| Exam Board | AQA |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2008 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Motion on a slope |
| Type | Coefficient of friction from motion |
| Difficulty | Moderate -0.3 This is a standard M1 mechanics question on forces on an inclined plane. Parts (a)-(c) follow routine procedures: drawing a force diagram, resolving perpendicular to find normal reaction, then resolving parallel with F=ma to find friction coefficient. Part (d) requires basic conceptual understanding. The calculations are straightforward with no novel problem-solving required, making it slightly easier than average. |
| Spec | 3.03e Resolve forces: two dimensions3.03v Motion on rough surface: including inclined planes |
| Answer | Marks | Guidance |
|---|---|---|
| Correct force diagram with labels and arrows. Accept components of the weight if shown in a different notation with the weight also shown. B0 if components are shown instead of the weight. | B1 | Correct force diagram with labels and arrows. Accept components of the weight if shown in a different notation with the weight also shown. B0 if components are shown instead of the weight. |
| Answer | Marks | Guidance |
|---|---|---|
| \((R =) 5 \times 9.8 \cos 40° = 37.5\) N | M1, A1 | Attempt at resolving perpendicular to the slope (eg 49sin40°). Correct value from correct working. |
| AG |
| Answer | Marks | Guidance |
|---|---|---|
| \(5 \times 0.8 = 5 \times 9.8 \sin 40° - \mu \times 5 \times 9.8 \cos 40°\) | B1, M1 | Use of \(F = \mu R\) at any stage and with any \(F\) but with \(R = 37.5\) OE. Three term equation of motion from resolving parallel to the slope with weight component, friction and \(ma\) term. |
| A1, A1 | Correct terms seen (may be as 31.5, 37.5μ (or F) and 4). Correct signs. | |
| \(\mu = \frac{5 \times 9.8 \sin 40° - 5 \times 0.8}{5 \times 9.8 \cos 40°} = 0.733\) | m1, A1 | Solving for \(\mu\). Al: Correct value for \(\mu\). Allow 0.732 but not \(\frac{11}{15}\) unless converted to a decimal. |
| Answer | Marks | Guidance |
|---|---|---|
| There is less friction so the coefficient of friction must be less. | B1, B1 | Less friction. Smaller coefficient of friction. If the answer and explanation contradict each other, award no marks. |
**6(a)**
| Correct force diagram with labels and arrows. Accept components of the weight if shown in a different notation with the weight also shown. B0 if components are shown instead of the weight. | B1 | Correct force diagram with labels and arrows. Accept components of the weight if shown in a different notation with the weight also shown. B0 if components are shown instead of the weight. |
**6(b)**
| $(R =) 5 \times 9.8 \cos 40° = 37.5$ N | M1, A1 | Attempt at resolving perpendicular to the slope (eg 49sin40°). Correct value from correct working. |
| | AG | |
**6(c)**
| $5 \times 0.8 = 5 \times 9.8 \sin 40° - \mu \times 5 \times 9.8 \cos 40°$ | B1, M1 | Use of $F = \mu R$ at any stage and with any $F$ but with $R = 37.5$ OE. Three term equation of motion from resolving parallel to the slope with weight component, friction and $ma$ term. |
| | A1, A1 | Correct terms seen (may be as 31.5, 37.5μ (or F) and 4). Correct signs. |
| $\mu = \frac{5 \times 9.8 \sin 40° - 5 \times 0.8}{5 \times 9.8 \cos 40°} = 0.733$ | m1, A1 | Solving for $\mu$. Al: Correct value for $\mu$. Allow 0.732 but not $\frac{11}{15}$ unless converted to a decimal. |
**6(d)**
| There is less friction so the coefficient of friction must be less. | B1, B1 | Less friction. Smaller coefficient of friction. If the answer and explanation contradict each other, award no marks. |6 A block, of mass 5 kg , slides down a rough plane inclined at $40 ^ { \circ }$ to the horizontal. When modelling the motion of the block, assume that there is no air resistance acting on it.
\begin{enumerate}[label=(\alph*)]
\item Draw and label a diagram to show the forces acting on the block.
\item Show that the magnitude of the normal reaction force acting on the block is 37.5 N , correct to three significant figures.
\item Given that the acceleration of the block is $0.8 \mathrm {~m} \mathrm {~s} ^ { - 2 }$, find the coefficient of friction between the block and the plane.
\item In reality, air resistance does act on the block. State how this would change your value for the coefficient of friction and explain why.
\end{enumerate}
\hfill \mbox{\textit{AQA M1 2008 Q6 [11]}}