R = 48\sqrt{5} \text{ N}$ :

\mu = \frac{F}{R} = \frac{24\sqrt{5} - 10}{48\sqrt{5}} = \frac{12 - \sqrt{5}}{24} \approx 0.407

Case 2: Finding $X$ when motion is impending up.

Now the block is about to move up, so friction reverses direction and acts down the plane. Taking upward as positive:

Forces up the plane: applied force $X$
Forces down the plane: weight component $12g \sin \alpha = 24\sqrt{5} \text{ N}$ and friction $F = 43.665 \text{ N}$

Equilibrium gives:

X = 12g \sin \alpha + F = 24\sqrt{5} + (24\sqrt{5} - 10) = 48\sqrt{5} - 10

X \approx 107.331 - 10 = 97.331 \approx 97.3 \text{ N}

Key Takeaways

In limiting equilibrium, friction is at its maximum: $F = \mu R$ .
The direction of friction reverses between the two limiting cases (impending down vs. impending up).
When on the point of moving down, friction acts up the plane; when on the point of moving up, friction acts down the plane.
The normal reaction $R$ and the coefficient of friction $\mu$ remain the same in both cases; only the applied force $X$ and the direction of friction change.
Setting up equilibrium equations ( $\sum F = 0$ ) is the correct approach when the block is on the point of moving.

Common Mistakes

Forgetting to reverse the direction of friction between the two cases. This is the most common error and will lead to incorrect equations.
Using Newton's second law ( $\sum F = ma$ ) instead of equilibrium ( $\sum F = 0$ ). Since the block is on the point of moving (not actually moving), acceleration is zero.
Calculating $R$ incorrectly. Remember $R = mg \cos \alpha$ , not $mg$ .
Not recognizing that $\mu$ is the same in both cases. The coefficient of friction is a property of the surfaces and does not change.
Sign errors when setting up the equilibrium equations. Clearly define a positive direction and stick to it.

Things to Be Careful About

The problem states $\mu \neq 0.15$ , which is a hint that the coefficient of friction is different from part (a). You must find the new value of $\mu$ from the first condition before using it in the second condition.
When resolving forces parallel to the plane, be consistent with your sign convention. Define one direction as positive and apply it throughout.
The exact answer for $\mu$ is $\frac{12 - \sqrt{5}}{24}$ , which simplifies to approximately $0.407$ . The exact answer for $X$ is $48\sqrt{5} - 10$ , which is approximately $97.3 \text{ N}$ .
The frictional force $F$ is the same in magnitude in both cases ( $F = \mu R$ ), but its direction changes. This is a key conceptual point.

Techniques used

set up equilibrium equation for impending motion down the planesolve for coefficient of frictionset up equilibrium equation for impending motion up the planesolve for required applied force

Q7MediumKinematics of Motion in a Straight Line

A particle moves in a straight line. The velocity $v\text{ms}^{-1}$ of the particle $t\text{s}$ after leaving a fixed point $O$ is given by $v = k(20 + pt - 6t^2)$ , where $k$ and $p$ are constants. The acceleration of the particle at $t = 1$ is $42\text{ms}^{-2}$ , and the displacement of the particle from $O$ at $t = 1$ is $93\text{m}$ .

(a)

Show that $k = 3$ and $p = 26$ .

6M Similar

DifficultyMedium

Worked solution

Approach

Differentiate the given expression for $v$ with respect to $t$ to obtain the acceleration $a$ . Substitute $t = 1$ and $a = 42$ to form a first equation in $k$ and $p$ . Integrate $v$ with respect to $t$ to obtain the displacement $s$ ; since the particle leaves $O$ at $t = 0$ , the integration constant is zero. Substitute $t = 1$ and $s = 93$ to form a second equation. Solve the two equations simultaneously for $k$ and $p$ .

