Edexcel M1 2004 January — Question 3 10 marks

Exam BoardEdexcel
ModuleM1 (Mechanics 1)
Year2004
SessionJanuary
Marks10
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicMotion on a slope
TypeMotion down rough slope
DifficultyModerate -0.3 This is a standard M1 mechanics problem requiring resolution of forces on an inclined plane and application of friction laws, followed by straightforward use of SUVAT equations. While it involves multiple steps (7 marks for part a), the techniques are routine and well-practiced. Part (c) tests conceptual understanding that acceleration is independent of mass. Slightly easier than average due to its textbook nature and standard approach.
Spec3.03r Friction: concept and vector form3.03v Motion on rough surface: including inclined planes

The tile on a roof becomes loose and slides from rest down the roof. The roof is modelled as a plane surface inclined at 30° to the horizontal. The coefficient of friction between the tile and the roof is 0.4. The tile is modelled as a particle of mass \(m\) kg.
  1. Find the acceleration of the tile as it slides down the roof. [7]
The tile moves a distance 3 m before reaching the edge of the roof.
  1. Find the speed of the tile as it reaches the edge of the roof. [2]
  2. Write down the answer to part (a) if the tile had mass \(2m\) kg. [1]

Part (a)
Answer/Working:
- \(R(\uparrow): R = mg \cos 30\)
- \(R(\rightarrow): ma = mg \sin 30 - F\)
- \(F = 0.4R\) used
- Eliminate \(R\): \(ma = mg \sin 30 - 0.4 \cdot mg \cos 30\)
- Solve: \(a = 4.9 - 0.4 \times 9.8 \times \frac{\sqrt{3}}{2} \approx 1.5\) or \(1.51 \text{ m s}^{-2}\)
Marks: B1, M1, A1, M1, A1 (7 marks)
Part (b)
Answer/Working: \(v^2 = 2 \times 1.51 \times 3 \Rightarrow v = 3\) or \(3.01 \text{ m s}^{-1}\)
Marks: M1, A1 (2 marks)
Part (c)
Answer/Working: \(1.5/1.51 \text{ m s}^{-2}\) (same as (a))
Marks: B1 (1 mark)
Guidance: Accept answer from part (a)
## Part (a)
**Answer/Working:**
- $R(\uparrow): R = mg \cos 30$
- $R(\rightarrow): ma = mg \sin 30 - F$
- $F = 0.4R$ used
- Eliminate $R$: $ma = mg \sin 30 - 0.4 \cdot mg \cos 30$
- Solve: $a = 4.9 - 0.4 \times 9.8 \times \frac{\sqrt{3}}{2} \approx 1.5$ or $1.51 \text{ m s}^{-2}$

**Marks:** B1, M1, A1, M1, A1 (7 marks)

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## Part (b)
**Answer/Working:** $v^2 = 2 \times 1.51 \times 3 \Rightarrow v = 3$ or $3.01 \text{ m s}^{-1}$

**Marks:** M1, A1 (2 marks)

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## Part (c)
**Answer/Working:** $1.5/1.51 \text{ m s}^{-2}$ (same as (a))

**Marks:** B1 (1 mark)

**Guidance:** Accept answer from part (a)

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The tile on a roof becomes loose and slides from rest down the roof. The roof is modelled as a plane surface inclined at 30° to the horizontal. The coefficient of friction between the tile and the roof is 0.4. The tile is modelled as a particle of mass $m$ kg.

\begin{enumerate}[label=(\alph*)]
\item Find the acceleration of the tile as it slides down the roof. [7]
\end{enumerate}

The tile moves a distance 3 m before reaching the edge of the roof.

\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{1}
\item Find the speed of the tile as it reaches the edge of the roof. [2]
\item Write down the answer to part (a) if the tile had mass $2m$ kg. [1]
\end{enumerate}

\hfill \mbox{\textit{Edexcel M1 2004 Q3 [10]}}