A particle $P$ of mass $0.2\text{kg}$ is projected vertically upwards from horizontal ground with speed $25\text{ms}^{-1}$. (a) Show that the speed of $P$ when it reaches $20\text{m}$ above the ground is $15\text{ms}^{-1}$. (b) When $P$ reaches $20\text{m}$ above the ground it collides with a second particle $Q$ of mass $0.1\text{kg}$ which is moving downwards at $20\text{ms}^{-1}$. $P$ is brought to instantaneous rest in the collision. Find the velocity of $Q$ immediately…

(a) 15 ms^{-1} · (b) 10 ms^{-1} upwards · (c) 5 m above the ground

9709/41 May/June 2024 — Mathematics Solved Past Paper

Q1Medium-EasyKinematics of Motion in a Straight Line

A car starts from rest and accelerates at $2\text{ms}^{-2}$ for $10\text{s}$ . It then travels at a constant speed for $30\text{s}$ . The car then uniformly decelerates to rest over a period of $20\text{s}$ .

(a)

Sketch a velocity-time graph for the motion of the car.

Fig. 1.1

2M Similar

DifficultyMedium-Easy

Worked solution

Approach

Find the maximum velocity reached during the acceleration phase using $v = u + at$ , then plot the three phases of motion on a velocity-time graph.

Working

During the first $10\text{ s}$ , the car accelerates from rest at $2\text{ m s}^{-2}$ :

v = u + at = 0 + 2 \times 10 = 20\text{ m s}^{-1}

The car then maintains this constant speed of $20\text{ m s}^{-1}$ for $30\text{ s}$ , from $t = 10$ to $t = 40$ .

Finally, it decelerates uniformly to rest over $20\text{ s}$ , from $t = 40$ to $t = 60$ .

The velocity-time graph is a trapezium with vertices at $(0, 0)$ , $(10, 20)$ , $(40, 20)$ , and $(60, 0)$ .

Answer

The velocity-time graph is a trapezium with vertices $(0,0)$ , $(10,20)$ , $(40,20)$ , $(60,0)$ .

Final answer

Trapezium with vertices (0,0), (10,20), (40,20), (60,0)

Detailed explanation

Walkthrough

First, we need to determine the key points on the velocity-time graph. The car starts from rest, so the graph begins at $(0, 0)$ .

During the first phase, the car accelerates at $2\text{ m s}^{-2}$ for $10\text{ s}$ . Using the constant acceleration formula $v = u + at$ , we find the velocity at $t = 10\text{ s}$ :

v = 0 + 2 \times 10 = 20\text{ m s}^{-1}

This gives the point $(10, 20)$ on the graph. The line from $(0, 0)$ to $(10, 20)$ has a positive gradient equal to the acceleration.

During the second phase, the car travels at a constant speed of $20\text{ m s}^{-1}$ for $30\text{ s}$ . This means the velocity does not change, so the graph is a horizontal line from $(10, 20)$ to $(40, 20)$ .

During the third phase, the car decelerates uniformly to rest over $20\text{ s}$ . The graph is a straight line from $(40, 20)$ down to $(60, 0)$ , with a negative gradient equal to the deceleration.

The overall shape is a trapezium with the parallel sides along the time axis (length $60$ ) and the horizontal segment (length $30$ ), and height $20$ .

Key Takeaways

The gradient of a velocity-time graph represents acceleration.
A horizontal line on a velocity-time graph indicates constant velocity.
The area under a velocity-time graph represents the total distance travelled.
Key coordinates on the graph are found using $v = u + at$ for each phase of motion.

Common Mistakes

Forgetting to calculate the maximum velocity before plotting the graph.
Plotting the wrong time intervals for each phase (e.g., using $30\text{ s}$ instead of the end time $40\text{ s}$ ).
Not labelling the axes or key coordinates on the graph.
Drawing curved lines instead of straight lines for uniform acceleration/deceleration.

Things to Be Careful About

Ensure the graph starts at the origin $(0, 0)$ since the car starts from rest.
The horizontal segment must span from $t = 10$ to $t = 40$ (a duration of $30\text{ s}$ ), not from $t = 0$ to $t = 30$ .
The final time is $10 + 30 + 20 = 60\text{ s}$ , so the graph must end at $(60, 0)$ .
Axis labels must be correct if used; if not labelled, assume $t$ is on the horizontal axis and $v$ on the vertical axis.

Techniques used

calculate final velocity from acceleration and timeplot velocity-time graphidentify key coordinates

(b)

Find the total distance travelled by the car.

2M Similar

DifficultyMedium-Easy

Worked solution

Approach

The total distance travelled equals the area under the velocity-time graph. Calculate this area using the trapezium formula or by splitting into simpler shapes (two triangles and a rectangle).

Working

The area under the graph is the area of the trapezium with parallel sides $30\text{ s}$ and $60\text{ s}$ and height $20\text{ m s}^{-1}$ :

\text{Area} = \frac{1}{2}(30 + 60) \times 20

= \frac{1}{2} \times 90 \times 20

= 900\text{ m}

Alternatively, splitting into two triangles and a rectangle:

\text{Area} = \frac{1}{2} \times 10 \times 20 + 30 \times 20 + \frac{1}{2} \times 20 \times 20

= 100 + 600 + 200

= 900\text{ m}

Answer

900\text{ m}

Final answer

900 m

Detailed explanation

Walkthrough

The total distance travelled by the car is equal to the area under the velocity-time graph. This is a fundamental principle of kinematics: since velocity is the rate of change of displacement, integrating velocity over time gives displacement (or distance, when velocity is always positive).

The velocity-time graph from part (a) is a trapezium. We can calculate its area in two ways:

Method 1: Trapezium formula
A trapezium has area $\frac{1}{2}(a + b) \times h$ , where $a$ and $b$ are the lengths of the parallel sides and $h$ is the perpendicular height between them.

