If any of my solutions look wrong, please refer to the mark scheme. You can exit full-screen mode for the question paper and mark scheme by clicking the icon in the bottom-right corner or by pressing Esc on your keyboard.

Edexcel A Level Mechanics (June 2019)

Mark Scheme Legend

M marks: Method marks for knowing a method and attempting to apply it.
A marks: Accuracy marks (dependent on M marks).
B marks: Unconditional accuracy marks (independent of M marks).
ft: Follow through marks.
cao: Correct answer only.

Question 1 (Kinematics – Calculus)
Question 2 (Kinematics – Constant Acceleration)
Question 3 (Forces – Inclined Plane)
Question 4 (Statics – Rod on Drum)
Question 5 (Projectiles)

Question 1 (6 marks)

[In this question position vectors are given relative to a fixed origin \(O\)]

At time \(t\) seconds, where \(t \geqslant 0\), a particle, \(P\), moves so that its velocity \(\mathbf{v} \, \mathrm{m\,s}^{-1}\) is given by

\[ \mathbf{v} = 6t\mathbf{i} – 5t^{\frac{3}{2}}\mathbf{j} \]

When \(t=0\), the position vector of \(P\) is \((-20\mathbf{i} + 20\mathbf{j}) \, \mathrm{m}\).

(a) Find the acceleration of \(P\) when \(t=4\).

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

Worked Solution

Part (a): Find the acceleration

Why we do this: Acceleration is the rate of change of velocity. To find the acceleration vector \(\mathbf{a}\) from the velocity vector \(\mathbf{v}\), we need to differentiate \(\mathbf{v}\) with respect to \(t\).

✏ Working:

Given: \(\mathbf{v} = 6t\mathbf{i} – 5t^{\frac{3}{2}}\mathbf{j}\)

Differentiate each component with respect to \(t\):

\[ \mathbf{a} = \frac{d\mathbf{v}}{dt} = \frac{d}{dt}(6t)\mathbf{i} – \frac{d}{dt}(5t^{\frac{3}{2}})\mathbf{j} \] \[ \mathbf{a} = 6\mathbf{i} – 5 \left(\frac{3}{2}t^{\frac{1}{2}}\right)\mathbf{j} \] \[ \mathbf{a} = 6\mathbf{i} – \frac{15}{2}t^{\frac{1}{2}}\mathbf{j} \]

✓ (M1 for differentiating, A1 for correct expression)

Substitute \(t=4\):

\[ \mathbf{a} = 6\mathbf{i} – \frac{15}{2}(4)^{\frac{1}{2}}\mathbf{j} \] \[ \mathbf{a} = 6\mathbf{i} – \frac{15}{2}(2)\mathbf{j} \] \[ \mathbf{a} = 6\mathbf{i} – 15\mathbf{j} \]

✓ (A1)

Part (b): Find the position vector

Why we do this: Position is found by integrating velocity. We also need to use the initial condition (\(t=0\)) to find the constant of integration.

✏ Working:

Integrate \(\mathbf{v}\) with respect to \(t\):

\[ \mathbf{r} = \int \mathbf{v} \, dt = \int (6t\mathbf{i} – 5t^{\frac{3}{2}}\mathbf{j}) \, dt \] \[ \mathbf{r} = \left(\frac{6t^2}{2}\right)\mathbf{i} – \left(\frac{5t^{\frac{5}{2}}}{\frac{5}{2}}\right)\mathbf{j} + \mathbf{C} \] \[ \mathbf{r} = 3t^2\mathbf{i} – 2t^{\frac{5}{2}}\mathbf{j} + \mathbf{C} \]

✓ (M1 for integrating, A1 for correct expression)

Use initial condition: When \(t=0\), \(\mathbf{r} = -20\mathbf{i} + 20\mathbf{j}\).

