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Pearson Edexcel A-Level Mechanics 2022

Pearson Edexcel A-Level Mechanics (Summer 2022)

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Question 1 (8 marks)

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

\[ \mathbf{v} = 3t^2\mathbf{i} – 6t^{\frac{1}{2}}\mathbf{j} \]

(a) Find the speed of \( P \) at time \( t = 2 \) seconds.

(b) Find an expression, in terms of \( t, \mathbf{i} \) and \( \mathbf{j} \), for the acceleration of \( P \) at time \( t \) seconds, where \( t > 0 \).

At time \( t = 4 \) seconds, the position vector of \( P \) is \( (\mathbf{i} – 4\mathbf{j}) \) m.

Part (a): Speed at \( t=2 \)

Step 1: Calculate velocity vector at \( t=2 \)

💡 What are we asked to find? We need the speed. Speed is the magnitude of the velocity vector. First, we must find the velocity vector when \( t=2 \).

✏ Working:

Substitute \( t=2 \) into the expression for \( \mathbf{v} \):

\[ \mathbf{v} = 3(2)^2\mathbf{i} – 6(2)^{\frac{1}{2}}\mathbf{j} \] \[ \mathbf{v} = 3(4)\mathbf{i} – 6\sqrt{2}\mathbf{j} \] \[ \mathbf{v} = 12\mathbf{i} – 6\sqrt{2}\mathbf{j} \]

Step 2: Calculate the magnitude

💡 How do we find speed? Use Pythagoras’ theorem on the vector components.

✏ Working:

\[ \text{Speed} = \sqrt{12^2 + (-6\sqrt{2})^2} \] \[ \text{Speed} = \sqrt{144 + (36 \times 2)} \] \[ \text{Speed} = \sqrt{144 + 72} = \sqrt{216} \]

Simplifying or calculating:

\[ \sqrt{216} = \sqrt{36 \times 6} = 6\sqrt{6} \approx 14.7 \text{ m s}^{-1} \]

Part (b): Acceleration

Step 1: Differentiate velocity

💡 Why calculate this? Acceleration is the rate of change of velocity. We differentiate \( \mathbf{v} \) with respect to \( t \).

✏ Working:

\[ \mathbf{v} = 3t^2\mathbf{i} – 6t^{\frac{1}{2}}\mathbf{j} \] \[ \mathbf{a} = \frac{d\mathbf{v}}{dt} = \frac{d}{dt}(3t^2)\mathbf{i} – \frac{d}{dt}(6t^{\frac{1}{2}})\mathbf{j} \] \[ \mathbf{a} = (3 \times 2t)\mathbf{i} – (6 \times \frac{1}{2}t^{-\frac{1}{2}})\mathbf{j} \] \[ \mathbf{a} = 6t\mathbf{i} – 3t^{-\frac{1}{2}}\mathbf{j} \]

Part (c): Position Vector

Step 1: Integrate velocity

💡 Why integrate? Velocity is the rate of change of position (\( \frac{d\mathbf{r}}{dt} = \mathbf{v} \)). To get position \( \mathbf{r} \) from velocity, we integrate. Don’t forget the constant of integration \( \mathbf{C} \)!

✏ Working:

\[ \mathbf{r} = \int (3t^2\mathbf{i} – 6t^{\frac{1}{2}}\mathbf{j}) \, dt \] \[ \mathbf{r} = \left(\frac{3t^3}{3}\right)\mathbf{i} – \left(\frac{6t^{\frac{3}{2}}}{\frac{3}{2}}\right)\mathbf{j} + \mathbf{C} \] \[ \mathbf{r} = t^3\mathbf{i} – (6 \times \frac{2}{3})t^{\frac{3}{2}}\mathbf{j} + \mathbf{C} \] \[ \mathbf{r} = t^3\mathbf{i} – 4t^{\frac{3}{2}}\mathbf{j} + \mathbf{C} \]

Step 2: Find constant C

💡 Use the given condition: At \( t=4 \), \( \mathbf{r} = \mathbf{i} – 4\mathbf{j} \).

