| Exam Board | Edexcel |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2007 |
| Session | January |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Work done and energy |
| Type | Motion on rough inclined plane |
| Difficulty | Moderate -0.8 This is a straightforward application of the work-energy principle requiring two standard calculations: (a) uses ΔKE = work done directly, and (b) uses work = friction force × distance with F = μR = μmg. Both parts follow textbook methods with no problem-solving insight needed, making it easier than average but not trivial due to the two-step structure. |
| Spec | 3.03v Motion on rough surface: including inclined planes6.02a Work done: concept and definition6.02d Mechanical energy: KE and PE concepts |
| Answer | Marks | Guidance |
|---|---|---|
| (a) \(\frac{1}{2} \times 0.8 \times (15^2 - 10^2) = 50\) (J) | M1 A1 | 2 marks |
| Answer | Marks | Guidance |
|---|---|---|
| \(\mu \approx 0.32\) | M1, M1 A1ft, A1 | accept 0.319; 6 marks total |
| Answer | Marks | Guidance |
|---|---|---|
| \(\mu \approx 0.32\) | M1, M1 A1ft, A1 | accept 0.319; 4 marks |
| Answer | Marks | Guidance |
|---|---|---|
| \(\mu \approx 0.32\) | M1, M1 A1 ft, A1 | 4 marks |
| Answer | Marks | Guidance |
|---|---|---|
| \(WD = F \times d \Rightarrow 2.5 \times 20 = 50J\) | (b)M1, (a)M1A1 | additional marks |
(a) $\frac{1}{2} \times 0.8 \times (15^2 - 10^2) = 50$ (J) | M1 A1 | 2 marks
(b) **Work-energy method:**
$F = \mu R = \mu 0.8g$
$\mu 0.8g \times 20 = 50$
$\mu \approx 0.32$ | M1, M1 A1ft, A1 | accept 0.319; 6 marks total
**Alternative for (b):**
$v^2 = u^2 + 2as \Rightarrow a = \frac{15^2 - 10^2}{2 \times 20} = 3.125$
N2L: $F = \mu mg = ma = 3.125m$
$\mu \approx 0.32$ | M1, M1 A1ft, A1 | accept 0.319; 4 marks
**Alternative for (b):**
$\text{WE: } F = \frac{50}{20} (= 2.5)$
$F = \mu R = \frac{50}{20} = \mu 0.8g$
$\mu \approx 0.32$ | M1, M1 A1 ft, A1 | 4 marks
**Alternative (first M1 for (b) could be scored in (a)):**
$v^2 = u^2 + 2as \Rightarrow 10^2 = 15^2 - 2 \times 20 \times (-a) \Rightarrow a = (-) \frac{125}{40}$
$F = ma \Rightarrow F = 2.5$
$WD = F \times d \Rightarrow 2.5 \times 20 = 50J$ | (b)M1, (a)M1A1 | additional marks
---A particle of mass 0.8 kg is moving in a straight line on a rough horizontal plane. The speed of the particle is reduced from 15 m s$^{-1}$ to 10 m s$^{-1}$ as the particle moves 20 m. Assuming that the only resistance to motion is the friction between the particle and the plane, find
\begin{enumerate}[label=(\alph*)]
\item the work done by friction in reducing the speed of the particle from 15 m s$^{-1}$ to 10 m s$^{-1}$, [2]
\item the coefficient of friction between the particle and the plane. [4]
\end{enumerate}
\hfill \mbox{\textit{Edexcel M2 2007 Q1 [6]}}