| Exam Board | OCR MEI |
|---|---|
| Module | M1 (Mechanics 1) |
| 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 SUVAT problem requiring two equations to find two unknowns (initial velocity and acceleration), followed by a straightforward proof using a quadratic discriminant. The algebraic manipulation is routine for M1 level, and the 'prove never at P' part is a common question type that students practice regularly. |
| 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 | M1 | Use of one relevant equation, including substitution |
| \(30 = \frac{1}{2} \times (u+9) \times 10\) | ||
| \(u = -3\) | A1 | |
| \(v = u + at\) | M1 | Use of a second relevant equation including substitution |
| \(9 = -3 + 10a\) | ||
| \(a = 1.2\) | 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 \(s = vt - \frac{1}{2}at^2\) | M1 | Use of one relevant equation, including substitution |
| \(\Rightarrow a = 1.2\) | A1 | |
| \(v = u + at\) | M1 | Use of a second relevant equation including substitution |
| \(\Rightarrow u = -3\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Either \(s = ut + \frac{1}{2}at^2\), solving for P: \(-5 = -3t + \frac{1}{2} \times 1.2t^2\) | M1 | Quadratic equation with \(s = -5\) |
| \(0.6t^2 - 3t + 5 = 0\) | ||
| Discriminant \(= 3^2 - 4 \times 0.6 \times 5 = -3\) | 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 \(v = 0\) | M1 | |
| \(v = u + at,\ v = 0 \Rightarrow t = 2.5\) | ||
| \(s = ut + \frac{1}{2}at^2\) 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 their \(u > 0\) and their \(a > 0\) | SC1 | "It is always going to the right" |
| SC1 | Demonstration that it is at \(-5\) for two negative times |
## Question 3:
### Part (i)
| Answer | Marks | Guidance |
|--------|-------|----------|
| **Either** $s = \frac{1}{2}(u+v)t$, take O as origin | M1 | Use of one relevant equation, including substitution |
| $30 = \frac{1}{2} \times (u+9) \times 10$ | | |
| $u = -3$ | A1 | |
| $v = u + at$ | M1 | Use of a second relevant equation including substitution |
| $9 = -3 + 10a$ | | |
| $a = 1.2$ | 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** $s = vt - \frac{1}{2}at^2$ | M1 | Use of one relevant equation, including substitution |
| $\Rightarrow a = 1.2$ | A1 | |
| $v = u + at$ | M1 | Use of a second relevant equation including substitution |
| $\Rightarrow u = -3$ | A1 | |
### Part (ii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| **Either** $s = ut + \frac{1}{2}at^2$, solving for P: $-5 = -3t + \frac{1}{2} \times 1.2t^2$ | M1 | Quadratic equation with $s = -5$ |
| $0.6t^2 - 3t + 5 = 0$ | | |
| Discriminant $= 3^2 - 4 \times 0.6 \times 5 = -3$ | 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 $v = 0$ | M1 | |
| $v = u + at,\ v = 0 \Rightarrow t = 2.5$ | | |
| $s = ut + \frac{1}{2}at^2$ 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 their $u > 0$ and their $a > 0$** | SC1 | "It is always going to the right" |
| | SC1 | Demonstration that it is at $-5$ for two negative times |
---3 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]{d5a09ed4-a32f-4ff7-aa08-6e54c2ab26a0-2_115_1169_1719_499}
\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 $9 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ 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 Q3 [7]}}