| Exam Board | Edexcel |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2020 |
| Session | June |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Constant acceleration (SUVAT) |
| Type | Constant acceleration vector (i and j) |
| Difficulty | Moderate -0.8 This is a straightforward M1 vector kinematics question requiring only routine techniques: substituting t=2 into the velocity vector, calculating magnitude for speed, using arctangent for angle, differentiating for acceleration, and solving simultaneous equations for parallel vectors. All parts are standard textbook exercises with no problem-solving insight required, making it easier than average. |
| Spec | 1.10h Vectors in kinematics: uniform acceleration in vector form3.02g Two-dimensional variable acceleration |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Put \(t = 2\) to give \(-3\mathbf{i} + 4\mathbf{j}\) | M1 | Allow column vectors |
| \(\sqrt{(-3)^2 + 4^2}\) (the \(-\) sign is not required) | M1 | Finding the magnitude of their \(\mathbf{v}\) |
| \(5\) (m s\(^{-1}\)) | A1 | Correct answer |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| e.g. \(\tan\theta = \frac{3}{4}\) | M1 | For a relevant trig equation |
| A correct equation | A1ft | A correct equation following through on their \(\mathbf{v}\) |
| \(37°\) or \(323°\) nearest degree | A1 | Correct answer (must be in degrees to nearest degree) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\mathbf{v} = (7-5t)\mathbf{i} + (12t-20)\mathbf{j} = (7\mathbf{i} - 20\mathbf{j}) + t(-5\mathbf{i} + 12\mathbf{j})\) | M1 | Collecting terms in \(t\) and constant terms (may be implied) |
| \(\frac{\mathbf{v} - (7\mathbf{i} - 20\mathbf{j})}{t} = (-5\mathbf{i} + 12\mathbf{j})\) | M1 A1 | Rearranging to required form; correct answer (isw if they find the magnitude) |
| OR: \(t=0\), \(\mathbf{v} = 7\mathbf{i} - 20\mathbf{j}\) | M1 | Finding the initial velocity or some other specific velocity |
| \(\frac{(-3\mathbf{i}+4\mathbf{j})-(7\mathbf{i}-20\mathbf{j})}{2} = (-5\mathbf{i}+12\mathbf{j})\) | M1 A1 | Use of \(\mathbf{a} = \frac{\mathbf{v}-\mathbf{u}}{t}\) with \(t=2\) (or possibly another appropriate value) |
| OR: Differentiate wrt \(t\): \(\frac{d\mathbf{v}}{dt} = \mathbf{a} = (-5\mathbf{i}+12\mathbf{j})\) | M2, A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\frac{(7-5t)}{(12t-20)} = \frac{-5}{8}\) | M1 A1 | Attempt at equation in \(t\) only, using ratio of components; allow reciprocal and a sign error; correct equation |
| Solve for \(t\) | M1 | Equation must have come from considering ratios |
| \(t = 2.2\) | A1 | Correct answer |
# Question 5:
## Part 5(a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Put $t = 2$ to give $-3\mathbf{i} + 4\mathbf{j}$ | M1 | Allow column vectors |
| $\sqrt{(-3)^2 + 4^2}$ (the $-$ sign is not required) | M1 | Finding the magnitude of their $\mathbf{v}$ |
| $5$ (m s$^{-1}$) | A1 | Correct answer |
## Part 5(b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| e.g. $\tan\theta = \frac{3}{4}$ | M1 | For a relevant trig equation |
| A correct equation | A1ft | A correct equation following through on their $\mathbf{v}$ |
| $37°$ or $323°$ nearest degree | A1 | Correct answer (must be in degrees to nearest degree) |
## Part 5(c):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\mathbf{v} = (7-5t)\mathbf{i} + (12t-20)\mathbf{j} = (7\mathbf{i} - 20\mathbf{j}) + t(-5\mathbf{i} + 12\mathbf{j})$ | M1 | Collecting terms in $t$ and constant terms (may be implied) |
| $\frac{\mathbf{v} - (7\mathbf{i} - 20\mathbf{j})}{t} = (-5\mathbf{i} + 12\mathbf{j})$ | M1 A1 | Rearranging to required form; correct answer (isw if they find the magnitude) |
| **OR:** $t=0$, $\mathbf{v} = 7\mathbf{i} - 20\mathbf{j}$ | M1 | Finding the initial velocity or some other specific velocity |
| $\frac{(-3\mathbf{i}+4\mathbf{j})-(7\mathbf{i}-20\mathbf{j})}{2} = (-5\mathbf{i}+12\mathbf{j})$ | M1 A1 | Use of $\mathbf{a} = \frac{\mathbf{v}-\mathbf{u}}{t}$ with $t=2$ (or possibly another appropriate value) |
| **OR:** Differentiate wrt $t$: $\frac{d\mathbf{v}}{dt} = \mathbf{a} = (-5\mathbf{i}+12\mathbf{j})$ | M2, A1 | |
## Part 5(d):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{(7-5t)}{(12t-20)} = \frac{-5}{8}$ | M1 A1 | Attempt at equation in $t$ only, using ratio of components; allow reciprocal and a sign error; correct equation |
| Solve for $t$ | M1 | Equation must have come from considering ratios |
| $t = 2.2$ | A1 | Correct answer |
---5. A particle $P$ is moving in a plane with constant acceleration. The velocity, $\mathbf { v } \mathrm { m } \mathrm { s } ^ { - 1 }$, of $P$ at time $t$ seconds is given by
$$\mathbf { v } = ( 7 - 5 t ) \mathbf { i } + ( 12 t - 20 ) \mathbf { j }$$
\begin{enumerate}[label=(\alph*)]
\item Find the speed of $P$ when $t = 2$
\item Find, to the nearest degree, the size of the angle between the direction of motion of $P$ and the vector $\mathbf { j }$, when $t = 2$
The constant acceleration of $P$ is a m s-2
\item Find $\mathbf { a }$ in terms of $\mathbf { i }$ and $\mathbf { j }$
\item Find the value of $t$ when $P$ is moving in the direction of the vector $( - 5 \mathbf { i } + 8 \mathbf { j } )$\\
\begin{center}
\end{center}
\end{enumerate}
\hfill \mbox{\textit{Edexcel M1 2020 Q5 [13]}}