| Exam Board | Edexcel |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2007 |
| Session | January |
| Marks | 9 |
| 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 -0.8 This is a straightforward mechanics question requiring only direct application of standard formulas: (a) uses a = Δv/Δt, (b) applies F = ma and Pythagoras, (c) uses v = u + at. All steps are routine calculations with no problem-solving insight needed, making it easier than average but not trivial due to vector notation and multiple parts. |
| Spec | 1.10a Vectors in 2D: i,j notation and column vectors1.10b Vectors in 3D: i,j,k notation3.02e Two-dimensional constant acceleration: with vectors3.03c Newton's second law: F=ma one dimension3.03d Newton's second law: 2D vectors |
| Answer | Marks | Guidance |
|---|---|---|
| (a) \(a = \frac{(15i - 4j) - (3i + 2j)}{4} = 3i - 1.5j\) | M1 A1 | 2 marks |
| (b) N2L: \(F = ma = 6i - 3j\) | M1 A1 | ft their \(a\) |
| (c) \(v_6 = (3i + 2j) + (3i - 1.5j) \times 6 = 21i - 7j\) (m s\(^{-1}\)) | M1 A1 | ft their \(a\) |
**(a)** $a = \frac{(15i - 4j) - (3i + 2j)}{4} = 3i - 1.5j$ | M1 A1 | 2 marks
**(b)** N2L: $F = ma = 6i - 3j$ | M1 A1 | ft their $a$ | $|F| = \sqrt{6^2 + 3^2} ≈ 6.71$ (N) | M1 A1 | accept $\sqrt{45}$, awrt 6.7 | 4 marks
**(c)** $v_6 = (3i + 2j) + (3i - 1.5j) \times 6 = 21i - 7j$ (m s$^{-1}$) | M1 A1 | ft their $a$ | 1 mark | 9 marks total
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A particle $P$ of mass 2 kg is moving under the action of a constant force $\mathbf{F}$ newtons. When $t = 0$, $P$ has velocity $(3\mathbf{i} + 2\mathbf{j})$ m s$^{-1}$ and at time $t = 4$ s, $P$ has velocity $(15\mathbf{i} - 4\mathbf{j})$ 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]
\item the velocity of $P$ at time $t = 6$ s. [3]
\end{enumerate}
\hfill \mbox{\textit{Edexcel M1 2007 Q3 [9]}}