June 2025 Paper 3 Mechanics Q3
3. [In this question, \(\mathbf{i}\) and \(\mathbf{j}\) are horizontal unit vectors due east and due north respectively.]
A particle \(P\) of mass 0.5 kg moves with constant acceleration \((2\mathbf{i} - 2.4\mathbf{j})\ \text{m s}^{-2}\) on a smooth horizontal plane under the action of a constant horizontal force \(\mathbf{F}\) N.
At time \(t = 0\), \(P\) is moving with velocity \((-7\mathbf{i} + 7.8\mathbf{j})\ \text{m s}^{-1}\)
At time \(t = 0\), \(P\) passes through the point \(O\).
At time \(t = 5\) seconds, \(P\) passes through the point \(A\).
| Scheme | Marks | AO |
|---|---|---|
| \((\mathbf{F} =)\ (\mathbf{i} - 1.2\mathbf{j})\) | B1 | 3.4 |
| (1) |
Notes
Accept column vectors throughout apart from in the final answer for (a) and (d) which must be in terms of \(\mathbf{i}\) and \(\mathbf{j}\), but only penalise ONCE for the whole question.
N.B. For final answers to (a), (b) and (d), penalise incorrect vector notation.
B1: Must be in terms of \(\mathbf{i}\) and \(\mathbf{j}\)
Do not accept \(0.5 \times (2\mathbf{i} - 2.4\mathbf{j})\)
Do not need \(\mathbf{F} =\)
| Scheme | Marks | AO |
|---|---|---|
| \(\mathbf{v} = (-7\mathbf{i} + 7.8\mathbf{j}) + 2(2\mathbf{i} - 2.4\mathbf{j})\) | M1 | 2.1 |
| \(= (-3\mathbf{i} + 3\mathbf{j})\ (\text{m s}^{-1})\) | A1 | 1.1b |
| (2) |
Notes
Accept column vectors throughout apart from in the final answer for (a) and (d) which must be in terms of \(\mathbf{i}\) and \(\mathbf{j}\), but only penalise ONCE for the whole question.
N.B. For final answers to (a), (b) and (d), penalise incorrect vector notation.
M1: Use of \(\mathbf{v} = \mathbf{u} + \mathbf{a}t\) or any other complete method, e.g. integration, to find an unsimplified expression for \(\mathbf{v}\) at \(t = 2\), condone sign errors.
N.B. If using integration, must have used \(t = 0\), \(\mathbf{v} = (-7\mathbf{i} + 7.8\mathbf{j})\) to find the constant of integration and included it in their expression for \(\mathbf{v}\) at \(t = 2\).
A1: cao. N.B. Ignore if they go on and find the speed.
| Scheme | Marks | AO |
|---|---|---|
| Use trig to find an equation in a relevant angle | M1 | 2.1 |
| e.g. \(\tan\alpha = 1\) or \(-1\) or \(\sin\alpha = \dfrac{3}{\sqrt{3^2 + 3^2}}\) or \(\dfrac{-3}{\sqrt{3^2 + 3^2}}\) or \(\cos\alpha = \dfrac{3}{\sqrt{3^2 + 3^2}}\) or \(\dfrac{-3}{\sqrt{3^2 + 3^2}}\) | A1 | 1.1b |
| \(315^\circ\) | A1 | 2.2a |
| (3) |
Notes
Accept column vectors throughout apart from in the final answer for (a) and (d) which must be in terms of \(\mathbf{i}\) and \(\mathbf{j}\), but only penalise ONCE for the whole question.
N.B. For final answers to (a), (b) and (d), penalise incorrect vector notation.
M1: Use trig to find an equation in a relevant angle for their \(\mathbf{v}\)
A1: Correct equation.
A1: Cao (with or without the degree sign), must come from a correct \(\mathbf{v}\)
N.B. \(\pm 45^\circ, \pm 135^\circ, \pm 225^\circ, -315^\circ\) with no working scores M1A1
\(315^\circ\) with no working scores all 3 marks.
| Scheme | Marks | AO |
|---|---|---|
| \(\overrightarrow{OA} = 5(-7\mathbf{i} + 7.8\mathbf{j}) + \dfrac{1}{2}(2\mathbf{i} - 2.4\mathbf{j}) \times 5^2\) | M1 | 3.1a |
| \(= (-10\mathbf{i} + 9\mathbf{j})\) (m) | A1 | 1.1b |
| (2) | ||
| (8 marks) |
Notes
Accept column vectors throughout apart from in the final answer for (a) and (d) which must be in terms of \(\mathbf{i}\) and \(\mathbf{j}\), but only penalise ONCE for the whole question.
N.B. For final answers to (a), (b) and (d), penalise incorrect vector notation.
M1: Use of \(\mathbf{r} = \mathbf{u}t + \dfrac{1}{2}\mathbf{a}t^2\) with \(t = 5\) or any other complete method, e.g. integration of their \(\mathbf{v} = f(t)\) and use of \(t = 5\), to find an unsimplified expression for \(\overrightarrow{OA}\) (No need to show that \(\mathbf{C} = \mathbf{0}\))
M0 if they use \(\mathbf{u} = \mathbf{0}\)
A1: Must be in terms of \(\mathbf{i}\) and \(\mathbf{j}\)
N.B. Ignore if they go on and find the length \(OA\)