The longer parallel side lies along the time axis from $t = 0$ to $t = 60$ , so $a = 60\text{ s}$ .
The shorter parallel side is the horizontal segment from $t = 10$ to $t = 40$ , so $b = 30\text{ s}$ .
The height is the maximum velocity, $h = 20\text{ m s}^{-1}$ .

\text{Area} = \frac{1}{2}(30 + 60) \times 20 = \frac{1}{2} \times 90 \times 20 = 900\text{ m}

Method 2: Split into simpler shapes

Triangle 1 (acceleration phase): base $= 10\text{ s}$ , height $= 20\text{ m s}^{-1}$ , area $= \frac{1}{2} \times 10 \times 20 = 100\text{ m}$
Rectangle (constant speed phase): width $= 30\text{ s}$ , height $= 20\text{ m s}^{-1}$ , area $= 30 \times 20 = 600\text{ m}$
Triangle 2 (deceleration phase): base $= 20\text{ s}$ , height $= 20\text{ m s}^{-1}$ , area $= \frac{1}{2} \times 20 \times 20 = 200\text{ m}$

Total distance $= 100 + 600 + 200 = 900\text{ m}$

Both methods give the same answer, confirming the result.

Key Takeaways

The area under a velocity-time graph represents the total distance travelled (when velocity is positive).
The area can be calculated using the trapezium formula or by decomposing into simpler shapes.
Units must be consistent: velocity in m/s and time in s gives distance in m.

Common Mistakes

Using the wrong parallel sides in the trapezium formula (e.g., using $10$ and $20$ instead of $30$ and $60$ ).
Forgetting to include all three phases when splitting into shapes.
Using an incorrect value for the maximum velocity (must use $20\text{ m s}^{-1}$ from part (a)).
Not showing the method (M1 mark) before the final answer (A1FT mark).

Things to Be Careful About

The trapezium formula requires the two parallel sides to be the horizontal lengths: $60\text{ s}$ (total time) and $30\text{ s}$ (constant speed duration).
The height is the vertical distance between the parallel sides: $20\text{ m s}^{-1}$ .
The final answer must carry correct units: metres (m), not seconds or m/s.
Follow-through marks depend on using the correct value of $20\text{ m s}^{-1}$ from part (a). Using $2$ instead of $20$ would give $0$ marks for the final answer.

Techniques used

calculate area under velocity-time graphuse trapezium area formula

Q2MediumForces and Equilibrium

Fig. 2.1

Two forces of magnitudes $20\text{N}$ and $F\text{N}$ act at a point $P$ in the directions shown in the diagram.

(a)

Given that the resultant force has no component in the $y$ -direction, calculate the value of $F$ .

2M Similar

DifficultyMedium-Easy

Worked solution

Approach

Resolve the forces into their vertical ( $y$ -direction) components. Since the resultant force has no component in the $y$ -direction, the sum of the vertical components must be zero.

Working

The $20\text{ N}$ force makes an angle of $60^\circ$ with the positive $x$ -axis, so its vertical component is $20\sin 60^\circ$ upwards. The force $F\text{ N}$ acts vertically downwards.

Setting the sum of vertical components to zero:

20\sin 60^\circ - F = 0

Solving for $F$ :

F = 20\sin 60^\circ = 20 \times \frac{\sqrt{3}}{2} = 10\sqrt{3}

F \approx 17.3\text{ N}

Answer

F = 17.3\text{ N}

Final answer

F = 17.3 N

Detailed explanation

Walkthrough

First, identify the vertical components of both forces. The $20\text{ N}$ force is at $60^\circ$ to the horizontal, so its upward vertical component is found using the sine ratio: $20\sin 60^\circ$ . The second force $F$ acts straight down along the negative $y$ -axis, contributing a downward component of $F$ . The problem states the resultant has no $y$ -component, meaning the upward and downward forces must balance exactly. We set $20\sin 60^\circ = F$ and evaluate to find $F = 10\sqrt{3} \approx 17.3\text{ N}$ .

Key Takeaways

When a resultant force has zero component in a specific direction, the sum of the force components in that direction must be zero. This allows you to solve for an unknown force magnitude directly.

Common Mistakes

Using the wrong trigonometric ratio (e.g., $20\cos 60^\circ$ instead of $20\sin 60^\circ$ for the vertical component). Forgetting that the vertical component of the $20\text{ N}$ force is upwards while $F$ is downwards, leading to an incorrect sign in the equation.

Things to Be Careful About

Ensure angles are measured correctly from the axis specified. Here, $60^\circ$ is from the positive $x$ -axis, so the vertical component uses $\sin 60^\circ$ . The mark scheme accepts $17.3$ or $10\sqrt{3}$ , so exact surd forms are preferred unless a decimal is explicitly required.

Techniques used

resolve forces into vertical componentsapply equilibrium condition in one direction

(b)

Given instead that $F = 10$ , find the magnitude and direction of the resultant force.

5M Similar

DifficultyMedium

Worked solution

Approach

With $F = 10$ , resolve both forces into horizontal ( $x$ ) and vertical ( $y$ ) components. Find the resultant components $X$ and $Y$ , then calculate the magnitude and direction of the resultant force.

