| Exam Board | OCR MEI |
|---|---|
| Module | M1 (Mechanics 1) |
| Marks | 18 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Constant acceleration (SUVAT) |
| Type | Read and interpret velocity-time graph |
| Difficulty | Moderate -0.3 This is a straightforward mechanics question testing basic calculus applications (differentiation for acceleration, integration for displacement) and graph interpretation. All parts involve standard techniques with no novel problem-solving required, making it slightly easier than average for A-level mechanics. |
| Spec | 1.02z Models in context: use functions in modelling3.02b Kinematic graphs: displacement-time and velocity-time3.02c Interpret kinematic graphs: gradient and area3.02d Constant acceleration: SUVAT formulae |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\frac{-20}{2} = -10\), \(-10\) m s\(^{-2}\) | M1 | Use of suitable triangle to attempt \(\Delta v / \Delta t\) for suitable interval. Accept wrong sign |
| A1 | cao. Allow both marks if correct answer seen |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Signed area under graph: \(\frac{1}{2} \times 2 \times 20 = 20\) | M1, A1 | Using the relevant area or other complete method |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Either using areas: Signed area \(2 \le t \le 5\): \(\frac{1}{2}\times((5-2)+(4.5-2.4))\times(-4) = -10.2\) | B1 | Allow \(+10.2\) |
| Signed area \(5 \le t \le 6\): \(\frac{1}{2}\times 1 \times 8 = 4\) | B1 | |
| Total displacement is \(13.8\) m | B1 | cao but FT from their 20 in part (A) |
| Or using suvat: \(t=0\) to \(t=2.4\): \(19.2\); \(t=4.5\) to \(t=6\): \(3.0\); \(t=2.4\) to \(t=4.5\): \(-8.4\); Total: \(13.8\) | B1, B1, B1 | Both required and both must be correct |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(a = 4t - 14\) | M1, A1 | Differentiate. Do not award for division by \(t\) |
| \(a(0.5) = -12\) so \(-12\) m s\(^{-2}\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Model A gives \(-4\) m s\(^{-1}\) | B1 | May be implied by other working |
| For model B we need \(v\) when \(a = 0\) | M1 | Using (iii) or argument based on symmetry/sketch that \(a=0\) when \(t=3.5\) |
| \(v\!\left(\tfrac{7}{2}\right) = -4.5\) | A1 | |
| So model B is \(0.5\) m s\(^{-1}\) less | F1 | Accept values without "more or less" |
## Question 4:
### Part (i)
| Answer | Mark | Guidance |
|--------|------|----------|
| $\frac{-20}{2} = -10$, $-10$ m s$^{-2}$ | M1 | Use of suitable triangle to attempt $\Delta v / \Delta t$ for suitable interval. Accept wrong sign |
| | A1 | cao. Allow both marks if correct answer seen |
### Part (ii)(A)
| Answer | Mark | Guidance |
|--------|------|----------|
| Signed area under graph: $\frac{1}{2} \times 2 \times 20 = 20$ | M1, A1 | Using the relevant area or other complete method |
### Part (ii)(B)
| Answer | Mark | Guidance |
|--------|------|----------|
| **Either** using areas: Signed area $2 \le t \le 5$: $\frac{1}{2}\times((5-2)+(4.5-2.4))\times(-4) = -10.2$ | B1 | Allow $+10.2$ |
| Signed area $5 \le t \le 6$: $\frac{1}{2}\times 1 \times 8 = 4$ | B1 | |
| Total displacement is $13.8$ m | B1 | cao but FT from their 20 in part (A) |
| **Or** using suvat: $t=0$ to $t=2.4$: $19.2$; $t=4.5$ to $t=6$: $3.0$; $t=2.4$ to $t=4.5$: $-8.4$; Total: $13.8$ | B1, B1, B1 | Both required and both must be correct |
### Part (iii)
| Answer | Mark | Guidance |
|--------|------|----------|
| $a = 4t - 14$ | M1, A1 | Differentiate. Do not award for division by $t$ |
| $a(0.5) = -12$ so $-12$ m s$^{-2}$ | A1 | |
### Part (iv)
| Answer | Mark | Guidance |
|--------|------|----------|
| Model A gives $-4$ m s$^{-1}$ | B1 | May be implied by other working |
| For model B we need $v$ when $a = 0$ | M1 | Using (iii) or argument based on symmetry/sketch that $a=0$ when $t=3.5$ |
| $v\!\left(\tfrac{7}{2}\right) = -4.5$ | A1 | |
| So model B is $0.5$ m s$^{-1}$ less | F1 | Accept values without "more or less" |4 A ring is moving on a straight wire. Its velocity is $v \mathrm {~m} \mathrm {~s} ^ { - 1 }$ at time $t$ seconds after passing a point Q .\\
Model A for the motion of the ring gives the velocity-time graph for $0 \leqslant t \leqslant 6$ shown in Fig. 7 .
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{196bd74f-c2b2-4cb3-b03c-8ecd9fce9c11-2_937_1414_325_404}
\captionsetup{labelformat=empty}
\caption{Fig. 7}
\end{center}
\end{figure}
Use model A to calculate the following.
\begin{enumerate}[label=(\roman*)]
\item The acceleration of the ring when $t = 0.5$.
\item The displacement of the ring from Q when\\
(A) $t = 2$,\\
(B) $t = 6$.
In an alternative model B , the velocity of the ring is given by $v = 2 t ^ { 2 } - 14 t + 20$ for $0 \leqslant t \leqslant 6$.
\item Calculate the acceleration of the ring at $t = 0.5$ as given by model B.
\item Calculate by how much the models differ in their values for the least $v$ in the time interval $0 \leqslant t \leqslant 6$.
\item Calculate the displacement of the ring from Q when $t = 6$ as given by model B .
\end{enumerate}
\hfill \mbox{\textit{OCR MEI M1 Q4 [18]}}