A Level - Edexcel - 2024 - Mechanics

Question 1 (3 marks)

Figure 1 shows a particle \( P \) of mass \( 0.5 \text{ kg} \) at rest on a rough horizontal plane.

(a) Find the magnitude of the normal reaction of the plane on \( P \).

(1)

The coefficient of friction between \( P \) and the plane is \( \frac{2}{7} \).

A horizontal force of magnitude \( X \) newtons is applied to \( P \).

Given that \( P \) is now in limiting equilibrium,

(b) find the value of \( X \).

(2)

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Question 2 (8 marks)

Figure 2 shows a speed-time graph for a model of the motion of an athlete running a \( 200 \text{ m} \) race in \( 24 \text{ s} \).

The athlete:

starts from rest at time \( t = 0 \) and accelerates at a constant rate, reaching a speed of \( 10 \text{ ms}^{-1} \) at \( t = 4 \).
then moves at a constant speed of \( 10 \text{ ms}^{-1} \) from \( t = 4 \) to \( t = 18 \).
then decelerates at a constant rate from \( t = 18 \) to \( t = 24 \), crossing the finishing line with speed \( U \text{ ms}^{-1} \).

Using the model,

(a) find the acceleration of the athlete during the first \( 4 \text{ s} \) of the race, stating the units of your answer.

(2)

(b) find the distance covered by the athlete during the first \( 18 \text{ s} \) of the race.

(3)

(c) find the value of \( U \).

(3)

✏️ Worked Solution

Part (a): Acceleration

💡 What are we being asked to find?

Acceleration is the rate of change of speed. On a speed-time graph, the acceleration is represented by the gradient of the line.

✏ Working:

We use the formula for gradient (slope):

\[ a = \frac{\Delta v}{\Delta t} = \frac{10 – 0}{4 – 0} \] \[ a = \frac{10}{4} = 2.5 \]

Don’t forget the units for acceleration:

\[ a = 2.5 \text{ ms}^{-2} \]

✓ (M1 A1)

Part (b): Distance (First 18s)

💡 What are we being asked to find?

The distance covered corresponds to the area under the speed-time graph. The motion from \( t=0 \) to \( t=18 \) consists of a triangle (acceleration phase) and a rectangle (constant speed phase).

✏ Working:

Area of triangle (0 to 4s):

\[ \text{Area}_1 = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 4 \times 10 = 20 \text{ m} \]

Area of rectangle (4 to 18s):

\[ \text{width} = 18 – 4 = 14 \] \[ \text{Area}_2 = \text{width} \times \text{height} = 14 \times 10 = 140 \text{ m} \]

Total distance:

\[ \text{Total} = 20 + 140 = 160 \text{ m} \]

✓ (M1 A1 A1)

Part (c): Finding U

💡 What are we being asked to find?

We know the total length of the race is \( 200 \text{ m} \). We have already calculated the distance for the first 18s (\( 160 \text{ m} \)). The remaining distance is covered in the trapezium from \( t=18 \) to \( t=24 \).

✏ Working:

Remaining distance:

\[ 200 – 160 = 40 \text{ m} \]

This distance corresponds to the area of the trapezium between \( t=18 \) and \( t=24 \). The parallel sides are height \( 10 \) and height \( U \). The width is \( 24 – 18 = 6 \).

\[ \text{Area} = \frac{1}{2} (a + b) h \] \[ 40 = \frac{1}{2} (10 + U) \times 6 \]

Simplify and solve for \( U \):

\[ 40 = 3(10 + U) \] \[ \frac{40}{3} = 10 + U \] \[ U = \frac{40}{3} – 10 \] \[ U = \frac{10}{3} \text{ or } 3.33 \text{ ms}^{-1} \]

✓ (M1 A1 A1)

🎯 Final Answers:

(a) \( 2.5 \text{ ms}^{-2} \)

(b) \( 160 \text{ m} \)

(c) \( U = \frac{10}{3} \) (or \( 3.33 \))

✓ Total: 8 marks

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Question 3 (7 marks)

A particle \( P \) of mass \( m \) is held at rest at a point on a rough inclined plane, as shown in Figure 3.

