A Level - Edexcel - 2020 - Mechanics

Question 1 (9 marks)

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

A brick \(P\) of mass \(m\) is placed on the plane.

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

Brick \(P\) is in equilibrium and on the point of sliding down the plane.

Brick \(P\) is modelled as a particle.

Using the model,

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

(2)

(b) show that \(\mu = \frac{3}{4}\)

(4)

For parts (c) and (d), you are not required to do any further calculations.

Brick \(P\) is now removed from the plane and a much heavier brick \(Q\) is placed on the plane.

The coefficient of friction between \(Q\) and the plane is also \(\frac{3}{4}\).

(c) Explain briefly why brick \(Q\) will remain at rest on the plane.

(1)

Brick \(Q\) is now projected with speed \(0.5 \text{ ms}^{-1}\) down a line of greatest slope of the plane.

Brick \(Q\) is modelled as a particle.

Using the model,

(d) describe the motion of brick \(Q\), giving a reason for your answer.

(2)

✏ Worked Solution

Part (a): Find the Normal Reaction R

💡 Strategy:

We resolve forces perpendicular to the inclined plane. The system is in equilibrium perpendicular to the plane, so the forces must balance.

First, we need \(\sin \alpha\) and \(\cos \alpha\) from \(\tan \alpha = \frac{3}{4}\).

✏ Working:

Given \(\tan \alpha = \frac{3}{4}\), this corresponds to a 3-4-5 triangle.

\[ \sin \alpha = \frac{3}{5}, \quad \cos \alpha = \frac{4}{5} \]

Resolving perpendicular to the plane (\(\nearrow\)):

\[ R – mg \cos \alpha = 0 \] \[ R = mg \cos \alpha \] \[ R = mg \left(\frac{4}{5}\right) \] \[ R = \frac{4}{5}mg \]

✓ (2 marks)

Part (b): Show Coefficient of Friction

💡 Strategy:

The brick is “on the point of sliding down”. This means it is in limiting equilibrium.

1. Friction is at its maximum: \(F = \mu R\).

2. Friction acts up the plane to oppose the tendency to slide down.

3. Forces parallel to the plane balance.

✏ Working:

Resolving parallel to the plane (\(\searrow\)):

\[ mg \sin \alpha – F = 0 \] \[ F = mg \sin \alpha \]

Using Limiting Friction \(F = \mu R\) and substituting our result for \(R\) from part (a):

\[ \mu \left(\frac{4}{5}mg\right) = mg \left(\frac{3}{5}\right) \]

Divide both sides by \(mg\):

\[ \frac{4}{5}\mu = \frac{3}{5} \] \[ \mu = \frac{3}{4} \]

✓ (4 marks)

Part (c): Heavier Brick Q

💡 Understanding:

We need to check if the equilibrium condition depends on mass. Look back at the equation: \(\mu R = mg \sin \alpha\).

✏ Working:

The condition for equilibrium is \(mg \sin \alpha = \mu (mg \cos \alpha)\).

The mass \(m\) cancels out from both sides of the equation.

Therefore, the equilibrium depends only on \(\alpha\) and \(\mu\), not on the mass.

✓ (1 mark)

Part (d): Motion of Q

💡 Strategy:

The brick is moving, so friction is limiting (\(F = \mu R\)). We need to find the resultant force down the plane to determine acceleration.

✏ Working:

Force down plane due to gravity: \(mg \sin \alpha = \frac{3}{5}mg\)

Resistive friction force (moving, so max friction): \(F_{max} = \mu R = \frac{3}{4}(\frac{4}{5}mg) = \frac{3}{5}mg\)

Resultant Force = Force down – Friction

\[ F_{net} = \frac{3}{5}mg – \frac{3}{5}mg = 0 \]

Since the resultant force is zero, acceleration is zero.

Conclusion: The brick moves with constant speed.

✓ (2 marks)

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Question 2 (8 marks)

A particle \(P\) moves with acceleration \((4\mathbf{i} – 5\mathbf{j}) \text{ ms}^{-2}\).

At time \(t = 0\), \(P\) is moving with velocity \((-2\mathbf{i} + 2\mathbf{j}) \text{ ms}^{-1}\).

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

(2)

At time \(t = 0\), \(P\) passes through the origin \(O\).

At time \(t = T\) seconds, where \(T > 0\), the particle \(P\) passes through the point \(A\).

