9709/42 Oct/Nov 2024 — Mathematics Solved Past Paper

Q1MediumKinematics of Motion in a Straight Line

Fig. 1

The velocity of a particle moving in a straight line at time $t$ seconds after leaving a fixed point $O$ is $v \text{ m s}^{-1}$ . The diagram shows a velocity-time graph which models the motion of the particle from $t = 0$ to $t = T$ . The graph consists of four straight line segments. The particle accelerates from rest to a speed of $V \text{ m s}^{-1}$ over a period of $4 \text{ s}$ , and then decelerates at $\frac{5}{3} \text{ m s}^{-2}$ to instantaneous rest over a period of $6 \text{ s}$ . The particle then travels back towards $O$ , reaching a maximum speed of $3 \text{ m s}^{-1}$ before coming to rest at time $t = T$ .

(a)

Find the value of $V$ .

2M Similar

DifficultyMedium-Easy

Worked solution

Approach

The particle decelerates from speed $V$ to rest over a period of 6 seconds (from $t = 4$ to $t = 10$ ) with a deceleration of $\frac{5}{3} \text{ m s}^{-2}$ . We can use the constant acceleration formula $v = u + at$ or the fact that acceleration is the gradient of the velocity-time graph.

Working

The time interval for deceleration is $10 - 4 = 6$ seconds.
Initial velocity $u = V$ , final velocity $v = 0$ , acceleration $a = -\frac{5}{3}$ .
Using $v = u + at$ :

0 = V - \frac{5}{3} \times 6

0 = V - 10

V = 10

Alternatively, using the gradient of the line from $(4, V)$ to $(10, 0)$ :

\frac{0 - V}{10 - 4} = -\frac{5}{3}

\frac{-V}{6} = -\frac{5}{3}

V = 10

Answer

V = 10

Final answer

10

Detailed explanation

Walkthrough

The problem gives a velocity-time graph with a deceleration phase from $t = 4$ to $t = 10$ . During this phase, the particle goes from speed $V$ to $0 \text{ m s}^{-1}$ in 6 seconds, with a known deceleration of $\frac{5}{3} \text{ m s}^{-2}$ . We apply the constant acceleration formula $v = u + at$ , where $v = 0$ , $u = V$ , $a = -\frac{5}{3}$ , and $t = 6$ . Solving this simple linear equation gives $V = 10$ . Alternatively, the gradient of the velocity-time graph represents acceleration, so $\frac{0 - V}{10 - 4} = -\frac{5}{3}$ , which also yields $V = 10$ .

Key Takeaways

The gradient of a velocity-time graph gives the acceleration.
The suvat equations can be applied to any phase of motion with constant acceleration.

Common Mistakes

Forgetting that deceleration means negative acceleration, leading to $0 = V + \frac{5}{3} \times 6$ and giving $V = -10$ .
Using the wrong time interval, e.g., using $t = 10$ instead of $t = 10 - 4 = 6$ .

Things to Be Careful About

Ensure the time elapsed is calculated correctly as $\Delta t = 10 - 4 = 6$ s, not $t = 10$ s.
The sign of acceleration must be negative since it is deceleration.

Techniques used

use velocity-time graph gradient for accelerationapply suvat equation v = u + at

(b)

Given that the total distance travelled by the particle from $t = 0$ to $t = T$ is $68 \text{ m}$ , find the value of $T$ .

3M Similar

DifficultyMedium

Worked solution

Approach

The total distance travelled is the total area under the velocity-time graph (treating all areas as positive, since distance is the integral of speed $|v|$ ). We calculate the area for the first 10 seconds, then add the area for the remaining time from $t = 10$ to $t = T$ , and set the sum equal to 68 m.

Working

Distance in the first 10 seconds:
The graph forms a triangle with base 10 and height $V = 10$ .

\text{Distance}_1 = \frac{1}{2} \times 10 \times 10 = 50 \text{ m}

Distance from $t = 10$ to $t = T$ :
The graph forms a triangle below the $t$ -axis with base $(T - 10)$ and height $|-3| = 3$ .

\text{Distance}_2 = \frac{1}{2} \times (T - 10) \times 3

Total distance:

\text{Distance}_1 + \text{Distance}_2 = 68

50 + \frac{3}{2}(T - 10) = 68

\frac{3}{2}(T - 10) = 18

T - 10 = 12

T = 22

Answer

T = 22

Final answer

22

Detailed explanation

Walkthrough

The total distance travelled is the total area under the velocity-time graph, treating all areas as positive because distance is the integral of speed (the magnitude of velocity). The graph from $t = 0$ to $t = 10$ is a triangle above the axis with base 10 and height $V = 10$ , giving an area of $\frac{1}{2} \times 10 \times 10 = 50$ m. From $t = 10$ to $t = T$ , the graph is a triangle below the axis with base $(T - 10)$ and height 3 (since the maximum speed in this direction is $3 \text{ m s}^{-1}$ ). The area of this triangle is $\frac{1}{2} \times (T - 10) \times 3$ . Adding these two areas and setting the sum to 68 m gives the equation $50 + \frac{3}{2}(T - 10) = 68$ . Solving this linear equation yields $T = 22$ .

Key Takeaways

Distance is the total area under a velocity-time graph, regardless of whether the velocity is positive or negative.
Areas below the time axis contribute positively to distance but negatively to displacement.

Common Mistakes

Treating the area below the $t$ -axis as negative and subtracting it from the first area, which would calculate displacement instead of distance.
Assuming the triangle from $t = 10$ to $t = T$ is isosceles without justification, which may not match the given maximum speed of $3 \text{ m s}^{-1}$ .
Assuming the particle returns to $O$ , which would mean displacement is zero, leading to an incorrect equation.

Things to Be Careful About

Always use the magnitude of velocity (speed) when calculating distance from a velocity-time graph.
Ensure the base of the second triangle is correctly identified as $(T - 10)$ rather than $T$ .

Techniques used

calculate area under velocity-time graph for distanceequate total area to given distancesolve linear equation for time

Q24MMediumEnergy, Work and Power

A block of mass $20 \text{ kg}$ is held at rest at the top of a plane inclined at $30^\circ$ to the horizontal. The block is projected with speed $5 \text{ m s}^{-1}$ down a line of greatest slope of the plane. There is a resistance force acting on the block. As the block moves $2 \text{ m}$ down the plane from its point of projection, the work done against this resistance force is $50 \text{ J}$ .