Working

Horizontal component of the resultant, $X$ :

X = 20\cos 60^\circ = 20 \times \frac{1}{2} = 10

Vertical component of the resultant, $Y$ :

Y = 20\sin 60^\circ - 10 = 10\sqrt{3} - 10 \approx 7.3205

Magnitude of the resultant force, $R$ :

R = \sqrt{X^2 + Y^2} = \sqrt{10^2 + (10\sqrt{3} - 10)^2}

R = \sqrt{100 + 300 - 200\sqrt{3} + 100} = \sqrt{500 - 200\sqrt{3}} \approx 12.393\text{ N}

Direction of the resultant force, $\theta$ (measured above the positive $x$ -axis):

\theta = \tan^{-1}\left(\frac{Y}{X}\right) = \tan^{-1}\left(\frac{10\sqrt{3} - 10}{10}\right) = \tan^{-1}(\sqrt{3} - 1) \approx 36.2^\circ

Answer

R = 12.4\text{ N}, \quad \theta = 36.2^\circ \text{ above the positive } x\text{-axis}

Final answer

Magnitude = 12.4 N, Direction = 36.2° above the positive x-axis

Detailed explanation

Walkthrough

First, resolve each force into its horizontal and vertical parts. The $20\text{ N}$ force contributes $20\cos 60^\circ = 10\text{ N}$ horizontally and $20\sin 60^\circ = 10\sqrt{3}\text{ N}$ vertically. The $10\text{ N}$ force contributes $0\text{ N}$ horizontally and $-10\text{ N}$ vertically. Adding these gives the resultant components: $X = 10$ and $Y = 10\sqrt{3} - 10 \approx 7.3205$ . The magnitude is found using Pythagoras' theorem: $R = \sqrt{X^2 + Y^2} \approx 12.393\text{ N}$ . The direction is found using the inverse tangent of the vertical component over the horizontal component: $\theta = \tan^{-1}(Y/X) \approx 36.2^\circ$ . Since both $X$ and $Y$ are positive, the resultant lies in the first quadrant, $36.2^\circ$ above the positive $x$ -axis.

Key Takeaways

To find the resultant of multiple forces, resolve each into perpendicular components, sum them to get the resultant components, then use Pythagoras' theorem and trigonometry to find the magnitude and direction.

Common Mistakes

Forgetting to include the sign of the vertical component for the downward force $F$ . Using $\cos$ instead of $\sin$ for the vertical component of the angled force. Calculating the angle from the wrong axis or misidentifying the quadrant.

Things to Be Careful About

Always specify the direction of the resultant relative to a clear reference (e.g., "above the positive $x$ -axis"). The mark scheme accepts $36.2^\circ$ or radians ( $0.632$ to 3sf). Ensure magnitude is given to 3 significant figures as $12.4\text{ N}$ .

Techniques used

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

Q34MMediumEnergy, Work and Power

A train of mass $180000\text{kg}$ ascends a straight hill of length $1.5\text{km}$ , inclined at an angle of $1.5^\circ$ to the horizontal. As it ascends the hill, the total work done to overcome the resistance to motion is $12000\text{kJ}$ and the speed of the train decreases from $45\text{ms}^{-1}$ to $40\text{ms}^{-1}$ .

Find the work done by the engine of the train as it ascends the hill, giving your answer in $\text{kJ}$ .

DifficultyMedium

Worked solution

Approach

Apply the work-energy principle: the work done by the engine is the total energy required to raise the train's gravitational potential energy, overcome resistance, and account for the change in kinetic energy.

Working

The hill is $1.5\text{ km} = 1500\text{ m}$ long, inclined at $1.5^\circ$ .

Gravitational potential energy gained:

\text{PE} = mgh = 180000 \times 10 \times 1500 \sin 1.5^\circ

\text{PE} = 70\,677\,760.4 \text{ J}

Initial kinetic energy:

\text{KE}_i = \frac{1}{2} \times 180000 \times 45^2 = 182\,250\,000 \text{ J}

Final kinetic energy:

\text{KE}_f = \frac{1}{2} \times 180000 \times 40^2 = 144\,000\,000 \text{ J}

Kinetic energy lost:

\text{KE}_i - \text{KE}_f = 182\,250\,000 - 144\,000\,000 = 38\,250\,000 \text{ J}

Work done against resistance $= 12000\text{ kJ} = 12\,000\,000 \text{ J}$ .

Work done by engine:

WD = \text{PE} + \text{work against resistance} - \text{KE lost}

WD = 70\,677\,760.4 + 12\,000\,000 - 38\,250\,000

WD = 44\,427\,760.4 \text{ J}

Converting to kJ:

WD = 44\,427.7604 \text{ kJ} \approx 44\,400 \text{ kJ}

Answer

44400 \text{ kJ}

Final answer

44400 kJ

Detailed explanation

Walkthrough

The train is moving up a hill. Three things happen: it gains height (so gains gravitational potential energy), it slows down (so loses kinetic energy), and it fights resistance (so energy is lost to resistance). The engine must supply the energy for the height gain and the resistance, but since the train is slowing down, some of the initial kinetic energy is converted to help, so the engine needs less. We use the work-energy principle: work done by engine equals PE gained plus work against resistance minus KE lost.

Step by step:

Convert 1.5 km to 1500 m.
Compute PE gained: $mgh$ where $h = 1500 \sin 1.5^\circ$ .
Compute initial and final KE.
Find KE lost.
Add PE gained and work against resistance, subtract KE lost.
Convert to kJ.

Key Takeaways

The work-energy principle links work done by forces to changes in kinetic energy.
When an object gains height, it gains gravitational potential energy.
When an object slows down, its kinetic energy decreases; this energy can be released to do work.
Always convert units consistently (kJ to J, km to m).

Common Mistakes

Using the vertical height instead of $1500 \sin 1.5^\circ$ .
Forgetting to convert 1.5 km to 1500 m.
Mixing up signs: the KE loss should be subtracted (or equivalently, added as a negative term).
Forgetting to convert the answer to kJ as required.
Using the wrong value of g (the mark scheme uses $g = 10\text{ m s}^{-2}$ ).

Things to Be Careful About

The mark scheme accepts $g = 10$ , giving PE = 70,677,760.4 J. If you use $g = 9.8$ , you get a slightly different value, but the final answer should be approximately 44400 kJ.
The work done by the engine is the work done by the driving force; it is positive.
The final answer must be in kJ.