✏ Working:

Substitute \( t=4 \) and \( \mathbf{r} = \mathbf{i} – 4\mathbf{j} \):

\[ \mathbf{i} – 4\mathbf{j} = (4^3)\mathbf{i} – 4(4)^{\frac{3}{2}}\mathbf{j} + \mathbf{C} \] \[ \mathbf{i} – 4\mathbf{j} = 64\mathbf{i} – 4(8)\mathbf{j} + \mathbf{C} \] \[ \mathbf{i} – 4\mathbf{j} = 64\mathbf{i} – 32\mathbf{j} + \mathbf{C} \]

Rearrange for \( \mathbf{C} \):

\[ \mathbf{C} = (\mathbf{i} – 4\mathbf{j}) – (64\mathbf{i} – 32\mathbf{j}) \] \[ \mathbf{C} = -63\mathbf{i} + 28\mathbf{j} \]

Step 3: Find position at t=1

✏ Working:

Complete expression for \( \mathbf{r} \):

\[ \mathbf{r} = t^3\mathbf{i} – 4t^{\frac{3}{2}}\mathbf{j} – 63\mathbf{i} + 28\mathbf{j} \]

At \( t=1 \):

\[ \mathbf{r} = (1)^3\mathbf{i} – 4(1)^{\frac{3}{2}}\mathbf{j} – 63\mathbf{i} + 28\mathbf{j} \] \[ \mathbf{r} = \mathbf{i} – 4\mathbf{j} – 63\mathbf{i} + 28\mathbf{j} \] \[ \mathbf{r} = -62\mathbf{i} + 24\mathbf{j} \]

🏁 Final Answers:

(a) \( 6\sqrt{6} \) or \( 14.7 \text{ m s}^{-1} \)

(b) \( 6t\mathbf{i} – 3t^{-\frac{1}{2}}\mathbf{j} \)

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Question 2 (10 marks)

A rough plane is inclined to the horizontal at an angle \( \alpha \), where \( \tan \alpha = \frac{3}{4} \).

A small block \( B \) of mass 5 kg is held in equilibrium on the plane by a horizontal force of magnitude \( X \) newtons, as shown in Figure 1. The force acts in a vertical plane which contains a line of greatest slope of the inclined plane.

The block \( B \) is modelled as a particle. The magnitude of the normal reaction of the plane on \( B \) is 68.6 N.

Using the model,

(a) (i) find the magnitude of the frictional force acting on \( B \),

(ii) state the direction of the frictional force acting on \( B \).

The horizontal force of magnitude \( X \) newtons is now removed and \( B \) moves down the plane.

Given that the coefficient of friction between \( B \) and the plane is 0.5,

(b) find the acceleration of \( B \) down the plane.

Part (a): Static Equilibrium

Step 1: Analyze trigonometry

💡 3-4-5 Triangle: We are given \( \tan \alpha = \frac{3}{4} \). This corresponds to a right-angled triangle with opposite 3, adjacent 4, and hypotenuse 5.

✏ Working:

\[ \sin \alpha = \frac{3}{5} = 0.6 \] \[ \cos \alpha = \frac{4}{5} = 0.8 \]

Step 2: Resolve forces perpendicular to the plane

💡 Why start here? We usually start by finding the unknown force \( X \). We know the normal reaction \( R = 68.6 \). By resolving forces perpendicular to the plane, we can link \( R \), weight, and \( X \).

