October 2021 Paper 3 Q13
13 In this question the unit vectors \(\mathbf{i}\) and \(\mathbf{j}\) are in the directions east and north respectively.
At time \(t\) seconds, where \(t \geqslant 0\), a particle \(P\) of mass \(2\,\mathrm{kg}\) is moving on a smooth horizontal surface under the action of a constant horizontal force \((-8\mathbf{i} - 54\mathbf{j})\,\mathrm{N}\) and a variable horizontal force \(\left(4t\mathbf{i} + 6(2t - 1)^2\mathbf{j}\right)\mathrm{N}\).
It is given that \(P\) is at rest when \(t = 0\).
| Scheme | Marks | AO |
|---|---|---|
| \(\mathbf{F} = (4t - 8)\mathbf{i} + 6\left((2t - 1)^2 - 9\right)\mathbf{j}\) | M1 | 3.1b |
| When \(t = 2\), the forces are in equilibrium | A1 | 1.1 |
| [2] |
Notes
M1: Combines given forces and considers either component equal to zero – allow M1 for either \(4t - 8 = 0\) or for \(6(2t - 1)^2 - 54 = 0\)
A1: \(t = 2\) only – need only consider \(\mathbf{i}\) or \(\mathbf{j}\) component but any contradictory working/answers scores A0
| Scheme | Marks | AO |
|---|---|---|
| \(m = 2 \Rightarrow \mathbf{a} = (2t - 4)\mathbf{i} + 3\left((2t - 1)^2 - 9\right)\mathbf{j}\) | B1 | 3.3 |
| \(\mathbf{v} = \left(t^2 - 4t\right)\mathbf{i} + 3\left(\dfrac{1}{6}(2t - 1)^3 - 9t\right)\mathbf{j}\,(+\mathbf{c})\) | M1* A1 | 3.1b 1.1 |
| \(t = 0, \mathbf{v} = \mathbf{0} \Rightarrow \mathbf{c} = \dfrac{1}{2}\mathbf{j}\) | M1dep* | 3.4 |
| Moving parallel to \(\mathbf{j} \Rightarrow \mathbf{i} = \mathbf{0}\) therefore \(t(t - 4) = 0\) | M1 | 3.1a |
| \(t = 4 \Rightarrow |\mathbf{v}| = 64\ \left(\mathrm{m\,s^{-1}}\right)\) | A1 | 2.2a |
| [6] |
Notes
B1: Using \(\mathbf{F} = 2\mathbf{a}\) correctly
Allow \(2\mathbf{a} = \ldots\)
M1*: Attempt to integrate \(\mathbf{a}\) (or \(\mathbf{F}\)) wrt \(t\) – two of their terms integrated correctly
M0 if only considering one force or one component for \(\mathbf{a}\) or \(\mathbf{F}\)
A1: Condone no \(+\mathbf{c}\) for this mark
\(\mathbf{v} = \left(t^2 - 4t\right)\mathbf{i} + \left(4t^3 - 6t^2 - 24t\right)\mathbf{j}\) (oe)
Allow \(2\mathbf{v} = \ldots\)
M1dep*: Uses correct initial conditions to find \(\mathbf{c}\) (or \(\mathbf{c} = \mathbf{0}\) if expanded version used)
M1: Sets \(\mathbf{i}\)-component of \(\mathbf{v}\) equal to 0 to obtain a quadratic equation in \(t\)
Dependent on first M mark
A1: Must have found \(+\mathbf{c}\) for this mark
| Scheme | Marks | AO |
|---|---|---|
| \(\mathbf{r} = \left(\dfrac{1}{3}t^3 - 2t^2\right)\mathbf{i} + 3\left(\dfrac{1}{48}(2t - 1)^4 - \dfrac{9}{2}t^2 + \dfrac{1}{6}t\right)\mathbf{j}\) | M1* A1 | 1.1 1.1 |
| \(t = 0 \Rightarrow \mathbf{r} = \dfrac{1}{16}\mathbf{j},\ t = 3 \Rightarrow \mathbf{r} = -9\mathbf{i} - \dfrac{1295}{16}\mathbf{j}\) | M1dep* | 1.1 |
| Dist. \(= \sqrt{(-9)^2 + \left(-\dfrac{1295}{16} - \dfrac{1}{16}\right)^2}\) | M1 | 1.1 |
| 81.5 (m) | A1 | 2.2a |
| [5] |
Notes
M1*: Attempt to integrate \(\mathbf{v}\) wrt \(t\) – two of their terms integrated correctly – dependent on first M mark in (b)
no vector constant of integration required in (c)
A1: \(\mathbf{r} = \left(\frac{1}{3}t^3 - 2t^2\right)\mathbf{i} + \left(t^4 - 2t^3 - 12t^2\right)\mathbf{j}\)
M1dep*: Attempt to find \(\mathbf{r}\) at both \(t = 0\) and \(t = 3\)
M1: Correct expression for the distance between given times (dependent on both previous M marks)
A1: (For reference: 81.49846624…)
\(\sqrt{6642} = 9\sqrt{82}\)