Techniques used

apply the work-energy principlecompute gravitational potential energy gaincompute kinetic energy changeconvert between joules and kilojoules

Approach

The driving force is found from the power equation $P = Fv$ . Once the driving force is known, apply Newton's second law separately to the car and the trailer, introducing the tension $T$ in the cable. Add or eliminate the tension to find the common acceleration, then substitute back to find the tension.

Working

The driving force $F$ is related to engine power $P$ and speed $v$ by $P = Fv$ , so

F = \frac{14000}{20} = 700\,\text{N}

Let the acceleration be $a$ and the tension in the cable be $T$ .

Apply Newton's second law to the car:

700 - 400 - T = 1700a

Apply Newton's second law to the trailer:

T - 150 = 300a

Apply Newton's second law to the whole system (car + trailer):

700 - 400 - 150 = (1700 + 300)a

Solve for the acceleration:

150 = 2000a

a = \frac{150}{2000} = \frac{3}{40} = 0.075\,\text{ms}^{-2}

Substitute into the trailer equation:

T - 150 = 300 \times 0.075 = 22.5

T = 150 + 22.5 = 172.5\,\text{N}

Answer

a = 0.075\,\text{ms}^{-2}, \qquad T = 172.5\,\text{N}

Final answer

Acceleration = 0.075 m s^-2; Tension = 172.5 N

Detailed explanation

Walkthrough

Start by finding the driving force of the engine. The engine delivers a power of $14000\,\text{W}$ while the car moves at $20\,\text{ms}^{-1}$ . The relation between power, driving force and speed is $P = Fv$ , so the driving force is

F = \frac{14000}{20} = 700\,\text{N}.

This driving force acts forwards on the car. The cable tension $T$ pulls the car backwards and pulls the trailer forwards, and the resistances oppose motion. For the car, the net forward force is driving force minus its resistance minus the tension:

700 - 400 - T = 1700a.

For the trailer, the only forward force is the tension, and the resistance opposes it:

T - 150 = 300a.

Adding these equations eliminates $T$ and gives a single equation for the whole system:

700 - 400 - 150 = 2000a,

so

150 = 2000a \quad \Rightarrow \quad a = 0.075\,\text{ms}^{-2}.

Finally, substitute this acceleration into the trailer equation to find $T$ :

T = 300(0.075) + 150 = 22.5 + 150 = 172.5\,\text{N}.

This is the tension in the cable while the car and trailer accelerate at $0.075\,\text{ms}^{-2}$ .

Key Takeaways

The key relationships used in this problem are:

$P = Fv$ : engine power equals driving force times speed.
Newton's second law $F_{\text{net}} = ma$ applied to each body.
Connected bodies moving together share the same acceleration.
The cable tension is an internal force when the car and trailer are treated as one system, so it cancels when applying Newton's second law to the combined system.

Common Mistakes

Using $14000\,\text{N}$ as the driving force instead of first dividing by the speed $20$ .
Getting the direction of the tension wrong: it acts backward on the car and forward on the trailer.
Forgetting one of the resistances when using the system equation.
Using different values of $T$ in the two separate equations; the tension is the same at both ends of a light cable.
Assuming the acceleration is zero because the speed is constant; a constant power can still produce acceleration.

Things to Be Careful About

Always calculate the driving force from power and speed before applying Newton's second law.
The system equation can be obtained either by adding the two body equations or directly by applying Newton's second law to the combined mass of $2000\,\text{kg}$ .
Since the road is horizontal, the weights of the car and trailer are perpendicular to the motion and play no part in the horizontal dynamics.
The mark scheme condones $173\,\text{N}$ as an acceptable rounding, but the exact value is $172.5\,\text{N}$ .

Techniques used

calculate driving force from power and speedapply Newton's second law to the car and trailer separatelyeliminate the cable tension to find the common accelerationsubstitute the acceleration back to find the tension

Approach

Take $g = 10\text{ m s}^{-2}$ . With no resistance, the only force along the slope is the component of the bobsled's weight down the slope. Use Newton's second law to find the acceleration, then apply the constant acceleration formula $v^2 = u^2 + 2as$ .

Working

Resolving along the slope:

mg\sin 12^\circ = ma

so

a = g\sin 12^\circ = 10\sin 12^\circ = 2.07911\ldots\text{ m s}^{-2}

Now use $v^2 = u^2 + 2as$ with $u = 5\text{ m s}^{-1}$ and $s = 60\text{ m}$ :

\begin{aligned} v^2 &= 5^2 + 2(2.07911\ldots)(60) \\ &= 25 + 249.493\ldots \\ &= 274.493\ldots \end{aligned}

v = \sqrt{274.493\ldots} = 16.5679\ldots\text{ m s}^{-1}

Answer

The speed at the bottom is $16.6\text{ m s}^{-1}$ (3 s.f.).

Final answer

16.6 m s^{-1}

Detailed explanation

Walkthrough

The bobsled slides down a straight slope of length $60\text{ m}$ inclined at $12^\circ$ . Since there is no resistance, the only force along the slope is the component of the weight pulling it downhill.

For a weight $mg$ on an incline at angle $\theta$ to the horizontal:

the component down the slope is $mg\sin\theta$ ,
the component perpendicular to the slope is $mg\cos\theta$ .

So resolving parallel to the slope gives $mg\sin 12^\circ = ma$ . The mass cancels, so $a = g\sin 12^\circ$ . Using $g = 10\text{ m s}^{-2}$ , this gives $a = 2.07911\ldots\text{ m s}^{-2}$ .

The acceleration is constant, so we can use the constant acceleration formula $v^2 = u^2 + 2as$ . Here $u = 5\text{ m s}^{-1}$ and $s = 60\text{ m}$ . Substituting gives $v^2 = 5^2 + 2(2.07911\ldots)(60) = 274.493\ldots$ , so $v = 16.5679\ldots\text{ m s}^{-1}$ .

