| Exam Board | AQA |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2008 |
| Session | January |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Newton's laws and connected particles |
| Type | Two connected particles, horizontal surface |
| Difficulty | Standard +0.3 This is a standard M1 connected particles problem with straightforward application of Newton's second law and friction formulae. Parts (a)-(c) involve routine calculations (F=μR, F=ma), while part (d) requires basic conceptual understanding that angled rope reduces normal reaction. Slightly easier than average due to 'show that' scaffolding and minimal problem-solving required. |
| Spec | 3.03k Connected particles: pulleys and equilibrium3.03o Advanced connected particles: and pulleys3.03r Friction: concept and vector form3.03t Coefficient of friction: F <= mu*R model |
| Answer | Marks | Guidance |
|---|---|---|
| \(F = 0.4 \times 1000 \times 9.8 = 3920\) | M1, A1 (Total: 2) | Use of \(F = \mu R\). Correct friction from correct working. Allow \(F = 0.4 \times 9800\). Allow verification. |
| Answer | Marks | Guidance |
|---|---|---|
| \(P - 3920 = 5000 \times 0.8\) → \(P = 7920 \text{ N}\) | M1, A1, A1 (Total: 3) | Three term equation of motion including an explicit 0.8. Correct equation. Correct force from correct working. Allow \(P = 5000 \times 0.8 + 3920\). |
| Answer | Marks | Guidance |
|---|---|---|
| \(T - 3920 = 1000 \times 0.8\) → \(T = 4720 \text{ N}\) or \(7920 - T = 4000 \times 0.8\) → \(T = 4720 \text{ N}\) | M1, A1, A1 (Total: 3) | Three term equation of motion. Correct equation. Correct tension. |
| Answer | Marks | Guidance |
|---|---|---|
| Friction is reduced because the normal reaction is reduced. | B1, E1 (Total: 2) | Friction reduced. Acceptable explanation. |
### Part (a)
$F = 0.4 \times 1000 \times 9.8 = 3920$ | M1, A1 (Total: 2) | Use of $F = \mu R$. Correct friction from correct working. Allow $F = 0.4 \times 9800$. Allow verification.
### Part (b)
$P - 3920 = 5000 \times 0.8$ → $P = 7920 \text{ N}$ | M1, A1, A1 (Total: 3) | Three term equation of motion including an explicit 0.8. Correct equation. Correct force from correct working. Allow $P = 5000 \times 0.8 + 3920$.
### Part (c)
$T - 3920 = 1000 \times 0.8$ → $T = 4720 \text{ N}$ or $7920 - T = 4000 \times 0.8$ → $T = 4720 \text{ N}$ | M1, A1, A1 (Total: 3) | Three term equation of motion. Correct equation. Correct tension.
### Part (d)
Friction is reduced because the normal reaction is reduced. | B1, E1 (Total: 2) | Friction reduced. Acceptable explanation.
---6 A tractor, of mass 4000 kg , is used to pull a skip, of mass 1000 kg , over a rough horizontal surface. The tractor is connected to the skip by a rope, which remains taut and horizontal throughout the motion, as shown in the diagram.\\
\includegraphics[max width=\textwidth, alt={}, center]{217f0e3e-9d1b-41f1-8299-f56d073fbbeb-4_243_880_477_571}
Assume that only two horizontal forces act on the tractor. One is a driving force, which has magnitude $P$ newtons and acts in the direction of motion. The other is the tension in the rope.
The coefficient of friction between the skip and the ground is 0.4 .\\
The tractor and the skip accelerate at $0.8 \mathrm {~ms} ^ { - 2 }$.
\begin{enumerate}[label=(\alph*)]
\item Show that the magnitude of the friction force acting on the skip is 3920 N .
\item Show that $P = 7920$.
\item Find the tension in the rope.
\item Suppose that, during the motion, the rope is not horizontal, but inclined at a small angle to the horizontal, with the higher end of the rope attached to the tractor, as shown in the diagram.\\
\includegraphics[max width=\textwidth, alt={}, center]{217f0e3e-9d1b-41f1-8299-f56d073fbbeb-4_241_880_1665_571}
How would the magnitude of the friction force acting on the skip differ from that found in part (a)?
Explain why.
\end{enumerate}
\hfill \mbox{\textit{AQA M1 2008 Q6 [10]}}