A Level - Edexcel - 2023 - Mechanics

Question 1 (3 marks)

A car is initially at rest on a straight horizontal road.

The car then accelerates along the road with a constant acceleration of \( 3.2 \, \text{m s}^{-2} \).

Find

(a) the speed of the car after \( 5 \, \text{s} \),

(b) the distance travelled by the car in the first \( 5 \, \text{s} \).

📝 Worked Solution

Part (a): Finding the Speed

💡 Why we do this:

Since the acceleration is constant, we can use the “suvat” equations of motion. We are given the initial velocity, acceleration, and time, and we need to find the final velocity.

✏ Working:

List the known variables:

\[ \begin{align} u &= 0 \, \text{m s}^{-1} \quad \text{(initially at rest)} \\ a &= 3.2 \, \text{m s}^{-2} \\ t &= 5 \, \text{s} \\ v &= ? \end{align} \]

Use the equation \( v = u + at \):

\[ v = 0 + 3.2 \times 5 \] \[ v = 16 \]

✓ (B1) Correct speed calculated

🔍 What this tells us:

The car is travelling at \( 16 \, \text{m s}^{-1} \) after 5 seconds.

Part (b): Finding the Distance

💡 Why we do this:

We need to find the displacement \( s \). We can use \( s = ut + \frac{1}{2}at^2 \) as it relates distance directly to the given information.

✏ Working:

Use the equation \( s = ut + \frac{1}{2}at^2 \):

\[ s = (0 \times 5) + \frac{1}{2} \times 3.2 \times 5^2 \] \[ s = 0 + 1.6 \times 25 \]

✓ (M1) Complete method to find s

\[ s = 40 \]

✓ (A1) Correct distance

🎯 Final Answer:

(a) \( 16 \, \text{m s}^{-1} \)

(b) \( 40 \, \text{m} \)

✓ Total: 3 marks

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Question 2 (4 marks)

A particle \( P \) has mass \( 5 \, \text{kg} \).

The particle is pulled along a rough horizontal plane by a horizontal force of magnitude \( 28 \, \text{N} \).

The only resistance to motion is a frictional force of magnitude \( F \) newtons, as shown in Figure 1.

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

The particle is accelerating along the plane at \( 1.4 \, \text{m s}^{-2} \).

(b) Find the value of \( F \).

The coefficient of friction between \( P \) and the plane is \( \mu \).

(c) Find the value of \( \mu \), giving your answer to 2 significant figures.

📝 Worked Solution

Part (a): Normal Reaction

💡 Why we do this:

The particle moves horizontally, meaning there is no vertical acceleration. Therefore, the vertical forces (forces up vs forces down) must be balanced.

✏ Working:

Resolve forces vertically:

\[ R = mg \]

Substitute \( m = 5 \, \text{kg} \) and \( g = 9.8 \, \text{m s}^{-2} \):

\[ R = 5 \times 9.8 \] \[ R = 49 \, \text{N} \]

✓ (B1) Correct reaction force

Part (b): Finding Friction F

💡 Why we do this:

We apply Newton’s Second Law (\( F_{\text{net}} = ma \)) horizontally. The pulling force acts forward, and friction acts backward.

✏ Working:

Equation of motion:

\[ 28 – F = ma \]

Substitute known values (\( m = 5 \), \( a = 1.4 \)):

\[ 28 – F = 5 \times 1.4 \]

✓ (M1) Correct equation of motion

\[ 28 – F = 7 \] \[ F = 28 – 7 \] \[ F = 21 \, \text{N} \]

✓ (A1) Correct friction force

Part (c): Coefficient of Friction

💡 Why we do this:

We use the definition of kinetic friction: \( F = \mu R \). We have calculated both \( F \) and \( R \).

✏ Working:

\[ \mu = \frac{F}{R} \] \[ \mu = \frac{21}{49} \]

Simplify fraction:

\[ \mu = \frac{3}{7} \]

Convert to decimal (2 sig figs required):

\[ \mu \approx 0.42857… \] \[ \mu = 0.43 \]

✓ (B1 ft) Correct value to 2 s.f.

