| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2023 |
| Session | November |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Motion on a slope |
| Type | Coefficient of friction from motion |
| Difficulty | Moderate -0.8 This is a straightforward mechanics problem requiring standard application of Newton's second law on an inclined plane with friction. Students resolve forces parallel and perpendicular to the plane, use F=ma with given acceleration to find μ, then apply a basic kinematics equation. All steps are routine textbook exercises with no problem-solving insight required. |
| Spec | 3.03v Motion on rough surface: including inclined planes |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Correct force diagram with 3 forces in the correct directions. | B1 | No labels required on the 3 forces; ignore wrong labels. Arrows needed. Allow either or both components of weight if fully labelled. Allow sin/cos mix. If forces are not connected to the block, the line of action of each force must go through the block. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(R = 8g\cos 30\ [= 40\sqrt{3} = 69.282\ldots]\) | B1 | Resolving perpendicular to the plane. |
| Resolving parallel to the plane and applying Newton's second law: \([8g\sin 30 - F = 8 \times 2.4 \Rightarrow F = 20.8]\) | M1* | 3 terms. Allow sign errors, sin/cos mix. Allow \(g\) missing, otherwise dimensionally correct. |
| Use \(F = \mu R\) to get equation in \(\mu\) only: \([8g\sin 30 - 8g\mu\cos 30 = 8\times 2.4 \quad 40 - 40\sqrt{3}\mu = 19.2]\) | DM1 | Allow \(g\) missing in either or both of \(F\) and \(R\). Allow sign errors, consistent sin/cos mix. \(R\) must be a single component of a force. Allow the 3 masses to be cancelled. |
| \(\mu = 0.3[0\ldots]\) — may first see \(\frac{20.8}{40\sqrt{3}}\) or \(\frac{20.8}{69.282\ldots}\) | A1 | Allow exact value \(\frac{13\sqrt{3}}{75}\) or \(\frac{104\sqrt{3}}{600}\) oe. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \([v^2 = 2\times 2.4\times 3 \Rightarrow \text{greatest speed} =]\ 3.79\ \text{ms}^{-1} = \frac{6\sqrt{10}}{5}\) | B1 | \(3.79473\ldots\) (3.8 without a more accurate value seen gets B0 and should be annotated SF). |
## Question 3(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Correct force diagram with 3 forces in the correct directions. | B1 | No labels required on the 3 forces; ignore wrong labels. Arrows needed. Allow either or both components of weight if fully labelled. Allow sin/cos mix. If forces are not connected to the block, the line of action of each force must go through the block. |
---
## Question 3(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $R = 8g\cos 30\ [= 40\sqrt{3} = 69.282\ldots]$ | B1 | Resolving perpendicular to the plane. |
| Resolving parallel to the plane and applying Newton's second law: $[8g\sin 30 - F = 8 \times 2.4 \Rightarrow F = 20.8]$ | M1* | 3 terms. Allow sign errors, sin/cos mix. Allow $g$ missing, otherwise dimensionally correct. |
| Use $F = \mu R$ to get equation in $\mu$ only: $[8g\sin 30 - 8g\mu\cos 30 = 8\times 2.4 \quad 40 - 40\sqrt{3}\mu = 19.2]$ | DM1 | Allow $g$ missing in either or both of $F$ and $R$. Allow sign errors, consistent sin/cos mix. $R$ must be a single component of a force. Allow the 3 masses to be cancelled. |
| $\mu = 0.3[0\ldots]$ — may first see $\frac{20.8}{40\sqrt{3}}$ or $\frac{20.8}{69.282\ldots}$ | A1 | Allow exact value $\frac{13\sqrt{3}}{75}$ or $\frac{104\sqrt{3}}{600}$ oe. |
---
## Question 3(c):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $[v^2 = 2\times 2.4\times 3 \Rightarrow \text{greatest speed} =]\ 3.79\ \text{ms}^{-1} = \frac{6\sqrt{10}}{5}$ | B1 | $3.79473\ldots$ (3.8 without a more accurate value seen gets B0 and should be annotated SF). |
---3 A block of mass 8 kg slides down a rough plane inclined at $30 ^ { \circ }$ to the horizontal, starting from rest. The coefficient of friction between the block and the plane is $\mu$. The block accelerates uniformly down the plane at $2.4 \mathrm {~m} \mathrm {~s} ^ { - 2 }$.
\begin{enumerate}[label=(\alph*)]
\item Draw a diagram showing the forces acting on the block.
\item Find the value of $\mu$.
\item Find the speed of the block after it has moved 3 m down the plane.
\end{enumerate}
\hfill \mbox{\textit{CAIE M1 2023 Q3 [6]}}