Find the speed of the block when it has moved $2 \text{ m}$ down the plane.

DifficultyMedium

Worked solution

Approach

Use the work-energy principle. The block loses gravitational potential energy as it moves down the plane, gains kinetic energy, and the resistance removes $50\,\text{J}$ of mechanical energy. Equate the increase in kinetic energy to the loss in gravitational potential energy minus the work done against the resistance. Take $g = 10\,\text{m s}^{-2}$ .

Working

Vertical drop after moving $2\,\text{m}$ down the plane is

h = 2\sin 30^\circ = 1\,\text{m}.

Loss in gravitational potential energy:

20g\sin 30^\circ \times 2 = 20(10)(0.5)(2) = 200\,\text{J}.

Initial kinetic energy:

\frac{1}{2}(20)(5^2) = 250\,\text{J}.

Let the final speed be $v$ . Final kinetic energy:

\frac{1}{2}(20)v^2 = 10v^2.

Work-energy equation:

\frac{1}{2}(20)v^2 - \frac{1}{2}(20)(5^2) = 20g\sin 30^\circ \times 2 - 50.

10v^2 - 250 = 200 - 50 = 150.

Solve for $v$ :

10v^2 = 400, \quad v^2 = 40.

Since speed is positive,

v = \sqrt{40} = 2\sqrt{10} \approx 6.32\,\text{m s}^{-1}.

Answer

6.32\,\text{m s}^{-1} \quad (\sqrt{40}\,\text{m s}^{-1}).

Final answer

6.32 m s^-1 (or sqrt(40) m s^-1)

Detailed explanation

Walkthrough

The block moves along the plane, but gravitational potential energy depends on the vertical height lost. Moving $2\,\text{m}$ down a plane inclined at $30^\circ$ gives a vertical drop

2\sin 30^\circ = 1\,\text{m}.

The potential energy lost by the block is its weight times this vertical drop, so

20(10)(1)=200\,\text{J}.

At the start the kinetic energy is

\frac12(20)(5^2)=250\,\text{J}.

At the end, with unknown speed $v$ , it is

\frac12(20)v^2=10v^2.

The work-energy principle states that the increase in kinetic energy equals the net work done on the block. Gravity does work equal to the loss in potential energy, $200\,\text{J}$ , while the resistance does $-50\,\text{J}$ of work. Therefore

10v^2-250=200-50.

This gives $10v^2=400$ , so $v^2=40$ and, because speed is positive,

v=\sqrt{40}=2\sqrt{10}\approx6.32\,\text{m s}^{-1}.

Key Takeaways

The work-energy principle connects changes in kinetic energy to work done by forces and changes in gravitational potential energy.
Gravitational potential energy depends only on vertical displacement, not distance along an incline.
When work is done against a resistance, that energy is removed from the mechanical energy of the system.
The final speed is found from an energy balance, avoiding the need to know the actual resistance force.

Common Mistakes

Using the whole distance $2\,\text{m}$ as the vertical height instead of $2\sin 30^\circ$ .
Writing the change in kinetic energy as $\frac12 m(v-5)^2$ ; the correct change is $\frac12 mv^2 - \frac12 m(5)^2$ .
Adding the $50\,\text{J}$ of work done against resistance instead of subtracting it.
Omitting one of the four energy terms in the equation.
Using the initial speed as $0$ or giving the final answer as a negative speed.

Things to Be Careful About

Adopt one consistent sign convention. Here the block descends, so potential energy is lost and contributes positively to the kinetic energy gain.
Use $g=10\,\text{m s}^{-2}$ as is standard in this syllabus.
The mark scheme requires the potential energy term to include the component of displacement vertically, so $mgh$ must use the vertical drop.
The answer may be given as $\sqrt{40}\,\text{m s}^{-1}$ , $2\sqrt{10}\,\text{m s}^{-1}$ , or $6.32\,\text{m s}^{-1}$ ; it should not be left as $v^2=40$ .

Techniques used

calculate vertical displacement from the incline anglecompute changes in gravitational potential energy and kinetic energyapply the work-energy principle including work done against resistancesolve for final speed from the energy balance

Approach

Use $P = Fv$ to find the driving force from the power and speed, then apply Newton's second law along the road to relate driving force, resistance and acceleration.

Working

The driving (pedalling) force is

F = \frac{P}{v} = \frac{250}{5} = 50\,\text{N}

Taking the direction of motion as positive, the resultant force is $F - R$ , where $R$ is the resistance. Newton's second law gives:

50 - R = 90(0.1) = 9

So

R = 50 - 9 = 41\,\text{N}

Answer

R = 41\,\text{N}

Final answer

41 N

Detailed explanation

Walkthrough

Start by finding the driving force. Since power is the rate of doing work and, for a constant force moving in the direction of motion, $P = Fv$ , the driving force is $F = \frac{250}{5} = 50\,\text{N}$ .

Next, apply Newton's second law. The cyclist is accelerating at $0.1\,\text{m s}^{-2}$ , so the resultant force on the cyclist must be $ma = 90 \times 0.1 = 9\,\text{N}$ . The driving force pushes forward and the resistance $R$ acts backward, so $50 - R = 9$ . Solving gives $R = 41\,\text{N}$ .

Key Takeaways

This part combines the power relation $P = Fv$ with Newton's second law. It shows that when an object accelerates, the resultant force is not zero; the driving force must overcome the resistance and still provide the acceleration.

Common Mistakes

Using $P = \frac{F}{v}$ instead of $P = Fv$ .
Forgetting the acceleration term and writing $50 = R$ .
Using the mass as the force directly, e.g. writing $90$ instead of $90 \times 0.1$ .
Sign errors: putting both forces on the same side without a consistent positive direction.

Things to Be Careful About

Choose a positive direction (forward) and keep it consistent. The resistance opposes motion, so it must be subtracted from the driving force. The mark scheme allows sign errors in the equation, but the final resistance must be positive. Also, the driving force found from $P = Fv$ must be used, not the power value $250$ itself.

Techniques used

use P = Fv to find the driving forceapply Newton's second law to the horizontal forcessolve the resulting linear equation for the resistance

(b)

The cyclist comes to the bottom of a hill inclined at $2^\circ$ to the horizontal.