🎯 Final Answer:

(a) \( 49 \, \text{N} \)

(b) \( 21 \, \text{N} \)

(c) \( \mu = 0.43 \)

✓ Total: 4 marks

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Question 3 (9 marks)

At time \( t \) seconds, where \( t \ge 0 \), a particle \( P \) has velocity \( \mathbf{v} \, \text{m s}^{-1} \) where

\[ \mathbf{v} = (t^2 – 3t + 7)\mathbf{i} + (2t^2 – 3)\mathbf{j} \]

Find

(a) the speed of \( P \) at time \( t = 0 \)

(b) the value of \( t \) when \( P \) is moving parallel to \( (\mathbf{i} + \mathbf{j}) \)

(c) the acceleration of \( P \) at time \( t \) seconds

(d) the value of \( t \) when the direction of the acceleration of \( P \) is perpendicular to \( \mathbf{i} \)

📝 Worked Solution

Part (a): Initial Speed

💡 Why we do this:

First, we find the velocity vector at \( t=0 \) by substitution. Then, speed is the magnitude of this velocity vector, calculated using Pythagoras’ theorem.

✏ Working:

Substitute \( t=0 \) into \( \mathbf{v} \):

\[ \mathbf{v} = (0 – 0 + 7)\mathbf{i} + (0 – 3)\mathbf{j} \] \[ \mathbf{v} = 7\mathbf{i} – 3\mathbf{j} \]

✓ (B1) Correct velocity vector

Calculate magnitude (speed):

\[ |\mathbf{v}| = \sqrt{7^2 + (-3)^2} \] \[ |\mathbf{v}| = \sqrt{49 + 9} = \sqrt{58} \]

✓ (M1) Use of Pythagoras

\[ \text{Speed} \approx 7.62 \, \text{m s}^{-1} \]

✓ (A1) Correct speed

Part (b): Parallel Motion

💡 Why we do this:

If a vector is parallel to \( \mathbf{i} + \mathbf{j} \), its \( \mathbf{i} \) component must equal its \( \mathbf{j} \) component (ratio 1:1).

✏ Working:

Equate the \( \mathbf{i} \) and \( \mathbf{j} \) components:

\[ t^2 – 3t + 7 = 2t^2 – 3 \]

✓ (M1) Equating components

Rearrange into a quadratic equation:

\[ 0 = t^2 + 3t – 10 \]

Factorise:

\[ (t + 5)(t – 2) = 0 \]

Solutions are \( t = -5 \) or \( t = 2 \). Since \( t \ge 0 \):

\[ t = 2 \]

✓ (A1) Correct value of t

Part (c): Acceleration

💡 Why we do this:

Acceleration is the rate of change of velocity. We need to differentiate the velocity vector \( \mathbf{v} \) with respect to time \( t \).

✏ Working:

\[ \mathbf{a} = \frac{d\mathbf{v}}{dt} \] \[ \mathbf{a} = \frac{d}{dt}(t^2 – 3t + 7)\mathbf{i} + \frac{d}{dt}(2t^2 – 3)\mathbf{j} \]

✓ (M1) Attempt to differentiate v

\[ \mathbf{a} = (2t – 3)\mathbf{i} + 4t\mathbf{j} \]

✓ (A1) Correct acceleration vector

Part (d): Direction of Acceleration

💡 Why we do this:

If the acceleration is perpendicular to \( \mathbf{i} \) (the horizontal unit vector), it must have no horizontal component. This means the \( \mathbf{i} \) component of acceleration is zero.

✏ Working:

Set the \( \mathbf{i} \) component of \( \mathbf{a} \) to zero:

\[ 2t – 3 = 0 \]

✓ (M1) Set i-component to zero

\[ 2t = 3 \] \[ t = 1.5 \]

✓ (A1) Correct value of t

🎯 Final Answer:

(a) \( \sqrt{58} \) or \( 7.6 \) \( \text{m s}^{-1} \)

(b) \( t = 2 \)

(c) \( (2t – 3)\mathbf{i} + 4t\mathbf{j} \)

(d) \( t = 1.5 \)

✓ Total: 9 marks

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Question 4 (10 marks)

[In this question, \( \mathbf{i} \) and \( \mathbf{j} \) are horizontal unit vectors and position vectors are given relative to a fixed origin \( O \)]

A particle \( P \) is moving on a smooth horizontal plane.

