| Exam Board | Edexcel |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2007 |
| Session | January |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable acceleration (1D) |
| Type | Velocity from acceleration by integration |
| Difficulty | Standard +0.3 This is a straightforward M2 mechanics question requiring standard techniques: applying F=ma to find acceleration from force, integrating to find velocity with given initial conditions, and using impulse-momentum theorem. All steps are routine applications of formulas with no conceptual challenges or novel problem-solving required. Slightly easier than average due to its mechanical nature. |
| Spec | 1.10a Vectors in 2D: i,j notation and column vectors3.03d Newton's second law: 2D vectors6.03f Impulse-momentum: relation |
| Answer | Marks | Guidance |
|---|---|---|
| \(\mathbf{a} = (3t^2 - 6)\mathbf{i} + 4t\mathbf{j}\) | M1, A1 | 2 marks |
| Answer | Marks | Guidance |
|---|---|---|
| \(t = 3\) \(\mathbf{v} = 9\mathbf{i} + 15\mathbf{j}\) (m s⁻¹) ★ | M1 A1, M1, A1, A1 | cso; 5 marks |
| Answer | Marks | Guidance |
|---|---|---|
| \( | Q | = 0.5\sqrt{5^2 + 12^2} = 6.5\) |
| Answer | Marks | Guidance |
|---|---|---|
| required angle is \(157°\) | M1 A1, A1 | awrt 157°, 203°; 13 marks total |
(a) N2L: $(1.5t^2 - 3)\mathbf{i} + 2t\mathbf{j} = 0.5\mathbf{a}$
$\mathbf{a} = (3t^2 - 6)\mathbf{i} + 4t\mathbf{j}$ | M1, A1 | 2 marks
(b) $\mathbf{v} = (t^3 - 6t)\mathbf{i} + 2t^2\mathbf{j}$ (+c)
$t = 2$ $-4\mathbf{i} + 5\mathbf{j} = -4\mathbf{i} + 8\mathbf{j} + c$ ($c = -3\mathbf{j}$)
$\mathbf{v} = (t^3 - 6t)\mathbf{i} + (2t^2 - 3)\mathbf{j}$ (m s⁻¹)
$t = 3$ $\mathbf{v} = 9\mathbf{i} + 15\mathbf{j}$ (m s⁻¹) ★ | M1 A1, M1, A1, A1 | cso; 5 marks
(c) $Q = 0.5(-3\mathbf{i} + 20\mathbf{j} - (9\mathbf{i} + 15\mathbf{j}))$ ($= 0.5(-12\mathbf{i} + 5\mathbf{j})$)
$|Q| = 0.5\sqrt{5^2 + 12^2} = 6.5$ | M1, M1 A1 | 3 marks
(d) acute angle is arctan $\frac{5}{12} \approx 23°$
or required angle is arctan $\frac{-5}{12}$
or acute angle is arccos $\frac{12}{13} \approx 23°$
or required angle is arccos $\frac{-12}{13}$
required angle is $157°$ | M1 A1, A1 | awrt 157°, 203°; 13 marks total
---A particle $P$ of mass 0.5 kg is moving under the action of a single force $\mathbf{F}$ newtons. At time $t$ seconds, $\mathbf{F} = (1.5t^2 - 3)\mathbf{i} + 2t\mathbf{j}$. When $t = 2$, the velocity of $P$ is $(-4\mathbf{i} + 5\mathbf{j})$ m s$^{-1}$.
\begin{enumerate}[label=(\alph*)]
\item Find the acceleration of $P$ at time $t$ seconds. [2]
\item Show that, when $t = 3$, the velocity of $P$ is $(9\mathbf{i} + 15\mathbf{j})$ m s$^{-1}$. [5]
\end{enumerate}
When $t = 3$, the particle $P$ receives an impulse $\mathbf{Q}$ N s. Immediately after the impulse the velocity of $P$ is $(-3\mathbf{i} + 20\mathbf{j})$ m s$^{-1}$. Find
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{2}
\item the magnitude of $\mathbf{Q}$, [3]
\item the angle between $\mathbf{Q}$ and $\mathbf{i}$. [3]
\end{enumerate}
\hfill \mbox{\textit{Edexcel M2 2007 Q6 [13]}}