| Exam Board | Edexcel |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2020 |
| Session | January |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors Introduction & 2D |
| Type | Resultant of two vector forces (direction/magnitude conditions) |
| Difficulty | Moderate -0.8 This is a straightforward M1 vectors question requiring standard techniques: (a) magnitude using Pythagoras, (b) angle between vectors using dot product or component ratios, (c) finding a constant using parallel vector conditions. All parts are routine textbook exercises with no problem-solving insight required, making it easier than average A-level questions. |
| Spec | 1.10c Magnitude and direction: of vectors3.03a Force: vector nature and diagrams |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Magnitude \(= \sqrt{10^2+1^2} = \sqrt{101}\) (N) | M1 A1 | Use of Pythagoras; A0 if only decimal given |
| Total | (2) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\tan\alpha = \frac{1}{10}\) | M1 | Any relevant trig ratio for \(\alpha\) or \((90°-\alpha)\) |
| \(45°\) | B1 | \(45°\) seen |
| Angle \(= 45° - \alpha = 39.289...°\) Accept \(39°\) or better | M1 A1 | Finding difference between \(45°\) and \(\alpha\) or \((90°-\alpha)\) and \(45°\) |
| Total | (4) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \((10\mathbf{i}+\mathbf{j})\cdot(\mathbf{i}+\mathbf{j}) = \sqrt{10^2+1^2}\cdot\sqrt{1^2+1^2}\cos\theta\) | M1 | |
| \((10\mathbf{i}+\mathbf{j})\cdot(\mathbf{i}+\mathbf{j}) = 11\) | B1 | |
| \(11 = \sqrt{10^2+1^2}\cdot\sqrt{1^2+1^2}\cos\theta\) | M1 | |
| \(\theta = 39°\) or better | A1 | (4) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \((10\mathbf{i}+\mathbf{j})-({\mathbf{i}+\mathbf{j}}) = 9\mathbf{i}\) or \((\mathbf{i}+\mathbf{j})-(10\mathbf{i}+\mathbf{j}) = -9\mathbf{i}\) | B1 | |
| \(9^2 = (10^2+1^2)+(1^2+1^2)-2\sqrt{10^2+1^2}\cdot\sqrt{1^2+1^2}\cos\theta\) | M1 | |
| \(\theta = 39°\) or better | A1 | (4) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \((10\mathbf{i}+\mathbf{j})+(-15\mathbf{i}+a\mathbf{j}) = -5\mathbf{i}+(a+1)\mathbf{j}\) | B1 | Adding two forces and collecting \(\mathbf{i}\)s and \(\mathbf{j}\)s |
| \(\frac{a+1}{-5} = \frac{-3}{2}\) | M1 A1 | Producing equation in \(a\) only using ratios; M0 if no resultant attempted or equation comes from equating resultant to \((2\mathbf{i}-3\mathbf{j})\); condone sign error but M0 if ratio upside down |
| Solve for \(a\) | M1 | Independent M mark; equation must come from ratio equation from resultant |
| \(a = 6.5\) | A1 | |
| Total | (5) |
# Question 6:
## Part 6(a)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Magnitude $= \sqrt{10^2+1^2} = \sqrt{101}$ (N) | M1 A1 | Use of Pythagoras; A0 if only decimal given |
| **Total** | **(2)** | |
## Part 6(b)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\tan\alpha = \frac{1}{10}$ | M1 | Any relevant trig ratio for $\alpha$ or $(90°-\alpha)$ |
| $45°$ | B1 | $45°$ seen |
| Angle $= 45° - \alpha = 39.289...°$ Accept $39°$ or better | M1 A1 | Finding difference between $45°$ and $\alpha$ or $(90°-\alpha)$ and $45°$ |
| **Total** | **(4)** | |
**Alternative 1 – Scalar Product:**
| Answer | Marks | Guidance |
|--------|-------|----------|
| $(10\mathbf{i}+\mathbf{j})\cdot(\mathbf{i}+\mathbf{j}) = \sqrt{10^2+1^2}\cdot\sqrt{1^2+1^2}\cos\theta$ | M1 | |
| $(10\mathbf{i}+\mathbf{j})\cdot(\mathbf{i}+\mathbf{j}) = 11$ | B1 | |
| $11 = \sqrt{10^2+1^2}\cdot\sqrt{1^2+1^2}\cos\theta$ | M1 | |
| $\theta = 39°$ or better | A1 | **(4)** |
**Alternative 2 – Cosine Rule:**
| Answer | Marks | Guidance |
|--------|-------|----------|
| $(10\mathbf{i}+\mathbf{j})-({\mathbf{i}+\mathbf{j}}) = 9\mathbf{i}$ or $(\mathbf{i}+\mathbf{j})-(10\mathbf{i}+\mathbf{j}) = -9\mathbf{i}$ | B1 | |
| $9^2 = (10^2+1^2)+(1^2+1^2)-2\sqrt{10^2+1^2}\cdot\sqrt{1^2+1^2}\cos\theta$ | M1 | |
| $\theta = 39°$ or better | A1 | **(4)** |
## Part 6(c)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $(10\mathbf{i}+\mathbf{j})+(-15\mathbf{i}+a\mathbf{j}) = -5\mathbf{i}+(a+1)\mathbf{j}$ | B1 | Adding two forces and collecting $\mathbf{i}$s and $\mathbf{j}$s |
| $\frac{a+1}{-5} = \frac{-3}{2}$ | M1 A1 | Producing equation in $a$ only using ratios; M0 if no resultant attempted or equation comes from equating resultant to $(2\mathbf{i}-3\mathbf{j})$; condone sign error but M0 if ratio upside down |
| Solve for $a$ | M1 | Independent M mark; equation must come from ratio equation from resultant |
| $a = 6.5$ | A1 | |
| **Total** | **(5)** | |
---6. A force $\mathbf { F }$ is given by $\mathbf { F } = ( 10 \mathbf { i } + \mathbf { j } ) \mathrm { N }$.
\begin{enumerate}[label=(\alph*)]
\item Find the exact value of the magnitude of $\mathbf { F }$.
\item Find, in degrees, the size of the angle between the direction of $\mathbf { F }$ and the direction of the vector $( \mathbf { i } + \mathbf { j } )$.
The resultant of the force $\mathbf { F }$ and the force $( - 15 \mathbf { i } + a \mathbf { j } ) \mathrm { N }$, where $a$ is a constant, is parallel to, but in the opposite direction to, the vector $( 2 \mathbf { i } - 3 \mathbf { j } )$.
\item Find the value of $a$.
\includegraphics[max width=\textwidth, alt={}, center]{916543cb-14f7-486c-ba3c-eda9be134045-19_104_59_2613_1886}
\end{enumerate}
\hfill \mbox{\textit{Edexcel M1 2020 Q6 [11]}}