A2 June 2022 Paper 2 Q6
6 A particle, \(P\), positioned at the origin, \(O\), is projected with a certain velocity along the \(x\)-axis. \(P\) is then acted on by a single force which varies in such a way that \(P\) moves backwards and forwards along the \(x\)-axis.
When the time after projection is \(t\) seconds, the displacement of \(P\) from the origin is \(x\) m and its velocity is \(v\) m s−1.
The motion of \(P\) is modelled using the differential equation \(\ddot{x} + \omega^2 x = 0\), where \(\omega\) rad s−1 is a positive constant.
\(D\) is the point where \(x = d\) for some positive constant, \(d\). When \(P\) reaches \(D\) it comes to instantaneous rest.
- \(x\)
- \(v\)
The quantity \(z\) is defined by \(z = \dfrac{1}{v}\).
One measure of the validity of the model is consideration of the value of \(z_m\). If \(z_m\) exceeds 8 then the model is considered to be valid.
The value of \(d\) is measured as 0.25 to 2 significant figures. The value of \(\omega\) is measured as \(0.75 \pm 0.02\).
| Scheme | Marks | AO |
|---|---|---|
| \(x = A\sin\omega t + B\cos\omega t\) or \(R\cos(\omega t + \phi)\) or \(R\sin(\omega t + \phi)\) | B1 | 1.2 |
| [1] |
Notes
B1: Correct form with 2 arbitrary constants. Must be “\(x =\)”.
Do not ISW; consider final answer as GS unless explicitly labelled otherwise.
Candidates may derive GS from, eg, auxiliary equation but GS must be in real form.
| Scheme | Marks | AO |
|---|---|---|
| \(t = 0,\ x = 0 \Rightarrow B = 0\) (so \(x = A\sin\omega t\)) | M1 | 3.3 |
| Stops when \(x = d \Rightarrow A = d\) so \(x = d\sin\omega t\) | A1 | 3.4 |
| \(v = \omega d\cos\omega t\) | A1 | 2.2a |
| [3] |
Notes
M1: Using one boundary condition (may be seen in (a)).
A1: Using other boundary condition
| Scheme | Marks | AO |
|---|---|---|
| \(\text{RHS} = \omega^2\left(d^2 - x^2\right) = \omega^2\left(d^2 - d^2\sin^2\omega t\right)\) \(= \omega^2 d^2\left(1 - \sin^2\omega t\right) = \omega^2 d^2\cos^2\omega t\) \(= (\omega d\cos\omega t)^2 = v^2 = \text{LHS}\) | B1 | 3.4 |
| [1] |
Notes
B1: AG. Sufficient working must be shown.
| Scheme | Marks | AO |
|---|---|---|
| \(\displaystyle z_m = \frac{1}{d - 0}\int_0^d z\,\mathrm{d}x = \frac{1}{d}\int_0^d \frac{1}{v}\,\mathrm{d}x\) | M1 | 3.3 |
| \(\displaystyle \begin{aligned} &= \frac{1}{d}\int_0^d \frac{1}{\omega\sqrt{d^2 - x^2}}\,\mathrm{d}x \\[6pt] &= \frac{1}{\omega d}\left[\sin^{-1}\frac{x}{d}\right]_0^d = \frac{1}{\omega d}\left[\sin^{-1}1 - \sin^{-1}0\right] \\[6pt] &= \frac{\pi}{2\omega d} \end{aligned}\) | A1 | 3.4 |
| [2] |
Notes
M1: Use of mean formula, over \(x\), with correct limits and \(z\) substituted
| Scheme | Marks | AO |
|---|---|---|
| Using \(d = 0.25\) and \(\omega = 0.75\) leads to \(z_m = 8.38\ldots\), which suggests that the model is valid, but \(d\) could be as high as 0.255 and \(\omega\) as high as 0.77 which would lead to \(z_m = 7.99998\). So it is highly likely that the model is valid (although just possible that it is not). | B1 | 2.2b |
| [1] |
Notes
B1: Indication that for most, but not all, of the possible combinations of values of \(d\) and \(\omega\) the value of \(z_m\) exceeds 8.
\(7.99998129\ldots \lt z_m \lt 8.782758327\ldots\)
| Scheme | Marks | AO |
|---|---|---|
| Initial velocity \(= \omega d\cos 0\) (or \(\omega d\)) | M1 | 3.4 |
| so \(u_{\min} = 0.73 \times 0.245 = 0.17885\) so \(0.18\text{ m s}^{-1}\) cao | A1 | 3.4 |
| [2] |
Notes
M1: Putting \(t = 0\) in expression for \(v\)
Or \(x = 0\) in expression for \(v^2\).
A1: Must include units.