It is given that:

the plane is inclined to the horizontal at an angle \( \alpha \), where \( \tan\alpha = \frac{5}{12} \)
the coefficient of friction between \( P \) and the plane is \( \mu \), where \( \mu < \frac{5}{12} \)

The particle \( P \) is released from rest and slides down the plane. Air resistance is modelled as being negligible.

(a) Find, in terms of \( m \) and \( g \), the magnitude of the normal reaction of the plane on \( P \).

(2)

(b) Show that, as \( P \) slides down the plane, the acceleration of \( P \) down the plane is \( \frac{1}{13}g(5 – 12\mu) \).

(4)

(c) State what would happen to \( P \) if it is released from rest but \( \mu \ge \frac{5}{12} \).

(1)

✏️ Worked Solution

Part (a): Normal Reaction

💡 Strategy:

First, we need to find \( \sin\alpha \) and \( \cos\alpha \) from the given \( \tan\alpha = \frac{5}{12} \). We can use a 5-12-13 Pythagorean triangle.

✏ Working:

Hypotenuse = \( \sqrt{5^2 + 12^2} = \sqrt{25 + 144} = \sqrt{169} = 13 \)

So, \( \sin\alpha = \frac{5}{13} \) and \( \cos\alpha = \frac{12}{13} \).

Resolving forces perpendicular to the plane:

\[ R = mg \cos\alpha \] \[ R = mg \left( \frac{12}{13} \right) \] \[ R = \frac{12}{13}mg \]

✓ (M1 A1)

Part (b): Acceleration

💡 Strategy:

We need to apply Newton’s Second Law (\( F_{net} = ma \)) down the plane. The forces acting parallel to the plane are the component of weight pulling it down (\( mg \sin\alpha \)) and friction resisting motion (\( F = \mu R \)) acting up the plane.

✏ Working:

Equation of motion down the plane:

\[ mg \sin\alpha – F = ma \]

Since the particle is sliding, friction is at its maximum:

\[ F = \mu R = \mu \left( \frac{12}{13}mg \right) \]

Substitute values into the motion equation:

\[ mg \left( \frac{5}{13} \right) – \mu \left( \frac{12}{13}mg \right) = ma \]

Divide through by \( m \) (mass cancels out):

\[ \frac{5}{13}g – \frac{12\mu}{13}g = a \]

Factor out \( \frac{1}{13}g \):

\[ a = \frac{1}{13}g(5 – 12\mu) \]

✓ (M1 A1 M1 A1)

Part (c): Condition for Motion

💡 Reasoning:

If \( \mu \ge \frac{5}{12} \), then \( 12\mu \ge 5 \), which means \( 5 – 12\mu \le 0 \). This implies the acceleration would be zero or negative (which is impossible as friction cannot cause backward motion from rest).

✏ Answer:

The particle would remain at rest (it would not slide down).

✓ (B1)

🎯 Final Answers:

(a) \( \frac{12}{13}mg \)

(b) Proof shown

(c) Particle would not move / remains at rest

✓ Total: 7 marks

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Question 4 (11 marks)

In this question you must show all stages of your working.

Solutions relying entirely on calculator technology are not acceptable.

[In this question, \( \mathbf{i} \) is a unit vector due east and \( \mathbf{j} \) is a unit vector due north. Position vectors are given relative to a fixed origin \( O \).]

At time \( t \) seconds, \( t \ge 1 \), the position vector of a particle \( P \) is \( \mathbf{r} \) metres, where

\[ \mathbf{r} = ct^{\frac{1}{2}}\mathbf{i} – \frac{3}{8}t^2\mathbf{j} \]

and \( c \) is a constant.

When \( t = 4 \), the bearing of \( P \) from \( O \) is \( 135^\circ \).

