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Pearson Edexcel A Level Mathematics: Mechanics Paper 32 (October 2021)
Exam Guide
- Time allowed: 1 hour 30 minutes
- Total marks: 50
- Advice: Read each question carefully. Check your answers if you have time at the end.
Click “Show Solution” to see the full worked answer with step-by-step explanations.
Table of Contents
Question 1 (4 marks)
A particle \( P \) moves with constant acceleration \( (2\mathbf{i} – 3\mathbf{j}) \, \text{m s}^{-2} \).
At time \( t = 0 \), \( P \) is moving with velocity \( 4\mathbf{i} \, \text{m s}^{-1} \).
(a) Find the velocity of \( P \) at time \( t = 2 \) seconds. (2)
At time \( t = 0 \), the position vector of \( P \) relative to a fixed origin \( O \) is \( (\mathbf{i} + \mathbf{j}) \, \text{m} \).
(b) Find the position vector of \( P \) relative to \( O \) at time \( t = 3 \) seconds. (2)
Worked Solution
Step 1: Part (a) – Finding Velocity
What are we asked to find? We need to find the velocity vector at a specific time.
What info do we have?
- Constant acceleration: \( \mathbf{a} = 2\mathbf{i} – 3\mathbf{j} \)
- Initial velocity (\( t=0 \)): \( \mathbf{u} = 4\mathbf{i} \) (which is \( 4\mathbf{i} + 0\mathbf{j} \))
- Time: \( t=2 \)
Strategy: Since acceleration is constant, we can use the vector equation \( \mathbf{v} = \mathbf{u} + \mathbf{a}t \).
Using \( \mathbf{v} = \mathbf{u} + \mathbf{a}t \):
\[ \mathbf{v} = (4\mathbf{i}) + (2\mathbf{i} – 3\mathbf{j}) \times 2 \] \[ \mathbf{v} = 4\mathbf{i} + 4\mathbf{i} – 6\mathbf{j} \] \[ \mathbf{v} = (4+4)\mathbf{i} – 6\mathbf{j} \] \[ \mathbf{v} = 8\mathbf{i} – 6\mathbf{j} \]✓ (M1 for method, A1 for answer)
Step 2: Part (b) – Finding Position
Strategy: To find the position vector \( \mathbf{r} \), we can use the constant acceleration formula \( \mathbf{r} = \mathbf{r}_0 + \mathbf{u}t + \frac{1}{2}\mathbf{a}t^2 \).
Note: \( \mathbf{r}_0 \) is the initial position at \( t=0 \), given as \( \mathbf{i} + \mathbf{j} \).
Given \( t = 3 \):
\[ \mathbf{r} = (\mathbf{i} + \mathbf{j}) + (4\mathbf{i})(3) + \frac{1}{2}(2\mathbf{i} – 3\mathbf{j})(3^2) \]Simplify each term:
\[ \mathbf{r} = (\mathbf{i} + \mathbf{j}) + 12\mathbf{i} + \frac{1}{2}(2\mathbf{i} – 3\mathbf{j})(9) \] \[ \mathbf{r} = \mathbf{i} + \mathbf{j} + 12\mathbf{i} + 4.5(2\mathbf{i} – 3\mathbf{j}) \] \[ \mathbf{r} = \mathbf{i} + \mathbf{j} + 12\mathbf{i} + 9\mathbf{i} – 13.5\mathbf{j} \]Collect \( \mathbf{i} \) and \( \mathbf{j} \) components:
\[ \mathbf{r} = (1 + 12 + 9)\mathbf{i} + (1 – 13.5)\mathbf{j} \] \[ \mathbf{r} = 22\mathbf{i} – 12.5\mathbf{j} \]✓ (M1 for substitution, A1 for answer)
Final Answer:
(a) \( \mathbf{v} = 8\mathbf{i} – 6\mathbf{j} \, \text{m s}^{-1} \)
(b) \( \mathbf{r} = 22\mathbf{i} – 12.5\mathbf{j} \, \text{m} \)
Question 2 (12 marks)
A small stone \( A \) of mass \( 3m \) is attached to one end of a string. A small stone \( B \) of mass \( m \) is attached to the other end of the string.
Initially \( A \) is held at rest on a fixed rough plane. The plane is inclined to the horizontal at an angle \( \alpha \), where \( \tan \alpha = \frac{3}{4} \).
