| Exam Board | AQA |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2007 |
| Session | January |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Constant acceleration (SUVAT) |
| Type | Sketch velocity-time graph |
| Difficulty | Standard +0.3 This is a straightforward three-stage motion problem requiring a velocity-time graph sketch, distance calculation from the graph (area under curve), and application of Newton's second law (F=ma) to find tension. All steps are standard M1 techniques with clear given values and no conceptual surprises, making it slightly easier than average for mechanics questions. |
| Spec | 3.02b Kinematic graphs: displacement-time and velocity-time3.02d Constant acceleration: SUVAT formulae3.03c Newton's second law: F=ma one dimension3.03f Weight: W=mg |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| Graph starts and finishes at rest | B1 | Starts and finishes at rest |
| Correct shape (trapezium) | B1 | Correct shape |
| Correct values on \(t\)-axis: 0, 4, 9, 12 | B1 | Correct values on \(t\)-axis |
| Correct values on \(v\)-axis: 2 | B1 | Total: 4 — Correct values on \(v\)-axis. Condone omission of the origin |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| \(s = \frac{1}{2}(5 + 12) \times 2\) | M1 | Use of the area under the graph (or equivalent) to find \(s\) |
| or \(s = \frac{1}{2} \times 2 \times 4 + 5 \times 2 + \frac{1}{2} \times 2 \times 3 = 17\) | ||
| \(= 17\) | A1 | Total: 2 — Correct distance. SC When 21 used instead of 12 allow full marks for \(s = 26\) |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| \(\max a = \frac{2}{4} = 0.5\) | B1 | Maximum acceleration |
| \(300 \times 0.5 = T - 300 \times 9.8\) | M1 | Three term equation of motion using their \(a\) |
| A1 | Correct equation using \(a = 0.5\) | |
| \(T = 2940 + 150 = 3090\) | A1 | Total: 4 — Correct tension |
## Question 2:
### Part (a)
| Working | Marks | Guidance |
|---------|-------|----------|
| Graph starts and finishes at rest | B1 | Starts and finishes at rest |
| Correct shape (trapezium) | B1 | Correct shape |
| Correct values on $t$-axis: 0, 4, 9, 12 | B1 | Correct values on $t$-axis |
| Correct values on $v$-axis: 2 | B1 | **Total: 4** — Correct values on $v$-axis. Condone omission of the origin |
### Part (b)
| Working | Marks | Guidance |
|---------|-------|----------|
| $s = \frac{1}{2}(5 + 12) \times 2$ | M1 | Use of the area under the graph (or equivalent) to find $s$ |
| **or** $s = \frac{1}{2} \times 2 \times 4 + 5 \times 2 + \frac{1}{2} \times 2 \times 3 = 17$ | | |
| $= 17$ | A1 | **Total: 2** — Correct distance. **SC** When 21 used instead of 12 allow full marks for $s = 26$ |
### Part (c)
| Working | Marks | Guidance |
|---------|-------|----------|
| $\max a = \frac{2}{4} = 0.5$ | B1 | Maximum acceleration |
| $300 \times 0.5 = T - 300 \times 9.8$ | M1 | Three term equation of motion using their $a$ |
| | A1 | Correct equation using $a = 0.5$ |
| $T = 2940 + 150 = 3090$ | A1 | **Total: 4** — Correct tension |
---2 A lift rises vertically from rest with a constant acceleration.\\
After 4 seconds, it is moving upwards with a velocity of $2 \mathrm {~ms} ^ { - 1 }$.\\
It then moves with a constant velocity for 5 seconds.\\
The lift then slows down uniformly, coming to rest after it has been moving for a total of 12 seconds.
\begin{enumerate}[label=(\alph*)]
\item Sketch a velocity-time graph for the motion of the lift.
\item Calculate the total distance travelled by the lift.
\item The lift is raised by a single vertical cable. The mass of the lift is 300 kg . Find the maximum tension in the cable during this motion.
\end{enumerate}
\hfill \mbox{\textit{AQA M1 2007 Q2 [10]}}