| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2020 |
| Session | November |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Motion on a slope |
| Type | Limiting equilibrium both directions |
| Difficulty | Standard +0.3 This is a standard two-part friction problem requiring resolution of forces parallel and perpendicular to the slope, with limiting equilibrium in both directions. Part (a) involves straightforward force resolution with friction acting down the slope, while part (b) requires resolving a horizontal force into components and finding the minimum coefficient. The calculations are routine for M1 level with no novel insight required, making it slightly easier than average. |
| Spec | 1.05g Exact trigonometric values: for standard angles3.03u Static equilibrium: on rough surfaces3.03v Motion on rough surface: including inclined planes |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(R = 5g\cos 30° \ [= 25\sqrt{3}]\) | B1 | |
| \(40 - 5g\sin 30° - F > 0\) | M1 | State that the net force up the plane is positive, 3 terms |
| \(F = \mu \times 5g\cos 30°\) | M1 | For using \(F = \mu R\) with \(R\) as a component of \(5g\) to obtain an equality/inequality in \(\mu\) only with 3 terms |
| \(\mu < \frac{1}{5}\sqrt{3}\) | A1 | AG |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(R = 5g\cos 30° \ [= 25\sqrt{3}]\) | B1 | |
| \(40 - 5g\sin 30° - F = 5a\) | M1 | Acceleration \(a > 0\) |
| \(F = \mu \times 5g\cos 30°\), \([40 - 5g\sin 30° - \mu \times 5g\cos 30° = 5a]\) | M1 | For using \(F = \mu R\) with \(R\) as a component of \(5g\) to obtain an equality in \(\mu\) and \(a\) |
| \(\mu < \frac{1}{5}\sqrt{3}\) | A1 | AG. From \(\mu = \frac{1}{5}\sqrt{3} = \frac{a}{g}\cos 30°\) with \(a > 0\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Attempt to resolve forces parallel to or perpendicular to the inclined plane, 3 relevant terms in either direction | M1 | |
| \(R = 5g\cos 30° + 40\sin 30° \ [= 20 + 25\sqrt{3} = 63.3]\) | A1 | |
| \(F = 40\cos 30° - 5g\sin 30° \ [= 20\sqrt{3} - 25 = 9.64]\) | A1 | |
| \(\mu \geqslant 0.152\) | B1 | AG. Using \(F \leqslant \mu R\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Attempt to resolve forces horizontally or vertically with 3 relevant terms in either direction | M1 | |
| \(40 = R\sin 30° + F\cos 30° \ [40 = \frac{1}{2}R + \frac{\sqrt{3}}{2}F]\) | A1 | |
| \(5g = R\cos 30° - F\sin 30° \ [5g = \frac{\sqrt{3}}{2}R - \frac{1}{2}F]\) | A1 | |
| \(\mu \geqslant 0.152\) | B1 | AG. Solve for \(R\) and \(F\) and use \(F \leqslant \mu R\) |
## Question 6(a):
| Answer | Mark | Guidance |
|--------|------|----------|
| $R = 5g\cos 30° \ [= 25\sqrt{3}]$ | B1 | |
| $40 - 5g\sin 30° - F > 0$ | M1 | State that the net force up the plane is positive, 3 terms |
| $F = \mu \times 5g\cos 30°$ | M1 | For using $F = \mu R$ with $R$ as a component of $5g$ to obtain an equality/inequality in $\mu$ only with 3 terms |
| $\mu < \frac{1}{5}\sqrt{3}$ | A1 | AG |
**Alternative scheme for 6(a):**
| Answer | Mark | Guidance |
|--------|------|----------|
| $R = 5g\cos 30° \ [= 25\sqrt{3}]$ | B1 | |
| $40 - 5g\sin 30° - F = 5a$ | M1 | Acceleration $a > 0$ |
| $F = \mu \times 5g\cos 30°$, $[40 - 5g\sin 30° - \mu \times 5g\cos 30° = 5a]$ | M1 | For using $F = \mu R$ with $R$ as a component of $5g$ to obtain an equality in $\mu$ and $a$ |
| $\mu < \frac{1}{5}\sqrt{3}$ | A1 | AG. From $\mu = \frac{1}{5}\sqrt{3} = \frac{a}{g}\cos 30°$ with $a > 0$ |
---
## Question 6(b):
| Answer | Mark | Guidance |
|--------|------|----------|
| Attempt to resolve forces parallel to or perpendicular to the inclined plane, 3 relevant terms in either direction | M1 | |
| $R = 5g\cos 30° + 40\sin 30° \ [= 20 + 25\sqrt{3} = 63.3]$ | A1 | |
| $F = 40\cos 30° - 5g\sin 30° \ [= 20\sqrt{3} - 25 = 9.64]$ | A1 | |
| $\mu \geqslant 0.152$ | B1 | AG. Using $F \leqslant \mu R$ |
**Alternative method for 6(b):**
| Answer | Mark | Guidance |
|--------|------|----------|
| Attempt to resolve forces horizontally or vertically with 3 relevant terms in either direction | M1 | |
| $40 = R\sin 30° + F\cos 30° \ [40 = \frac{1}{2}R + \frac{\sqrt{3}}{2}F]$ | A1 | |
| $5g = R\cos 30° - F\sin 30° \ [5g = \frac{\sqrt{3}}{2}R - \frac{1}{2}F]$ | A1 | |
| $\mu \geqslant 0.152$ | B1 | AG. Solve for $R$ and $F$ and use $F \leqslant \mu R$ |
---6 A block of mass 5 kg is placed on a plane inclined at $30 ^ { \circ }$ to the horizontal. The coefficient of friction between the block and the plane is $\mu$.
\begin{enumerate}[label=(\alph*)]
\item \begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{fcc3d739-5c36-48ad-9c34-f69b28a06dba-10_424_709_392_760}
\captionsetup{labelformat=empty}
\caption{Fig. 6.1}
\end{center}
\end{figure}
When a force of magnitude 40 N is applied to the block, acting up the plane parallel to a line of greatest slope, the block begins to slide up the plane (see Fig. 6.1).
Show that $\mu < \frac { 1 } { 5 } \sqrt { 3 }$.
\item \begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{fcc3d739-5c36-48ad-9c34-f69b28a06dba-11_422_727_264_749}
\captionsetup{labelformat=empty}
\caption{Fig. 6.2}
\end{center}
\end{figure}
When a force of magnitude 40 N is applied horizontally, in a vertical plane containing a line of greatest slope, the block does not move (see Fig. 6.2).
Show that, correct to 3 decimal places, the least possible value of $\mu$ is 0.152 .
\end{enumerate}
\hfill \mbox{\textit{CAIE M1 2020 Q6 [8]}}