\[ (-20\mathbf{i} + 20\mathbf{j}) = 3(0)^2\mathbf{i} – 2(0)^{\frac{5}{2}}\mathbf{j} + \mathbf{C} \] \[ \mathbf{C} = -20\mathbf{i} + 20\mathbf{j} \]

Substitute \(\mathbf{C}\) back into the expression for \(\mathbf{r}\) and evaluate at \(t=4\):

\[ \mathbf{r} = (3(4)^2 – 20)\mathbf{i} + (-2(4)^{\frac{5}{2}} + 20)\mathbf{j} \] \[ \mathbf{r} = (3(16) – 20)\mathbf{i} + (-2(32) + 20)\mathbf{j} \] \[ \mathbf{r} = (48 – 20)\mathbf{i} + (-64 + 20)\mathbf{j} \] \[ \mathbf{r} = 28\mathbf{i} – 44\mathbf{j} \]

Final Answers:

(a) \( \mathbf{a} = (6\mathbf{i} – 15\mathbf{j}) \, \mathrm{m\,s}^{-2} \)

(b) \( \mathbf{r} = (28\mathbf{i} – 44\mathbf{j}) \, \mathrm{m} \)

✓ Total: 6 marks

↑ Back to Top

Question 2 (8 marks)

A particle, \(P\), moves with constant acceleration \((2\mathbf{i} – 3\mathbf{j}) \, \mathrm{m\,s}^{-2}\).

At time \(t=0\), the particle is at the point \(A\) and is moving with velocity \((-\mathbf{i} + 4\mathbf{j}) \, \mathrm{m\,s}^{-1}\).

At time \(t=T\) seconds, \(P\) is moving in the direction of vector \((3\mathbf{i} – 4\mathbf{j})\).

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

At time \(t=4\) seconds, \(P\) is at the point \(B\).

(b) Find the distance \(AB\).

Worked Solution

Part (a): Find the time T

Why we do this: Since acceleration is constant, we can use the vector equation \(\mathbf{v} = \mathbf{u} + \mathbf{a}t\). “Moving in the direction of \((3\mathbf{i} – 4\mathbf{j})\)” means the velocity vector is parallel to this vector. This implies the ratio of the \(\mathbf{j}\) component to the \(\mathbf{i}\) component is the same, or \(\mathbf{v} = k(3\mathbf{i} – 4\mathbf{j})\).

✏ Working:

Given: \(\mathbf{u} = -\mathbf{i} + 4\mathbf{j}\), \(\mathbf{a} = 2\mathbf{i} – 3\mathbf{j}\).

At time \(T\):

\[ \mathbf{v} = \mathbf{u} + \mathbf{a}T \] \[ \mathbf{v} = (-\mathbf{i} + 4\mathbf{j}) + (2\mathbf{i} – 3\mathbf{j})T \] \[ \mathbf{v} = (-1 + 2T)\mathbf{i} + (4 – 3T)\mathbf{j} \]

✓ (M1 for using v=u+at, A1 for correct expression)

This velocity is parallel to \(3\mathbf{i} – 4\mathbf{j}\). We can equate the ratios of components:

\[ \frac{\mathbf{j}\text{-component}}{\mathbf{i}\text{-component}} = \frac{-4}{3} \] \[ \frac{4 – 3T}{-1 + 2T} = \frac{-4}{3} \]

✓ (M1 for ratio)

Solve for \(T\):

\[ 3(4 – 3T) = -4(-1 + 2T) \] \[ 12 – 9T = 4 – 8T \] \[ 12 – 4 = 9T – 8T \] \[ 8 = T \]

✓ (A1)

Part (b): Find distance AB

Why we do this: The distance \(AB\) is the magnitude of the displacement vector \(\mathbf{s}\) at \(t=4\). We use the equation \(\mathbf{s} = \mathbf{u}t + \frac{1}{2}\mathbf{a}t^2\).

