June 2022 Paper 3 Q12
12 In this question the unit vectors \(\mathbf{i}\) and \(\mathbf{j}\) are in the directions east and north respectively.
A particle \(P\) is moving on a smooth horizontal surface under the action of a single force \(\mathbf{F}\) N. At time \(t\) seconds, where \(t \geqslant 0\), the velocity \(\mathbf{v}\,\mathrm{m\,s^{-1}}\) of \(P\), relative to a fixed origin \(O\), is given by
\(\mathbf{v} = (1 - 2t)\mathbf{i} + (2t^2 + t - 13)\mathbf{j}\).
The mass of \(P\) is 0.5 kg.
When \(t = 1\), \(P\) is at the point with position vector \(\frac{1}{6}\mathbf{j}\).
| Scheme | Marks | AO |
|---|---|---|
| \(\mathbf{v} = (1 - 2t)\mathbf{i} + (2t^2 + t - 13)\mathbf{j}\) If \(P\) is stationary, then \(1 - 2t = 0\) and \(2t^2 + t - 13 = 0\) | M1 | 3.1b |
| \(\mathbf{i}: 1 - 2t = 0 \Rightarrow t = \frac{1}{2}\) \(\mathbf{j}: 2t^2 + t - 13 = 0 \Rightarrow t = 2.3117\ldots, -2.8117\ldots\) No value of \(t\) is common to both components, so \(P\) is never stationary | A1 | 2.2a |
| [2] |
Notes
M1: Considers either the i or j component equal to zero or forms a five-term quartic equation for \(|\mathbf{v}|^2 = 0\) (oe)
\((4t^4 + 4t^3 - 47t^2 - 30t + 170 = 0)\)
A1: BC – need not see the negative value of \(t\) or for substituting \(t = 0.5\) into quadratic expression for \(\mathbf{j}\) and showing this gives a non-zero answer (oe) with correct working and conclusion
A1 for the correct quartic equation with roots stated as \(2.3 \pm 0.35\mathrm{i}\), \(-2.8 \pm 0.65\mathrm{i}\) + correct conclusion
| Scheme | Marks | AO |
|---|---|---|
| \(\Rightarrow \mathbf{a} = -2\mathbf{i} + (4t + 1)\mathbf{j}\) \((\mathrm{m\,s^{-2}})\) | B1 | 1.1 |
| [1] |
Notes
B1: Correct derivative – or as a column vector
Brackets must be around the \(4t + 1\)
| Scheme | Marks | AO |
|---|---|---|
| \(-2(2t^2 + t - 13) = 1(1 - 2t)\) | M1* | 3.1b |
| \(4t^2 - 25 = 0 \Rightarrow t = 2.5\) | M1dep* | 1.1 |
| \(\mathbf{F} = m\mathbf{a} \Rightarrow \mathbf{F} = 0.5\{-2\mathbf{i} + (4t + 1)\mathbf{j}\}\) | M1* | 3.4 |
| \(|\mathbf{F}| = \sqrt{(-1)^2 + 5.5^2}\) | M1dep* | 1.1 |
| \(|\mathbf{F}| = 5.59\) (N) | A1 | 1.1 |
| [5] |
Notes
M1*: Setting up a quadratic equation in \(t\) only – allow sign errors (including on the 1 and 2) and the 1 and \(-2\) on the wrong side
Or multiples of 1 and \(-2\)
M1dep*: Solves their (two or three term) quadratic and selects their positive value of \(t\)
Check unsupported solutions if incorrect quadratic equation
M1*: Substitute their \(\mathbf{a}\) into \(\mathbf{F} = 0.5\mathbf{a}\) or their \(|\mathbf{a}|\) into \(|\mathbf{F}| = 0.5|\mathbf{a}|\). If \(\mathbf{F}\) not stated in terms of \(t\) then one component must be correct following through from their \(\mathbf{a}\) (and possibly \(t\))
Must use correct value of 0.5 for \(m\) but can be in terms of \(t\)
M1dep*: Dependent on previous M mark only
From a value of \(t \gt 0\)
A1: awrt 5.59 (exact: \(\dfrac{5\sqrt{5}}{2}\))
5.590169…
| Scheme | Marks | AO |
|---|---|---|
| \(\mathbf{s} = (t - t^2)\mathbf{i} + \left(\dfrac{2}{3}t^3 + \dfrac{1}{2}t^2 - 13t\right)\mathbf{j}\ (+\mathbf{c})\) | M1* | 1.1 |
| \(t = 1\), \(\mathbf{s} = \dfrac{1}{6}\mathbf{j} \Rightarrow \mathbf{c} = (0\mathbf{i} +)\,12\mathbf{j}\) | A1 | 3.4 |
| When \(t = 1.5\), \(\mathbf{s} = -\dfrac{3}{4}\mathbf{i} - \dfrac{33}{8}\mathbf{j}\) | M1dep* | 1.1 |
| \(\tan^{-1}\left(\dfrac{\pm 3/4}{\pm 33/8}\right)\) or \(\tan^{-1}\left(\dfrac{\pm 33/8}{\pm 3/4}\right)\) | M1 | 3.1b |
| Bearing \(= 180 + \tan^{-1}\left(\dfrac{3/4}{33/8}\right) = 190^\circ\) | A1 | 3.2a |
| [5] |
Notes
M1*: Integrates \(\mathbf{v}\) wrt \(t\) – at least three terms correct
Allow without \(+\mathbf{c}\)
A1: Uses given conditions to find correct \(\mathbf{c}\) – dependent on a completely correct integrated expression for \(\mathbf{s}\)
www
M1dep*: Substitute \(t = 1.5\) into their \(\mathbf{s}\)
M1: Attempt to find a relevant angle using the components of their \(\mathbf{s}\) (allow use of sin/cos with the magnitude of \(\mathbf{s}\))
Dependent on both previous M marks.
Written in terms of arctan is sufficient
A1: awrt 190 (or from \(270 - \tan^{-1}\left(\frac{33/8}{3/4}\right)\))
190.3048465…