The position vector of \(A\) is \((\lambda\mathbf{i} – 4.5\mathbf{j}) \text{ m}\) relative to \(O\), where \(\lambda\) is a constant.

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

(4)

(c) Hence find the value of \(\lambda\).

(2)

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Question 3 (12 marks)

(i) At time \(t\) seconds, where \(t \ge 0\), a particle \(P\) moves so that its acceleration \(\mathbf{a} \text{ ms}^{-2}\) is given by

\[ \mathbf{a} = (1 – 4t)\mathbf{i} + (3 – t^2)\mathbf{j} \]

At the instant when \(t = 0\), the velocity of \(P\) is \(36\mathbf{i} \text{ ms}^{-1}\).

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

(3)

(b) Find the value of \(t\) at the instant when \(P\) is moving in a direction perpendicular to \(\mathbf{i}\)

(3)

(ii) At time \(t\) seconds, where \(t \ge 0\), a particle \(Q\) moves so that its position vector \(\mathbf{r}\) metres, relative to a fixed origin \(O\), is given by

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

Find the value of \(t\) at the instant when the speed of \(Q\) is \(5 \text{ ms}^{-1}\)

(6)

✏ Worked Solution

Part (i)(a): Velocity at t=4

💡 Strategy:

Acceleration is a function of time, so we must integrate to find velocity.

\(\mathbf{v} = \int \mathbf{a} \, dt\).

Don’t forget the constant of integration \(\mathbf{C}\), which we find using the initial condition at \(t=0\).

✏ Working:

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

At \(t = 0\), \(\mathbf{v} = 36\mathbf{i}\). Substituting this in:

\[ 36\mathbf{i} = (0)\mathbf{i} + (0)\mathbf{j} + \mathbf{C} \implies \mathbf{C} = 36\mathbf{i} \]

So the full expression for velocity is:

\[ \mathbf{v} = (t – 2t^2 + 36)\mathbf{i} + (3t – \frac{1}{3}t^3)\mathbf{j} \]

Find \(\mathbf{v}\) at \(t = 4\):

\[ \mathbf{v} = (4 – 2(4)^2 + 36)\mathbf{i} + (3(4) – \frac{1}{3}(4)^3)\mathbf{j} \] \[ \mathbf{v} = (4 – 32 + 36)\mathbf{i} + (12 – \frac{64}{3})\mathbf{j} \] \[ \mathbf{v} = 8\mathbf{i} + \left(\frac{36}{3} – \frac{64}{3}\right)\mathbf{j} \] \[ \mathbf{v} = 8\mathbf{i} – \frac{28}{3}\mathbf{j} \text{ ms}^{-1} \]

✓ (3 marks)

Part (i)(b): Moving Perpendicular to i

💡 Strategy:

“Perpendicular to \(\mathbf{i}\)” means the particle is moving entirely in the \(\mathbf{j}\) direction (vertical).

This implies the \(\mathbf{i}\) component of velocity is zero.

✏ Working:

Set the \(\mathbf{i}\) component of \(\mathbf{v}\) to 0:

\[ t – 2t^2 + 36 = 0 \]

Rearrange for a positive \(t^2\):

\[ 2t^2 – t – 36 = 0 \]

Factorise (or use formula):

\[ (2t – 9)(t + 4) = 0 \]

Since \(t \ge 0\), ignore \(t = -4\).

\[ 2t = 9 \implies t = 4.5 \text{ s} \]

✓ (3 marks)

Part (ii): Speed of Q

💡 Strategy:

We are given position \(\mathbf{r}\). We need to:

1. Differentiate position to get velocity \(\mathbf{v} = \frac{d\mathbf{r}}{dt}\).

2. Find speed, which is the magnitude of velocity \(|\mathbf{v}|\).

3. Set magnitude equal to 5 and solve for \(t\).

✏ Working:

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

Differentiate w.r.t \(t\):

\[ \mathbf{v} = (2t – 1)\mathbf{i} + 3\mathbf{j} \]

Speed is magnitude:

\[ |\mathbf{v}| = \sqrt{(2t – 1)^2 + 3^2} \]

We are told speed = 5:

\[ \sqrt{(2t – 1)^2 + 9} = 5 \]

Square both sides:

\[ (2t – 1)^2 + 9 = 25 \] \[ (2t – 1)^2 = 16 \]

Square root both sides:

\[ 2t – 1 = \pm 4 \]

Case 1: \(2t – 1 = 4 \implies 2t = 5 \implies t = 2.5\)

Case 2: \(2t – 1 = -4 \implies 2t = -3 \implies t = -1.5\) (Reject as \(t \ge 0\))

Final Answer: \(t = 2.5 \text{ s}\)

✓ (6 marks)

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Question 4 (10 marks)

A ladder \(AB\) has mass \(M\) and length \(6a\).