Given that the power and resistance to motion are unchanged, find the steady speed which the cyclist could maintain when riding up the hill.

2M Similar

DifficultyMedium

Worked solution

Approach

At steady speed up the hill the acceleration is zero, so the resultant force along the slope is zero. Express the driving force as $\frac{250}{v}$ , add the resistance and the component of weight down the slope, then solve for $v$ .

Working

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

90g\sin 2^\circ = 900\sin 2^\circ \approx 31.4\,\text{N}

The driving force up the slope is

\frac{250}{v}

For steady speed, the resultant force along the slope is zero:

\frac{250}{v} - 41 - 900\sin 2^\circ = 0

So

v = \frac{250}{41 + 900\sin 2^\circ} \approx \frac{250}{72.4} \approx 3.45\,\text{m s}^{-1}

Answer

v = 3.45\,\text{m s}^{-1}

Final answer

3.45 m s^-1

Detailed explanation

Walkthrough

At steady speed up the hill, acceleration is zero, so the resultant force along the slope is zero. The power is unchanged, so the driving force up the slope is $\frac{250}{v}$ .

Down the slope there are two forces: the constant resistance $41\,\text{N}$ and the component of the weight acting down the slope, $mg\sin 2^\circ = 90g\sin 2^\circ$ . Using $g = 10\,\text{m s}^{-2}$ , this component is $90 \times 10 \times \sin 2^\circ \approx 31.4\,\text{N}$ .

For steady speed, the driving force must balance the total force down the slope:

\frac{250}{v} = 41 + 90g\sin 2^\circ

Therefore

v = \frac{250}{41 + 900\sin 2^\circ} \approx 3.45\,\text{m s}^{-1}

Key Takeaways

Steady speed means zero acceleration, hence zero resultant force. On an incline, only the component of weight parallel to the slope matters, $mg\sin\theta$ . The power relation $P = Fv$ links the driving force to the speed.

Common Mistakes

Using the full weight $mg$ instead of its component $mg\sin\theta$ .
Using $mg\cos\theta$ instead of $mg\sin\theta$ .
Forgetting to include the resistance $41\,\text{N}$ .
Setting acceleration equal to something other than zero; steady speed means $a=0$ .
Using $g = 9.8$ instead of $g = 10$ ; the mark scheme expects $3.45\,\text{m s}^{-1}$ .

Things to Be Careful About

The mark scheme requires a component of weight, not the mass. It allows a sin/cos mix in the equation but the correct component is $mg\sin\theta$ . Use the value of $R$ from part (a). Keep the final answer to 3 significant figures: $3.45\,\text{m s}^{-1}$ .

Techniques used

use P = Fv to express the driving force in terms of vresolve the component of weight along the slopeset the resultant force along the slope to zero for steady speedsolve for the steady speed

Q4MediumForces and Equilibrium

Fig. 2

The diagram shows two particles, $A$ and $B$ , of masses $0.2 \text{ kg}$ and $0.1 \text{ kg}$ respectively. The particles are suspended below a horizontal ceiling by two strings, $AP$ and $BQ$ , attached to fixed points $P$ and $Q$ on the ceiling. The particles are connected by a horizontal string, $AB$ . Angle $APQ = 45^\circ$ and $BQP = \theta^\circ$ . Each string is light and inextensible. The particles are in equilibrium.

(a)

Find the value of the tension in the string $AB$ .

2M Similar

DifficultyMedium-Easy

Worked solution

Approach

Focus on particle $A$ and apply the equilibrium condition. Resolve the forces acting on $A$ horizontally and vertically to eliminate the unknown tension $T_{AP}$ and solve for $T_{AB}$ .

Working

The forces acting on particle $A$ are:

Its weight $W_A = 0.2g$ acting vertically downwards.
The tension $T_{AB}$ in the horizontal string $AB$ acting to the right.
The tension $T_{AP}$ in the string $AP$ acting at $45^\circ$ above the horizontal to the left.

Resolving vertically at $A$ :

\begin{aligned} T_{AP} \sin 45^\circ &= 0.2g \end{aligned}

Resolving horizontally at $A$ :

\begin{aligned} T_{AP} \cos 45^\circ &= T_{AB} \end{aligned}

Divide the vertical equation by the horizontal equation to eliminate $T_{AP}$ :

\frac{T_{AP} \sin 45^\circ}{T_{AP} \cos 45^\circ} = \frac{0.2g}{T_{AB}}

\tan 45^\circ = \frac{0.2g}{T_{AB}}

Using $g = 10 \text{ m/s}^2$ and $\tan 45^\circ = 1$ :

1 = \frac{0.2 \times 10}{T_{AB}} = \frac{2}{T_{AB}}

T_{AB} = 2 \text{ N}

Answer

T_{AB} = 2 \text{ N}

Final answer

2 N

Detailed explanation

Walkthrough

First, isolate particle $A$ and identify all forces acting on it. There are three forces: the weight $0.2g$ pulling straight down, the tension $T_{AB}$ pulling horizontally to the right along string $AB$ , and the tension $T_{AP}$ pulling up and to the left along string $AP$ at an angle of $45^\circ$ to the ceiling.

Because the particle is in equilibrium, the net force in both the horizontal and vertical directions must be zero. We resolve the forces vertically to get an equation involving $T_{AP}$ and the weight, and horizontally to get an equation involving $T_{AP}$ and $T_{AB}$ . By dividing the vertical equation by the horizontal equation, the unknown $T_{AP}$ cancels out, leaving a direct relationship between $T_{AB}$ and the known weight. Substituting $g = 10 \text{ m/s}^2$ and $\tan 45^\circ = 1$ gives $T_{AB} = 2 \text{ N}$ .

Key Takeaways

When a particle is in equilibrium, the sum of forces in any direction is zero. Resolving horizontally and vertically is the standard approach.
Dividing two resolved equations is a powerful technique to eliminate an unknown tension or normal reaction.
Always check which direction each force acts before writing the equations; signs matter.

Common Mistakes

Using the wrong trigonometric ratio (e.g., using $\sin 45^\circ$ for the horizontal component instead of $\cos 45^\circ$ ). Remember that the angle $45^\circ$ is with the horizontal ceiling, so the horizontal component of $T_{AP}$ uses $\cos$ and the vertical uses $\sin$ .
Forgetting that $g = 10 \text{ m/s}^2$ is used in this syllabus unless otherwise stated, leading to $T_{AB} = 1.96 \text{ N}$ instead of $2 \text{ N}$ .
Assuming $T_{AB} = T_{AP}$ ; the tensions in different strings are generally not equal.