A useful check is energy conservation: the loss in gravitational potential energy is $mg(60\sin 12^\circ)$ , and this equals the gain in kinetic energy $\frac{1}{2}m(v^2 - 5^2)$ . Cancelling $m$ gives the same value of $v$ .

Key Takeaways

On a frictionless incline, the acceleration is $g\sin\theta$ , independent of mass.
The weight component along the slope is $mg\sin\theta$ ; the component perpendicular to the slope is $mg\cos\theta$ .
Constant acceleration problems can be solved with the suvat equations.
Energy methods can be used as an alternative when no work is done against resistance.

Common Mistakes

Using $mg\cos 12^\circ$ as the force down the slope instead of $mg\sin 12^\circ$ .
Forgetting the initial speed $5\text{ m s}^{-1}$ and using $v^2 = 2as$ .
Using the wrong suvat equation, e.g. $v = u + at$ without knowing $t$ .
In the energy method, using $60$ as the vertical height instead of $60\sin 12^\circ$ , or forgetting the initial kinetic energy.

Things to Be Careful About

The mark scheme uses $g = 10\text{ m s}^{-2}$ ; with $g = 9.8\text{ m s}^{-2}$ the answer would be slightly different.
Keep units consistent: distances in metres, speeds in $\text{m s}^{-1}$ .
Give the final speed to 3 significant figures: $16.6\text{ m s}^{-1}$ .

Techniques used

resolve the weight component along the slopeapply Newton's second law to find the accelerationuse the suvat equation v^2 = u^2 + 2as

(b)

It is given instead that the coefficient of friction between the bobsled and the slope is $0.03$ .

Find the time that it takes for the bobsled to reach the bottom of the slope.

5M Similar

DifficultyMedium

Worked solution

Approach

Resolve perpendicular to the slope to find the normal reaction $R$ . Use the friction model $F = \mu R$ with $\mu = 0.03$ . Apply Newton's second law parallel to the slope to find the acceleration, then use $s = ut + \frac{1}{2}at^2$ to find the time.

Working

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

Perpendicular to the slope:

R = mg\cos 12^\circ

Friction:

F = \mu R = 0.03mg\cos 12^\circ

Parallel to the slope, taking down the slope as positive:

mg\sin 12^\circ - F = ma

Substitute $F$ :

mg\sin 12^\circ - 0.03mg\cos 12^\circ = ma

Cancel $m$ :

a = g\sin 12^\circ - 0.03g\cos 12^\circ

a = 10\sin 12^\circ - 0.3\cos 12^\circ = 2.07911\ldots - 0.29343\ldots = 1.78568\ldots\text{ m s}^{-2}

Use $s = ut + \frac{1}{2}at^2$ with $s = 60$ , $u = 5$ :

60 = 5t + \frac{1}{2}(1.78568\ldots)t^2

0.89284\ldots t^2 + 5t - 60 = 0

Solve using the quadratic formula, taking the positive root:

\begin{aligned} t &= \frac{-5 + \sqrt{5^2 - 4(0.89284\ldots)(-60)}}{2(0.89284\ldots)} \\ &= \frac{-5 + \sqrt{239.281\ldots}}{1.78568\ldots} \\ &= \frac{-5 + 15.4694\ldots}{1.78568\ldots} \\ &= 5.86260\ldots\text{ s} \end{aligned}

Answer

The time taken is $5.86\text{ s}$ (3 s.f.).

Final answer

5.86 s

Detailed explanation

Walkthrough

Now there is friction, so we first need the normal reaction force $R$ . The bobsled does not accelerate perpendicular to the slope, so the perpendicular component of weight $mg\cos 12^\circ$ is balanced by $R$ :

R = mg\cos 12^\circ

The friction force is $F = \mu R = 0.03mg\cos 12^\circ$ . It acts up the slope because the bobsled is moving down the slope.

Take down the slope as positive. The weight component down the slope is $mg\sin 12^\circ$ , and friction opposes motion, so Newton's second law gives

mg\sin 12^\circ - F = ma

Substituting $F$ and cancelling $m$ :

a = g\sin 12^\circ - 0.03g\cos 12^\circ

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

a = 10\sin 12^\circ - 0.3\cos 12^\circ = 1.78568\ldots\text{ m s}^{-2}

Because the acceleration is constant, use $s = ut + \frac{1}{2}at^2$ with $s = 60$ , $u = 5$ :

60 = 5t + \frac{1}{2}(1.78568\ldots)t^2

This is the quadratic $0.89284\ldots t^2 + 5t - 60 = 0$ . Solving with the quadratic formula and taking the positive root gives $t = 5.86260\ldots\text{ s}$ .

An energy alternative: work done by gravity minus work done by friction equals the change in kinetic energy. This gives the final speed $v = 15.4687\ldots\text{ m s}^{-1}$ , then $60 = \frac{1}{2}(5 + v)t$ gives the same time.

Key Takeaways

On an incline, the normal reaction is $R = mg\cos\theta$ , not $mg$ .
The friction force is $F = \mu R$ and always opposes the motion.
Newton's second law is applied separately parallel and perpendicular to the slope.
Constant acceleration allows the suvat equations; solving for time may require the quadratic formula.

Common Mistakes

Using $R = mg$ instead of $mg\cos 12^\circ$ .
Using $F = \mu mg\sin 12^\circ$ or $F = \mu mg$ instead of $F = \mu mg\cos 12^\circ$ .
Taking friction down the slope instead of up the slope.
Using $mg\cos 12^\circ$ as the force down the slope.
Losing the mass when applying Newton's second law.
Forgetting to discard the negative root of the quadratic.
In the energy method, using the slope length as the vertical height for gravitational potential energy.

