| Exam Board | Edexcel |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2022 |
| Session | January |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Forces, equilibrium and resultants |
| Type | Forces in vector form: kinematics extension |
| Difficulty | Moderate -0.3 This is a straightforward M1 vector mechanics question requiring standard application of Newton's second law (F=ma) to find constants, basic vector angle calculation, and constant acceleration kinematics. All steps are routine textbook procedures with no novel insight required, making it slightly easier than average. |
| Spec | 1.10b Vectors in 3D: i,j,k notation3.02e Two-dimensional constant acceleration: with vectors3.03d Newton's second law: 2D vectors |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \((p\mathbf{i} + q\mathbf{j}) + (2q\mathbf{i} + p\mathbf{j}) = 2(\mathbf{i} - \mathbf{j})\) | M1 | Use of \(\mathbf{F} = m\mathbf{a}\) with \(m=2\), correct no. of terms, must attempt to add two forces |
| Equating coefficients of \(\mathbf{i}\) or \(\mathbf{j}\) | M1 | Must have equation in \(p\) and \(q\) only (no vectors); available if \(m\) omitted |
| \(p + 2q = 2\) | A1 | Correct equation in any form |
| \(q + p = -2\) | A1 | Two correct equations in any form |
| \(p = -6;\ q = 4\) | A1 | cao |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\tan\alpha = \pm 1\); e.g. \(45°\) or \(\frac{\pi}{4}\) | M1 | Use of trig to find a relevant angle |
| Angle is \(135°\) or \(225°\) or \(\frac{3\pi}{4}\) or \(\frac{5\pi}{4}\) | A1 | cao, accept radians or degrees |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\mathbf{v} = (3\mathbf{i} - 4\mathbf{j}) + T(\mathbf{i} - \mathbf{j})\) | M1 | Use of \(\mathbf{v} = \mathbf{u} + \mathbf{a}T\) to obtain a velocity vector |
| \(\frac{3+T}{-4-T} = \frac{11}{-13}\) | M1A1 | Use of ratios using their \(\mathbf{v}\) (must be a velocity) to produce equation in \(T\); condone sign error but must be correct way up |
| Solve for \(T\) | DM1 | Dependent on previous M mark |
| \(T = 2.5\) | A1 | cao |
## Question 6:
### Part 6(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $(p\mathbf{i} + q\mathbf{j}) + (2q\mathbf{i} + p\mathbf{j}) = 2(\mathbf{i} - \mathbf{j})$ | M1 | Use of $\mathbf{F} = m\mathbf{a}$ with $m=2$, correct no. of terms, must attempt to add two forces |
| Equating coefficients of $\mathbf{i}$ or $\mathbf{j}$ | M1 | Must have equation in $p$ and $q$ only (no vectors); available if $m$ omitted |
| $p + 2q = 2$ | A1 | Correct equation in any form |
| $q + p = -2$ | A1 | Two correct equations in any form |
| $p = -6;\ q = 4$ | A1 | cao |
### Part 6(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\tan\alpha = \pm 1$; e.g. $45°$ or $\frac{\pi}{4}$ | M1 | Use of trig to find a relevant angle |
| Angle is $135°$ or $225°$ or $\frac{3\pi}{4}$ or $\frac{5\pi}{4}$ | A1 | cao, accept radians or degrees |
### Part 6(c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\mathbf{v} = (3\mathbf{i} - 4\mathbf{j}) + T(\mathbf{i} - \mathbf{j})$ | M1 | Use of $\mathbf{v} = \mathbf{u} + \mathbf{a}T$ to obtain a velocity vector |
| $\frac{3+T}{-4-T} = \frac{11}{-13}$ | M1A1 | Use of ratios using their $\mathbf{v}$ (must be a velocity) to produce equation in $T$; condone sign error but must be correct way up |
| Solve for $T$ | DM1 | Dependent on previous M mark |
| $T = 2.5$ | A1 | cao |
---\begin{enumerate}
\item \hspace{0pt} [In this question $\mathbf { i }$ and $\mathbf { j }$ are horizontal unit vectors.]
\end{enumerate}
A particle $P$ of mass 2 kg moves under the action of two forces, $( p \mathbf { i } + q \mathbf { j } ) \mathrm { N }$ and $( 2 q \mathbf { i } + p \mathbf { j } ) \mathrm { N }$, where $p$ and $q$ are constants.
Given that the acceleration of $P$ is $( \mathbf { i } - \mathbf { j } ) \mathrm { ms } ^ { - 2 }$\\
(a) find the value of $p$ and the value of $q$.\\
(b) Find the size of the angle between the direction of the acceleration and the vector $\mathbf { j }$.
At time $t = 0$, the velocity of $P$ is $( 3 \mathbf { i } - 4 \mathbf { j } ) \mathrm { m } \mathrm { s } ^ { - 1 }$\\
At $t = T$ seconds, $P$ is moving in the direction of the vector $( 11 \mathbf { i } - 13 \mathbf { j } )$.\\
(c) Find the value of $T$.
\hfill \mbox{\textit{Edexcel M1 2022 Q6 [12]}}