Things to Be Careful About

The angle $45^\circ$ is given between the string $AP$ and the horizontal ceiling $PQ$ . This means it is also the angle between $T_{AP}$ and the horizontal, making $\cos 45^\circ$ the horizontal component and $\sin 45^\circ$ the vertical component.
The mark scheme accepts condoning $T_{AB} = 0.2g$ directly from $\tan 45^\circ = 1$ , but showing the resolution steps is safer to earn method marks.

Techniques used

isolate particle A as a free bodyresolve forces horizontally and verticallyapply equilibrium conditionuse trigonometric ratios to eliminate unknown tension

(b)

Find the value of $\theta$ and the tension in the string $BQ$ .

4M Similar

DifficultyMedium

Worked solution

Approach

Focus on particle $B$ and apply the equilibrium condition. Resolve the forces acting on $B$ horizontally and vertically to find $\theta$ and $T_{BQ}$ .

Working

The forces acting on particle $B$ are:

Its weight $W_B = 0.1g$ acting vertically downwards.
The tension $T_{AB} = 2 \text{ N}$ in the horizontal string $AB$ acting to the left.
The tension $T_{BQ}$ in the string $BQ$ acting at angle $\theta$ above the horizontal to the right.

Resolving vertically at $B$ :

\begin{aligned} T_{BQ} \sin \theta &= 0.1g \end{aligned}

Resolving horizontally at $B$ :

\begin{aligned} T_{BQ} \cos \theta &= T_{AB} = 2 \end{aligned}

Using $g = 10 \text{ m/s}^2$ , the vertical equation becomes:

T_{BQ} \sin \theta = 0.1 \times 10 = 1

Divide the vertical equation by the horizontal equation to eliminate $T_{BQ}$ :

\frac{T_{BQ} \sin \theta}{T_{BQ} \cos \theta} = \frac{1}{2}

\tan \theta = 0.5

\theta = \tan^{-1}(0.5) = 26.565...^\circ \approx 26.6^\circ

To find $T_{BQ}$ , square both resolved equations and add them:

(T_{BQ} \sin \theta)^2 + (T_{BQ} \cos \theta)^2 = 1^2 + 2^2

T_{BQ}^2 (\sin^2 \theta + \cos^2 \theta) = 1 + 4

T_{BQ}^2 = 5

T_{BQ} = \sqrt{5} \approx 2.24 \text{ N}

Answer

\theta = 26.6^\circ, \quad T_{BQ} = \sqrt{5} \text{ N} \approx 2.24 \text{ N}

Final answer

θ = 26.6°, T_BQ = √5 N (or 2.24 N)

Detailed explanation

Walkthrough

Isolate particle $B$ and identify the three forces acting on it: its weight $0.1g$ downwards, the known tension $T_{AB} = 2 \text{ N}$ pulling to the left, and the unknown tension $T_{BQ}$ pulling up and to the right at angle $\theta$ to the horizontal.

Apply the equilibrium condition by resolving vertically and horizontally. Vertically, the upward component of $T_{BQ}$ must balance the weight: $T_{BQ} \sin \theta = 0.1g = 1$ . Horizontally, the rightward component of $T_{BQ}$ must balance $T_{AB}$ : $T_{BQ} \cos \theta = 2$ .

Dividing the vertical equation by the horizontal gives $\tan \theta = 1/2 = 0.5$ , from which $\theta = 26.6^\circ$ . To find $T_{BQ}$ , square both equations and add them, using the identity $\sin^2 \theta + \cos^2 \theta = 1$ , yielding $T_{BQ}^2 = 1^2 + 2^2 = 5$ , so $T_{BQ} = \sqrt{5} \approx 2.24 \text{ N}$ .

Key Takeaways

When two unknowns remain after resolving forces, divide the equations to find an angle, or square and add them to find the resultant magnitude.
The tension in the connecting string $AB$ is the same on both sides; it pulls particle $B$ to the left with $2 \text{ N}$ .
Pythagorean theorem applies directly to the resolved components of a single tension force.

Common Mistakes

Assuming $T_{AB} = T_{BQ}$ ; the tensions are different because the angles and weights are different.
Using the wrong angle for the components of $T_{BQ}$ . The angle $\theta$ is with the horizontal, so the horizontal component is $T_{BQ} \cos \theta$ and the vertical is $T_{BQ} \sin \theta$ .
Forgetting to use $g = 10 \text{ m/s}^2$ , which gives $0.1g = 1$ instead of $0.98$ .
Not providing both values for $\theta$ and $T_{BQ}$ as required by the question.

Things to Be Careful About

The mark scheme allows alternative methods such as resolving parallel and perpendicular to $BQ$ , using a triangle of forces, or applying Lami's theorem. All must yield the same final values.
When using Lami's theorem at $B$ , the angles between the forces are $90^\circ$ (between weight and $T_{AB}$ ), $90^\circ + \theta$ (between $T_{AB}$ and $T_{BQ}$ ), and $180^\circ - \theta$ (between $T_{BQ}$ and weight).
Always carry forward the correct value of $T_{AB} = 2 \text{ N}$ from part (a); using an incorrect value will cascade errors.

Techniques used

isolate particle B as a free bodyresolve forces horizontally and verticallyapply equilibrium conditionuse Pythagorean theorem and inverse tangent to find unknowns

Q5Medium-HardKinematics of Motion in a Straight LineMomentum

Two particles, $P$ and $Q$ , of masses $2m \text{ kg}$ and $m \text{ kg}$ respectively, are held at rest in the same vertical line. The heights of $P$ and $Q$ above horizontal ground are $1 \text{ m}$ and $2 \text{ m}$ respectively. $P$ is projected vertically upwards with speed $2 \text{ m s}^{-1}$ . At the same instant, $Q$ is released from rest.

(a)

Find the speed of each particle immediately before they collide.

4M Similar

DifficultyMedium

Worked solution

Approach

Take upward as positive. Write the displacement of each particle from its own starting height using $s = ut + \frac{1}{2}at^2$ . At the collision the particles are at the same height, so equate their heights. Then use $v = u + at$ to find the velocities just before impact.