✏ Working:

Using \(t=4\):

\[ \mathbf{s} = \mathbf{u}t + \frac{1}{2}\mathbf{a}t^2 \] \[ \mathbf{s} = (-\mathbf{i} + 4\mathbf{j})(4) + \frac{1}{2}(2\mathbf{i} – 3\mathbf{j})(4^2) \] \[ \mathbf{s} = (-4\mathbf{i} + 16\mathbf{j}) + 8(2\mathbf{i} – 3\mathbf{j}) \] \[ \mathbf{s} = (-4\mathbf{i} + 16\mathbf{j}) + (16\mathbf{i} – 24\mathbf{j}) \] \[ \mathbf{s} = 12\mathbf{i} – 8\mathbf{j} \]

✓ (M1 for using suvat, A1 for correct vector)

Now find the magnitude of \(\mathbf{s}\) (distance):

\[ |\mathbf{s}| = \sqrt{12^2 + (-8)^2} \] \[ |\mathbf{s}| = \sqrt{144 + 64} \] \[ |\mathbf{s}| = \sqrt{208} \] \[ |\mathbf{s}| = \sqrt{16 \times 13} = 4\sqrt{13} \approx 14.4 \]

✓ (M1 for Pythagoras, A1 for answer)

Final Answers:

(a) \( T = 8 \)

(b) \( 4\sqrt{13} \, \mathrm{m} \) or \( 14.4 \, \mathrm{m} \)

✓ Total: 8 marks

↑ Back to Top

Question 3 (12 marks)

Two blocks, \(A\) and \(B\), of masses \(2m\) and \(3m\) respectively, are attached to the ends of a light string. Initially \(A\) is held at rest on a fixed rough plane. The plane is inclined at angle \(\alpha\) to the horizontal ground, where \(\tan \alpha = \frac{5}{12}\).

The string passes over a small smooth pulley, \(P\), fixed at the top of the plane. The part of the string from \(A\) to \(P\) is parallel to a line of greatest slope of the plane. Block \(B\) hangs freely below \(P\), as shown in Figure 1.

The coefficient of friction between \(A\) and the plane is \(\frac{2}{3}\).

The blocks are released from rest with the string taut and \(A\) moves up the plane. The tension in the string immediately after the blocks are released is \(T\).

(a) Show that \(T = \frac{12mg}{5}\).

After \(B\) reaches the ground, \(A\) continues to move up the plane until it comes to rest before reaching \(P\).

(b) Determine whether \(A\) will remain at rest, carefully justifying your answer.

Worked Solution

Part (a): Find Tension T

Why we do this: We need to apply Newton’s Second Law (\(F=ma\)) to both blocks separately. Because they are connected by an inextensible string, they share the same acceleration magnitude \(a\) and tension \(T\).

Key Trig Values: Since \(\tan \alpha = \frac{5}{12}\), this is a 5-12-13 triangle.

\[ \sin \alpha = \frac{5}{13}, \quad \cos \alpha = \frac{12}{13} \]

✏ Working:

For Block A (on the plane):

Resolve forces perpendicular to plane to find Reaction \(R\):

\[ R = 2mg \cos \alpha = 2mg \left(\frac{12}{13}\right) = \frac{24mg}{13} \]

Calculate Friction \(F_r\). Since moving, \(F_r = \mu R\):

\[ F_r = \frac{2}{3} R = \frac{2}{3} \left(\frac{24mg}{13}\right) = \frac{16mg}{13} \]

Equation of motion parallel to plane (upwards):

\[ T – F_r – 2mg \sin \alpha = 2ma \] \[ T – \frac{16mg}{13} – 2mg\left(\frac{5}{13}\right) = 2ma \] \[ T – \frac{16mg}{13} – \frac{10mg}{13} = 2ma \] \[ T – \frac{26mg}{13} = 2ma \] \[ T – 2mg = 2ma \quad \text{— (1)} \]

✓ (M1 A1)

For Block B (hanging):

Equation of motion (downwards):

\[ 3mg – T = 3ma \quad \text{— (2)} \]

✓ (M1 A1)

Solve Simultaneous Equations:

From (1): \(ma = \frac{T – 2mg}{2}\). Substitute into (2):

\[ 3mg – T = 3\left(\frac{T – 2mg}{2}\right) \]

Multiply by 2:

\[ 6mg – 2T = 3(T – 2mg) \] \[ 6mg – 2T = 3T – 6mg \] \[ 12mg = 5T \] \[ T = \frac{12mg}{5} \]

✓ (A1 Given Answer)

Part (b): Will A remain at rest?