Things to Be Careful About

Use $g = 10\text{ m s}^{-2}$ to match the mark scheme values.
The acceleration is smaller than in part (a) because friction opposes motion.
The quadratic has two roots; time must be positive.
Give the final answer to 3 significant figures: $5.86\text{ s}$ .
The mark scheme awards marks for resolving perpendicular (B1), applying Newton's second law (M1), using $F = \mu R$ (DM1), and solving the suvat equation (DM1), so show each stage clearly.

Techniques used

resolve forces perpendicular and parallel to the slopeapply the friction model F = μRapply Newton's second law along the slopesolve the suvat quadratic for time

Q6Medium-HardKinematics of Motion in a Straight Line

A particle moves in a straight line, starting from a point $O$ . The velocity of the particle at time $t\text{s}$ after leaving $O$ is $v\text{ms}^{-1}$ . It is given that $v = kt^{\frac{1}{2}} - 2t - 8$ , where $k$ is a positive constant. The maximum velocity of the particle is $4.5\text{ms}^{-1}$ .

(a)

Show that $k = 10$ .

5M Similar

DifficultyMedium

Worked solution

Approach

At a maximum value of $v$ , the acceleration $\frac{dv}{dt}$ is zero. Differentiate $v$ , solve $\frac{dv}{dt}=0$ for $t$ in terms of $k$ , substitute into $v$ with the known maximum $4.5\text{ ms}^{-1}$ , and solve for $k$ .

Working

The velocity is

v = kt^{\frac{1}{2}} - 2t - 8

Differentiate with respect to $t$ :

\frac{dv}{dt} = \frac{1}{2}kt^{-\frac{1}{2}} - 2

At the maximum, $\frac{dv}{dt}=0$ :

\frac{1}{2}kt^{-\frac{1}{2}} - 2 = 0

\frac{1}{2}kt^{-\frac{1}{2}} = 2

kt^{-\frac{1}{2}} = 4

t = \left(\frac{k}{4}\right)^2 = \frac{k^2}{16}

Substitute this $t$ into $v$ :

v = k\left(\frac{k^2}{16}\right)^{\frac{1}{2}} - 2\left(\frac{k^2}{16}\right) - 8

= k\left(\frac{k}{4}\right) - \frac{k^2}{8} - 8 = \frac{k^2}{4} - \frac{k^2}{8} - 8 = \frac{k^2}{8} - 8

Since the maximum velocity is $4.5$ ,

\frac{k^2}{8} - 8 = 4.5

\frac{k^2}{8} = 12.5

k^2 = 100

Because $k>0$ ,

k = 10

Answer

k = 10

Final answer

k = 10

Detailed explanation

Walkthrough

The velocity is given as a function of time, so its maximum occurs where its derivative, the acceleration, is zero. Differentiate term by term:

\frac{d}{dt}\left(kt^{\frac{1}{2}}\right) = \frac{1}{2}kt^{-\frac{1}{2}}, \quad \frac{d}{dt}(-2t) = -2, \quad \frac{d}{dt}(-8) = 0.

Setting the derivative equal to zero gives $kt^{-\frac{1}{2}} = 4$ , so the maximum occurs at $t = \frac{k^2}{16}$ . Substitute this into the original expression for $v$ :

v = k\left(\frac{k}{4}\right) - 2\left(\frac{k^2}{16}\right) - 8 = \frac{k^2}{8} - 8.

Because the maximum velocity is $4.5$ , solve $\frac{k^2}{8} - 8 = 4.5$ . This gives $\frac{k^2}{8} = 12.5$ , so $k^2 = 100$ . Since $k$ is positive, $k = 10$ .

Key Takeaways

A stationary point of $v(t)$ occurs when $\frac{dv}{dt}=0$ .
Fractional powers differentiate using $\frac{d}{dt}(t^n)=nt^{n-1}$ .
A known value at a stationary point can be used to find an unknown parameter.

Common Mistakes

Forgetting that the derivative of $-8$ is $0$ .
Losing the $-2$ when differentiating $-2t$ .
Taking $k = -10$ instead of rejecting it because $k>0$ .
Trying boundary values such as $t=0$ instead of solving $\frac{dv}{dt}=0$ .

Things to Be Careful About

This is an "answer given" question, so any error in the working loses the final accuracy mark.
The derivative of $kt^{\frac{1}{2}}$ is $\frac{1}{2}kt^{-\frac{1}{2}}$ , not $kt^{-\frac{1}{2}}$ .
Reject the negative root because the constant $k$ is stated to be positive.

Techniques used

differentiate velocity to obtain accelerationset acceleration to zero to locate maximum velocitysolve for the time of maximum velocity in terms of ksubstitute into the velocity equation and solve for k

(b)

6M

(i)

Verify that $v = 0$ when $t = 1$ and $t = 16$ .

1M Similar

(ii)

Find the distance travelled by the particle in the first $16\text{s}$ .

5M Similar

DifficultyMedium-Hard

Worked solution

Approach

Distance travelled is the integral of speed, not signed velocity. Since $v$ changes sign at $t=1$ , split the interval $0 \leq t \leq 16$ into $[0,1]$ and $[1,16]$ . On $[0,1]$ , $v<0$ ; on $[1,16]$ , $v>0$ . Integrate over each interval, take the absolute value of the negative displacement, and add.

Working

Integrate $v$ :

s(t) = \int \left(10t^{\frac{1}{2}} - 2t - 8\right)\,dt = \frac{20}{3}t^{\frac{3}{2}} - t^2 - 8t + C

The constant $C$ cancels in definite integrals.

On $[0,1]$ , the displacement is

s(1) - s(0) = \left(\frac{20}{3}(1)^{\frac{3}{2}} - 1^2 - 8(1)\right) - 0 = \frac{20}{3} - 9 = -\frac{7}{3}.

The distance on this interval is

\left|-\frac{7}{3}\right| = \frac{7}{3}.