The string passes over a pulley \( P \) that is 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. Stone \( B \) hangs freely below \( P \).
The coefficient of friction between \( A \) and the plane is \( \frac{1}{6} \). Stone \( A \) is released from rest and begins to move down the plane.
(a) Write down an equation of motion for \( A \). (2)
(b) Show that the acceleration of \( A \) is \( \frac{1}{10}g \). (7)
(c) Sketch a velocity-time graph for the motion of \( B \), from the instant when \( A \) is released from rest to the instant just before \( B \) reaches the pulley, explaining your answer. (2)
(d) In reality, the string is not light. State how this would affect the working in part (b). (1)
Worked Solution
Step 1: Understanding the Forces & Geometry
Geometry: Given \( \tan \alpha = \frac{3}{4} \). This corresponds to a 3-4-5 triangle.
\( \sin \alpha = \frac{3}{5} = 0.6 \)
\( \cos \alpha = \frac{4}{5} = 0.8 \)
Motion: \( A \) moves down the plane (given). This means friction acts up the plane. \( B \) moves up.
Forces on A (mass 3m):
- Weight component down plane: \( 3mg \sin \alpha \)
- Tension \( T \) up plane
- Friction \( F \) up plane (opposing motion)
- Equation: \( F_{net} = ma \)
Step 2: Part (a) – Equation of Motion for A
Applying Newton’s Second Law parallel to the plane (downwards is positive):
\[ 3mg \sin \alpha – F – T = 3ma \]This is the required equation.
✓ (M1 for terms, A1 for correct equation)
Step 3: Part (b) – Finding Acceleration
We need to find \( F \) (Friction) and eliminate \( T \) (Tension). To find \( F \), we need the Reaction force \( R \).
Resolve perpendicular to plane for A: \( R = 3mg \cos \alpha \)
Friction Law: \( F = \mu R \)
Equation for B (mass m): B moves UP. Forces are Tension (up) and Weight (down).
1. Find Friction \( F \):
\[ R = 3mg(0.8) = 2.4mg \] \[ F = \frac{1}{6}R = \frac{1}{6}(2.4mg) = 0.4mg \]2. Equation for B (moving up):
\[ T – mg = ma \implies T = m(g + a) \]3. Substitute \( T \) and \( F \) into A’s equation:
\[ 3mg \sin \alpha – F – T = 3ma \] \[ 3mg(0.6) – 0.4mg – m(g + a) = 3ma \] \[ 1.8mg – 0.4mg – mg – ma = 3ma \]Divide by \( m \):
\[ 1.8g – 0.4g – g = 4a \] \[ 0.4g = 4a \] \[ a = \frac{0.4g}{4} = 0.1g \] \[ a = \frac{1}{10}g \]✓ Correctly shown
Step 4: Part (c) – Velocity-Time Graph
The acceleration is constant (\( 0.1g \)).
Since \( A \) starts from rest, velocity starts at 0.
For constant acceleration starting from rest, \( v = at \), which is a straight line through the origin.
Explanation: The acceleration is constant, so the gradient is constant. The particle starts from rest, so the line passes through the origin.
Step 5: Part (d) – Heavy String
If the string is not light, it has mass. As the string moves over the pulley, the amount of string (and thus mass) on each side changes.
The tension would be different at A and B, and the acceleration would not be constant (it would vary with time/position).
Answer: The tension in the string would not be constant throughout its length (or tensions at A and B would be different).
Final Answer:
(a) \( 3mg \sin \alpha – F – T = 3ma \)
(b) \( a = \frac{1}{10}g \)
(c) Straight line starting from origin with positive gradient.
(d) Tensions would be different at each end.
Question 3 (10 marks)
A beam \( AB \) has mass \( m \) and length \( 2a \). The beam rests in equilibrium with \( A \) on rough horizontal ground and with \( B \) against a smooth vertical wall.
The beam is inclined to the horizontal at an angle \( \theta \), as shown in Figure 2.
The coefficient of friction between the beam and the ground is \( \mu \).
The beam is modelled as a uniform rod resting in a vertical plane that is perpendicular to the wall.