The particle has constant acceleration \( (2.4\mathbf{i} + \mathbf{j}) \, \text{m s}^{-2} \)

At time \( t = 0 \), \( P \) passes through the point \( A \).

At time \( t = 5 \, \text{s} \), \( P \) passes through the point \( B \).

The velocity of \( P \) as it passes through \( A \) is \( (-16\mathbf{i} – 3\mathbf{j}) \, \text{m s}^{-1} \)

(a) Find the speed of \( P \) as it passes through \( B \).

The position vector of \( A \) is \( (44\mathbf{i} – 10\mathbf{j}) \, \text{m} \).

At time \( t = T \) seconds, where \( T > 5 \), \( P \) passes through the point \( C \).

The position vector of \( C \) is \( (4\mathbf{i} + c\mathbf{j}) \, \text{m} \).

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

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

📝 Worked Solution

Part (a): Speed at B

💡 Why we do this:

Acceleration is constant, so we can use vector suvat equation \( \mathbf{v} = \mathbf{u} + \mathbf{a}t \). We know \( \mathbf{u} \) (velocity at A), \( \mathbf{a} \), and \( t=5 \).

✏ Working:

\[ \mathbf{v}_B = \mathbf{u} + \mathbf{a}t \] \[ \mathbf{v}_B = (-16\mathbf{i} – 3\mathbf{j}) + 5(2.4\mathbf{i} + \mathbf{j}) \]

✓ (M1) Use of v = u + at

\[ \mathbf{v}_B = (-16 + 12)\mathbf{i} + (-3 + 5)\mathbf{j} \] \[ \mathbf{v}_B = -4\mathbf{i} + 2\mathbf{j} \]

✓ (A1) Correct velocity vector

Calculate speed (magnitude):

\[ |\mathbf{v}_B| = \sqrt{(-4)^2 + 2^2} = \sqrt{16 + 4} = \sqrt{20} \]

✓ (M1) Pythagoras

\[ \text{Speed} \approx 4.47 \, \text{m s}^{-1} \]

✓ (A1) Correct speed

Part (b): Finding T

💡 Why we do this:

We need the position vector \( \mathbf{r} \) at time \( T \). We use \( \mathbf{r} = \mathbf{r}_0 + \mathbf{u}t + \frac{1}{2}\mathbf{a}t^2 \), where \( \mathbf{r}_0 \) is the initial position at A. We can then equate the \( \mathbf{i} \) component to 4 to find \( T \).

✏ Working:

\[ \mathbf{r}_C = \mathbf{r}_A + \mathbf{u}T + \frac{1}{2}\mathbf{a}T^2 \] \[ 4\mathbf{i} + c\mathbf{j} = (44\mathbf{i} – 10\mathbf{j}) + (-16\mathbf{i} – 3\mathbf{j})T + \frac{1}{2}(2.4\mathbf{i} + \mathbf{j})T^2 \]

✓ (M1) Vector equation set up

Equate \( \mathbf{i} \) components:

\[ 4 = 44 – 16T + \frac{1}{2}(2.4)T^2 \] \[ 4 = 44 – 16T + 1.2T^2 \]

Rearrange to form a quadratic:

\[ 1.2T^2 – 16T + 40 = 0 \]

✓ (A1) Correct quadratic in T

Multiply by 10 to clear decimals:

\[ 12T^2 – 160T + 400 = 0 \]

Divide by 4:

\[ 3T^2 – 40T + 100 = 0 \]

Factorise:

\[ (3T – 10)(T – 10) = 0 \]

So \( T = \frac{10}{3} \) or \( T = 10 \).

Since \( T > 5 \):

\[ T = 10 \]

✓ (A1) Correct value of T

Part (c): Finding c

💡 Why we do this:

Now that we know \( T = 10 \), we substitute this into the \( \mathbf{j} \) component equation to find \( c \).