Working

Let upward be positive and take $g = 10 \text{ m s}^{-2}$ .

For $P$ , starting height $1 \text{ m}$ , $u = 2$ , $a = -g$ :

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

For $Q$ , starting height $2 \text{ m}$ , $u = 0$ , $a = -g$ :

s_Q = -\frac{1}{2}gt^2

At collision the heights are equal:

1 + s_P = 2 + s_Q

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

2t = 1 \Rightarrow t = \frac{1}{2} \text{ s}

Velocities just before collision:

v_P = 2 - gt = 2 - 10\left(\frac{1}{2}\right) = -3 \text{ m s}^{-1}

v_Q = 0 - gt = -10\left(\frac{1}{2}\right) = -5 \text{ m s}^{-1}

Speeds are the magnitudes of these velocities:

|v_P| = 3 \text{ m s}^{-1}, \quad |v_Q| = 5 \text{ m s}^{-1}

Answer

Speed of $P$ = $3 \text{ m s}^{-1}$ ; speed of $Q$ = $5 \text{ m s}^{-1}$ .

Final answer

Speed of P = 3 m s^-1; speed of Q = 5 m s^-1

Detailed explanation

Walkthrough

We need the speeds just before the particles collide. The most reliable way is to write down where each particle is at time $t$ and find the time when those positions are the same.

Choose upward as positive. $P$ starts $1 \text{ m}$ above the ground with initial velocity $+2 \text{ m s}^{-1}$ , so its displacement from its starting point is $s_P = 2t - \frac{1}{2}gt^2$ . $Q$ starts $2 \text{ m}$ above the ground and is dropped from rest, so its displacement from its starting point is $s_Q = -\frac{1}{2}gt^2$ (negative because it moves downward).

At the collision, the height of $P$ above the ground, $1 + s_P$ , equals the height of $Q$ above the ground, $2 + s_Q$ . Substituting gives $1 + 2t - \frac{1}{2}gt^2 = 2 - \frac{1}{2}gt^2$ . The quadratic terms cancel, leaving $2t = 1$ , so $t = \frac{1}{2} \text{ s}$ .

Now use $v = u + at$ for each particle. For $P$ , $v_P = 2 - gt = 2 - 10 \times \frac{1}{2} = -3 \text{ m s}^{-1}$ . For $Q$ , $v_Q = 0 - gt = -5 \text{ m s}^{-1}$ . The negative signs mean both are moving downward. Speeds are positive magnitudes, so the speeds are $3$ and $5 \text{ m s}^{-1}$ .

Key Takeaways

Use $s = ut + \frac{1}{2}at^2$ to describe position when acceleration is constant.
At a collision, the particles are at the same position, so equate their heights.
Velocity is a vector: a negative answer means downward. Speed is the magnitude and is always positive.

Common Mistakes

Using the wrong sign for $g$ : if upward is positive, $g$ is $-10$ in the acceleration.
Forgetting the initial heights when equating positions, e.g. writing $s_P = s_Q$ instead of $1 + s_P = 2 + s_Q$ .
Giving $-3$ and $-5$ as speeds instead of taking magnitudes.
Using $u = 0$ for $P$ ; $P$ is projected upwards with $u = 2$ .

Things to Be Careful About

The collision time is found from positions, not from velocities.
The quadratic terms in $t^2$ cancel because both particles have the same acceleration $g$ .
If the mark scheme asks for speeds, final answers must be positive.

Techniques used

use suvat displacement equation for each particleequate heights to find collision timeuse suvat velocity equation and take magnitudes

(b)

It is given that immediately after the collision the downward speed of $Q$ is $3.5 \text{ m s}^{-1}$ .

Find the speed of $P$ at the instant that it reaches the ground.

5M Similar

DifficultyMedium-Hard

Worked solution

Approach

Use conservation of linear momentum during the collision to find the velocity of $P$ immediately after impact. Then find the height of $P$ at the collision from part (a), and use $v^2 = u^2 + 2as$ to find its speed when it reaches the ground.

Working

Take downward as positive for the collision.

Before impact:

$P$ : mass $2m$ , speed $3 \text{ m s}^{-1}$ downward, momentum $6m$ .
$Q$ : mass $m$ , speed $5 \text{ m s}^{-1}$ downward, momentum $5m$ .

After impact, $Q$ has downward speed $3.5 \text{ m s}^{-1}$ , so its momentum is $3.5m$ . Let $P$ have downward velocity $v$ after impact.

Conservation of momentum:

6m + 5m = 2mv + 3.5m

11m = 2mv + 3.5m

7.5m = 2mv

v = 3.75 \text{ m s}^{-1} \text{ downward}

Height of $P$ at the collision:

s_P = 2\left(\frac{1}{2}\right) - \frac{1}{2}(10)\left(\frac{1}{2}\right)^2 = 1 - 1.25 = -0.25 \text{ m}

So $P$ is $1 - 0.25 = 0.75 \text{ m}$ above the ground.

Now $P$ falls $0.75 \text{ m}$ with initial downward speed $3.75 \text{ m s}^{-1}$ and acceleration $g = 10 \text{ m s}^{-2}$ downward:

v^2 = u^2 + 2as = 3.75^2 + 2(10)(0.75)

v^2 = 14.0625 + 15 = 29.0625 = \frac{465}{16}

v = \frac{\sqrt{465}}{4} \approx 5.39 \text{ m s}^{-1}

Answer

Speed of $P$ when it reaches the ground = $\frac{\sqrt{465}}{4} \approx 5.39 \text{ m s}^{-1}$ .

Final answer

5.39 m s^-1 (exact sqrt(465)/4)

Detailed explanation

Walkthrough

First use momentum conservation. Choose downward as positive. Before the collision both particles are moving downward, so their momenta are $2m \times 3 = 6m$ and $m \times 5 = 5m$ . After the collision $Q$ still moves downward at $3.5 \text{ m s}^{-1}$ , so its momentum is $3.5m$ . Let $P$ 's downward velocity after the collision be $v$ . Conservation of momentum gives $6m + 5m = 2mv + 3.5m$ , so $2mv = 7.5m$ and $v = 3.75 \text{ m s}^{-1}$ downward.