(a) Show that \( \mu \ge \frac{1}{2} \cot \theta \). (5)
A horizontal force of magnitude \( kmg \), where \( k \) is a constant, is now applied to the beam at \( A \). This force acts in a direction that is perpendicular to the wall and towards the wall.
Given that \( \tan \theta = \frac{5}{4} \), \( \mu = \frac{1}{2} \) and the beam is now in limiting equilibrium,
(b) use the model to find the value of \( k \). (5)
Worked Solution
Step 1: Part (a) – Forces and Moments
Let’s identify the forces acting on the beam:
- At A: Normal reaction \( R \) (up), Friction \( F \) (towards wall, resisting slip away).
- At B: Normal reaction \( S \) (horizontal, from wall). Wall is smooth, so no friction.
- Weight \( mg \): Acting downwards at the midpoint (uniform rod).
For equilibrium, forces balance and moments balance.
Resolving Forces:
(\( \uparrow \)): \( R = mg \)
(\( \rightarrow \)): \( F = S \)
Taking Moments about A:
Weight \( mg \) at distance \( a \). Perpendicular distance = \( a \cos \theta \).
Reaction \( S \) at B at distance \( 2a \). Perpendicular distance = \( 2a \sin \theta \).
\[ mg(a \cos \theta) = S(2a \sin \theta) \]Divide by \( a \):
\[ mg \cos \theta = 2S \sin \theta \] \[ S = \frac{mg \cos \theta}{2 \sin \theta} = \frac{1}{2}mg \cot \theta \]Since \( F = S \), we have \( F = \frac{1}{2}mg \cot \theta \).
Friction Condition: \( F \le \mu R \)
\[ \frac{1}{2}mg \cot \theta \le \mu (mg) \] \[ \frac{1}{2} \cot \theta \le \mu \] \[ \mu \ge \frac{1}{2} \cot \theta \]✓ Shown
Step 2: Part (b) – New Force Applied
Situation:
- Force \( kmg \) applied at A towards the wall.
- \( \tan \theta = \frac{5}{4} \), so \( \cot \theta = \frac{4}{5} \), \( \sin \theta = \frac{5}{\sqrt{41}} \), \( \cos \theta = \frac{4}{\sqrt{41}} \).
- \( \mu = \frac{1}{2} \).
- Limiting equilibrium: Friction is at maximum \( F_{max} = \mu R \).
Direction of slipping? The applied force pushes towards the wall. This tends to make the ladder slip up the wall (or A moves towards wall). However, usually, with a push, we check moments.
Actually, adding a push at A towards the wall makes it likely to slip up the wall if the push is strong, or it prevents it from slipping down. The question says “limiting equilibrium”.
Standard case: Before push, it tends to slip down. The push resists this. Or the push is strong enough to make it slip up?
Let’s consider the Moments equation again. The moment of Weight tries to rotate it down. The moment of \( S \) and horizontal forces balance it.
Let’s look at the forces at A. Horizontal: Push \( P = kmg \) and Friction \( F \). Vertical: \( R \).
Wait, if we push A towards the wall, friction at A will oppose this motion, so Friction acts away from the wall.
Resolving (\( \rightarrow \)): \( P – F = S \) (if slipping towards wall) OR \( P + F = S \) (if slipping away)?
Actually, let’s stick to the physical intuition. Usually, without the force, it wants to slip away. The force \( kmg \) helps push it in. If it’s in limiting equilibrium now, it could be on the verge of slipping up (over-pushed) or still on the verge of slipping down (under-pushed but helped).
Let’s test the “verge of slipping up” scenario (Friction acts away from wall). And “verge of slipping down” (Friction acts towards wall).
Given \( k \) is a value we need to find. Let’s assume it’s on the verge of slipping up/inwards because we applied an extra force.
Direction of motion A: Towards wall. -> Friction \( F \) acts Away from wall.
Direction of motion B: Up wall.
Assumption: Slipping UP/IN. Friction at A acts AWAY from wall.
(\( \uparrow \)): \( R = mg \)
(\( \rightarrow \)): Force \( kmg \) towards wall. Reaction \( S \) from wall (left). Friction \( F \) away from wall (left).
\[ kmg = S + F \]Moments about A (same as before as forces at A don’t create moments):
\[ mg a \cos \theta = S (2a \sin \theta) \] \[ S = \frac{1}{2}mg \cot \theta \]Substitute values: \( \cot \theta = 0.8 \).
\[ S = \frac{1}{2}mg (0.8) = 0.4mg \]Limiting Friction: \( F = \mu R = 0.5 (mg) = 0.5mg \).