✏ Working:

Equate \( \mathbf{j} \) components:

\[ c = -10 – 3T + \frac{1}{2}(1)T^2 \]

✓ (M1) Equation for c

Substitute \( T = 10 \):

\[ c = -10 – 3(10) + 0.5(100) \] \[ c = -10 – 30 + 50 \] \[ c = 10 \]

✓ (A1) Correct value of c

🎯 Final Answer:

(a) \( 4.5 \, \text{m s}^{-1} \) (or \( 2\sqrt{5} \))

(b) \( T = 10 \)

(c) \( c = 10 \)

✓ Total: 10 marks

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Question 5 (11 marks)

A small ball is projected with speed \( 28 \, \text{m s}^{-1} \) from a point \( O \) on horizontal ground.

After moving for \( T \) seconds, the ball passes through the point \( A \).

The point \( A \) is \( 40 \, \text{m} \) horizontally and \( 20 \, \text{m} \) vertically from the point \( O \), as shown in Figure 2.

Given that the ball is projected at an angle \( \alpha \) to the ground, use the model to

(a) show that \( T = \frac{10}{7 \cos \alpha} \)

(b) show that \( \tan^2 \alpha – 4 \tan \alpha + 3 = 0 \)

(c) find the greatest possible height, in metres, of the ball above the ground as the ball moves from \( O \) to \( A \).

The model does not include air resistance.

(d) State one other limitation of the model.

📝 Worked Solution

Part (a): Horizontal Motion

💡 Why we do this:

Horizontal acceleration is zero. We use \( \text{distance} = \text{speed} \times \text{time} \) with the horizontal component of velocity \( u \cos \alpha \).

✏ Working:

Horizontal component of velocity: \( u_x = 28 \cos \alpha \)

Horizontal distance: \( x = 40 \)

\[ x = u_x T \] \[ 40 = (28 \cos \alpha) T \]

✓ (M1) Equation for horizontal motion

Rearrange for \( T \):

\[ T = \frac{40}{28 \cos \alpha} \]

Simplify fraction (divide by 4):

\[ T = \frac{10}{7 \cos \alpha} \]

✓ (A1*) Result shown correctly

Part (b): Vertical Motion & Trigonometry

💡 Why we do this:

We use the vertical displacement equation \( s = ut + \frac{1}{2}at^2 \). Then we substitute the expression for \( T \) derived in part (a) and use the identity \( \sec^2 \alpha = 1 + \tan^2 \alpha \) to form an equation solely in \( \tan \alpha \).

✏ Working:

Vertical motion (upwards positive, \( a = -g \)):

\[ 20 = (28 \sin \alpha)T – \frac{1}{2}g T^2 \]

Substitute \( T = \frac{10}{7 \cos \alpha} \) and \( g = 9.8 \):

\[ 20 = 28 \sin \alpha \left( \frac{10}{7 \cos \alpha} \right) – 4.9 \left( \frac{10}{7 \cos \alpha} \right)^2 \]

✓ (M1) Substitute T into vertical equation

Simplify terms:

\[ 20 = 40 \frac{\sin \alpha}{\cos \alpha} – 4.9 \left( \frac{100}{49 \cos^2 \alpha} \right) \] \[ 20 = 40 \tan \alpha – 10 \frac{1}{\cos^2 \alpha} \] \[ 20 = 40 \tan \alpha – 10 \sec^2 \alpha \]

Divide by 10:

\[ 2 = 4 \tan \alpha – \sec^2 \alpha \]

Use \( \sec^2 \alpha = 1 + \tan^2 \alpha \):

\[ 2 = 4 \tan \alpha – (1 + \tan^2 \alpha) \]

✓ (M1) Use trig identity

\[ 2 = 4 \tan \alpha – 1 – \tan^2 \alpha \]

Rearrange to form quadratic:

\[ \tan^2 \alpha – 4 \tan \alpha + 3 = 0 \]

✓ (A1*) Result shown

Part (c): Greatest Height

💡 Why we do this:

First, we solve the quadratic to find possible angles \( \alpha \). The greatest height for projectile motion is given by \( H = \frac{u^2 \sin^2 \alpha}{2g} \). We should test which angle gives the larger height.