(a) Show that \( c = 3 \).

(3)

(b) Find the speed of \( P \) when \( t = 4 \).

(4)

When \( t = T \), \( P \) is accelerating in the direction of \( (-\mathbf{i} – 27\mathbf{j}) \).

(c) Find the value of \( T \).

(4)

✏️ Worked Solution

Part (a): Finding Constant c

💡 What this tells us:

A bearing of \( 135^\circ \) means the particle is South-East. In terms of coordinates, this means the vector makes an angle of \( 45^\circ \) with the positive x-axis (East) in the clockwise direction (or \( -45^\circ \)).

Mathematically, if the position is \( x\mathbf{i} + y\mathbf{j} \), then \( \tan(45^\circ) = \frac{|y|}{|x|} = 1 \). Since it is South-East, \( x \) is positive and \( y \) is negative, so \( y = -x \).

✏ Working:

Substitute \( t = 4 \) into \( \mathbf{r} \):

\[ \mathbf{r} = c(4)^{\frac{1}{2}}\mathbf{i} – \frac{3}{8}(4)^2\mathbf{j} \] \[ \mathbf{r} = c(2)\mathbf{i} – \frac{3}{8}(16)\mathbf{j} \] \[ \mathbf{r} = 2c\mathbf{i} – 6\mathbf{j} \]

Since the bearing is \( 135^\circ \) (South-East):

\[ \frac{\text{component } \mathbf{j}}{\text{component } \mathbf{i}} = \tan(-45^\circ) = -1 \] \[ \frac{-6}{2c} = -1 \] \[ -6 = -2c \] \[ 2c = 6 \] \[ c = 3 \]

✓ (B1 M1 A1)

Part (b): Finding Speed

💡 Strategy:

Speed is the magnitude of velocity. To get velocity \( \mathbf{v} \), we differentiate the position vector \( \mathbf{r} \) with respect to \( t \).

✏ Working:

Using \( c=3 \), \( \mathbf{r} = 3t^{\frac{1}{2}}\mathbf{i} – \frac{3}{8}t^2\mathbf{j} \).

Differentiate to find \( \mathbf{v} \):

\[ \mathbf{v} = \frac{d\mathbf{r}}{dt} = \frac{1}{2}(3)t^{-\frac{1}{2}}\mathbf{i} – 2\left(\frac{3}{8}\right)t\mathbf{j} \] \[ \mathbf{v} = \frac{3}{2}t^{-\frac{1}{2}}\mathbf{i} – \frac{3}{4}t\mathbf{j} \]

Substitute \( t=4 \):

\[ \mathbf{v} = \frac{3}{2}(4)^{-\frac{1}{2}}\mathbf{i} – \frac{3}{4}(4)\mathbf{j} \] \[ \mathbf{v} = \frac{3}{2}\left(\frac{1}{2}\right)\mathbf{i} – 3\mathbf{j} \] \[ \mathbf{v} = \frac{3}{4}\mathbf{i} – 3\mathbf{j} \]

Calculate speed (magnitude):

\[ |\mathbf{v}| = \sqrt{\left(\frac{3}{4}\right)^2 + (-3)^2} \] \[ |\mathbf{v}| = \sqrt{\frac{9}{16} + 9} \] \[ |\mathbf{v}| = \sqrt{\frac{9 + 144}{16}} = \sqrt{\frac{153}{16}} \] \[ |\mathbf{v}| = \frac{\sqrt{153}}{4} \approx 3.09 \text{ ms}^{-1} \]

✓ (M1 A1 M1 A1)

Part (c): Acceleration Direction

💡 Strategy:

First, find acceleration \( \mathbf{a} \) by differentiating \( \mathbf{v} \). Then, set the direction of \( \mathbf{a} \) equal to the direction of \( (-\mathbf{i} – 27\mathbf{j}) \). This means the ratio of their components must be equal.