Substitute into horizontal equation:
\[ kmg = 0.4mg + 0.5mg \] \[ k = 0.9 \]Check: Is this the only possibility? If it were slipping down, \( F \) acts towards wall. \( kmg + F = S \). \( kmg = 0.4mg – 0.5mg \). This gives negative \( k \), impossible. So it must be slipping up.
✓ k = 0.9
Final Answer:
(a) \( \mu \ge \frac{1}{2} \cot \theta \)
(b) \( k = 0.9 \)
Question 4 (10 marks)
A small stone is projected with speed \( 65 \, \text{m s}^{-1} \) from a point \( O \) at the top of a vertical cliff.
Point \( O \) is \( 70 \, \text{m} \) vertically above the point \( N \). Point \( N \) is on horizontal ground.
The stone is projected at an angle \( \alpha \) above the horizontal, where \( \tan \alpha = \frac{5}{12} \).
The stone hits the ground at the point \( A \), as shown in Figure 3.
The acceleration due to gravity is modelled as having magnitude \( 10 \, \text{m s}^{-2} \).
(a) Find the time taken for the stone to travel from \( O \) to \( A \). (4)
(b) Find the speed of the stone at the instant just before it hits the ground at \( A \). (5)
(c) State one other limitation of the model that could affect the reliability of your answers. (1)
Worked Solution
Step 1: Part (a) – Time of Flight
Initial Values:
\( u = 65 \), \( \tan \alpha = \frac{5}{12} \).
Triangle 5-12-13. \( \sin \alpha = \frac{5}{13} \), \( \cos \alpha = \frac{12}{13} \).
Initial Vertical Velocity \( u_y = 65 \sin \alpha = 65 \times \frac{5}{13} = 25 \, \text{m s}^{-1} \) (Upwards).
Displacement \( s = -70 \, \text{m} \) (since it lands below O).
Acceleration \( a = -10 \, \text{m s}^{-2} \) (gravity).
Use \( s = ut + \frac{1}{2}at^2 \).
Rearrange to quadratic:
\[ 5t^2 – 25t – 70 = 0 \]Divide by 5:
\[ t^2 – 5t – 14 = 0 \]Factorise:
\[ (t – 7)(t + 2) = 0 \]Since \( t > 0 \), \( t = 7 \) seconds.
✓ t = 7s
Step 2: Part (b) – Final Speed
We need the magnitude of the velocity vector at \( t = 7 \).
Horizontal Component: Constant.
\( v_x = u_x = 65 \cos \alpha = 65 \times \frac{12}{13} = 60 \, \text{m s}^{-1} \).
Vertical Component: Use \( v = u + at \).
Speed (Magnitude):
\[ V = \sqrt{v_x^2 + v_y^2} \] \[ V = \sqrt{60^2 + (-45)^2} \] \[ V = \sqrt{3600 + 2025} \] \[ V = \sqrt{5625} \] \[ V = 75 \, \text{m s}^{-1} \]✓ 75 m/s
Step 3: Part (c) – Limitations
The question asks for a limitation other than air resistance.
Possible answers:
- The stone is modelled as a particle (rotational effects/dimensions ignored).
- Wind effects.
- Value of g is assumed constant (though for 70m this is reasonable).
Selected Answer: Dimensions of the stone are ignored.
Final Answer:
(a) 7 seconds
(b) \( 75 \, \text{m s}^{-1} \)
(c) Stone size/dimensions ignored.
Question 5 (14 marks)
At time \( t \) seconds, a particle \( P \) has velocity \( \mathbf{v} \, \text{m s}^{-1} \), where
\[ \mathbf{v} = 3t^{\frac{1}{2}}\mathbf{i} – 2t\mathbf{j} \quad t > 0 \]
(a) Find the acceleration of \( P \) at time \( t \) seconds. (2)
(b) Find the value of \( t \) at the instant when \( P \) is moving in the direction of \( \mathbf{i} – \mathbf{j} \). (3)
At time \( t \) seconds, the position vector of \( P \) relative to \( O \) is \( \mathbf{r} \) metres. When \( t = 1 \), \( \mathbf{r} = -\mathbf{j} \).
