| Exam Board | AQA |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2005 |
| Session | January |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Newton's laws and connected particles |
| Type | Block on rough horizontal surface – accelerating (finding acceleration or applied force) |
| Difficulty | Moderate -0.3 This is a standard M1 connected particles question with friction requiring resolution of forces in two directions and application of F=ma. All parts are routine calculations following well-practiced methods (force diagrams, resolving horizontally/vertically, friction coefficient formula), with parts (b) and (c) being 'show that' questions that guide students to the answer. Part (e) requires minimal conceptual reasoning about how angle affects normal reaction and hence friction. |
| Spec | 3.03d Newton's second law: 2D vectors3.03e Resolve forces: two dimensions3.03u Static equilibrium: on rough surfaces3.03v Motion on rough surface: including inclined planes |
| Answer | Marks | Guidance |
|---|---|---|
| Diagram showing \(F\), \(T\), \(mg\) | B1 (1) | Correct diagram |
| Answer | Marks | Guidance |
|---|---|---|
| \(40\cos30° - F = 25\times0.1\) | M1 | Three term equation of motion |
| \(F = 40\cos30° - 2.5 = 32.1 \text{ N}\) | A1, A1 (3) | Correct equation; AG correct force from correct working |
| Answer | Marks | Guidance |
|---|---|---|
| \(R + 40\sin30° = 25\times9.8\) | M1 | Resolving vertically |
| \(R = 225 \text{ N}\) | A1, A1 (3) | Correct equation; AG correct force from correct working |
| Answer | Marks | Guidance |
|---|---|---|
| \(32.1 = 225\mu\) | M1 | Use of \(F = \mu R\) |
| \(\mu = \frac{32.1}{225} = 0.143\) | A1 (2) | Correct \(\mu\) |
| Answer | Marks | Guidance |
|---|---|---|
| Friction will decrease as normal reaction decreases | B1, B1 (2) | Decrease in friction; Normal reaction decreases |
## Question 3:
### Part (a)
Diagram showing $F$, $T$, $mg$ | B1 (1) | Correct diagram
### Part (b)
$40\cos30° - F = 25\times0.1$ | M1 | Three term equation of motion
$F = 40\cos30° - 2.5 = 32.1 \text{ N}$ | A1, A1 (3) | Correct equation; AG correct force from correct working
### Part (c)
$R + 40\sin30° = 25\times9.8$ | M1 | Resolving vertically
$R = 225 \text{ N}$ | A1, A1 (3) | Correct equation; AG correct force from correct working
### Part (d)
$32.1 = 225\mu$ | M1 | Use of $F = \mu R$
$\mu = \frac{32.1}{225} = 0.143$ | A1 (2) | Correct $\mu$
### Part (e)
Friction will decrease as normal reaction decreases | B1, B1 (2) | Decrease in friction; Normal reaction decreases
---3 The diagram shows a rope that is attached to a box of mass 25 kg , which is being pulled along rough horizontal ground. The rope is at an angle of $30 ^ { \circ }$ to the ground. The tension in the rope is 40 N . The box accelerates at $0.1 \mathrm {~ms} ^ { - 2 }$.\\
\includegraphics[max width=\textwidth, alt={}, center]{eb1f2470-aeeb-4b1d-a6c0-bdeb7048edd5-3_214_729_504_644}
\begin{enumerate}[label=(\alph*)]
\item Draw a diagram to show all of the forces acting on the box.
\item Show that the magnitude of the friction force acting on the box is 32.1 N , correct to three significant figures.
\item Show that the magnitude of the normal reaction force that the ground exerts on the box is 225 N .
\item Find the coefficient of friction between the box and the ground.
\item State what would happen to the magnitude of the friction force if the angle between the rope and the horizontal were increased. Give a reason for your answer.
\end{enumerate}
\hfill \mbox{\textit{AQA M1 2005 Q3 [11]}}