Why we do this: When \(B\) hits the ground, the string goes slack, so \(T=0\). \(A\) moves up, stops, and then we must check if the component of weight down the slope is enough to overcome the maximum possible friction. If gravity down slope > max friction, it slides down.

✏ Working:

Forces acting down the plane:

\[ \text{Weight component} = 2mg \sin \alpha = 2mg \left(\frac{5}{13}\right) = \frac{10mg}{13} \]

Maximum available friction (acting up the plane to prevent sliding down):

\[ F_{\text{max}} = \mu R = \frac{16mg}{13} \]

Compare forces:

\[ \frac{16mg}{13} > \frac{10mg}{13} \] \[ \text{Max Friction} > \text{Weight Component} \]

Since the maximum friction is greater than the force trying to pull it down the slope, friction can match the weight component.

Conclusion: A will not move.

✓ (M1 A1)

Part (c): Refinements

Why we do this: Real-life physics involves factors ignored in simple models.

Possible refinements:

Include the mass/weight of the string.
Consider the friction at the pulley (pulley not smooth).
Include air resistance.
Do not model blocks as particles (consider dimensions/rotation).
String is extensible.

✓ (B1 B1)

Final Answers:

(a) Proof shown.

(b) It will remain at rest.

✓ Total: 12 marks

↑ Back to Top

Question 4 (11 marks)

A ramp, \(AB\), of length \(8 \, \mathrm{m}\) and mass \(20 \, \mathrm{kg}\), rests in equilibrium with the end \(A\) on rough horizontal ground.

The ramp rests on a smooth solid cylindrical drum which is partly under the ground. The drum is fixed with its axis at the same horizontal level as \(A\).

The point of contact between the ramp and the drum is \(C\), where \(AC = 5 \, \mathrm{m}\), as shown in Figure 2.

The ramp is resting in a vertical plane which is perpendicular to the axis of the drum, at an angle \(\theta\) to the horizontal, where \(\tan \theta = \frac{7}{24}\).

The ramp is modelled as a uniform rod.

(a) Explain why the reaction from the drum on the ramp at point \(C\) acts in a direction which is perpendicular to the ramp.

(b) Find the magnitude of the resultant force acting on the ramp at \(A\).

The ramp is still in equilibrium in the position shown in Figure 2 but the ramp is not now modelled as being uniform.

Given that the centre of mass of the ramp is assumed to be closer to \(A\) than to \(B\),

Worked Solution

Part (a): Direction of Reaction

Why we do this: The direction of a reaction force depends on friction. If a surface is smooth, it cannot exert a force parallel to the surface (friction).

The drum is smooth (no friction), so the reaction force must be normal (perpendicular) to the tangent of the surface at the point of contact.

✓ (B1)

Part (b): Resultant Force at A

Why we do this: We need to find the ground reaction components (Normal \(N_A\) and Friction \(F_A\)) at \(A\). Since the rod is in equilibrium, the sum of moments is zero and the sum of forces is zero. Taking moments about \(A\) eliminates the unknown forces at \(A\) and allows us to find the reaction at \(C\).

Trig Values: \(\tan \theta = \frac{7}{24}\). This is a 7-24-25 triangle.

\[ \sin \theta = \frac{7}{25}, \quad \cos \theta = \frac{24}{25} \]

✏ Working:

Let \(R\) be the reaction at \(C\) (perpendicular to rod). Mass \(20 \, \mathrm{kg}\) acts at midpoint (\(4 \, \mathrm{m}\) from \(A\)).