The components perpendicular to the plane are:

\( R \) (outwards)
\( 5g \cos \alpha \) (component of weight inwards)
\( X \sin \alpha \) (component of horizontal force inwards)

✏ Working:

\[ R = 5g \cos \alpha + X \sin \alpha \]

Substitute known values (\( R=68.6, g=9.8 \)):

\[ 68.6 = 5(9.8)(0.8) + X(0.6) \] \[ 68.6 = 39.2 + 0.6X \] \[ 0.6X = 68.6 – 39.2 \] \[ 0.6X = 29.4 \] \[ X = \frac{29.4}{0.6} = 49 \text{ N} \]

Step 3: Resolve forces parallel to the plane

💡 Finding Friction: Now we resolve parallel to the plane to find the friction \( F \). We don’t know the direction yet, so let’s calculate the driving forces.

Forces up the plane: \( X \cos \alpha \)
Forces down the plane: \( 5g \sin \alpha \)

✏ Working:

Upward force component: \( 49 \times 0.8 = 39.2 \text{ N} \)

Downward force component: \( 5 \times 9.8 \times 0.6 = 29.4 \text{ N} \)

Since the upward force (39.2) is greater than the downward force (29.4), friction must act down the plane to maintain equilibrium.

Equilibrium Equation:

\[ F_{\text{up}} = F_{\text{down}} \] \[ X \cos \alpha = 5g \sin \alpha + F \] \[ 39.2 = 29.4 + F \] \[ F = 39.2 – 29.4 = 9.8 \text{ N} \]

Part (b): Dynamics (Acceleration)

Step 1: New setup

💡 What changed? Force \( X \) is removed. The block moves down the plane. Friction opposes motion, so it acts up the plane.

We need to recalculate \( R \) because \( X \) is gone.

✏ Working:

Resolve perpendicular to plane:

\[ R = 5g \cos \alpha \] \[ R = 5(9.8)(0.8) = 39.2 \text{ N} \]

Calculate Friction (\( F = \mu R \)):

\[ F = 0.5 \times 39.2 = 19.6 \text{ N} \]

Step 2: Equation of Motion

💡 Newton’s Second Law: \( F_{net} = ma \)

Forces down the plane (positive direction): Component of weight.

Forces up the plane (negative direction): Friction.

✏ Working:

\[ 5g \sin \alpha – F = 5a \] \[ (5 \times 9.8 \times 0.6) – 19.6 = 5a \] \[ 29.4 – 19.6 = 5a \] \[ 9.8 = 5a \] \[ a = \frac{9.8}{5} = 1.96 \text{ m s}^{-2} \]

🏁 Final Answers:

(a)(i) \( 9.8 \text{ N} \)

(a)(ii) Down the plane

(b) \( 1.96 \text{ m s}^{-2} \)

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Question 3 (9 marks)

A particle \( P \) of mass 4 kg is at rest at the point \( A \) on a smooth horizontal plane.

At time \( t = 0 \), two forces, \( \mathbf{F}_1 = (4\mathbf{i} – \mathbf{j}) \text{ N} \) and \( \mathbf{F}_2 = (\lambda\mathbf{i} + \mu\mathbf{j}) \text{ N} \), where \( \lambda \) and \( \mu \) are constants, are applied to \( P \).

Given that \( P \) moves in the direction of the vector \( (3\mathbf{i} + \mathbf{j}) \),

(a) show that \( \lambda – 3\mu + 7 = 0 \).

At time \( t = 4 \) seconds, \( P \) passes through the point \( B \).

Given that \( \lambda = 2 \),

(b) find the length of \( AB \).

Part (a): Vector Ratios

Step 1: Find the resultant force

✏ Working:

\[ \mathbf{R} = \mathbf{F}_1 + \mathbf{F}_2 \] \[ \mathbf{R} = (4\mathbf{i} – \mathbf{j}) + (\lambda\mathbf{i} + \mu\mathbf{j}) \] \[ \mathbf{R} = (4 + \lambda)\mathbf{i} + (\mu – 1)\mathbf{j} \]

Step 2: Use the direction property

💡 Parallel Vectors: The particle moves in direction \( 3\mathbf{i} + \mathbf{j} \). This means the resultant force (and acceleration) must be a multiple of this vector.

