| Exam Board | Edexcel |
|---|---|
| Module | M1 (Mechanics 1) |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Newton's laws and connected particles |
| Type | Block on rough horizontal surface – equilibrium (finding friction, normal reaction, or coefficient of friction) |
| Difficulty | Standard +0.3 This is a straightforward M1 friction problem requiring resolution of forces in two directions and application of F=μR. The 'show that' format guides students to the answer, and the angle/coefficient values are chosen to give clean arithmetic. Part (b) is a simple qualitative reasoning question worth minimal marks. |
| Spec | 3.03e Resolve forces: two dimensions3.03m Equilibrium: sum of resolved forces = 03.03v Motion on rough surface: including inclined planes |
| Answer | Marks |
|---|---|
| Resolve \(\uparrow\): \(R - P\cos 60° - 0.5g = 0 \therefore R = 0.5g + P\cos 60°\) | M1 A1 |
| Resolve \(\rightarrow\): \(P\sin 60° - F = 0\) | M1 |
| \(F = \mu R = \frac{1}{\sqrt{3}}(0.5g + 0.5P)\) | M1 A1 |
| Sub. in giving \(\frac{\sqrt{3}}{2}P - \frac{1}{\sqrt{3}}(0.5g + 0.5P) = 0\) | M1 |
| \(3P - P - g = 0 \therefore 2P = g\) so \(P = \frac{g}{2}\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Brush is moved slowly so very little air resistance | B1 | (8) |
**Part (a)**
Resolve $\uparrow$: $R - P\cos 60° - 0.5g = 0 \therefore R = 0.5g + P\cos 60°$ | M1 A1 |
Resolve $\rightarrow$: $P\sin 60° - F = 0$ | M1 |
$F = \mu R = \frac{1}{\sqrt{3}}(0.5g + 0.5P)$ | M1 A1 |
Sub. in giving $\frac{\sqrt{3}}{2}P - \frac{1}{\sqrt{3}}(0.5g + 0.5P) = 0$ | M1 |
$3P - P - g = 0 \therefore 2P = g$ so $P = \frac{g}{2}$ | A1 |
**Part (b)**
Brush is moved slowly so very little air resistance | B1 | (8)2. A monk uses a small brush to clean the stone floor of a monastery by pushing the brush with a force of $P$ Newtons at an angle of $60 ^ { \circ }$ to the vertical. He moves the brush at a constant speed. The mass of the brush is 0.5 kg and the coefficient of friction between the brush and the floor is $\frac { 1 } { \sqrt { 3 } }$. The brush is modelled as a particle and air resistance is ignored.
\begin{enumerate}[label=(\alph*)]
\item Show that $P = \frac { g } { 2 }$ Newtons.
\item Explain why it is reasonable to ignore air resistance in this situation.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M1 Q2 [8]}}