(c) Find an expression for \( \mathbf{r} \) in terms of \( t \). (3)
(d) Find the exact distance of \( P \) from \( O \) at the instant when \( P \) is moving with speed \( 10 \, \text{m s}^{-1} \). (6)
Worked Solution
Step 1: Part (a) – Acceleration
Acceleration is the derivative of velocity with respect to time.
\[ \mathbf{a} = \frac{d\mathbf{v}}{dt} \]Differentiate each component:
\( \frac{d}{dt}(3t^{1/2}) = 3 \times \frac{1}{2}t^{-1/2} = \frac{3}{2}t^{-1/2} \)
\( \frac{d}{dt}(-2t) = -2 \)
\[ \mathbf{a} = \frac{3}{2}t^{-\frac{1}{2}}\mathbf{i} – 2\mathbf{j} \]Step 2: Part (b) – Direction of Motion
Moving in the direction of \( \mathbf{i} – \mathbf{j} \) means the velocity vector is parallel to \( \mathbf{i} – \mathbf{j} \).
The ratio of the \( \mathbf{j} \) component to the \( \mathbf{i} \) component must be the same as in \( 1\mathbf{i} – 1\mathbf{j} \), which is \( -1 \).
Alternatively: \( \frac{v_y}{v_x} = \frac{-1}{1} = -1 \).
Divide by \( t^{1/2} \) (since \( t>0 \)):
\[ 2t^{1/2} = 3 \] \[ t^{1/2} = 1.5 \] \[ t = 1.5^2 = 2.25 \]Fraction: \( t = \frac{9}{4} \)
Step 3: Part (c) – Position Vector
Position is the integral of velocity. Don’t forget the constant of integration \( \mathbf{C} \).
\[ \mathbf{r} = \int \mathbf{v} \, dt \]Use boundary condition: \( t=1, \mathbf{r} = -\mathbf{j} \).
\[ -\mathbf{j} = 2(1)^{3/2}\mathbf{i} – (1)^2\mathbf{j} + \mathbf{C} \] \[ -\mathbf{j} = 2\mathbf{i} – \mathbf{j} + \mathbf{C} \] \[ \mathbf{C} = -2\mathbf{i} \]So,
\[ \mathbf{r} = (2t^{3/2} – 2)\mathbf{i} – t^2\mathbf{j} \]Step 4: Part (d) – Distance at Speed 10
First, find the time \( t \) when speed is 10.
Speed = \( |\mathbf{v}| = \sqrt{v_x^2 + v_y^2} \).
Square both sides:
\[ 9t + 4t^2 = 100 \] \[ 4t^2 + 9t – 100 = 0 \]Solve quadratic for \( t \):
\[ t = \frac{-9 \pm \sqrt{81 – 4(4)(-100)}}{8} \] \[ t = \frac{-9 \pm \sqrt{81 + 1600}}{8} = \frac{-9 \pm \sqrt{1681}}{8} \] \[ \sqrt{1681} = 41 \] \[ t = \frac{-9 + 41}{8} = \frac{32}{8} = 4 \](Ignore negative time).
Now find distance \( |\mathbf{r}| \) at \( t=4 \).
Substitute \( t=4 \) into \( \mathbf{r} \):
\[ \mathbf{r} = (2(4)^{3/2} – 2)\mathbf{i} – (4)^2\mathbf{j} \] \[ 4^{3/2} = (\sqrt{4})^3 = 8 \] \[ \mathbf{r} = (2(8) – 2)\mathbf{i} – 16\mathbf{j} \] \[ \mathbf{r} = 14\mathbf{i} – 16\mathbf{j} \]Distance is magnitude:
\[ d = \sqrt{14^2 + (-16)^2} \] \[ d = \sqrt{196 + 256} = \sqrt{452} \] \[ d = \sqrt{4 \times 113} = 2\sqrt{113} \]✓ 2√113 m
Final Answer:
(a) \( \frac{3}{2}t^{-1/2}\mathbf{i} – 2\mathbf{j} \)
(b) \( t = 2.25 \)
(c) \( \mathbf{r} = (2t^{3/2} – 2)\mathbf{i} – t^2\mathbf{j} \)
(d) \( 2\sqrt{113} \, \text{m} \)