| Exam Board | Edexcel |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2017 |
| Session | October |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Work done and energy |
| Type | Motion on rough inclined plane |
| Difficulty | Standard +0.3 This is a straightforward M2 mechanics question requiring standard application of friction force calculation (R = mg cos θ, F = μR) and the work-energy principle. All steps are routine: find normal reaction, calculate friction work, then apply work-energy equation with given values. Slightly above average difficulty due to being M2 content and requiring careful sign conventions, but no novel problem-solving or geometric insight needed. |
| Spec | 3.03v Motion on rough surface: including inclined planes6.02d Mechanical energy: KE and PE concepts6.02i Conservation of energy: mechanical energy principle |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Notes |
| \(N = 4g\cos 40°\) \((= 30.028\ldots)\) | ||
| \(F = 0.5 \times 4g\cos 40°\) | M1 | Use of \(F = \mu N\) where \(N\) is a resolved component of \(4g\). Condone sin/cos confusion |
| Work done \(= 12 \times 0.5 \times 4g\cos 40°\) | M1 | Their \(F \times 12\) |
| \((= 180.17\ldots) = 180\) J | A1 | Max 3 s.f. |
| Total: (3) |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Notes |
| Work done \(+\) Final KE \(=\) Initial KE \(+\) GPE | M1 | Must be using W.E. Need all terms. Condone sign errors. Terms must be dimensionally correct. Condone sin/cos confusion |
| Their WD \(+ \dfrac{1}{2}\times 4 \times 24^2 = \dfrac{1}{2}\times 4u^2 + 4g\times 12\sin 40°\) follow their WD | A1ft | At most one error. Incorrect sign(s) is one error. \((4g\times 12\sin 40 = 302.367\ldots)\) |
| A1ft | Correct unsimplified (for their WD) | |
| \(u (= 22.691\ldots) = 23\) or \(22.7 \text{ m s}^{-1}\) | A1 | Max 3 s.f. |
| Total: (4) [7] |
# Question 3(a):
| Working/Answer | Marks | Notes |
|---|---|---|
| $N = 4g\cos 40°$ $(= 30.028\ldots)$ | | |
| $F = 0.5 \times 4g\cos 40°$ | M1 | Use of $F = \mu N$ where $N$ is a resolved component of $4g$. Condone sin/cos confusion |
| Work done $= 12 \times 0.5 \times 4g\cos 40°$ | M1 | Their $F \times 12$ |
| $(= 180.17\ldots) = 180$ J | A1 | Max 3 s.f. |
| **Total: (3)** | | |
# Question 3(b):
| Working/Answer | Marks | Notes |
|---|---|---|
| Work done $+$ Final KE $=$ Initial KE $+$ GPE | M1 | Must be using W.E. Need all terms. Condone sign errors. Terms must be dimensionally correct. Condone sin/cos confusion |
| Their WD $+ \dfrac{1}{2}\times 4 \times 24^2 = \dfrac{1}{2}\times 4u^2 + 4g\times 12\sin 40°$ follow their WD | A1ft | At most one error. Incorrect sign(s) is one error. $(4g\times 12\sin 40 = 302.367\ldots)$ |
| | A1ft | Correct unsimplified (for their WD) |
| $u (= 22.691\ldots) = 23$ or $22.7 \text{ m s}^{-1}$ | A1 | Max 3 s.f. |
| **Total: (4) [7]** | | |
---3. A particle $P$ of mass 4 kg moves from point $A$ to point $B$ down a line of greatest slope of a fixed rough plane. The plane is inclined at $40 ^ { \circ }$ to the horizontal and $A B = 12 \mathrm {~m}$. The coefficient of friction between $P$ and the plane is 0.5
\begin{enumerate}[label=(\alph*)]
\item Find the work done against friction as $P$ moves from $A$ to $B$.
Given that the speed of $P$ at $B$ is $24 \mathrm {~m} \mathrm {~s} ^ { - 1 }$
\item use the work-energy principle to find the speed of $P$ at $A$.
\begin{center}
\end{center}
\end{enumerate}
\hfill \mbox{\textit{Edexcel M2 2017 Q3 [7]}}