| Exam Board | Edexcel |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2016 |
| Session | October |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Advanced work-energy problems |
| Type | Rough inclined plane work-energy |
| Difficulty | Standard +0.3 This is a standard M2 work-energy problem with two parts: (a) uses energy conservation to find work done against friction (result given, so verification only), and (b) applies work-energy principle for the return journey. The calculations are straightforward with clear setup, though students must correctly account for friction acting in different directions on ascent vs descent. Slightly above average due to the two-stage nature and careful sign consideration required. |
| 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 | Mark | Notes |
| Form work-energy equation | M1 | Need both terms on RHS. Must be dimensionally correct. Condone sign errors |
| Change in energy \(= \pm\left(\frac{1}{2} \times 4 \times 6^2 - 4g \times 10\sin\alpha\right)\) | A2 | \(-1\) each error |
| \(= 72 - 40g \times \frac{1}{7} = 16\) (J) given answer | A1 | \(-16\) is A0. Condone \(-16\) becoming \(+16\) |
| (4 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Notes |
| Complete strategy using suvat and N2L to find work done | M1 | |
| \(v^2 = u^2 + 2as \Rightarrow 36 = -20a\) \((a = -1.8)\) | A1 | |
| \(Fr + 4g\sin\theta = 4 \times (\text{their } 1.8)\), \((Fr = 1.6)\) | A1 | |
| Work Done \(= 1.6 \times 10 = 16\) (J) given answer | A1 | |
| (4 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Notes |
| Considering whole journey: \(\frac{1}{2} \times 4v^2 = \frac{1}{2} \times 4 \times 36 - 2 \times 16\) | M1 | Requires all 3 terms. Must be dimensionally correct. Condone sign errors |
| A1 | Correct unsimplified equation | |
| \(v^2 = 20\), \(v = 4.47\) (m s\(^{-1}\)) (4.5) | A1 | Accept \(2\sqrt{5}\) |
| (3 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Notes |
| \(\frac{1}{2} \times 4 \times v^2 + 16 = 40g\sin\alpha\) | M1 | Requires all 3 terms. Must be dimensionally correct. Condone sign errors |
| A1 | Correct unsimplified equation | |
| \(v^2 = 20\), \(v = 4.47\) (m s\(^{-1}\)) (4.5) | A1 | Accept \(2\sqrt{5}\) |
| (3 marks total, 7 marks for Q3) |
# Question 3:
## Part 3a:
| Working | Mark | Notes |
|---------|------|-------|
| Form work-energy equation | M1 | Need both terms on RHS. Must be dimensionally correct. Condone sign errors |
| Change in energy $= \pm\left(\frac{1}{2} \times 4 \times 6^2 - 4g \times 10\sin\alpha\right)$ | A2 | $-1$ each error |
| $= 72 - 40g \times \frac{1}{7} = 16$ (J) **given answer** | A1 | $-16$ is A0. Condone $-16$ becoming $+16$ |
| **(4 marks)** | | |
**Alt 3a:**
| Working | Mark | Notes |
|---------|------|-------|
| Complete strategy using suvat and N2L to find work done | M1 | |
| $v^2 = u^2 + 2as \Rightarrow 36 = -20a$ $(a = -1.8)$ | A1 | |
| $Fr + 4g\sin\theta = 4 \times (\text{their } 1.8)$, $(Fr = 1.6)$ | A1 | |
| Work Done $= 1.6 \times 10 = 16$ (J) **given answer** | A1 | |
| **(4 marks)** | | |
## Part 3b:
| Working | Mark | Notes |
|---------|------|-------|
| Considering whole journey: $\frac{1}{2} \times 4v^2 = \frac{1}{2} \times 4 \times 36 - 2 \times 16$ | M1 | Requires all 3 terms. Must be dimensionally correct. Condone sign errors |
| | A1 | Correct unsimplified equation |
| $v^2 = 20$, $v = 4.47$ (m s$^{-1}$) (4.5) | A1 | Accept $2\sqrt{5}$ |
| **(3 marks)** | | |
**Alt 3b:** Working from $B$ to $A$:
| Working | Mark | Notes |
|---------|------|-------|
| $\frac{1}{2} \times 4 \times v^2 + 16 = 40g\sin\alpha$ | M1 | Requires all 3 terms. Must be dimensionally correct. Condone sign errors |
| | A1 | Correct unsimplified equation |
| $v^2 = 20$, $v = 4.47$ (m s$^{-1}$) (4.5) | A1 | Accept $2\sqrt{5}$ |
| **(3 marks total, 7 marks for Q3)** | | |
---3. A particle $P$ of mass 4 kg is projected with speed $6 \mathrm {~ms} ^ { - 1 }$ up a line of greatest slope of a fixed rough inclined plane. The plane is inclined at angle $\alpha$ to the horizontal, where $\sin \alpha = \frac { 1 } { 7 }$. The particle is projected from the point $A$ on the plane and comes to instantaneous rest at the point $B$ on the plane, where $A B = 10 \mathrm {~m}$.
\begin{enumerate}[label=(\alph*)]
\item Show that the work done against friction as $P$ moves from $A$ to $B$ is 16 joules.
After coming to instantaneous rest at $B$, the particle slides back down the plane.
\item Use the work-energy principle to find the speed of $P$ at the instant it returns to $A$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M2 2016 Q3 [7]}}