A2 June 2019 Paper 2 Q5
5. An engineer is investigating the motion of a sprung diving board at a swimming pool.
Let \(E\) be the position of the end of the diving board when it is at rest in its equilibrium position and when there is no diver standing on the diving board.
A diver jumps from the diving board.
The vertical displacement, \(h\) cm, of the end of the diving board above \(E\) is modelled by the differential equation
where \(t\) seconds is the time after the diver jumps.
When \(t = 0\), the end of the diving board is 20 cm below \(E\) and is moving upwards with a speed of 55 cm s−1.
| Scheme | Marks | AO |
|---|---|---|
| \(4m^2 + 4m + 37 = 0 \Rightarrow m = -\dfrac{1}{2} \pm 3\mathrm{i}\) | M1 | 1.1b |
| \(h = \mathrm{e}^{-0.5t}(A\cos 3t + B\sin 3t)\) | A1 | 1.1b |
| (2) |
Notes
M1: Uses the model to form and solve the auxiliary equation \(4m^2 + 4m + 37 = 0\)
See General Guidance for awarding this mark. This can be implied by correct values for \(m\) (from calculator)
A1: Correct general solution including “\(h =\)”
| Scheme | Marks | AO |
|---|---|---|
| \(t = 0,\ h = -20 \Rightarrow A = -20\) | M1 | 3.4 |
| \(\dfrac{\mathrm{d}h}{\mathrm{d}t} = -0.5\mathrm{e}^{-0.5t}(A\cos 3t + B\sin 3t) + \mathrm{e}^{-0.5t}(-3A\sin 3t + 3B\cos 3t)\) \(t = 0,\ \dfrac{\mathrm{d}h}{\mathrm{d}t} = 55 \Rightarrow B = \ldots\) (NB \(B = 15\)) | M1 | 3.4 |
| \((h =)\,\mathrm{e}^{-0.5t}(15\sin 3t - 20\cos 3t)\) | A1 | 1.1b |
| \(-0.5\mathrm{e}^{-0.5t}(15\sin 3t - 20\cos 3t) + \mathrm{e}^{-0.5t}(60\sin 3t + 45\cos 3t) = 0\) or e.g. \(-0.5\mathrm{e}^{-0.5t}(15\sin 3t - 20\cos 3t) + \dfrac{25\sqrt{37}}{2}\mathrm{e}^{-0.5t}\sin\left(3t + \arctan\dfrac{22}{21}\right) = 0\) \(\Rightarrow t = \ldots\) | M1 | 3.1b |
| \(\tan 3t = -\dfrac{22}{21}\) or e.g. \(3t + \tan^{-1}\dfrac{22}{21} = 0\) | A1 M1 on ePEN | 2.1 |
| \(t = 0.778\) s | A1 | 1.1b |
| \(h = \mathrm{e}^{-0.5 \times \text{"}0.778\text{"}}\bigl(15\sin(3 \times \text{"}0.778\text{"}) - 20\cos(3 \times \text{"}0.778\text{"})\bigr)\) | dM1 | 1.1b |
| = 16.7 cm | A1 | 3.2a |
| (8) |
Notes
M1: Uses the model and the initial conditions to establish the value of “\(A\)”. Need to see \(t = 0\) and \(h = \pm 20\) leading to a value for “\(A\)”. This may be implied by \(A = -20\) or \(A = 20\).
M1: Differentiates their model using the product rule and uses the initial conditions, \(t = 0\) with \(\mathrm{d}h/\mathrm{d}t = \pm 55\), to establish the value of “\(B\)”
A1: Correct particular solution or correct values for \(A\) and \(B\)
M1: Uses their solution to the model with a correct strategy to obtain a value for \(t\) e.g. differentiates or uses their derivative from earlier, sets equal to zero and solves for \(t\)
A1(M1 on ePEN): Correct equation for \(t\)
A1: Correct value for \(t\) (allow awrt 0.778 if necessary) but this value may be implied.
dM1: Uses the model and their positive value for \(t\) to find the maximum displacement - if their \(t\) is incorrect there must be some indication that they are using their \(h\) and not just a number written down. E.g. must see substitution into their \(h\) or they re-state their \(h\) and obtain a value for \(h\).
Dependent on all the previous method marks
A1: Correct value (awrt 16.7 (units not needed))
| Scheme | Marks | AO |
|---|---|---|
| E.g. considers large values of \(t\) in the model for \(h\) or states that for large values of \(t\), \(h\) becomes smaller or becomes zero | M1 | 3.4 |
E.g.
| A1 B1 on ePEN | 3.2b |
| (2) | ||
| (12 marks) |
Notes
M1: Considers the model for large values of \(t\) either by substituting values or by considering the expression and commenting on its behaviour for large values of \(t\). E.g. as \(t \to \infty,\ h \to 0\) or as \(t \to \infty,\ e^{-0.5t} \to 0\) or as \(t \to \infty\) the oscillations become smaller etc.
A1: Makes a suitable comment – see scheme for examples