| Exam Board | Edexcel |
|---|---|
| Module | FM1 AS (Further Mechanics 1 AS) |
| Year | 2021 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Motion on a slope |
| Type | Energy methods on slope |
| Difficulty | Standard +0.3 This is a standard two-part mechanics problem combining forces on a slope with energy methods. Part (a) requires resolving perpendicular to the plane (routine), while part (b) applies work-energy principle with friction over two stages (sliding then projectile). The setup is straightforward with given values, requiring systematic application of standard techniques rather than problem-solving insight. Slightly easier than average due to clear structure and standard methods. |
| Spec | 3.03r Friction: concept and vector form3.03v Motion on rough surface: including inclined planes6.02a Work done: concept and definition6.02b Calculate work: constant force, resolved component |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Resolve perpendicular to the plane | M1 | Allow sin/cos confusion |
| \(R = \frac{4}{5}mg\) | A1 | cao |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Work done against friction \(= 0.4R \times 0.5\) \((= 0.16mg)\) | M1 | Correct form for work done against friction \((1.568m)\) |
| PE Loss \(= mg \times 0.5\sin\alpha + 0.8mg\) \((= 1.1mg)\) | M1 | Correct no. of terms, dimensionally correct, condone sin/cos confusion \((10.78m)\) |
| Using work-energy principle | M1 | Correct number of terms (using their WD and PE for the whole journey to the floor) |
| \(1.1mg = 0.16mg + \frac{1}{2}mv^2\) | A1 | Correct unsimplified equation |
| \(v = 4.3\) or \(4.29\ (\text{m s}^{-1})\) | A1 | Either of the two possible answers (as \(g = 9.8\) has been used) |
## Question 1:
**Part (a):**
| Answer/Working | Mark | Guidance |
|---|---|---|
| Resolve perpendicular to the plane | M1 | Allow sin/cos confusion |
| $R = \frac{4}{5}mg$ | A1 | cao |
**Part (b):**
| Answer/Working | Mark | Guidance |
|---|---|---|
| Work done against friction $= 0.4R \times 0.5$ $(= 0.16mg)$ | M1 | Correct form for work done against friction $(1.568m)$ |
| PE Loss $= mg \times 0.5\sin\alpha + 0.8mg$ $(= 1.1mg)$ | M1 | Correct no. of terms, dimensionally correct, condone sin/cos confusion $(10.78m)$ |
| Using work-energy principle | M1 | Correct number of terms (using their WD and PE for the **whole journey** to the floor) |
| $1.1mg = 0.16mg + \frac{1}{2}mv^2$ | A1 | Correct unsimplified equation |
| $v = 4.3$ or $4.29\ (\text{m s}^{-1})$ | A1 | Either of the two possible answers (as $g = 9.8$ has been used) |
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\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{05f6f654-05e5-41d5-a6e4-11cd91a6df83-02_826_700_244_550}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{center}
\end{figure}
A small book of mass $m$ is held on a rough straight desk lid which is inclined at an angle $\alpha$ to the horizontal, where $\tan \alpha = \frac { 3 } { 4 }$. The book is released from rest at a distance of 0.5 m from the edge of the desk lid, as shown in Figure 1. The book slides down the desk lid and then hits the floor that is 0.8 m below the edge of the desk lid. The coefficient of friction between the book and the desk lid is 0.4
The book is modelled as a particle which, after leaving the desk lid, is assumed to move freely under gravity.
\begin{enumerate}[label=(\alph*)]
\item Find, in terms of $m$ and $g$, the magnitude of the normal reaction on the book as it slides down the desk lid.
\item Use the work-energy principle to find the speed of the book as it hits the floor.
\end{enumerate}
\hfill \mbox{\textit{Edexcel FM1 AS 2021 Q1 [7]}}