If vector \( \mathbf{A} \) is parallel to vector \( \mathbf{B} \), the ratio of their components must be equal: \( \frac{x_A}{x_B} = \frac{y_A}{y_B} \).

✏ Working:

\[ \frac{4 + \lambda}{3} = \frac{\mu – 1}{1} \]

Cross multiply:

\[ 1(4 + \lambda) = 3(\mu – 1) \] \[ 4 + \lambda = 3\mu – 3 \] \[ \lambda – 3\mu + 4 + 3 = 0 \] \[ \lambda – 3\mu + 7 = 0 \]

(Shown)

Part (b): Kinematics

Step 1: Find acceleration vector

💡 Find inputs: We are given \( \lambda = 2 \). First, find \( \mu \) using the equation from (a), then find the force, then acceleration (\( F=ma \)).

✏ Working:

Using \( \lambda = 2 \):

\[ 2 – 3\mu + 7 = 0 \] \[ 3\mu = 9 \implies \mu = 3 \]

Resultant Force:

\[ \mathbf{R} = (4 + 2)\mathbf{i} + (3 – 1)\mathbf{j} = 6\mathbf{i} + 2\mathbf{j} \text{ N} \]

Newton’s Second Law (\( m = 4 \)):

\[ \mathbf{F} = m\mathbf{a} \] \[ 6\mathbf{i} + 2\mathbf{j} = 4\mathbf{a} \] \[ \mathbf{a} = 1.5\mathbf{i} + 0.5\mathbf{j} \text{ m s}^{-2} \]

Step 2: Calculate distance

💡 SUVAT: Start from rest (\( u=0 \)). Constant acceleration. Time \( t=4 \). We want displacement \( s \).

✏ Working:

\[ \mathbf{s} = \mathbf{u}t + \frac{1}{2}\mathbf{a}t^2 \] \[ \mathbf{s} = \mathbf{0}(4) + \frac{1}{2}(1.5\mathbf{i} + 0.5\mathbf{j})(4^2) \] \[ \mathbf{s} = 0 + 8(1.5\mathbf{i} + 0.5\mathbf{j}) \] \[ \mathbf{s} = 12\mathbf{i} + 4\mathbf{j} \]

Length of AB is the magnitude of \( \mathbf{s} \):

\[ |\mathbf{s}| = \sqrt{12^2 + 4^2} \] \[ |\mathbf{s}| = \sqrt{144 + 16} = \sqrt{160} \] \[ \sqrt{160} = \sqrt{16 \times 10} = 4\sqrt{10} \approx 12.6 \text{ m} \]

🏁 Final Answer:

\( 4\sqrt{10} \text{ m} \) (or 12.6 m)

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Question 4 (11 marks)

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

A particle of mass \( 2M \) is attached to the rod at the point \( C \), where \( AC = 1.5a \).

The rod rests with its end \( A \) on rough horizontal ground.

The rod is held in equilibrium at an angle \( \theta \) to the ground by a light string that is attached to the end \( B \) of the rod.

The string is perpendicular to the rod, as shown in Figure 2.

(a) Explain why the frictional force acting on the rod at \( A \) acts horizontally to the right on the diagram.

The tension in the string is \( T \).

(b) Show that \( T = 2Mg \cos \theta \).

Given that \( \cos \theta = \frac{3}{5} \),

The coefficient of friction between the rod and the ground is \( \mu \). Given that the rod is in limiting equilibrium,

(d) show that \( \mu = \frac{8}{19} \).

Part (a): Friction Direction

💡 Resolve Horizontally: To stay in equilibrium, horizontal forces must balance.

The tension \( T \) pulls perpendicular to the rod (up and to the left). Therefore, \( T \) has a horizontal component to the left.

To balance this, friction must act to the right.

Part (b): Moments about A

Step 1: Identify all moments

💡 Why moments about A? Taking moments about A eliminates the unknown reaction forces (Friction and Normal Reaction) at A. We only need to deal with Weight and Tension.