Next find where $P$ is when the collision happens. From part (a), at $t = \frac{1}{2}$ , $s_P = 2 \times \frac{1}{2} - \frac{1}{2}g\left(\frac{1}{2}\right)^2 = -0.25 \text{ m}$ . Since $P$ started $1 \text{ m}$ above the ground, it is now $1 - 0.25 = 0.75 \text{ m}$ above the ground.

Finally, $P$ travels $0.75 \text{ m}$ downward to the ground with initial speed $3.75 \text{ m s}^{-1}$ and acceleration $10 \text{ m s}^{-2}$ downward. Using $v^2 = u^2 + 2as$ :

v^2 = 3.75^2 + 2(10)(0.75) = 14.0625 + 15 = 29.0625 = \frac{465}{16}

So $v = \frac{\sqrt{465}}{4} \approx 5.39 \text{ m s}^{-1}$ .

Key Takeaways

Momentum is conserved in a direct collision between two particles, provided no external impulse acts during the collision.
Momentum is a vector; choose a positive direction and apply it consistently.
After a collision, a particle may continue with a new velocity; use suvat equations from the new position and new velocity.

Common Mistakes

Using $v_P = 2$ or $v_Q = 0$ in the momentum equation instead of the speeds just before impact from part (a).
Forgetting $Q$ 's momentum after the collision.
Using the wrong height for $P$ at the collision, e.g. $1 \text{ m}$ or $2 \text{ m}$ instead of $0.75 \text{ m}$ .
Taking $P$ 's velocity after the collision as upward when conservation of momentum gives it downward.
Using $u = 0$ in the final suvat calculation; $P$ already has speed $3.75$ downward after the collision.

Things to Be Careful About

The height at collision can be found from either particle; both give $0.75 \text{ m}$ .
In the final fall, downward is positive, so $a = +10$ and $s = +0.75$ .
The exact answer is $\frac{\sqrt{465}}{4}$ ; the mark scheme accepts AWRT $5.39$ .

Techniques used

apply conservation of momentum in a direct impactdetermine collision height from part (a)use v^2 = u^2 + 2as for the final fall

Q6Medium-HardKinematics of Motion in a Straight Line

A particle, $P$ , travels in a straight line, starting from a point $O$ with velocity $6 \text{ m s}^{-1}$ . The acceleration of $P$ at time $t \text{ s}$ after leaving $O$ is $a \text{ m s}^{-2}$ , where

\begin{aligned} a &= -1.5t^{\frac{1}{2}} && \text{for } 0 \le t \le 1, \\ a &= 1.5t^{\frac{1}{2}} - 3t^{-\frac{1}{2}} && \text{for } t > 1. \end{aligned}

(a)

Find the velocity of $P$ at $t = 1$ .

3M Similar

DifficultyMedium-Easy

Worked solution

Approach

Integrate the acceleration for $0 \le t \le 1$ to obtain the velocity, then use the initial condition $v = 6$ when $t = 0$ to find the constant of integration. Finally substitute $t = 1$ .

Working

For $0 \le t \le 1$ ,

v = \int -\frac{3}{2} t^{1/2} \, dt = -t^{3/2} + c.

Use $v = 6$ when $t = 0$ :

6 = -0 + c \Rightarrow c = 6.

So

v = 6 - t^{3/2}.

At $t = 1$ :

v = 6 - 1^{3/2} = 5.

Answer

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

Final answer

5 m s^-1

Detailed explanation

Walkthrough

The acceleration is the rate of change of velocity, so to go from acceleration to velocity we integrate with respect to time. Integrating $-\frac{3}{2}t^{1/2}$ gives $-t^{3/2}$ because increasing the power by 1 gives $t^{3/2}$ and multiplying by the reciprocal of the new power, $\frac{2}{3}$ , cancels the $\frac{3}{2}$ . The constant $c$ is found from the fact that the particle starts at $O$ with velocity $6$ when $t = 0$ . Substituting $t=1$ then gives the required velocity.

Key Takeaways

Integration of acceleration produces velocity.
The initial condition fixes the arbitrary constant.
Evaluating at a specific time completes the problem.

Common Mistakes

Forgetting the constant of integration.
Using $v = at$ instead of integrating; the mark scheme awards no method mark for this.
Incorrectly integrating $t^{1/2}$ : the new power is $3/2$ and the coefficient becomes $2/3$ .

Things to Be Careful About

The initial condition is at $t = 0$ , not at $t = 1$ .
Keep the expression for $v$ for the correct interval; this part only uses $0 \le t \le 1$ .

Techniques used

integrate acceleration with respect to timeapply initial condition to determine constant of integrationevaluate velocity at a given time

(b)

Given that there is no change in the velocity of $P$ when $t = 1$ , find an expression for the velocity of $P$ for $t > 1$ .

3M Similar

DifficultyMedium

Worked solution

Approach

Integrate the acceleration expression valid for $t > 1$ . Use the fact that the velocity is continuous at $t = 1$ : the value from part (a), $v = 5$ , must also be obtained by the new expression when $t = 1$ . This determines the new constant.

Working

For $t > 1$ ,

v = \int \left( \frac{3}{2} t^{1/2} - 3 t^{-1/2} \right) dt = t^{3/2} - 6 t^{1/2} + c.

Since there is no change in velocity at $t = 1$ , $v = 5$ when $t = 1$ :

5 = 1^{3/2} - 6(1^{1/2}) + c = 1 - 6 + c.

So

c = 10.

Therefore, for $t > 1$ ,

v = t^{3/2} - 6 t^{1/2} + 10.

Answer

v = t^{3/2} - 6 t^{1/2} + 10

Final answer

v = t^(3/2) - 6 t^(1/2) + 10

Detailed explanation

Walkthrough

For $t > 1$ the acceleration changes, so we integrate the new expression. The antiderivative of $\frac{3}{2}t^{1/2}$ is $t^{3/2}$ , and the antiderivative of $-3t^{-1/2}$ is $-6t^{1/2}$ . Because the velocity does not change at $t = 1$ , the new expression must give $v = 5$ when $t = 1$ . Substituting $t = 1$ and $v = 5$ gives $5 = 1 - 6 + c$ , so $c = 10$ .

Key Takeaways

A piecewise acceleration needs a separate velocity expression for each interval.
Continuity of velocity at the boundary determines the new constant of integration.

Common Mistakes

Using a velocity other than $5$ at $t = 1$ .
Forgetting the constant of integration.
Wrong antiderivative of $t^{-1/2}$ : it becomes $2t^{1/2}$ , so $-3t^{-1/2}$ becomes $-6t^{1/2}$ .