✏ Working:

Differentiate \( \mathbf{v} = \frac{3}{2}t^{-\frac{1}{2}}\mathbf{i} – \frac{3}{4}t\mathbf{j} \):

\[ \mathbf{a} = \frac{d\mathbf{v}}{dt} = -\frac{1}{2}\left(\frac{3}{2}\right)t^{-\frac{3}{2}}\mathbf{i} – \frac{3}{4}\mathbf{j} \] \[ \mathbf{a} = -\frac{3}{4}t^{-\frac{3}{2}}\mathbf{i} – \frac{3}{4}\mathbf{j} \]

The vector \( \mathbf{a} \) is parallel to \( -\mathbf{i} – 27\mathbf{j} \). Ratio of components must match:

\[ \frac{\text{j-component}}{\text{i-component}} = \frac{-27}{-1} = 27 \]

Using our acceleration vector at \( t=T \):

\[ \frac{-\frac{3}{4}}{-\frac{3}{4}T^{-\frac{3}{2}}} = 27 \] \[ \frac{1}{T^{-\frac{3}{2}}} = 27 \] \[ T^{\frac{3}{2}} = 27 \]

Solve for \( T \):

\[ T = 27^{\frac{2}{3}} \] \[ T = (3^3)^{\frac{2}{3}} = 3^2 = 9 \]

✓ (M1 A1 M1 A1)

🎯 Final Answers:

(a) Proof shown

(b) \( 3.09 \text{ ms}^{-1} \) (or \( \frac{3\sqrt{17}}{4} \))

(c) \( T = 9 \)

✓ Total: 11 marks

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Question 5 (12 marks)

At time \( t = 0 \), a small stone is projected with velocity \( 35 \text{ ms}^{-1} \) from a point \( O \) on horizontal ground.

The stone is projected at an angle \( \alpha \) to the horizontal, where \( \tan\alpha = \frac{3}{4} \).

In an initial model:

the stone is modelled as a particle \( P \) moving freely under gravity
the stone hits the ground at the point \( A \)

For the motion of \( P \) from \( O \) to \( A \):

at time \( t \) seconds, the horizontal distance of \( P \) from \( O \) is \( x \) metres
at time \( t \) seconds, the vertical distance of \( P \) above the ground is \( y \) metres

(a) Using the model, show that

\[ y = \frac{3}{4}x – \frac{1}{160}x^2 \]

(6)

(b) Use the answer to (a), or otherwise, to find the length \( OA \).

(2)

Using the model, the greatest height of the stone above the ground is found to be \( H \) metres.

(c) Use the answer to (a), or otherwise, to find the value of \( H \).

(2)

The model is refined to include air resistance. Using this new model, the greatest height is found to be \( K \) metres.

(d) State which is greater, \( H \) or \( K \), justifying your answer.

(1)

(e) State one limitation of this refined model.

(1)

✏️ Worked Solution

Part (a): Equation of Trajectory

💡 Strategy:

We need to set up the equations for horizontal motion (\( x \)) and vertical motion (\( y \)) in terms of \( t \). Then we eliminate \( t \) to find the relationship between \( y \) and \( x \).

First, find \( \sin\alpha \) and \( \cos\alpha \) from \( \tan\alpha = \frac{3}{4} \) (a 3-4-5 triangle).

✏ Working:

Using 3-4-5 triangle: \( \sin\alpha = \frac{3}{5} \), \( \cos\alpha = \frac{4}{5} \).