Forces creating clockwise moments:

Weight of rod (\( Mg \)) at distance \( a \). Perpendicular distance = \( a \cos \theta \).
Weight of particle (\( 2Mg \)) at distance \( 1.5a \). Perpendicular distance = \( 1.5a \cos \theta \).

Force creating anticlockwise moment:

Tension (\( T \)) at distance \( 2a \). Since the string is perpendicular to the rod, the distance is just \( 2a \).

✏ Working:

\[ \text{Clockwise} = \text{Anticlockwise} \] \[ Mg(a \cos \theta) + 2Mg(1.5a \cos \theta) = T(2a) \] \[ Mga \cos \theta + 3Mga \cos \theta = 2Ta \] \[ 4Mga \cos \theta = 2Ta \]

Divide by \( 2a \):

\[ 2Mg \cos \theta = T \] \[ T = 2Mg \cos \theta \]

(Shown)

Part (c): Vertical Force at A

Step 1: Resolve Vertically

💡 Identify vertical forces:

Upwards: Vertical reaction \( R \) at A + Vertical component of \( T \).
Downwards: Total weight \( Mg + 2Mg = 3Mg \).

Angle of string: The rod is at \( \theta \) to horizontal. The string is \( 90^\circ \) to rod. The angle of the string to the vertical is \( \theta \). The angle to the horizontal is \( 90 – \theta \).

Vertical component of \( T \): \( T \sin(90-\theta) \) or \( T \cos \theta \).

Wait, let’s check the geometry. Rod is angle \( \theta \) to horizontal. String is perpendicular. Angle of string to vertical is \( \theta \). So vertical component is \( T \cos \theta \).

✏ Working:

\[ R + T \cos \theta = 3Mg \]

We know \( T = 2Mg \cos \theta \), so substitute this in:

\[ R + (2Mg \cos \theta) \cos \theta = 3Mg \] \[ R = 3Mg – 2Mg \cos^2 \theta \]

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

\[ R = 3Mg – 2Mg \left(\frac{3}{5}\right)^2 \] \[ R = 3Mg – 2Mg \left(\frac{9}{25}\right) \] \[ R = 3Mg – \frac{18Mg}{25} \] \[ R = \frac{75Mg}{25} – \frac{18Mg}{25} \] \[ R = \frac{57Mg}{25} \]

(Shown)

Part (d): Coefficient of Friction

Step 1: Find Friction Force F

💡 Resolve Horizontally: \( F = \text{Horizontal component of } T \).

Horizontal component of \( T \) is \( T \sin \theta \).

First, if \( \cos \theta = \frac{3}{5} \), then \( \sin \theta = \frac{4}{5} \).

✏ Working:

\[ F = T \sin \theta \] \[ F = (2Mg \cos \theta) \sin \theta \] \[ F = 2Mg \left(\frac{3}{5}\right) \left(\frac{4}{5}\right) \] \[ F = \frac{24Mg}{25} \]

Step 2: Use Limiting Friction Formula

✏ Working:

\[ F = \mu R \] \[ \frac{24Mg}{25} = \mu \left( \frac{57Mg}{25} \right) \]

Cancel \( Mg \) and \( 25 \):

\[ 24 = 57\mu \] \[ \mu = \frac{24}{57} \]

Divide top and bottom by 3:

\[ \mu = \frac{8}{19} \]

(Shown)

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Question 5 (12 marks)

A golf ball is at rest at the point \( A \) on horizontal ground.

The ball is hit and initially moves at an angle \( \alpha \) to the ground.

The ball first hits the ground at the point \( B \), where \( AB = 120 \text{ m} \), as shown in Figure 3.

The motion of the ball is modelled as that of a particle, moving freely under gravity, whose initial speed is \( U \, \text{m s}^{-1} \).

Using this model,

(a) show that \( U^2 \sin \alpha \cos \alpha = 588 \).

