| Exam Board | OCR MEI |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2013 |
| Session | January |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Constant acceleration (SUVAT) |
| Type | SUVAT simultaneous equations: find u and a |
| Difficulty | Standard +0.3 This is a standard two-part SUVAT question requiring systematic application of kinematic equations to find unknowns, followed by a straightforward proof using a quadratic discriminant. The algebraic manipulation is routine for M1 level, though part (ii) requires recognizing that no real solution means the particle never reaches P. |
| Spec | 3.02d Constant acceleration: SUVAT formulae |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Either: \(s = \frac{1}{2}(u+v)t\), take O as origin: \(30 = \frac{1}{2}\times(u+9)\times10\) | M1 | Use of one relevant equation, including substitution |
| \(u = -3\) | A1 | |
| \(v = u + at\): \(9 = -3 + 10a\) | M1 | Use of a second relevant equation including substitution |
| \(a = 1.2\) | A1 | |
| Or: \(v = u + at \Rightarrow u + 10a = 9\) | M1 | Use of one relevant equation, including substitution |
| \(s = ut + \frac{1}{2}at^2 \Rightarrow u + 5a = 3\) | M1 | Use of a second relevant equation including substitution |
| Solving simultaneously: \(a = 1.2\) | A1 | |
| \(u = -3\) | A1 | |
| Or: \(s = vt - \frac{1}{2}at^2 \Rightarrow a = 1.2\) | M1, A1 | Use of one relevant equation |
| \(v = u + at \Rightarrow u = -3\) | M1, A1 | Use of a second relevant equation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Either: \(s = ut + \frac{1}{2}at^2\); solving for P: \(-5 = -3t + \frac{1}{2}\times1.2t^2\) | M1 | Quadratic equation with \(s = -5\) |
| \(0.6t^2 - 3t + 5 = 0\) | ||
| Discriminant \(= 3^2 - 4\times0.6\times5 = -3\) | M1 | Considering the discriminant or equivalent |
| No real roots for \(t\) (\(\Rightarrow\) Particle is never at P) | E1 | Cao without wrong working in the whole question |
| Or: Find when \(v=0\): \(v = u + at\), \(v=0 \Rightarrow t = 2.5\) | M1 | |
| \(s = ut + \frac{1}{2}at^2\) and \(t = 2.5\) | M1 | Or use \(v^2 = u^2 + 2as\) |
| \(\Rightarrow s = -3.75 > -5\) | E1 | Cao without wrong working. Comparison necessary |
| Special cases when \(u > 0\) and \(a > 0\): "It is always going to the right" | SC1 | |
| Demonstration that it is at \(-5\) for two negative times | SC1 |
## Question 4:
**Part (i)**
| Answer | Marks | Guidance |
|--------|-------|----------|
| **Either:** $s = \frac{1}{2}(u+v)t$, take O as origin: $30 = \frac{1}{2}\times(u+9)\times10$ | M1 | Use of one relevant equation, including substitution |
| $u = -3$ | A1 | |
| $v = u + at$: $9 = -3 + 10a$ | M1 | Use of a second relevant equation including substitution |
| $a = 1.2$ | A1 | |
| **Or:** $v = u + at \Rightarrow u + 10a = 9$ | M1 | Use of one relevant equation, including substitution |
| $s = ut + \frac{1}{2}at^2 \Rightarrow u + 5a = 3$ | M1 | Use of a second relevant equation including substitution |
| Solving simultaneously: $a = 1.2$ | A1 | |
| $u = -3$ | A1 | |
| **Or:** $s = vt - \frac{1}{2}at^2 \Rightarrow a = 1.2$ | M1, A1 | Use of one relevant equation |
| $v = u + at \Rightarrow u = -3$ | M1, A1 | Use of a second relevant equation |
**Part (ii)**
| Answer | Marks | Guidance |
|--------|-------|----------|
| **Either:** $s = ut + \frac{1}{2}at^2$; solving for P: $-5 = -3t + \frac{1}{2}\times1.2t^2$ | M1 | Quadratic equation with $s = -5$ |
| $0.6t^2 - 3t + 5 = 0$ | | |
| Discriminant $= 3^2 - 4\times0.6\times5 = -3$ | M1 | Considering the discriminant or equivalent |
| No real roots for $t$ ($\Rightarrow$ Particle is never at P) | E1 | Cao without wrong working in the whole question |
| **Or:** Find when $v=0$: $v = u + at$, $v=0 \Rightarrow t = 2.5$ | M1 | |
| $s = ut + \frac{1}{2}at^2$ and $t = 2.5$ | M1 | Or use $v^2 = u^2 + 2as$ |
| $\Rightarrow s = -3.75 > -5$ | E1 | Cao without wrong working. Comparison necessary |
| **Special cases when $u > 0$ and $a > 0$:** "It is always going to the right" | SC1 | |
| Demonstration that it is at $-5$ for two negative times | SC1 | |
---4 A particle is travelling along a straight line with constant acceleration. $\mathrm { P } , \mathrm { O }$ and Q are points on the line, as illustrated in Fig. 4. The distance from P to O is 5 m and the distance from O to Q is 30 m .
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{13f555cc-d506-48e5-a0e4-225cae4251dc-4_113_1173_1576_447}
\captionsetup{labelformat=empty}
\caption{Fig. 4}
\end{center}
\end{figure}
Initially the particle is at O . After 10 s , it is at Q and its velocity is $9 \mathrm {~ms} ^ { - 1 }$ in the direction $\overrightarrow { \mathrm { OQ } }$.\\
(i) Find the initial velocity and the acceleration of the particle.\\
(ii) Prove that the particle is never at P .
\hfill \mbox{\textit{OCR MEI M1 2013 Q4 [7]}}