| Exam Board | AQA |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2005 |
| Session | January |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Constant acceleration (SUVAT) |
| Type | SUVAT single equation: straightforward find |
| Difficulty | Moderate -0.8 This is a straightforward M1 question testing basic SUVAT equations and Newton's second law with given values. Part (a) involves simple substitution into v=u+at and s=ut+½at², while part (b) requires F=ma with resistance forces—all standard textbook exercises requiring only routine application of formulas with no problem-solving insight. |
| Spec | 3.02d Constant acceleration: SUVAT formulae3.03c Newton's second law: F=ma one dimension |
| Answer | Marks | Guidance |
|---|---|---|
| \(40 = 12 + 100a\) | M1 | Use of constant acceleration equation to form equation for \(a\) |
| \(a = \frac{40-12}{100} = 0.28 \text{ ms}^{-2}\) AG | A1 (2) | AG; correct answer from correct working |
| Answer | Marks | Guidance |
|---|---|---|
| \(s = \frac{1}{2}(12+40)\times100\) | M1 | Expression for distance, using \(t=100\) |
| \(= 2600 \text{ m}\) | A1 (2) | Correct final distance |
| Answer | Marks | Guidance |
|---|---|---|
| \(F - 40000 = 200\times1000\times0.28\) | M1 | Three term equation of motion |
| \(F = 40000 + 56000 = 96000 \text{ N}\) | A1, A1 (3) | Correct equation; Correct force |
## Question 1:
### Part (a)(i)
$40 = 12 + 100a$ | M1 | Use of constant acceleration equation to form equation for $a$
$a = \frac{40-12}{100} = 0.28 \text{ ms}^{-2}$ AG | A1 (2) | AG; correct answer from correct working
### Part (a)(ii)
$s = \frac{1}{2}(12+40)\times100$ | M1 | Expression for distance, using $t=100$
$= 2600 \text{ m}$ | A1 (2) | Correct final distance
### Part (c)
$F - 40000 = 200\times1000\times0.28$ | M1 | Three term equation of motion
$F = 40000 + 56000 = 96000 \text{ N}$ | A1, A1 (3) | Correct equation; Correct force
---1 A train travels along a straight horizontal track. It is travelling at a speed of $12 \mathrm {~ms} ^ { - 1 }$ when it begins to accelerate uniformly. It reaches a speed of $40 \mathrm {~ms} ^ { - 1 }$ after accelerating for 100 seconds.
\begin{enumerate}[label=(\alph*)]
\item \begin{enumerate}[label=(\roman*)]
\item Show that the acceleration of the train is $0.28 \mathrm {~m} \mathrm {~s} ^ { - 2 }$.
\item Find the distance that the train travelled in the 100 seconds.
\end{enumerate}\item The mass of the train is 200 tonnes and a resistance force of 40000 N acts on the train. Find the magnitude of the driving force produced by the engine that acts on the train as it accelerates.
\end{enumerate}
\hfill \mbox{\textit{AQA M1 2005 Q1 [7]}}