Things to Be Careful About

The phrase no change in velocity means the two velocity expressions agree at $t = 1$ .
Use the exact value $v = 5$ from part (a), not a rounded or invented value.

Techniques used

integrate acceleration on the second intervalapply continuity of velocity at t = 1solve for constant of integration

(c)

Given that the velocity of $P$ is positive for $t \le 4$ , find the total distance travelled between $t = 0$ and $t = 4$ .

4M Similar

DifficultyMedium-Hard

Worked solution

Approach

Since the velocity is positive for $0 \le t \le 4$ , the particle never reverses direction, so the total distance is the total displacement. Integrate the velocity expression piecewise: use $v = 6 - t^{3/2}$ on $0 \le t \le 1$ and $v = t^{3/2} - 6 t^{1/2} + 10$ on $1 \le t \le 4$ . Add the two displacements.

Working

For $0 \le t \le 1$ :

s_1 = \int_0^1 \left(6 - t^{3/2}\right) dt = \left[6t - \frac{2}{5}t^{5/2}\right]_0^1.

s_1 = 6 - \frac{2}{5} = \frac{28}{5}.

For $1 \le t \le 4$ :

s_2 = \int_1^4 \left(t^{3/2} - 6 t^{1/2} + 10\right) dt = \left[\frac{2}{5}t^{5/2} - 4t^{3/2} + 10t\right]_1^4.

At $t = 4$ :

\frac{2}{5}(4^{5/2}) - 4(4^{3/2}) + 10(4) = \frac{64}{5} - 32 + 40 = \frac{104}{5}.

At $t = 1$ :

\frac{2}{5} - 4 + 10 = \frac{32}{5}.

So

s_2 = \frac{104}{5} - \frac{32}{5} = \frac{72}{5}.

Total distance:

s = s_1 + s_2 = \frac{28}{5} + \frac{72}{5} = 20.

Answer

20 \text{ m}

Final answer

20 m

Detailed explanation

Walkthrough

The particle moves in a straight line and its velocity is stated to be positive up to $t = 4$ , so it never turns back. Therefore distance travelled equals displacement. We integrate each velocity expression over its interval. The first displacement uses limits $0$ and $1$ ; the second uses limits $1$ and $4$ . Finally add the two positive displacements.

Key Takeaways

When velocity keeps one sign, distance equals displacement.
Piecewise-defined velocity requires integrating each piece separately.
Definite integrals with correct limits give displacements for each interval.

Common Mistakes

Using $s = vt$ instead of integrating; the mark scheme awards no method mark for this.
Not increasing powers or not changing coefficients when integrating.
Using incorrect limits, such as $0$ to $4$ for the whole motion without splitting at $t = 1$ .
Forgetting to add the two displacements.
If a displacement expression has an incorrect constant of integration, the final answer is not awarded even if $20$ m is seen.

Things to Be Careful About

Use $v = 6 - t^{3/2}$ only on $0 \le t \le 1$ .
Use $v = t^{3/2} - 6 t^{1/2} + 10$ only on $1 \le t \le 4$ .
Evaluate $s_2$ at $t = 4$ and at $t = 1$ , then subtract.
The two displacements are $\frac{28}{5}$ and $\frac{72}{5}$ , giving total $20$ m.

Techniques used

integrate velocity piecewiseevaluate definite integrals with correct limitsadd displacements to find total distance

Q7MediumNewton's Laws of MotionKinematics of Motion in a Straight Line

Fig. 3

Two particles, $A$ and $B$ , of masses $0.2 \text{ kg}$ and $0.3 \text{ kg}$ respectively, are attached to the ends of a light inextensible string. The string passes over a small fixed smooth pulley which is attached to the bottom of a rough plane inclined at an angle $\theta$ to the horizontal where $\sin \theta = 0.6$ . Particle $A$ lies on the plane, and particle $B$ hangs vertically below the pulley, $0.25 \text{ m}$ above horizontal ground. The string between $A$ and the pulley is parallel to a line of greatest slope of the plane (see diagram). The coefficient of friction between $A$ and the plane is $1.125$ . Particle $A$ is released from rest.

(a)

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

7M Similar

DifficultyMedium

Worked solution

Approach

Resolve the forces acting on particle $A$ perpendicular and parallel to the inclined plane to find the normal reaction $R$ and the limiting friction $F$ . Then apply Newton's second law to both particles to form a system of equations in the tension $T$ and acceleration $a$ . Solve these simultaneously.

Working

For particle $A$ , resolve perpendicular to the plane:

R = 0.2g \cos \theta

Given $\sin \theta = 0.6$ , we have $\cos \theta = 0.8$ . Using $g = 10 \text{ m s}^{-2}$ :

R = 0.2 \times 10 \times 0.8 = 1.6 \text{ N}

The limiting friction is:

F = \mu R = 1.125 \times 1.6 = 1.8 \text{ N}

Apply Newton's second law to particle $B$ (moving downwards):

0.3g - T = 0.3a \implies 3 - T = 0.3a \quad \text{...(1)}

Apply Newton's second law to particle $A$ (moving down the plane):

T + 0.2g \sin \theta - F = 0.2a \implies T + 0.2 \times 10 \times 0.6 - 1.8 = 0.2a \implies T - 0.6 = 0.2a \quad \text{...(2)}

Add equations (1) and (2):

(3 - T) + (T - 0.6) = 0.3a + 0.2a

2.4 = 0.5a

a = 4.8 \text{ m s}^{-2}

Substitute $a = 4.8$ into equation (1) to find $T$ :

3 - T = 0.3 \times 4.8

3 - T = 1.44

T = 1.56 \text{ N}

Answer

T = 1.56 \text{ N}, \quad a = 4.8 \text{ m s}^{-2}

Final answer

Tension = 1.56 N, acceleration = 4.8 m s^-2

Detailed explanation

Walkthrough

First, we resolve the forces on particle $A$ perpendicular to the inclined plane to find the normal reaction force $R$ . Since $\sin \theta = 0.6$ , we know $\cos \theta = 0.8$ . Using $g = 10 \text{ m s}^{-2}$ , we calculate $R = 0.2 \times 10 \times 0.8 = 1.6 \text{ N}$ . Next, we use the coefficient of friction $\mu = 1.125$ to find the limiting friction $F = \mu R = 1.125 \times 1.6 = 1.8 \text{ N}$ .