On $[1,16]$ :

s(16) - s(1) = \left(\frac{20}{3}(16)^{\frac{3}{2}} - 16^2 - 8(16)\right) - \left(\frac{20}{3}(1)^{\frac{3}{2}} - 1^2 - 8(1)\right)

Since $16^{\frac{3}{2}} = 64$ and the value at $t=1$ is $-\frac{7}{3}$ :

= \left(\frac{20}{3}\cdot 64 - 256 - 128\right) - \left(-\frac{7}{3}\right)

= \frac{1280}{3} - 384 + \frac{7}{3} = \frac{128}{3} + \frac{7}{3} = 45.

Since $v>0$ on this interval, this is also the distance.

Total distance:

\frac{7}{3} + 45 = \frac{7}{3} + \frac{135}{3} = \frac{142}{3}\text{ m}.

Answer

\frac{142}{3}\text{ m} \approx 47.3\text{ m}

Final answer

142/3 m (approximately 47.3 m)

Detailed explanation

Walkthrough

Distance is total length travelled, so integrating $v$ over the whole interval would only give net displacement. A particle changes direction exactly where $v=0$ . From part (b)(i), $v=0$ at $t=1$ and $t=16$ . A quick sign check, for example $v(0)=-8<0$ and $v(4)=4>0$ , shows the particle moves backwards for $0<t<1$ and forwards for $1<t<16$ .

Integrate the velocity term by term:

\int \left(10t^{\frac{1}{2}} - 2t - 8\right)\,dt = \frac{20}{3}t^{\frac{3}{2}} - t^2 - 8t + C.

The displacement from $0$ to $1$ is

\frac{20}{3}(1) - 1 - 8 = -\frac{7}{3}.

Because this interval is backwards, the distance is $\frac{7}{3}$ .

The displacement from $1$ to $16$ is

\left(\frac{20}{3}(64) - 256 - 128\right) - \left(-\frac{7}{3}\right) = \frac{128}{3} + \frac{7}{3} = 45.

Add the two distances:

\frac{7}{3} + 45 = \frac{142}{3} \approx 47.3\text{ m}.

Key Takeaways

The integral of velocity gives displacement, not distance.
To find distance, split the motion at times when velocity is zero, integrate over each interval, take absolute values, and sum.
$16^{\frac{3}{2}} = 64$ is essential for the numerical evaluation.

Common Mistakes

Integrating from $0$ to $16$ and treating the result as distance; the net displacement is actually different from the distance.
Forgetting to take the absolute value of $-\frac{7}{3}$ .
Evaluating $16^{\frac{3}{2}}$ incorrectly.
Sign errors when subtracting the value at $t=1$ , especially missing the double negative.

Things to Be Careful About

Use limits $1$ and $16$ for the second interval, not $0$ and $16$ .
The first interval contributes positive distance $\frac{7}{3}$ even though its displacement is negative.
The mark scheme accepts $47.3$ or better as the final numerical answer.
Keep the unit metres in the final answer.

Techniques used

integrate velocity to obtain displacementsplit the interval where velocity changes signevaluate definite integrals on each intervaladd absolute displacements to find total distance

Q7Medium-HardKinematics of Motion in a Straight LineMomentum

A particle $P$ of mass $0.2\text{kg}$ is projected vertically upwards from horizontal ground with speed $25\text{ms}^{-1}$ .

(a)

Show that the speed of $P$ when it reaches $20\text{m}$ above the ground is $15\text{ms}^{-1}$ .

2M Similar

DifficultyMedium-Easy

Worked solution

Approach

Use the constant acceleration formula connecting initial speed, acceleration, displacement and final speed. Take upward as positive, so $u = 25$ , $a = -g = -10$ and $s = 20$ .

Working

v^2 = u^2 + 2as = 25^2 + 2(-10)(20)

v^2 = 625 - 400 = 225

v = 15

Since the particle is still moving upwards at this height, the speed is $15\text{ ms}^{-1}$ .

Answer

$15\text{ ms}^{-1}$

Final answer

15 ms^{-1}

Detailed explanation

Walkthrough

We know the initial speed, the displacement and the acceleration, and we want the final speed. The suvat equation $v^2 = u^2 + 2as$ is ideal because it does not involve time. Taking upward as positive, the acceleration is $-g = -10$ . Substituting $u = 25$ , $a = -10$ and $s = 20$ gives $v^2 = 225$ , so $v = 15$ . Because the particle is still moving upwards at that height, the speed is $15\text{ ms}^{-1}$ .

Key Takeaways

This part tests the ability to choose the correct constant acceleration formula and to substitute values with a consistent sign convention. It also shows that speed is the magnitude of velocity, so the positive root is chosen.

Common Mistakes

Quoting the answer without showing an intermediate equation loses the method mark.
Using $a = +g$ instead of $-g$ for upward motion gives $v^2 = 625 + 400$ , which is wrong.
Mixing up $u$ and $v$ , or using $s = -20$ , can produce an incorrect sign.

Things to Be Careful About

The mark scheme requires at least one intermediate step from the equation of motion to the given result; a bare statement of $v = 15$ is not enough. Also, if you use the energy method, you must show the energy equation with all three terms.

Techniques used

apply the suvat equation v^2 = u^2 + 2assubstitute given values with a consistent sign conventionsolve for the speed

(b)

When $P$ reaches $20\text{m}$ above the ground it collides with a second particle $Q$ of mass $0.1\text{kg}$ which is moving downwards at $20\text{ms}^{-1}$ . $P$ is brought to instantaneous rest in the collision.

Find the velocity of $Q$ immediately after the collision.

2M Similar

DifficultyMedium-Easy

Worked solution

Approach

Use conservation of linear momentum during the collision. Take upward as positive. Before the collision $P$ has velocity $+15$ and $Q$ has velocity $-20$ ; after the collision $P$ has velocity $0$ . Let $v$ be the velocity of $Q$ immediately after the collision.