Horizontal motion (constant velocity):

\[ x = (u \cos\alpha)t \] \[ x = 35 \left(\frac{4}{5}\right) t = 28t \implies t = \frac{x}{28} \]

Vertical motion (gravity \( -g \)):

\[ y = (u \sin\alpha)t – \frac{1}{2}gt^2 \] \[ y = 35 \left(\frac{3}{5}\right) t – \frac{1}{2}(9.8)t^2 \] \[ y = 21t – 4.9t^2 \]

Substitute \( t = \frac{x}{28} \) into the \( y \) equation:

\[ y = 21\left(\frac{x}{28}\right) – 4.9\left(\frac{x}{28}\right)^2 \] \[ y = \frac{21}{28}x – 4.9\left(\frac{x^2}{784}\right) \] \[ y = \frac{3}{4}x – \frac{4.9}{784}x^2 \]

Simplify \( \frac{4.9}{784} \): Note that \( \frac{4.9}{784} = \frac{49}{7840} = \frac{1}{160} \).

\[ y = \frac{3}{4}x – \frac{1}{160}x^2 \]

✓ (M1 A1 M1 A1 DM1 A1)

Part (b): Horizontal Range OA

💡 Strategy:

At point A (on the ground), the vertical height \( y = 0 \). We solve the equation from part (a) for \( x \).

✏ Working:

\[ 0 = \frac{3}{4}x – \frac{1}{160}x^2 \] \[ 0 = x \left(\frac{3}{4} – \frac{1}{160}x\right) \]

\( x = 0 \) (Point O) or:

\[ \frac{3}{4} – \frac{1}{160}x = 0 \] \[ \frac{3}{4} = \frac{x}{160} \] \[ x = \frac{3 \times 160}{4} = 3 \times 40 = 120 \] \[ OA = 120 \text{ m} \]

✓ (M1 A1)

Part (c): Greatest Height H

💡 Strategy:

The trajectory is a parabola. The maximum height occurs at the vertex. Due to symmetry (assuming flat ground), the vertex x-coordinate is half the range. Alternatively, complete the square or differentiate.

✏ Working:

Axis of symmetry is at \( x = \frac{120}{2} = 60 \).

Substitute \( x = 60 \) into equation for \( y \):

\[ H = \frac{3}{4}(60) – \frac{1}{160}(60)^2 \] \[ H = 45 – \frac{3600}{160} \] \[ H = 45 – 22.5 \] \[ H = 22.5 \text{ m} \]

✓ (M1 A1)

Parts (d) & (e): Model Refinement

(d) Comparison:

\( H > K \). Air resistance opposes motion and removes energy from the system, thus reducing the maximum height reached.

✓ (B1)

(e) Limitation:

Examples: Wind effects are not considered, or the stone’s dimensions/spin are ignored (modelled as a particle).

✓ (B1)

🎯 Final Answers:

(a) Proof shown

(b) \( 120 \text{ m} \)

(c) \( 22.5 \text{ m} \)

(d) \( H > K \)

✓ Total: 12 marks

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Question 6 (9 marks)

Figure 5 shows a uniform rod \( AB \) of mass \( M \) and length \( 2a \).

the rod has its end \( A \) on rough horizontal ground
the rod rests in equilibrium against a small smooth fixed horizontal peg \( P \)
the point \( C \) on the rod, where \( AC = 1.5a \), is the point of contact between the rod and the peg
the rod is at an angle \( \theta \) to the ground, where \( \tan\theta = \frac{4}{3} \)

The rod lies in a vertical plane perpendicular to the peg.

The magnitude of the normal reaction of the peg on the rod at \( C \) is \( S \).

(a) Show that \( S = \frac{2}{5}Mg \).

(3)

The coefficient of friction between the rod and the ground is \( \mu \).

Given that the rod is in limiting equilibrium,

(b) find the value of \( \mu \).

(6)

✏️ Worked Solution

Part (a): Moments to find S

💡 Strategy:

To eliminate the unknown forces at the ground (Reaction \( R \) and Friction \( F \)), we take moments about point \( A \).

The forces creating moments are:

The weight \( Mg \) acting downwards at the center of the rod (distance \( a \) from \( A \)).
The reaction \( S \) acting perpendicular to the rod at \( C \) (distance \( 1.5a \) from \( A \)).

First, find \( \cos\theta \) using the 3-4-5 triangle property from \( \tan\theta = \frac{4}{3} \).