The ball reaches a maximum height of 10 m above the ground.

(b) Show that \( U^2 = 1960 \).

In a refinement to the model, the effect of air resistance is included. The motion of the ball, from \( A \) to \( B \), is now modelled as that of a particle whose initial speed is \( V \, \text{m s}^{-1} \). This refined model is used to calculate a value for \( V \).

(d) State one further refinement to the model that would make the model more realistic.

Part (a): Horizontal Range

Step 1: Write expressions for Horizontal and Vertical Motion

💡 Strategy: Time of flight is determined by vertical motion (starting and ending at \( s_y = 0 \)). Range is determined by horizontal velocity \(\times\) time of flight.

Initial components:

\( u_x = U \cos \alpha \)
\( u_y = U \sin \alpha \)

✏ Working:

Vertical Motion (\( s = ut + \frac{1}{2}at^2 \)):

\[ 0 = (U \sin \alpha)T – \frac{1}{2}gT^2 \]

Since \( T \neq 0 \), we can divide by \( T \):

\[ 0 = U \sin \alpha – \frac{1}{2}gT \] \[ T = \frac{2U \sin \alpha}{g} \]

Horizontal Motion (Constant velocity):

\[ \text{Range} = (U \cos \alpha) \times T \] \[ 120 = (U \cos \alpha) \times \frac{2U \sin \alpha}{g} \] \[ 120 = \frac{2U^2 \sin \alpha \cos \alpha}{g} \]

Step 2: Rearrange to show result

✏ Working:

\[ 120g = 2U^2 \sin \alpha \cos \alpha \] \[ 60g = U^2 \sin \alpha \cos \alpha \]

Take \( g = 9.8 \):

\[ 60 \times 9.8 = U^2 \sin \alpha \cos \alpha \] \[ 588 = U^2 \sin \alpha \cos \alpha \]

(Shown)

Part (b): Max Height

Step 1: Use Vertical Energy/Suvat

💡 Max Height Condition: At max height, \( v_y = 0 \). Displacement \( s = 10 \).

✏ Working:

\[ v^2 = u^2 + 2as \] \[ 0 = (U \sin \alpha)^2 – 2g(10) \] \[ U^2 \sin^2 \alpha = 20g \] \[ U^2 \sin^2 \alpha = 20(9.8) = 196 \]

Step 2: Solve Simultaneous Equations

💡 We have two equations:

\( U^2 \sin \alpha \cos \alpha = 588 \)
\( U^2 \sin^2 \alpha = 196 \)

If we divide (2) by (1), the \( U^2 \) and one \( \sin \alpha \) cancel out, leaving \( \tan \alpha \).

✏ Working:

\[ \frac{U^2 \sin^2 \alpha}{U^2 \sin \alpha \cos \alpha} = \frac{196}{588} \] \[ \frac{\sin \alpha}{\cos \alpha} = \frac{196}{588} \] \[ \tan \alpha = \frac{1}{3} \]

Now we need \( U^2 \). From \( \tan \alpha = \frac{1}{3} \), we can find \( \sin \alpha \).

Imagine a triangle with opp=1, adj=3. Hypotenuse \( = \sqrt{1^2 + 3^2} = \sqrt{10} \).

\[ \sin \alpha = \frac{1}{\sqrt{10}} \] \[ \sin^2 \alpha = \frac{1}{10} \]

Substitute back into equation 2:

\[ U^2 \left(\frac{1}{10}\right) = 196 \] \[ U^2 = 1960 \]

(Shown)

Parts (c) & (d): Modelling

(c) Air Resistance: Air resistance opposes motion and removes energy from the system. To achieve the same distance (120m) against air resistance, you would need to hit the ball harder (faster) initially.

Therefore, \( V > U \).

(d) Refinement: Consider factors ignored by the particle model:

Size of the ball (dimensions).
Spin of the ball (Magnus effect).
Wind effects.

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