| Exam Board | OCR MEI |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2012 |
| Session | June |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Constant acceleration (SUVAT) |
| Type | Find acceleration from distances/times |
| Difficulty | Moderate -0.3 This is a straightforward two-part SUVAT question requiring standard application of kinematic equations with given values. Part (i) uses s=ut+½at² directly, and part (ii) uses v²=u²+2as then v=u+at. No problem-solving insight needed, just systematic substitution into familiar formulas. |
| Spec | 3.02d Constant acceleration: SUVAT formulae |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| At C: \(s = ut + \frac{1}{2}at^2\) | ||
| \(500 = 5\times20 + 0.5\times a\times20^2\) | M1 | M1 for a method which if correctly applied would give \(a\) |
| \(a = 2\ (\text{ms}^{-2})\) | A1 | Cao. Special case: If 800 used for \(s\) giving \(a=3.5\), treat as misread SC SC, give M1 A0 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| At B: \(v^2 - u^2 = 2as\) | M1 | M1 for method giving either \(v\) or \(t\). Apply FT from incorrect \(a\) for M mark only |
| \(v^2 - 5^2 = 2\times2\times300\) | ||
| \(v = 35\ \ \) Speed is \(35\ \text{ms}^{-1}\) | A1 | Cao. No FT except SC1 for 46.2 following \(a=3.5\) after use of \(s=800\) |
| At B: \(v = u + at\) | ||
| \(35 = 5 + 2\times t\) | ||
| \(t = 15\ \) Time is \(15\ \text{s}\) | A1 | Cao. No FT except SC1 for 11.7 following \(a=3.5\) after use of \(s=800\) |
# Question 4:
## Part (i)
| Answer | Marks | Guidance |
|--------|-------|----------|
| At C: $s = ut + \frac{1}{2}at^2$ | | |
| $500 = 5\times20 + 0.5\times a\times20^2$ | M1 | M1 for a method which if correctly applied would give $a$ |
| $a = 2\ (\text{ms}^{-2})$ | A1 | Cao. Special case: If 800 used for $s$ giving $a=3.5$, treat as misread SC SC, give M1 A0 |
## Part (ii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| At B: $v^2 - u^2 = 2as$ | M1 | M1 for method giving either $v$ or $t$. Apply FT from incorrect $a$ for M mark only |
| $v^2 - 5^2 = 2\times2\times300$ | | |
| $v = 35\ \ $ Speed is $35\ \text{ms}^{-1}$ | A1 | Cao. No FT except SC1 for 46.2 following $a=3.5$ after use of $s=800$ |
| At B: $v = u + at$ | | |
| $35 = 5 + 2\times t$ | | |
| $t = 15\ $ Time is $15\ \text{s}$ | A1 | Cao. No FT except SC1 for 11.7 following $a=3.5$ after use of $s=800$ |
---4 Fig. 4 illustrates points $\mathrm { A } , \mathrm { B }$ and C on a straight race track. The distance AB is 300 m and AC is 500 m .\\
A car is travelling along the track with uniform acceleration.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{076ad371-b029-4d57-aa0f-8a78ed03ccf3-3_65_1324_897_372}
\captionsetup{labelformat=empty}
\caption{Fig. 4}
\end{center}
\end{figure}
Initially the car is at A and travelling in the direction AB with speed $5 \mathrm {~ms} ^ { - 1 }$. After 20 s it is at C .\\
(i) Find the acceleration of the car.\\
(ii) Find the speed of the car at B and how long it takes to travel from A to B .
\hfill \mbox{\textit{OCR MEI M1 2012 Q4 [5]}}