✏ Working:

Solve \( \tan^2 \alpha – 4 \tan \alpha + 3 = 0 \):

\[ (\tan \alpha – 3)(\tan \alpha – 1) = 0 \] \[ \tan \alpha = 3 \quad \text{or} \quad \tan \alpha = 1 \]

For max height, we want the larger vertical component, so larger \( \alpha \). \( \tan \alpha = 3 \) gives a larger angle than \( \tan \alpha = 1 \).

If \( \tan \alpha = 3 \), construct triangle: opp=3, adj=1, hyp=\( \sqrt{10} \). So \( \sin \alpha = \frac{3}{\sqrt{10}} \).

Formula for max height:

\[ H = \frac{u^2 \sin^2 \alpha}{2g} \] \[ H = \frac{28^2 \times \left(\frac{3}{\sqrt{10}}\right)^2}{2 \times 9.8} \]

✓ (M1) Use max height formula or conservation of energy

\[ H = \frac{784 \times 0.9}{19.6} \] \[ H = \frac{705.6}{19.6} \] \[ H = 36 \, \text{m} \]

✓ (A1) Correct height

Part (d): Limitations

Any one of:

The ball is not a particle (it has size/dimensions).
Spin of the ball is ignored.
Wind effects are ignored.
Variable gravity is ignored (though negligible here).

✓ (B1) Valid limitation

🎯 Final Answer:

(c) 36 m

(d) e.g. Dimensions of the ball are ignored

✓ Total: 11 marks

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Question 6 (13 marks)

A rod \( AB \) has mass \( M \) and length \( 2a \).

The rod has its end \( A \) on rough horizontal ground and its end \( B \) against a smooth vertical wall.

The rod makes an angle \( \theta \) with the ground, as shown in Figure 3.

The rod is at rest in limiting equilibrium.

(a) State the direction (left or right on Figure 3 above) of the frictional force acting on the rod at \( A \). Give a reason for your answer.

The magnitude of the normal reaction of the wall on the rod at \( B \) is \( S \).

In an initial model, the rod is modelled as being uniform.

Use this initial model to answer parts (b), (c) and (d).

(b) By taking moments about \( A \), show that

\[ S = \frac{1}{2} Mg \cot \theta \]

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

Given that \( \tan \theta = \frac{3}{4} \)

(c) find the value of \( \mu \)

(d) find, in terms of \( M \) and \( g \), the magnitude of the resultant force acting on the rod at \( A \).

In a new model, the rod is modelled as being non-uniform, with its centre of mass closer to \( B \) than it is to \( A \).

A new value for \( S \) is calculated using this new model, with \( \tan \theta = \frac{3}{4} \)

(e) State whether this new value for \( S \) is larger, smaller or equal to the value that \( S \) would take using the initial model. Give a reason for your answer.

📝 Worked Solution

Part (a): Direction of Friction

💡 Why we do this:

Friction always opposes the direction of motion or the tendency of motion. If the wall is smooth, the rod would naturally slip down and the bottom end A would slide to the left (away from the wall).

✏ Working:

Right (towards the wall).

Reason: The rod tends to slip to the left, so friction opposes this motion.

✓ (B1) Correct direction and reason

Part (b): Moments about A

💡 Why we do this:

Taking moments about A eliminates the unknown reaction force \( R \) and friction \( F \) at A. We consider the weight \( Mg \) acting at the midpoint (uniform rod) and the reaction \( S \) at B.

✏ Working:

Clockwise Moment = Anticlockwise Moment

Weight \( Mg \) acts at distance \( a \) from A. The perpendicular distance is \( a \cos \theta \).