Working

v = k(20 + pt - 6t^2)

Differentiating:

a = \frac{dv}{dt} = k(p - 12t)

At $t = 1$ , $a = 42$ :

42 = k(p - 12) \tag{1}

Integrating:

s = \int v\,dt = k\left(20t + \frac{p}{2}t^2 - 2t^3\right) + c

Since $s = 0$ when $t = 0$ , $c = 0$ . At $t = 1$ , $s = 93$ :

93 = k\left(20 + \frac{p}{2} - 2\right) = k\left(18 + \frac{p}{2}\right) \tag{2}

From (1):

k = \frac{42}{p - 12}

Substituting into (2):

93 = \frac{42\left(18 + \frac{p}{2}\right)}{p - 12}

93(p - 12) = 42\left(18 + \frac{p}{2}\right)

93p - 1116 = 756 + 21p

72p = 1872

p = 26

Then:

k = \frac{42}{26 - 12} = \frac{42}{14} = 3

Answer

k = 3, \quad p = 26

Final answer

k = 3, p = 26

Detailed explanation

Walkthrough

We are given the velocity $v = k(20 + pt - 6t^2)$ with two unknown constants $k$ and $p$ , and two pieces of information at $t = 1$ : $a = 42$ and $s = 93$ . Since $a = \frac{dv}{dt}$ , we differentiate term by term: the constant $20$ gives $0$ , $pt$ gives $p$ , and $-6t^2$ gives $-12t$ , so $a = k(p - 12t)$ . Substituting $t = 1$ and $a = 42$ gives equation (1): $42 = k(p - 12)$ .

Next, since $s = \int v\,dt$ , we integrate term by term: $\int 20\,dt = 20t$ , $\int pt\,dt = \frac{p}{2}t^2$ , and $\int -6t^2\,dt = -2t^3$ , giving $s = k\left(20t + \frac{p}{2}t^2 - 2t^3\right) + c$ . The particle leaves $O$ at $t = 0$ , so $s(0) = 0$ and therefore $c = 0$ . Using $s = 93$ at $t = 1$ gives equation (2): $93 = k\left(20 + \frac{p}{2} - 2\right) = k\left(18 + \frac{p}{2}\right)$ .

Now solve the two equations. From (1), $k = \frac{42}{p - 12}$ . Substituting into (2) and multiplying by $p - 12$ eliminates $k$ and gives a linear equation in $p$ : $93p - 1116 = 756 + 21p$ . Collecting terms, $72p = 1872$ , so $p = 26$ . Finally, $k = \frac{42}{26 - 12} = 3$ , confirming both values.

Key Takeaways

Differentiation and integration connect $v$ , $a$ and $s$ : $a = \frac{dv}{dt}$ and $s = \int v\,dt$ .
The constant of integration is fixed by the initial condition that the particle is at $O$ when $t = 0$ .
Two unknown constants need two equations; substitute each condition to build the system.
Solving the simultaneous system requires careful algebraic manipulation, especially clearing a fraction.

Common Mistakes

Using $a = v/t$ or $s = vt$ instead of differentiating or integrating — the mark scheme awards these no method marks.
Dropping the constant of integration, or failing to set $c = 0$ from the initial condition.
Sign errors when expanding $93(p - 12)$ or $42\left(18 + \frac{p}{2}\right)$ .
Failing to show at least one line of working after eliminating $k$ or $p$ ; the simultaneous-equation mark requires visible working.

Things to Be Careful About

The method marks require the power of $t$ to change by exactly 1 with a coefficient change in at least one term.
Because the question asks to "Show that $k = 3$ and $p = 26$ ", the values are given; any error in the working makes the final answer wrong (A0).
Make sure both equations are correctly formed before solving — a wrong equation can still produce a consistent-looking algebra path but gives wrong constants.

Techniques used

differentiate velocity to obtain accelerationintegrate velocity to obtain displacementapply initial conditions to fix the constant of integrationform and solve simultaneous equations

(b)

Find the distance moved by the particle between the time at which its acceleration is zero and the time at which its velocity is zero.

5M Similar

DifficultyMedium

Worked solution

Approach

Find the time at which the acceleration is zero by setting $a = 3(26 - 12t) = 0$ and solving for $t$ . Find the time at which the velocity is zero by setting $v = 3(20 + 26t - 6t^2) = 0$ , solving the quadratic, and keeping the positive root. Since the velocity remains positive throughout the interval, the distance travelled equals the change in displacement. Integrate $v$ to obtain $s$ , and evaluate $s$ at the two times, subtracting the smaller from the larger.