✏ Working:

Given \( \tan\theta = \frac{4}{3} \), we have \( \sin\theta = \frac{4}{5} \) and \( \cos\theta = \frac{3}{5} \).

Take moments about \( A \):

\[ \text{Clockwise Moment} = \text{Anticlockwise Moment} \] \[ Mg \times (a \cos\theta) = S \times (1.5a) \]

Substitute \( \cos\theta = \frac{3}{5} \):

\[ Mg \cdot a \cdot \frac{3}{5} = S \cdot \frac{3}{2}a \]

Divide both sides by \( a \) and simplify:

\[ \frac{3}{5}Mg = \frac{3}{2}S \] \[ S = \frac{3}{5}Mg \times \frac{2}{3} \] \[ S = \frac{2}{5}Mg \]

✓ (M1 A1 A1)

Part (b): Coefficient of Friction

💡 Strategy:

We need to find the friction \( F \) and ground reaction \( R \). We do this by resolving forces horizontally and vertically. The reaction \( S \) acts perpendicular to the rod, so we must resolve \( S \) into horizontal and vertical components.

Since the rod makes angle \( \theta \) with the horizontal, the force \( S \) (perpendicular to rod) makes angle \( \theta \) with the vertical.

Horizontal component of \( S \): \( S \sin\theta \) (to the left)
Vertical component of \( S \): \( S \cos\theta \) (upwards)

✏ Working:

Resolve Horizontally (\( \rightarrow \)):

Friction \( F \) opposes the tendency to slip (rod tends to slip A to the right? No, check diagram. The rod leans right. Gravity pulls it down. The bottom A tends to slip away from the peg, i.e., to the left. So friction acts to the right? Let’s check the horizontal force of the peg. The peg pushes up-left on the rod. So the horizontal component of \( S \) pushes the rod left. To balance this, Friction \( F \) must act to the right.)

\[ F = S \sin\theta \] \[ F = \left(\frac{2}{5}Mg\right) \left(\frac{4}{5}\right) = \frac{8}{25}Mg \]

✓ (M1 A1)

Resolve Vertically (\( \uparrow \)):

\[ R + S \cos\theta = Mg \] \[ R + \left(\frac{2}{5}Mg\right) \left(\frac{3}{5}\right) = Mg \] \[ R + \frac{6}{25}Mg = Mg \] \[ R = Mg – \frac{6}{25}Mg = \frac{19}{25}Mg \]

✓ (M1 A1)

Limiting Equilibrium:

\[ F = \mu R \] \[ \frac{8}{25}Mg = \mu \left( \frac{19}{25}Mg \right) \] \[ \mu = \frac{8/25}{19/25} = \frac{8}{19} \]

Decimal approximation: \( \mu \approx 0.421 \)

✓ (M1 A1)

🎯 Final Answers:

(a) Proof shown

(b) \( \mu = \frac{8}{19} \) (or \( 0.421 \))

✓ Total: 9 marks

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Pearson Edexcel A Level Mechanics (June 2024)

Mark Scheme Legend

Table of Contents

Question 1 (3 marks)

✏️ Worked Solution

Part (a): Normal Reaction

Part (b): Limiting Friction

Question 2 (8 marks)

✏️ Worked Solution

Part (a): Acceleration

Part (b): Distance (First 18s)

Part (c): Finding U

Question 3 (7 marks)

✏️ Worked Solution

Part (a): Normal Reaction

Part (b): Acceleration

Part (c): Condition for Motion

Question 4 (11 marks)

✏️ Worked Solution

Part (a): Finding Constant c

Part (b): Finding Speed

Part (c): Acceleration Direction

Question 5 (12 marks)

✏️ Worked Solution

Part (a): Equation of Trajectory

Part (b): Horizontal Range OA

Part (c): Greatest Height H

Parts (d) & (e): Model Refinement

Question 6 (9 marks)

✏️ Worked Solution

Part (a): Moments to find S

Part (b): Coefficient of Friction