1. Take Moments about A:

\[ R \times 5 = 20g \times (4 \cos \theta) \] \[ 5R = 80g \left(\frac{24}{25}\right) \] \[ 5R = \frac{1920g}{25} = 76.8g \] \[ R = 15.36g \]

✓ (M1 A1 for moments equation)

2. Resolve Forces:

Let \(H\) be horizontal component and \(V\) be vertical component of force at \(A\).

Resolve Horizontally: \(R\) acts at \(90^\circ\) to rod, so angle with vertical is \(\theta\). Horizontal component of \(R\) is \(R \sin \theta\).

\[ H = R \sin \theta \] \[ H = 15.36g \left(\frac{7}{25}\right) \approx 4.3g \approx 42.1 \, \mathrm{N} \]

Resolve Vertically: Vertical component of \(R\) is \(R \cos \theta\).

\[ V + R \cos \theta = 20g \] \[ V = 20g – 15.36g \left(\frac{24}{25}\right) \] \[ V = 20g – 14.7456g = 5.2544g \approx 51.5 \, \mathrm{N} \]

✓ (M1 A1 for resolving)

3. Resultant Magnitude:

\[ \text{Resultant} = \sqrt{H^2 + V^2} \] \[ \text{Resultant} = \sqrt{(R \sin \theta)^2 + (20g – R \cos \theta)^2} \]

Alternatively, using vector subtraction:

\[ \text{Force at A} + \mathbf{R} + \mathbf{W} = 0 \] \[ |Force at A| = |-(\mathbf{R} + \mathbf{W})| \]

Calculation:

\[ H = 42.15 \] \[ V = 51.49 \] \[ \text{Resultant} = \sqrt{42.15^2 + 51.49^2} \approx \sqrt{1776 + 2651} \approx \sqrt{4427} \] \[ \text{Resultant} \approx 66.5 \, \mathrm{N} \]

✓ (M1 A1)

Part (c): Effect of non-uniform mass

Why we do this: If the centre of mass moves closer to \(A\), the distance in the moments equation decreases.

The moment of the weight about \(A\) is \(20g \times d \cos \theta\).

If \(d\) (distance to CoM) decreases, the clockwise moment decreases.

Since \(5R = \text{Clockwise Moment}\), \(R\) must decrease.

✓ (B1)

Final Answers:

(a) Smooth surface implies no friction.

(b) \( 66.5 \, \mathrm{N} \) (or \( 67 \, \mathrm{N} \))

✓ Total: 11 marks

↑ Back to Top

Question 5 (13 marks)

The points \(A\) and \(B\) lie \(50 \, \mathrm{m}\) apart on horizontal ground.

At time \(t=0\) two small balls, \(P\) and \(Q\), are projected in the vertical plane containing \(AB\).

Ball \(P\) is projected from \(A\) with speed \(20 \, \mathrm{m\,s}^{-1}\) at \(30^\circ\) to \(AB\).

Ball \(Q\) is projected from \(B\) with speed \(u \, \mathrm{m\,s}^{-1}\) at angle \(\theta\) to \(BA\), as shown in Figure 3.

At time \(t=2\) seconds, \(P\) and \(Q\) collide.

Until they collide, the balls are modelled as particles moving freely under gravity.

(a) Find the velocity of \(P\) at the instant before it collides with \(Q\).

(b) Find (i) the size of angle \(\theta\), (ii) the value of \(u\).

Worked Solution

Part (a): Velocity of P at t=2

Why we do this: The velocity has two components: horizontal (constant) and vertical (affected by gravity). We calculate both at \(t=2\).

✏ Working:

Initial velocity of P:

\[ u_x = 20 \cos 30^\circ = 10\sqrt{3} \] \[ u_y = 20 \sin 30^\circ = 10 \]

Velocity at \(t=2\):

Horizontal component \(v_x = u_x = 10\sqrt{3} \approx 17.32\).