Working

Setting $a = 0$ :

3(26 - 12t) = 0 \implies t = \frac{26}{12} = \frac{13}{6}

Setting $v = 0$ :

3(20 + 26t - 6t^2) = 0

6t^2 - 26t - 20 = 0

3t^2 - 13t - 10 = 0

(3t + 2)(t - 5) = 0

t = 5 \quad \text{or} \quad t = -\frac{2}{3}

Since $t \geq 0$ , only $t = 5$ is valid. For $\frac{13}{6} \leq t \leq 5$ the velocity is positive, so distance equals change in displacement.

s = \int v\,dt = 3\left(20t + 13t^2 - 2t^3\right)

At $t = 5$ :

s(5) = 3(100 + 325 - 250) = 525

At $t = \frac{13}{6}$ :

s\left(\frac{13}{6}\right) = 3\left(20 \cdot \frac{13}{6} + 13 \cdot \frac{169}{36} - 2 \cdot \frac{2197}{216}\right)

= 3\left(\frac{130}{3} + \frac{2197}{36} - \frac{2197}{108}\right) = 3\left(\frac{4537}{54}\right) = \frac{4537}{18}

Distance travelled:

525 - \frac{4537}{18} = \frac{9450}{18} - \frac{4537}{18} = \frac{4913}{18} \approx 273\,\text{m}

Answer

\frac{4913}{18}\,\text{m} \approx 273\,\text{m}

Final answer

4913/18 m (≈273 m)

Detailed explanation

Walkthrough

From part (a), $k = 3$ and $p = 26$ , so $v = 3(20 + 26t - 6t^2)$ and $a = 3(26 - 12t)$ .

Step 1 — when is the acceleration zero? Set $a = 0$ : $3(26 - 12t) = 0$ , so $t = \frac{26}{12} = \frac{13}{6}$ s. This is the start of the interval.

Step 2 — when is the velocity zero? Set $v = 0$ : $3(20 + 26t - 6t^2) = 0$ . Divide by 3, multiply by $-1$ and reorder to $6t^2 - 26t - 20 = 0$ , then divide by 2: $3t^2 - 13t - 10 = 0$ . Factor as $(3t + 2)(t - 5) = 0$ to get $t = 5$ or $t = -\frac{2}{3}$ . Since time cannot be negative here, we keep $t = 5$ .

Step 3 — check the direction of motion. The velocity $v = 3(20 + 26t - 6t^2)$ is a downward-opening quadratic whose roots are $-\frac{2}{3}$ and $5$ ; it is therefore positive for $t$ between those roots. The interval $\left[\frac{13}{6}, 5\right]$ lies inside, so the particle never reverses; distance equals the change in displacement.

Step 4 — compute displacement. $s = \int v\,dt = 3\left(20t + 13t^2 - 2t^3\right)$ . At $t = 5$ , $s(5) = 525$ . At $t = \frac{13}{6}$ , $s\left(\frac{13}{6}\right) = \frac{4537}{18} \approx 252.06$ .

Step 5 — distance $= s(5) - s\left(\frac{13}{6}\right) = 525 - \frac{4537}{18} = \frac{9450}{18} - \frac{4537}{18} = \frac{4913}{18} \approx 273$ m.

Key Takeaways

Setting $a = 0$ and $v = 0$ locates the instants that delimit the interval.
Quadratic equations for $t$ often have a negative root to reject on physical grounds.
When $v$ keeps one sign, distance travelled equals the change in displacement.
Definite integration of $v$ gives displacement; subtracting the end values gives the distance.

Common Mistakes

Forgetting to divide out the factor 3 when solving $20 + 26t - 6t^2 = 0$ , or mishandling the signs when rearranging to $6t^2 - 26t - 20 = 0$ .
Keeping both quadratic roots and integrating from the negative one, which has no physical meaning.
Assuming the particle changes direction without checking the sign of $v$ ; here it stays positive so distance equals displacement.
Dropping the factor 3 in the antiderivative, giving answers off by a factor of 3.
Using the limits in the wrong order, which would produce a negative distance.

Things to Be Careful About

The method mark for $a = 0$ requires a positive $t$ ; the method mark for $v = 0$ requires at least one $t$ value from solving the quadratic.
The distance evaluation must use the student's own integrated $s$ with positive limits; a bare numerical answer loses the final method mark.
The exact distance is $\frac{4913}{18}$ m; writing $273$ m is a rounded approximation and is allowed.
Double check the fractional arithmetic at $t = \frac{13}{6}$ — a single slip in the common denominator can change the answer.

Techniques used

solve for time when acceleration is zerosolve a quadratic for time when velocity is zerointegrate velocity to obtain displacementevaluate the difference of displacements to find distance

Mathematics 9709/42 — February/March 2025

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