The end \(A\) of the ladder is on rough horizontal ground.

The ladder rests against a fixed smooth horizontal rail at the point \(C\).

The point \(C\) is at a vertical height \(4a\) above the ground.

The vertical plane containing \(AB\) is perpendicular to the rail.

The ladder is inclined to the horizontal at an angle \(\alpha\), where \(\sin \alpha = \frac{4}{5}\).

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

The ladder rests in limiting equilibrium.

The ladder is modelled as a uniform rod.

Using the model,

(a) show that the magnitude of the force exerted on the ladder by the rail at \(C\) is \(\frac{9Mg}{25}\)

(3)

(b) Hence, or otherwise, find the value of \(\mu\).

(7)

✏ Worked Solution

Part (a): Find Reaction at C

💡 Strategy:

We take moments about A. This eliminates the unknown reaction \(R\) and friction \(F\) at the ground.

We need to know distances:

1. Ladder length \(6a\), Uniform weight acts at centre \(3a\).

2. Distance \(AC\). We know height of C is \(4a\) and \(\sin \alpha = \frac{4}{5}\).

Note: The force at the rail \(N\) acts perpendicular to the ladder because the rail is smooth.

✏ Working:

First, find distance \(AC\):

\[ \sin \alpha = \frac{\text{Opp}}{\text{Hyp}} = \frac{4a}{AC} \] \[ \frac{4}{5} = \frac{4a}{AC} \implies AC = 5a \]

Now take moments about A (\(\circlearrowleft = \circlearrowright\)):

The weight \(Mg\) acts at distance \(3a\) along ladder. Perpendicular distance = \(3a \cos \alpha\).

The reaction \(N\) acts at distance \(5a\) along ladder. It is already perpendicular to the ladder.

\[ N \times 5a = Mg \times (3a \cos \alpha) \]

Since \(\sin \alpha = \frac{4}{5}\), we have a 3-4-5 triangle, so \(\cos \alpha = \frac{3}{5}\).

\[ 5a N = Mg (3a) \left(\frac{3}{5}\right) \] \[ 5a N = \frac{9a}{5} Mg \]

Cancel \(a\) and divide by 5:

\[ N = \frac{9Mg}{25} \]

✓ (3 marks)

Part (b): Find Friction Coefficient μ

💡 Strategy:

We need to resolve forces horizontally and vertically.

1. Resolve \(\rightarrow\): Friction \(F\) must balance horizontal component of \(N\).

2. Resolve \(\uparrow\): Reaction \(R\) plus vertical component of \(N\) balances weight.

3. Use Limiting Friction: \(F = \mu R\).

✏ Working:

Resolve Horizontally (\(\rightarrow\)):

\[ F = N \sin \alpha \] \[ F = \left(\frac{9Mg}{25}\right) \left(\frac{4}{5}\right) = \frac{36Mg}{125} \]

Resolve Vertically (\(\uparrow\)):

\[ R + N \cos \alpha = Mg \] \[ R = Mg – N \cos \alpha \] \[ R = Mg – \left(\frac{9Mg}{25}\right) \left(\frac{3}{5}\right) \] \[ R = Mg – \frac{27Mg}{125} \] \[ R = \frac{125 – 27}{125}Mg = \frac{98Mg}{125} \]

Using \(F = \mu R\):

\[ \frac{36Mg}{125} = \mu \left(\frac{98Mg}{125}\right) \]

Cancel \(Mg/125\):

\[ 36 = 98 \mu \] \[ \mu = \frac{36}{98} = \frac{18}{49} \]

✓ (7 marks)

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Question 5 (11 marks)

A small ball is projected with speed \(U \text{ ms}^{-1}\) from a point \(O\) at the top of a vertical cliff.

The point \(O\) is \(25 \text{ m}\) vertically above the point \(N\) which is on horizontal ground.

The ball is projected at an angle of \(45^\circ\) above the horizontal.

The ball hits the ground at a point \(A\), where \(AN = 100 \text{ m}\).

