| Exam Board | OCR |
|---|---|
| Module | H240/03 (Pure Mathematics and Mechanics) |
| Year | 2018 |
| Session | March |
| Marks | 14 |
| Topic | Vectors Introduction & 2D |
| Type | Closest approach of two objects |
| Difficulty | Standard +0.8 This is a multi-part kinematics question requiring differentiation to find velocities, equating speeds (involving square roots and solving a quartic), deriving a distance formula, then minimizing using calculus. Part (iii) requires completing the square or substitution on a quartic expression. While the individual techniques are standard M1 content, the combination of steps, algebraic manipulation of the quartic, and the 14-mark length make this moderately challenging—above average but not requiring exceptional insight. |
| Spec | 1.10h Vectors in kinematics: uniform acceleration in vector form3.02a Kinematics language: position, displacement, velocity, acceleration3.02f Non-uniform acceleration: using differentiation and integration |
| Answer | Marks | Guidance |
|---|---|---|
| \(\mathbf{F}_A = 2\mathbf{i} + 3\mathbf{j}\) | B1 | |
| \(\mathbf{F}_B = -4\mathbf{i} + (3 - 4t)\mathbf{j}\) | B1 | |
| \((2t)^2 + 9 = (-4t)^2 + (3 - 4t)^2\) | M1 | \( |
| \(7t^2 - 6t = 0 \Rightarrow t = \ldots\) | M1 | Expand and attempt to solve quadratic in \(t\) (to obtain two solutions) |
| \(t = 0\) or \(t = \frac{6}{7}\) | A1 | Both values of \(t\) must be given |
| Answer | Marks | Guidance |
|---|---|---|
| \(\mathbf{F}_A - \mathbf{F}_B = (3t^2 - 1)\mathbf{i} + (-1 + 2t^2)\mathbf{j}\) | *M1 | Consider \(\pm(\mathbf{F}_A - \mathbf{F}_B)\); Condone one sign error |
| \(d^2 = (3t^2 - 1)^2 + (-1 + 2t^2)^2\) | dep*M1 | Use of \(d^2 = |
| \(= 9t^4 - 6t^2 + 1 + 4t^4 - 4t^2 + 1 = 13t^4 - 10t^2 + 2\) | A1 | AG Expand correctly to given answer; Must show at least one intermediate step |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{d}{dt}(d^2) = 52t^3 - 20t\) | B1 | |
| \(\frac{d}{dt}(d^2) = 0 \Rightarrow t = \ldots\) | *M1 | Set their derivative \(= 0\) and solve for \(t\) |
| \(t = 0\) and \(t = \sqrt{\frac{5}{13}}\) | A1 | Both values correct; accept 0.620... |
| Test nature of stationary point with correct value(s) of \(t\) | B1 | e.g. \(\frac{d^2}{dt^2}(d^2) = 156t^2 - 20 > 0\) when \(t^2 = \frac{5}{13}\) so minimum; Or any other valid method |
| Substitute their non-zero \(t\) into \(d\) or \(d^2\) | dep*M1 | |
| \(d = \frac{1}{\sqrt{13}}\) or 0.277 | A1 | Dependent on all previous marks; 0.277 350... |
## Part (i):
$\mathbf{F}_A = 2\mathbf{i} + 3\mathbf{j}$ | B1 |
$\mathbf{F}_B = -4\mathbf{i} + (3 - 4t)\mathbf{j}$ | B1 |
$(2t)^2 + 9 = (-4t)^2 + (3 - 4t)^2$ | M1 | $|\mathbf{F}_A| = |\mathbf{F}_B|$ with/without square root
$7t^2 - 6t = 0 \Rightarrow t = \ldots$ | M1 | Expand and attempt to solve quadratic in $t$ (to obtain two solutions)
$t = 0$ or $t = \frac{6}{7}$ | A1 | Both values of $t$ must be given
## Part (ii):
$\mathbf{F}_A - \mathbf{F}_B = (3t^2 - 1)\mathbf{i} + (-1 + 2t^2)\mathbf{j}$ | *M1 | Consider $\pm(\mathbf{F}_A - \mathbf{F}_B)$; Condone one sign error
$d^2 = (3t^2 - 1)^2 + (-1 + 2t^2)^2$ | dep*M1 | Use of $d^2 = |\mathbf{F}_A - \mathbf{F}_B|^2$
$= 9t^4 - 6t^2 + 1 + 4t^4 - 4t^2 + 1 = 13t^4 - 10t^2 + 2$ | A1 | AG Expand correctly to given answer; Must show at least one intermediate step
## Part (iii):
$\frac{d}{dt}(d^2) = 52t^3 - 20t$ | B1 |
$\frac{d}{dt}(d^2) = 0 \Rightarrow t = \ldots$ | *M1 | Set their derivative $= 0$ and solve for $t$
$t = 0$ and $t = \sqrt{\frac{5}{13}}$ | A1 | Both values correct; accept 0.620...
Test nature of stationary point with correct value(s) of $t$ | B1 | e.g. $\frac{d^2}{dt^2}(d^2) = 156t^2 - 20 > 0$ when $t^2 = \frac{5}{13}$ so minimum; Or any other valid method
Substitute their non-zero $t$ into $d$ or $d^2$ | dep*M1 |
$d = \frac{1}{\sqrt{13}}$ or 0.277 | A1 | Dependent on all previous marks; 0.277 350...Two particles $A$ and $B$ have position vectors $\mathbf{r}_A$ metres and $\mathbf{r}_B$ metres at time $t$ seconds, where
$$\mathbf{r}_A = t^2\mathbf{i} + (3t - 1)\mathbf{j} \quad \text{and} \quad \mathbf{r}_B = (1 - 2t^2)\mathbf{i} + (3t - 2t^2)\mathbf{j}, \quad \text{for } t \geqslant 0.$$
\begin{enumerate}[label=(\roman*)]
\item Find the values of $t$ when $A$ and $B$ are moving with the same speed. [5]
\item Show that the distance, $d$ metres, between $A$ and $B$ at time $t$ satisfies
$$d^2 = 13t^4 - 10t^2 + 2.$$ [3]
\item Hence find the shortest distance between $A$ and $B$ in the subsequent motion. [6]
\end{enumerate}
\hfill \mbox{\textit{OCR H240/03 2018 Q9 [14]}}