| Exam Board | OCR MEI |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2016 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Constant acceleration (SUVAT) |
| Type | SUVAT simultaneous equations: find u and a |
| Difficulty | Moderate -0.3 This is a standard SUVAT problem requiring students to form two simultaneous equations from given conditions and solve them, followed by finding a turning point. While it involves algebraic manipulation and understanding of constant acceleration, it's a routine mechanics question with no novel insight required—slightly easier than average due to its straightforward structure. |
| Spec | 3.02d Constant acceleration: SUVAT formulae |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(s = ut + \frac{1}{2}at^2\) | The final mark scheme will include commonly used alternative methods. | |
| \(t = 2 \Rightarrow 2u + 2a = 12\) | B1 | Allow one equation with \(v = 0\) and \(t = 4\) |
| \(t = 6 \Rightarrow 6u + 18a = 12\) | B1 | |
| Solving the simultaneous equations | M1 | Attempt to solve non-trivial simultaneous equations in \(u\) and \(a\) |
| \(u = 8, a = -2\) | A1 | CAO |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| At B, \(v^2 - u^2 = 2as\) | Follow through for their values of \(u\) and \(a\). | |
| \(\Rightarrow 0^2 - 8^2 = 2 \times (-2) \times s\) | M1 | Allow the use of \(s = ut + \frac{1}{2}at^2\) with \(t = 4\). |
| \(s = 16\) | A1 | |
| AB is 4 m. | A1 | CAO |
**(i)**
| Answer | Mark | Guidance |
|--------|------|----------|
| $s = ut + \frac{1}{2}at^2$ | | The final mark scheme will include commonly used alternative methods. |
| $t = 2 \Rightarrow 2u + 2a = 12$ | B1 | Allow one equation with $v = 0$ and $t = 4$ |
| $t = 6 \Rightarrow 6u + 18a = 12$ | B1 | |
| Solving the simultaneous equations | M1 | Attempt to solve non-trivial simultaneous equations in $u$ and $a$ |
| $u = 8, a = -2$ | A1 | CAO |
**(ii)**
| Answer | Mark | Guidance |
|--------|------|----------|
| At B, $v^2 - u^2 = 2as$ | | Follow through for their values of $u$ and $a$. |
| $\Rightarrow 0^2 - 8^2 = 2 \times (-2) \times s$ | M1 | Allow the use of $s = ut + \frac{1}{2}at^2$ with $t = 4$. |
| $s = 16$ | A1 | |
| AB is 4 m. | A1 | CAO |
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\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{4c8c96cf-5184-46e4-9c45-a8a80d0a6ff8-2_117_1162_1486_438}
\captionsetup{labelformat=empty}
\caption{Fig. 2}
\end{center}
\end{figure}
A particle moves on the straight line shown in Fig. 2. The positive direction is indicated on the diagram. The time, $t$, is measured in seconds. The particle has constant acceleration, $a \mathrm {~ms} ^ { - 2 }$.
Initially it is at the point O and has velocity $u \mathrm {~ms} ^ { - 1 }$.\\
When $t = 2$, the particle is at A where OA is 12 m . The particle is also at A when $t = 6$.\\
(i) Write down two equations in $u$ and $a$ and solve them.\\
(ii) The particle changes direction when it is at B .
Find the distance AB .
\hfill \mbox{\textit{OCR MEI M1 2016 Q2 [7]}}