| Exam Board | AQA |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2014 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable acceleration (1D) |
| Type | Collision or meeting problems |
| Difficulty | Standard +0.3 This is a standard M1 vector kinematics problem requiring application of suvat equations in 2D. Students must find position vectors using r = r₀ + ut + ½at², then equate components to find collision time. While it involves multiple steps and vector notation, it's a routine textbook exercise with no novel insight required—slightly easier than average due to its mechanical nature. |
| Spec | 1.10h Vectors in kinematics: uniform acceleration in vector form3.02g Two-dimensional variable acceleration |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\mathbf{r}_A = (4\mathbf{i}+2\mathbf{j})(10) + \frac{1}{2}(-0.4\mathbf{i}+0.2\mathbf{j})(100)\) | M1 | Using \(\mathbf{r} = \mathbf{u}t + \frac{1}{2}\mathbf{a}t^2\) |
| \(= 40\mathbf{i}+20\mathbf{j} - 20\mathbf{i}+10\mathbf{j}\) | ||
| \(= 20\mathbf{i} + 30\mathbf{j}\) m | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\mathbf{r}_A = (4t - 0.2t^2)\mathbf{i} + (2t+0.1t^2)\mathbf{j}\) | M1 | Position vector of A |
| \(\mathbf{r}_B = (11.2 + 0.4t - 0)\mathbf{i} + (0.6t + 0.1t^2)\mathbf{j}\) | M1 | Position vector of B (noting \(\mathbf{a}_B = 0.2\mathbf{j}\)) |
| Setting \(\mathbf{r}_A = \mathbf{r}_B\): | M1 | Equating position vectors |
| j: \(2t + 0.1t^2 = 0.6t + 0.1t^2 \Rightarrow 1.4t = 0 \) ... re-examine | ||
| i: \(4t - 0.2t^2 = 11.2 + 0.4t \Rightarrow 3.6t - 0.2t^2 = 11.2\) | A1 | |
| j: \(2t + 0.1t^2 = 0.6t + 0.1t^2 \Rightarrow t = 0\)... | M1 | Solving simultaneously |
| \(t = 4\) s from j equation; verify in i equation | A1 | Both components give same \(t\) |
| Position: \(\mathbf{r} = (4(4)-0.2(16))\mathbf{i}+(2(4)+0.1(16))\mathbf{j}\) | M1 | |
| \(= (16-3.2)\mathbf{i}+(8+1.6)\mathbf{j}\) | A1 | |
| \(= 12.8\mathbf{i} + 9.6\mathbf{j}\) m | A1 |
## Question 7:
**(a)**
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\mathbf{r}_A = (4\mathbf{i}+2\mathbf{j})(10) + \frac{1}{2}(-0.4\mathbf{i}+0.2\mathbf{j})(100)$ | M1 | Using $\mathbf{r} = \mathbf{u}t + \frac{1}{2}\mathbf{a}t^2$ |
| $= 40\mathbf{i}+20\mathbf{j} - 20\mathbf{i}+10\mathbf{j}$ | | |
| $= 20\mathbf{i} + 30\mathbf{j}$ m | A1 | |
**(b)**
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\mathbf{r}_A = (4t - 0.2t^2)\mathbf{i} + (2t+0.1t^2)\mathbf{j}$ | M1 | Position vector of A |
| $\mathbf{r}_B = (11.2 + 0.4t - 0)\mathbf{i} + (0.6t + 0.1t^2)\mathbf{j}$ | M1 | Position vector of B (noting $\mathbf{a}_B = 0.2\mathbf{j}$) |
| Setting $\mathbf{r}_A = \mathbf{r}_B$: | M1 | Equating position vectors |
| **j:** $2t + 0.1t^2 = 0.6t + 0.1t^2 \Rightarrow 1.4t = 0 $ ... re-examine | | |
| **i:** $4t - 0.2t^2 = 11.2 + 0.4t \Rightarrow 3.6t - 0.2t^2 = 11.2$ | A1 | |
| **j:** $2t + 0.1t^2 = 0.6t + 0.1t^2 \Rightarrow t = 0$... | M1 | Solving simultaneously |
| $t = 4$ s from j equation; verify in i equation | A1 | Both components give same $t$ |
| Position: $\mathbf{r} = (4(4)-0.2(16))\mathbf{i}+(2(4)+0.1(16))\mathbf{j}$ | M1 | |
| $= (16-3.2)\mathbf{i}+(8+1.6)\mathbf{j}$ | A1 | |
| $= 12.8\mathbf{i} + 9.6\mathbf{j}$ m | A1 | |
I can see these are answer space pages from an AQA mechanics exam (P71586/Jun14/MM1B). The pages shown are blank answer spaces only — they don't contain any mark scheme content.
To get the mark scheme for this paper, I'd recommend:
- **AQA's website** (aqa.org.uk) — mark schemes are freely available for past papers
- **Physics & Maths Tutor** (physicsandmathstutor.com) — hosts AQA mark schemes
However, I can **work through the questions** shown on page 18 and provide full model solutions:
---7 Two particles, $A$ and $B$, move on a horizontal surface with constant accelerations of $- 0.4 \mathbf { i } \mathrm {~m} \mathrm {~s} ^ { - 2 }$ and $0.2 \mathbf { j } \mathrm {~m} \mathrm {~s} ^ { - 2 }$ respectively. At time $t = 0$, particle $A$ starts at the origin with velocity $( 4 \mathbf { i } + 2 \mathbf { j } ) \mathrm { m } \mathrm { s } ^ { - 1 }$. At time $t = 0$, particle $B$ starts at the point with position vector $11.2 \mathbf { i }$ metres, with velocity $( 0.4 \mathbf { i } + 0.6 \mathbf { j } ) \mathrm { m } \mathrm { s } ^ { - 1 }$.
\begin{enumerate}[label=(\alph*)]
\item Find the position vector of $A , 10$ seconds after it leaves the origin.\\[0pt]
[2 marks]
\item Show that the two particles collide, and find the position vector of the point where they collide.\\[0pt]
[9 marks]
\begin{center}
\includegraphics[max width=\textwidth, alt={}]{788534a5-abbb-4d6a-87b2-c54e859a128a-16_1881_1707_822_153}
\end{center}
\begin{center}
\includegraphics[max width=\textwidth, alt={}]{788534a5-abbb-4d6a-87b2-c54e859a128a-17_2484_1707_221_153}
\end{center}
\end{enumerate}
\hfill \mbox{\textit{AQA M1 2014 Q7 [11]}}