The motion of the ball is modelled as that of a particle moving freely under gravity.

Using this initial model,

(a) show that \(U = 28\)

(6)

(b) find the greatest height of the ball above the horizontal ground \(NA\).

(3)

In a refinement to the model of the motion of the ball from \(O\) to \(A\), the effect of air resistance is included.

This refined model is used to find a new value of \(U\).

(c) How would this new value of \(U\) compare with 28, the value given in part (a)?

(1)

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

(1)

✏ Worked Solution

Part (a): Show U = 28

💡 Strategy:

We analyse horizontal and vertical motion separately.

1. Horizontal: Constant velocity. \(x = (U \cos 45^\circ) t\).

2. Vertical: Constant acceleration \(g\). \(y = (U \sin 45^\circ) t – \frac{1}{2}gt^2\).

We set the origin at \(O\). When it hits the ground at \(A\):

\(x = 100\)

\(y = -25\) (since it falls 25m below launch point).

✏ Working:

Horizontal motion:

\[ 100 = U \cos 45^\circ t \] \[ 100 = U \frac{1}{\sqrt{2}} t \implies t = \frac{100\sqrt{2}}{U} \]

Vertical motion (using \(s = ut + \frac{1}{2}at^2\) with downwards negative):

\[ -25 = U \sin 45^\circ t – \frac{1}{2}(9.8)t^2 \]

Substitute the expression for \(t\):

\[ -25 = U \left(\frac{1}{\sqrt{2}}\right) \left(\frac{100\sqrt{2}}{U}\right) – 4.9 \left(\frac{100\sqrt{2}}{U}\right)^2 \]

Simplify terms:

\[ -25 = 100 – 4.9 \left(\frac{20000}{U^2}\right) \]

Rearrange to solve for \(U^2\):

\[ 4.9 \left(\frac{20000}{U^2}\right) = 125 \] \[ \frac{98000}{U^2} = 125 \] \[ U^2 = \frac{98000}{125} \] \[ U^2 = 784 \] \[ U = \sqrt{784} = 28 \text{ ms}^{-1} \]

✓ (6 marks)

Part (b): Greatest Height above Ground

💡 Strategy:

At max height, vertical velocity \(v_y = 0\).

We find the max height above projection point O first, then add the 25m cliff height.

Use \(v^2 = u^2 + 2as\).

✏ Working:

Vertical initial velocity \(u_y = 28 \sin 45^\circ = 14\sqrt{2}\).

At max height \(v_y = 0\). Let height above O be \(h\).

\[ 0^2 = (14\sqrt{2})^2 – 2(9.8)h \] \[ 19.6h = 196 \times 2 \] \[ 19.6h = 392 \] \[ h = \frac{392}{19.6} = 20 \text{ m} \]

Total height above ground \(NA\):

\[ H = 20 + 25 = 45 \text{ m} \]

✓ (3 marks)

Part (c): Effect of Air Resistance

💡 Reasoning:

Air resistance opposes motion, slowing the ball down horizontally and vertically. To reach the same distance (100m) against this resistance, you would need to throw it harder.

✏ Answer:

The new value of \(U\) would be greater than 28.

✓ (1 mark)

Part (d): Further Refinement

✏ Answer:

Any one of:

Include the dimensions/size of the ball (do not model as a particle).
Include the effect of wind.
Include the effect of spin (Magnus effect).
Use a more accurate value for \(g\).

✓ (1 mark)

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A Level Mechanics 2020 – Paper 32

💡 Key to Solutions

Table of Contents

Question 1 (9 marks)

✏ Worked Solution

Part (a): Find the Normal Reaction R

Part (b): Show Coefficient of Friction

Part (c): Heavier Brick Q

Part (d): Motion of Q

Question 2 (8 marks)

✏ Worked Solution

Part (a): Velocity at t=2

Part (b): Find T

Part (c): Find Lambda

Question 3 (12 marks)

✏ Worked Solution

Part (i)(a): Velocity at t=4

Part (i)(b): Moving Perpendicular to i

Part (ii): Speed of Q

Question 4 (10 marks)

✏ Worked Solution

Part (a): Find Reaction at C

Part (b): Find Friction Coefficient μ

Question 5 (11 marks)

✏ Worked Solution

Part (a): Show U = 28

Part (b): Greatest Height above Ground

Part (c): Effect of Air Resistance

Part (d): Further Refinement