Reaction \( S \) acts horizontally at B. The perpendicular vertical distance is \( 2a \sin \theta \).

\[ S \times (2a \sin \theta) = Mg \times (a \cos \theta) \]

✓ (M1) Moments equation

Divide by \( a \) and rearrange for \( S \):

\[ 2S \sin \theta = Mg \cos \theta \] \[ S = \frac{Mg \cos \theta}{2 \sin \theta} \]

Use \( \cot \theta = \frac{\cos \theta}{\sin \theta} \):

\[ S = \frac{1}{2} Mg \cot \theta \]

✓ (A1*) Result shown

Part (c): Finding \( \mu \)

💡 Why we do this:

We need to find the friction force \( F \) and the normal reaction \( R \) at A. We resolve forces horizontally and vertically. Since equilibrium is limiting, \( F = \mu R \).

✏ Working:

Resolve Vertically (\( \uparrow = \downarrow \)):

\[ R = Mg \]

✓ (B1) Vertical resolution

Resolve Horizontally (\( \leftarrow = \rightarrow \)):

\[ F = S \]

✓ (B1) Horizontal resolution

Using the result from (b):

\[ F = \frac{1}{2} Mg \cot \theta \]

Given \( \tan \theta = \frac{3}{4} \), so \( \cot \theta = \frac{4}{3} \):

\[ F = \frac{1}{2} Mg \left( \frac{4}{3} \right) = \frac{2}{3} Mg \]

Using limiting friction \( F = \mu R \):

\[ \frac{2}{3} Mg = \mu (Mg) \] \[ \mu = \frac{2}{3} \]

✓ (A1) Correct value

Part (d): Resultant Force at A

💡 Why we do this:

The total force at A is the vector sum of the normal reaction \( R \) (vertical) and friction \( F \) (horizontal). We use Pythagoras.

✏ Working:

\[ \text{Resultant} = \sqrt{R^2 + F^2} \]

Substitute \( R = Mg \) and \( F = \frac{2}{3} Mg \):

\[ \text{Resultant} = \sqrt{(Mg)^2 + \left( \frac{2}{3} Mg \right)^2} \] \[ \text{Resultant} = \sqrt{(Mg)^2 + \frac{4}{9} (Mg)^2} \] \[ \text{Resultant} = Mg \sqrt{1 + \frac{4}{9}} \] \[ \text{Resultant} = Mg \sqrt{\frac{13}{9}} \] \[ \text{Resultant} = \frac{\sqrt{13}}{3} Mg \]

✓ (A1) Correct magnitude

Part (e): New Model

💡 Why we do this:

We analyze the moments equation again. If the centre of mass moves closer to B, the distance from A (let’s call it \( d \)) increases (\( d > a \)).

✏ Working:

Moments equation becomes: \( S \times 2a \sin \theta = Mg \times d \cos \theta \)

\[ S = \frac{Mg \, d \cot \theta}{2a} \]

Since the centre of mass is closer to B, \( d > a \).

Therefore, \( S \) will be larger.

Reason: The moment of the weight about A increases.

✓ (B1) Correct conclusion and reason

🎯 Final Answer:

(c) \( \mu = \frac{2}{3} \) (or 0.67)

(d) \( \frac{\sqrt{13}}{3} Mg \)

(e) Larger

✓ Total: 13 marks

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Pearson Edexcel GCE A Level Mathematics Paper 32 Mechanics (Summer 2023)

💡 Mark Scheme Legend

📋 Table of Contents

Question 1 (3 marks)

📝 Worked Solution

Part (a): Finding the Speed

Part (b): Finding the Distance

Question 2 (4 marks)

📝 Worked Solution

Part (a): Normal Reaction

Part (b): Finding Friction F

Part (c): Coefficient of Friction

Question 3 (9 marks)

📝 Worked Solution

Part (a): Initial Speed

Part (b): Parallel Motion

Part (c): Acceleration

Part (d): Direction of Acceleration

Question 4 (10 marks)

📝 Worked Solution

Part (a): Speed at B

Part (b): Finding T

Part (c): Finding c

Question 5 (11 marks)

📝 Worked Solution

Part (a): Horizontal Motion

Part (b): Vertical Motion & Trigonometry

Part (c): Greatest Height

Part (d): Limitations

Question 6 (13 marks)

📝 Worked Solution

Part (a): Direction of Friction

Part (b): Moments about A

Part (c): Finding \( \mu \)

Part (d): Resultant Force at A

Part (e): New Model