| Exam Board | Edexcel |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2003 |
| Session | November |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors Introduction & 2D |
| Type | Newton's second law with vector forces (find acceleration or force) |
| Difficulty | Moderate -0.3 This is a straightforward M1 mechanics question testing basic vector kinematics and Newton's second law. Part (a) requires simple calculation of acceleration from change in velocity; part (b) applies F=ma directly; part (c) involves finding velocity at t=6s then using constant velocity motion. All steps are routine applications of standard formulas with no problem-solving insight required, making it slightly easier than average. |
| Spec | 1.10h Vectors in kinematics: uniform acceleration in vector form3.03d Newton's second law: 2D vectors |
| Answer | Marks | Guidance |
|---|---|---|
| \(a = \frac{1}{4}[(5i + 11j) - (3i - 5j)] = -2i + 4j\) | M1 | A1 |
| (2) |
| Answer | Marks | Guidance |
|---|---|---|
| \(\mathbf{F} = m\mathbf{a} = -6i + 12j\) | M1 | A1 |
| \( | \mathbf{F} | = \sqrt{180} \approx 13.4\) N (AWRT) |
| \([\text{OR } | \mathbf{a} | = \sqrt{20} \approx 4.47 \Rightarrow |
| Answer | Marks | Guidance |
|---|---|---|
| \(t = 6\) \(v = 3i - 5j + 6(-2i + 4j) = [-9i + 19j]\) | M1 | A1 ✓ |
| At B: \(r = (6i - 29j) + 3(-9i + 19j) = [-21i + 28j]\) | M1 | A1 ✓ |
| \(OB = \sqrt{(21)^2 + (28)^2} = \underline{35 \text{ m}}\) | M1 | A1 ✓ |
| (6) |
## (a)
$a = \frac{1}{4}[(5i + 11j) - (3i - 5j)] = -2i + 4j$ | M1 | A1
| | (2)
## (b)
$\mathbf{F} = m\mathbf{a} = -6i + 12j$ | M1 | A1
$|\mathbf{F}| = \sqrt{180} \approx 13.4$ N (AWRT) | M1 | A1
$[\text{OR } |\mathbf{a}| = \sqrt{20} \approx 4.47 \Rightarrow |\mathbf{F}| = 3 \times 4.47 \approx 13.4 \text{ N}]$ | | (4)
## (c)
$t = 6$ $v = 3i - 5j + 6(-2i + 4j) = [-9i + 19j]$ | M1 | A1 ✓
At B: $r = (6i - 29j) + 3(-9i + 19j) = [-21i + 28j]$ | M1 | A1 ✓
$OB = \sqrt{(21)^2 + (28)^2} = \underline{35 \text{ m}}$ | M1 | A1 ✓
| | (6)
**Total: 12 marks**
---A particle $P$ of mass 3 kg is moving under the action of a constant force $\mathbf{F}$ newtons. At $t = 0$, $P$ has velocity $(3\mathbf{i} - 5\mathbf{j})$ m s$^{-1}$. At $t = 4$ s, the velocity of $P$ is $(-5\mathbf{i} + 11\mathbf{j})$ m s$^{-1}$. Find
\begin{enumerate}[label=(\alph*)]
\item the acceleration of $P$, in terms of $\mathbf{i}$ and $\mathbf{j}$. [2]
\item the magnitude of $\mathbf{F}$. [4]
\end{enumerate}
At $t = 6$ s, $P$ is at the point $A$ with position vector $(6\mathbf{i} - 29\mathbf{j})$ m relative to a fixed origin $O$. At this instant the force $\mathbf{F}$ newtons is removed and $P$ then moves with constant velocity. Three seconds after the force has been removed, $P$ is at the point $B$.
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{2}
\item Calculate the distance of $B$ from $O$. [6]
\end{enumerate}
\hfill \mbox{\textit{Edexcel M1 2003 Q5 [12]}}