Working

0.2(15) + 0.1(-20) = 0.2(0) + 0.1v

3 - 2 = 0.1v

v = 10

The positive sign means upwards.

Answer

$10\text{ ms}^{-1}\text{ upwards}$

Final answer

10 ms^{-1} upwards

Detailed explanation

Walkthrough

During a collision, if no external impulse acts, total momentum is conserved. Choose upward as positive. Before the collision $P$ has momentum $0.2 \times 15$ and $Q$ has momentum $0.1 \times (-20)$ because it is moving downwards. After the collision $P$ is at rest, so its momentum is zero. If $v$ is the velocity of $Q$ after the collision, conservation gives $0.2(15) + 0.1(-20) = 0.1v$ . Solving gives $v = 10$ , so $Q$ moves upwards at $10\text{ ms}^{-1}$ .

Key Takeaways

Momentum is a vector, so a positive direction must be chosen and velocities substituted with their signs. In a direct impact, total momentum before equals total momentum after.

Common Mistakes

Using the projection speed $25$ instead of the speed at collision $15$ is a common error and is not awarded the method mark.
Forgetting the negative sign on $Q$ 's downward velocity.
Writing momentum as $mgv$ instead of $mv$ ; the mark scheme gives at most M1 A0 for that.
Giving only a magnitude without stating the direction; the accuracy mark requires the direction.

Things to Be Careful About

The mark scheme allows sign errors in the momentum equation for the method mark, but the final answer must include the direction. Use the speed found in part (a), not the initial projection speed.

Techniques used

choose a positive directionapply conservation of linear momentumsolve for the unknown velocity and state its direction

(c)

When $P$ reaches the ground it rebounds back directly upwards with half of the speed that it had immediately before hitting the ground.

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

6M Similar

DifficultyMedium-Hard

Worked solution

Approach

Track the motion in stages. First find how long $P$ takes to fall from $20\text{ m}$ to the ground and its speed just before impact. Then find where $Q$ is at that same instant. Finally, after $P$ rebounds, set the upward distance travelled by $P$ plus the downward distance travelled by $Q$ equal to the $20\text{ m}$ separation.

Working

Stage 1: $P$ falls from rest at $20\text{ m}$ .

20 = \frac{1}{2}(10)t^2 \quad \Rightarrow \quad t = 2

Speed just before ground:

v = 10(2) = 20

So $P$ rebounds upwards with speed $10$ .

At $t = 2$ , $Q$ has been moving for $2\text{ s}$ with initial velocity $10$ upwards and acceleration $-10$ :

s_Q = 10(2) - \frac{1}{2}(10)(2)^2 = 20 - 20 = 0

so $Q$ is back at $20\text{ m}$ above the ground, moving downwards at

v_Q = 10 - 10(2) = -10

i.e. $10\text{ ms}^{-1}$ downwards.

Stage 2: Let $t$ be the time after $P$ rebounds. Upward displacement of $P$ :

s_P = 10t - \frac{1}{2}(10)t^2 = 10t - 5t^2

Downward displacement of $Q$ from $20\text{ m}$ :

s_Q = 10t + \frac{1}{2}(10)t^2 = 10t + 5t^2

They collide when:

s_P + s_Q = 20

(10t - 5t^2) + (10t + 5t^2) = 20

20t = 20 \quad \Rightarrow \quad t = 1

Height above ground:

s_P = 10(1) - 5(1)^2 = 5

Answer

$5\text{ m}\text{ above the ground}$

Final answer

5 m above the ground

Detailed explanation

Walkthrough

After the first collision, $P$ is instantaneously at rest at $20\text{ m}$ , while $Q$ moves upwards at $10\text{ ms}^{-1}$ (from part (b)). $P$ then falls. From $20 = \frac12 g t^2$ with $g=10$ , it reaches the ground after $2\text{ s}$ , and its speed just before impact is $v = gt = 20$ . It rebounds with half this speed, so it leaves the ground at $10\text{ ms}^{-1}$ upwards.

At that same instant, $Q$ has been moving for $2\text{ s}$ with initial velocity $10$ upwards and acceleration $-10$ . Its displacement is $10(2) - 5(2)^2 = 0$ , so it is also back at $20\text{ m}$ , now moving downwards at $10\text{ ms}^{-1}$ .

Now take $t = 0$ at the instant $P$ rebounds. $P$ rises from the ground with upward displacement $s_P = 10t - 5t^2$ . $Q$ falls from $20\text{ m}$ with downward displacement $s_Q = 10t + 5t^2$ . They meet when the sum of these displacements equals the original $20\text{ m}$ separation: $s_P + s_Q = 20$ . This gives $20t = 20$ , so $t = 1$ . The height is then $s_P = 10(1) - 5(1)^2 = 5\text{ m}$ .

Key Takeaways

This is a multi-stage kinematics problem. The key is to identify a convenient time origin, here the instant $P$ rebounds, and to write consistent displacement equations for both particles. The collision condition is that the two displacements add to the initial separation.

Common Mistakes

Using $25$ or $15$ for the speed of $P$ at ground impact instead of $20$ .
Forgetting that $Q$ also moves during the $2\text{ s}$ that $P$ is falling.
Using the same sign convention for both particles without adjusting for their different directions, leading to $s_P - s_Q = 20$ instead of $s_P + s_Q = 20$ .
Forgetting to halve the impact speed when $P$ rebounds.
Stopping at $t = 1$ without converting to height.

Things to Be Careful About

The mark scheme awards separate marks for the speed at impact, the time to reach the ground, the two displacement equations, the time of collision and the final height. It uses 'their 10' from part (b), so a wrong value in part (b) can still gain method marks in part (c) if the working is consistent. The final height must be correct with no earlier errors (CWO).

Techniques used

find the time of fall using a suvat equationcompute the speed at impact and the rebound speedtrack the displacements of both particles after the reboundsolve the collision condition

Mathematics 9709/41 — May/June 2024

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