| Exam Board | Edexcel |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2002 |
| Session | June |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Forces, equilibrium and resultants |
| Type | Forces in vector form: resultant and acceleration |
| Difficulty | Moderate -0.3 This is a straightforward M1 mechanics question testing standard vector kinematics with constant acceleration. Parts (a)-(d) involve routine calculations: finding angles using arctangent, computing acceleration from velocity change, applying F=ma, and using v=u+at. Part (e) requires recognizing when velocity components are equal, but this is a standard technique. The multi-part structure and 13 total marks indicate moderate length, but each step follows directly from basic formulas without requiring problem-solving insight or novel approaches. |
| Spec | 1.10d Vector operations: addition and scalar multiplication3.02a Kinematics language: position, displacement, velocity, acceleration3.03d Newton's second law: 2D vectors |
| Answer | Marks | Guidance |
|---|---|---|
| \(\tan \theta = \frac{1}{2} \Rightarrow \theta = 26.6°\) | M1 A1 | |
| angle required \(= 153.4°\) | A1 | (3) |
| Answer | Marks | Guidance |
|---|---|---|
| \(\mathbf{a} = \frac{1}{4}[(\mathbf{i} - 2\mathbf{j}) - (-5\mathbf{i} + 7\mathbf{j})]\) | M1 | |
| \(= (2\mathbf{i} - 3\mathbf{j}) \text{ m s}^{-2}\) | A1 | (2) |
| Answer | Marks | Guidance |
|---|---|---|
| \(\mathbf{F} = m\mathbf{a} = 4\mathbf{i} - 6\mathbf{j}\) | M1 | |
| \( | \mathbf{F} | = \sqrt{(16 + 36)} = 7.21 \text{ N}\) |
| Answer | Marks | Guidance |
|---|---|---|
| \(\mathbf{v} = (-5 + 2t)\mathbf{i} + (7 - 3t)\mathbf{j}\) | M1 A1ft | (2) |
| Answer | Marks | Guidance |
|---|---|---|
| \(\mathbf{v}\) parallel to \(\mathbf{i} + \mathbf{j} \Rightarrow \frac{-5 + 2t}{7 - 3t} = 1\) | M1 | |
| \(\Rightarrow t = 2.4 \text{ s}\) | M1 A1 | (3) |
## (a)
$\tan \theta = \frac{1}{2} \Rightarrow \theta = 26.6°$ | M1 A1 |
angle required $= 153.4°$ | A1 | (3) |
## (b)
$\mathbf{a} = \frac{1}{4}[(\mathbf{i} - 2\mathbf{j}) - (-5\mathbf{i} + 7\mathbf{j})]$ | M1 |
$= (2\mathbf{i} - 3\mathbf{j}) \text{ m s}^{-2}$ | A1 | (2) |
## (c)
$\mathbf{F} = m\mathbf{a} = 4\mathbf{i} - 6\mathbf{j}$ | M1 |
$|\mathbf{F}| = \sqrt{(16 + 36)} = 7.21 \text{ N}$ | M1 A1 | (3) |
## (d)
$\mathbf{v} = (-5 + 2t)\mathbf{i} + (7 - 3t)\mathbf{j}$ | M1 A1ft | (2) |
## (e)
$\mathbf{v}$ parallel to $\mathbf{i} + \mathbf{j} \Rightarrow \frac{-5 + 2t}{7 - 3t} = 1$ | M1 |
$\Rightarrow t = 2.4 \text{ s}$ | M1 A1 | (3) |
**Total: (13 marks)**
---A particle $P$ of mass $2 \text{ kg}$ moves in a plane under the action of a single constant force $\mathbf{F}$ newtons. At time $t$ seconds, the velocity of $P$ is $\mathbf{v} \text{ m s}^{-1}$. When $t = 0$, $\mathbf{v} = (-5\mathbf{i} + 7\mathbf{j})$ and when $t = 3$, $\mathbf{v} = (\mathbf{i} - 2\mathbf{j})$.
\begin{enumerate}[label=(\alph*)]
\item Find in degrees the angle between the direction of motion of $P$ when $t = 3$ and the vector $\mathbf{j}$. [3]
\item Find the acceleration of $P$. [2]
\item Find the magnitude of $\mathbf{F}$. [3]
\item Find in terms of $t$ the velocity of $P$. [2]
\item Find the time at which $P$ is moving parallel to the vector $\mathbf{i} + \mathbf{j}$. [3]
\end{enumerate}
\hfill \mbox{\textit{Edexcel M1 2002 Q5 [13]}}