| Exam Board | AQA |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2008 |
| Session | January |
| Marks | 6 |
| 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 requiring standard SUVAT application and Newton's second law with vertical motion. Part (a) is a 'show that' using s=ut+½at² with given values, part (b) applies F=ma with weight, and part (c) is trivial (distance/time). All steps are routine textbook exercises with no problem-solving insight required, making it easier than average. |
| Spec | 3.02d Constant acceleration: SUVAT formulae3.03c Newton's second law: F=ma one dimension3.03f Weight: W=mg |
| Answer | Marks | Guidance |
|---|---|---|
| \(8 = \frac{1}{2}a \times 5^2\) → \(a = \frac{2 \times 8}{25} = 0.64 \text{ ms}^{-2}\) | M1, A1, A1 (Total: 2) | Use of constant acceleration equation with \(u = 0\) to find \(a\). Correct answer from correct working, showing evidence of solving for \(a\). Allow verification/substitution. |
| Answer | Marks | Guidance |
|---|---|---|
| \(T - 70 \times 9.8 = 70 \times 0.64\) → \(T = 730.8 = 731 \text{ N to 3 sf}\) | M1, A1, A1 (Total: 3) | Three term equation of motion for crate. Correct equation. Correct tension. |
| Answer | Marks | Guidance |
|---|---|---|
| \(v = \frac{8}{5} = 1.6 \text{ ms}^{-1}\) | B1 (Total: 1) | Correct average speed. Accept \(\frac{8}{5}\). Allow \(\frac{3.2 + 0}{2} = 1.6 \text{ ms}^{-1}\). |
### Part (a)
$8 = \frac{1}{2}a \times 5^2$ → $a = \frac{2 \times 8}{25} = 0.64 \text{ ms}^{-2}$ | M1, A1, A1 (Total: 2) | Use of constant acceleration equation with $u = 0$ to find $a$. Correct answer from correct working, showing evidence of solving for $a$. Allow verification/substitution.
### Part (b)
$T - 70 \times 9.8 = 70 \times 0.64$ → $T = 730.8 = 731 \text{ N to 3 sf}$ | M1, A1, A1 (Total: 3) | Three term equation of motion for crate. Correct equation. Correct tension.
### Part (c)
$v = \frac{8}{5} = 1.6 \text{ ms}^{-1}$ | B1 (Total: 1) | Correct average speed. Accept $\frac{8}{5}$. Allow $\frac{3.2 + 0}{2} = 1.6 \text{ ms}^{-1}$.
---1 A crane is used to lift a crate, of mass 70 kg , vertically upwards. As the crate is lifted, it accelerates uniformly from rest, rising 8 metres in 5 seconds.
\begin{enumerate}[label=(\alph*)]
\item Show that the acceleration of the crate is $0.64 \mathrm {~m} \mathrm {~s} ^ { - 2 }$.
\item The crate is attached to the crane by a single cable. Assume that there is no resistance to the motion of the crate.
Find the tension in the cable.
\item Calculate the average speed of the crate during these 5 seconds.
\end{enumerate}
\hfill \mbox{\textit{AQA M1 2008 Q1 [6]}}