| Exam Board | Edexcel |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2018 |
| Session | Specimen |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Forces, equilibrium and resultants |
| Type | Forces in vector form: resultant and acceleration |
| Difficulty | Moderate -0.8 This is a straightforward M1 vector mechanics question requiring basic vector addition and understanding of parallel vectors (equal component ratios). Part (a) involves simple algebraic manipulation to show a given result, and part (b) applies F=ma with magnitude calculation. Both parts are routine textbook exercises with no problem-solving insight required. |
| Spec | 1.10d Vector operations: addition and scalar multiplication3.03d Newton's second law: 2D vectors |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \((2\mathbf{i} - 3\mathbf{j}) + (p\mathbf{i} + q\mathbf{j}) = (p+2)\mathbf{i} + (q-3)\mathbf{j}\) | M1 | Resultant force \(= \mathbf{F}_1 + \mathbf{F}_2\) in the form \(a\mathbf{i} + b\mathbf{j}\) |
| \(\frac{p+2}{q-3} = \frac{1}{2}\) or \(\begin{cases} p+2=n \\ q-3=2n \end{cases}\) for \(n \neq 1\) | M1 | Use parallel vector to form a scalar equation in \(p\) and \(q\) |
| Correct equation (accept any equivalent form) | A1 | |
| \(4 + 2p = -3 + q\) | DM1 | Dependent on no errors seen in comparing the vectors. Rearrange to obtain given answer. At least one stage of working between the fraction and the given answer |
| \(2p - q + 7 = 0\) | A1 | Given Answer |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(q = 11 \Rightarrow p = 2\) | B1 | |
| \(\mathbf{R} = 4\mathbf{i} + 8\mathbf{j}\) | M1 | \((2+p)\mathbf{i} + 8\mathbf{j}\) for their \(p\) |
| \(4\mathbf{i} + 8\mathbf{j} = 2\mathbf{a} \quad (\mathbf{a} = 2\mathbf{i} + 4\mathbf{j})\) | M1 | Use of \(\mathbf{F} = m\mathbf{a}\) |
| \( | \mathbf{a} | = \sqrt{2^2 + 4^2}\) |
| \(= \sqrt{20} = 4.5\) or \(4.47\) or better \((\text{m s}^{-2})\), i.e. \(2\sqrt{5}\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \( | \mathbf{F} | = \sqrt{16 + 64} \left(= \sqrt{80}\right)\) |
| \(\sqrt{80} = 2 \times | \mathbf{a} | \) |
## Question 5:
### Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $(2\mathbf{i} - 3\mathbf{j}) + (p\mathbf{i} + q\mathbf{j}) = (p+2)\mathbf{i} + (q-3)\mathbf{j}$ | M1 | Resultant force $= \mathbf{F}_1 + \mathbf{F}_2$ in the form $a\mathbf{i} + b\mathbf{j}$ |
| $\frac{p+2}{q-3} = \frac{1}{2}$ or $\begin{cases} p+2=n \\ q-3=2n \end{cases}$ for $n \neq 1$ | M1 | Use parallel vector to form a scalar equation in $p$ and $q$ |
| Correct equation (accept any equivalent form) | A1 | |
| $4 + 2p = -3 + q$ | DM1 | Dependent on no errors seen in comparing the vectors. Rearrange to obtain given answer. At least one stage of working between the fraction and the given answer |
| $2p - q + 7 = 0$ | A1 | **Given Answer** |
### Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $q = 11 \Rightarrow p = 2$ | B1 | |
| $\mathbf{R} = 4\mathbf{i} + 8\mathbf{j}$ | M1 | $(2+p)\mathbf{i} + 8\mathbf{j}$ for their $p$ |
| $4\mathbf{i} + 8\mathbf{j} = 2\mathbf{a} \quad (\mathbf{a} = 2\mathbf{i} + 4\mathbf{j})$ | M1 | Use of $\mathbf{F} = m\mathbf{a}$ |
| $|\mathbf{a}| = \sqrt{2^2 + 4^2}$ | DM1 | Correct method for $|\mathbf{a}|$. Dependent on preceding M1 |
| $= \sqrt{20} = 4.5$ or $4.47$ or better $(\text{m s}^{-2})$, i.e. $2\sqrt{5}$ | A1 | |
**Alternative for last two M marks:**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $|\mathbf{F}| = \sqrt{16 + 64} \left(= \sqrt{80}\right)$ | M1 | Correct method for $|\mathbf{F}|$ |
| $\sqrt{80} = 2 \times |\mathbf{a}|$ | DM1 | Use of $|\mathbf{F}| = m|\mathbf{a}|$. Dependent on preceding M1 |\begin{enumerate}
\item Two forces, $\mathbf { F } _ { 1 }$ and $\mathbf { F } _ { 2 }$, act on a particle $A$.\\
$\mathbf { F } _ { 1 } = ( 2 \mathbf { i } - 3 \mathbf { j } ) \mathrm { N }$ and $\mathbf { F } _ { 2 } = ( p \mathbf { i } + q \mathbf { j } ) \mathrm { N }$, where $p$ and $q$ are constants.\\
Given that the resultant of $\mathbf { F } _ { 1 }$ and $\mathbf { F } _ { 2 }$ is parallel to ( $\mathbf { i } + 2 \mathbf { j }$ ),\\
(a) show that $2 p - q + 7 = 0$
\end{enumerate}
Given that $q = 11$ and that the mass of $A$ is 2 kg , and that $\mathbf { F } _ { 1 }$ and $\mathbf { F } _ { 2 }$ are the only forces acting on $A$,\\
(b) find the magnitude of the acceleration of $A$.\\
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\hfill \mbox{\textit{Edexcel M1 2018 Q5 [10]}}