Then, we apply Newton's second law ( $F_{net} = ma$ ) to both particles. For particle $B$ , which moves vertically downwards, the net force is its weight minus the tension: $0.3g - T = 0.3a$ . For particle $A$ , which moves down the plane, the net force is the tension plus the component of its weight down the plane, minus the friction opposing the motion: $T + 0.2g \sin \theta - F = 0.2a$ . Substituting the known values gives two simultaneous equations in $T$ and $a$ . Adding these equations eliminates $T$ , allowing us to solve for $a = 4.8 \text{ m s}^{-2}$ . Substituting this back into the equation for $B$ gives the tension $T = 1.56 \text{ N}$ .

Key Takeaways

Resolving forces perpendicular to an inclined plane is essential for finding the normal reaction and friction.
Connected particles over a smooth pulley share the same magnitude of acceleration and tension.
Newton's second law must be applied separately to each particle, taking care with the direction of the net force for each.

Common Mistakes

Forgetting to include the component of weight down the plane for particle $A$ in the equation of motion.
Using the wrong value for $\cos \theta$ or mixing up $\sin$ and $\cos$ .
Sign errors when setting up the equations of motion (e.g., adding friction instead of subtracting it).

Things to Be Careful About

Ensure $g = 10 \text{ m s}^{-2}$ is used as implied by the mark scheme values.
Friction always opposes the direction of motion; since $A$ moves down the plane, friction acts up the plane.
The tension is the same throughout the string because the pulley is smooth and the string is light and inextensible.

Techniques used

resolve forces perpendicular to planecalculate limiting frictionapply Newton's second law to each particlesolve simultaneous equations

(b)

When $B$ reaches the ground, it comes to rest.

Find the total distance that $A$ travels down the plane from when it is released until it comes to rest. You may assume that $A$ does not reach the pulley.

4M Similar

DifficultyMedium

Worked solution

Approach

The motion of particle $A$ occurs in two stages. In the first stage, both particles accelerate together until $B$ hits the ground. Use $v^2 = u^2 + 2as$ to find the velocity of $A$ at this point. In the second stage, the string becomes slack, so tension is zero. Calculate the new acceleration of $A$ using Newton's second law, then use $v^2 = u^2 + 2as$ again to find the additional distance $A$ travels before coming to rest.

Working

Stage 1: Both particles move together

Particle $B$ falls a distance $s_1 = 0.25 \text{ m}$ with initial velocity $u = 0$ and acceleration $a = 4.8 \text{ m s}^{-2}$ . Find the velocity $v$ when $B$ hits the ground:

v^2 = u^2 + 2as_1

v^2 = 0^2 + 2 \times 4.8 \times 0.25 = 2.4

Stage 2: Particle $A$ decelerates after the string goes slack

When $B$ hits the ground, the string becomes slack, so the tension $T = 0$ . Particle $A$ continues to move down the plane but is now decelerating due to friction and the component of weight acting up the plane (relative to its motion).

Apply Newton's second law to $A$ (taking down the plane as positive):

0.2g \sin \theta - F = 0.2a'

0.2 \times 10 \times 0.6 - 1.8 = 0.2a'

1.2 - 1.8 = 0.2a' \implies -0.6 = 0.2a' \implies a' = -3 \text{ m s}^{-2}

Now use $v^2 = u^2 + 2a's_2$ to find the additional distance $s_2$ traveled by $A$ before coming to rest ( $v_{final} = 0$ ):

0^2 = 2.4 + 2 \times (-3) \times s_2

0 = 2.4 - 6s_2 \implies 6s_2 = 2.4 \implies s_2 = 0.4 \text{ m}

Total distance

The total distance $A$ travels down the plane is the sum of the distances in both stages:

\text{Total distance} = s_1 + s_2 = 0.25 + 0.4 = 0.65 \text{ m}

Answer

\text{Total distance} = 0.65 \text{ m}

Final answer

0.65 m

Detailed explanation

Walkthrough

The problem involves two distinct phases of motion. In the first phase, particles $A$ and $B$ are connected and accelerate together. Particle $B$ falls $0.25 \text{ m}$ , so we use the constant acceleration formula $v^2 = u^2 + 2as$ with $u = 0$ , $a = 4.8 \text{ m s}^{-2}$ , and $s = 0.25 \text{ m}$ to find $v^2 = 2.4$ . This is the square of the velocity of particle $A$ at the moment $B$ hits the ground.

In the second phase, $B$ is at rest, so the string goes slack and the tension drops to zero. Particle $A$ is still moving down the plane, but now the only forces acting along the plane are the component of its weight ( $0.2g \sin \theta = 1.2 \text{ N}$ ) pulling it down and the friction ( $F = 1.8 \text{ N}$ ) opposing the motion. The net force is $1.2 - 1.8 = -0.6 \text{ N}$ , giving a deceleration of $a' = -3 \text{ m s}^{-2}$ . We use $v^2 = u^2 + 2a's$ again, with final velocity $0$ , initial $v^2 = 2.4$ , and $a' = -3$ , to find the additional distance $s_2 = 0.4 \text{ m}$ .

Adding the two distances gives the total distance traveled by $A$ : $0.25 + 0.4 = 0.65 \text{ m}$ .

Key Takeaways

When a connected system is broken (e.g., a string goes slack), the motion must be analyzed in separate stages.
The velocity at the end of the first stage becomes the initial velocity for the second stage.
Always re-evaluate the forces and acceleration for each stage, as the tension may change or disappear.

Common Mistakes

Assuming the acceleration remains $4.8 \text{ m s}^{-2}$ after $B$ hits the ground.
Forgetting to add the distance traveled during the first stage to the distance traveled during the second stage.
Sign errors when calculating the new acceleration (e.g., getting $+3$ instead of $-3$ ).

Things to Be Careful About

Ensure you use the correct initial velocity for the second stage; it is the velocity reached at the end of the first stage.
The acceleration in the second stage is negative (deceleration) because friction is greater than the component of weight down the plane.
The question asks for the total distance, not just the distance after the string goes slack.

Techniques used

apply suvat equation to find velocity at impactdetermine new acceleration when string slackensapply suvat equation again to find stopping distance

Mathematics 9709/42 — October/November 2024

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