Vertical component \(v_y = u_y – gt = 10 – 9.8(2)\):

\[ v_y = 10 – 19.6 = -9.6 \, \mathrm{m\,s}^{-1} \]

✓ (M1 A1)

Magnitude (Speed):

\[ v = \sqrt{(10\sqrt{3})^2 + (-9.6)^2} = \sqrt{300 + 92.16} = \sqrt{392.16} \approx 19.8 \, \mathrm{m\,s}^{-1} \]

Direction:

\[ \alpha = \tan^{-1}\left(\frac{9.6}{10\sqrt{3}}\right) \approx 29^\circ \text{ below horizontal} \]

Velocity is \(19.8 \, \mathrm{m\,s}^{-1}\) at \(29^\circ\) below horizontal.

✓ (M1 A1)

Part (b): Find u and theta

Why we do this: For a collision to occur at \(t=2\), the sum of horizontal distances travelled must equal \(50 \, \mathrm{m}\), and the vertical heights must be equal.

✏ Working:

Horizontal Motion:

\[ x_P + x_Q = 50 \] \[ (20 \cos 30^\circ \times 2) + (u \cos \theta \times 2) = 50 \] \[ 40 \left(\frac{\sqrt{3}}{2}\right) + 2u \cos \theta = 50 \] \[ 20\sqrt{3} + 2u \cos \theta = 50 \] \[ 2u \cos \theta = 50 – 20\sqrt{3} \] \[ u \cos \theta = 25 – 10\sqrt{3} \approx 7.68 \quad \text{— (1)} \]

✓ (M1 A1)

Vertical Motion:

Heights are equal at \(t=2\):

\[ (20 \sin 30^\circ \times 2 – \frac{1}{2}g(2^2)) = (u \sin \theta \times 2 – \frac{1}{2}g(2^2)) \]

The \(-\frac{1}{2}gt^2\) terms cancel out (standard property of projectiles colliding at same time):

\[ 20 \sin 30^\circ \times 2 = u \sin \theta \times 2 \] \[ 20(0.5) = u \sin \theta \] \[ 10 = u \sin \theta \quad \text{— (2)} \]

✓ (M1 A1)

Solve (1) and (2):

Divide (2) by (1) to find \(\theta\):

\[ \tan \theta = \frac{10}{25 – 10\sqrt{3}} \] \[ \tan \theta = \frac{10}{7.679…} \approx 1.302 \] \[ \theta \approx 52.5^\circ \]

Find \(u\) using (2):

\[ u = \frac{10}{\sin 52.5^\circ} \approx 12.6 \, \mathrm{m\,s}^{-1} \]

✓ (M1 A1)

Part (c): Limitation

Limitations include:

Size of the balls (modelled as particles).
Spin of the balls.
Wind effects.

✓ (B1)

Final Answers:

(a) \( 19.8 \, \mathrm{m\,s}^{-1} \) at \( 29^\circ \) below horizontal.

(b) (i) \( \theta = 52.5^\circ \), (ii) \( u = 12.6 \)

✓ Total: 13 marks

↑ Back to Top

Edexcel A Level Mechanics (June 2019)

Mark Scheme Legend

Table of Contents

Question 1 (6 marks)

Worked Solution

Part (a): Find the acceleration

Part (b): Find the position vector

Question 2 (8 marks)

Worked Solution

Part (a): Find the time T

Part (b): Find distance AB

Question 3 (12 marks)

Worked Solution

Part (a): Find Tension T

Part (b): Will A remain at rest?

Part (c): Refinements

Question 4 (11 marks)

Worked Solution

Part (a): Direction of Reaction

Part (b): Resultant Force at A

Part (c): Effect of non-uniform mass

Question 5 (13 marks)

Worked Solution

Part (a): Velocity of P at t=2

Part (b): Find u and theta

Part (c): Limitation