| Exam Board | Edexcel |
|---|---|
| Module | M2 (Mechanics 2) |
| Session | Specimen |
| Marks | 8 |
| 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 standard M2 work-energy question requiring straightforward application of the work-energy principle and friction formula. Part (a) uses energy conservation (PE lost = KE gained + work against friction), and part (b) applies work = friction force × distance with F = μR. Both parts follow textbook methods with no novel problem-solving required, making it slightly easier than average. |
| 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 component6.02c Work by variable force: using integration |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Notes |
| K.E. gained \(= \frac{1}{2} \times 0.6 \times 4^2\) | M1 | |
| P.E. lost \(= 0.6 \times g \times (12\sin 30)\) | A1 | |
| Change in energy \(= 0.6 \times g \times 12\sin 30 - \frac{1}{2} \times 0.6 \times 4^2\) | A1 | |
| \(= 30.48\), Work done against friction \(= 30\) or \(30.5\) J | A1 | (4 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Notes |
| \(R = 0.6g\cos 30\) | B1 | |
| \(F = \frac{30.48}{12}\) | B1ft | |
| \(F = \mu R\) | M1 | |
| \(\mu = \frac{30.48}{12 \times 0.6g\cos 30}\) | ||
| \(\mu = 0.499\) or \(0.50\) | A1 | (4 marks) |
# Question 2:
## Part (a):
| Working | Marks | Notes |
|---------|-------|-------|
| K.E. gained $= \frac{1}{2} \times 0.6 \times 4^2$ | M1 | |
| P.E. lost $= 0.6 \times g \times (12\sin 30)$ | A1 | |
| Change in energy $= 0.6 \times g \times 12\sin 30 - \frac{1}{2} \times 0.6 \times 4^2$ | A1 | |
| $= 30.48$, Work done against friction $= 30$ or $30.5$ J | A1 | (4 marks) |
## Part (b):
| Working | Marks | Notes |
|---------|-------|-------|
| $R = 0.6g\cos 30$ | B1 | |
| $F = \frac{30.48}{12}$ | B1ft | |
| $F = \mu R$ | M1 | |
| $\mu = \frac{30.48}{12 \times 0.6g\cos 30}$ | | |
| $\mu = 0.499$ or $0.50$ | A1 | (4 marks) |
---2 A particle $P$ of mass 0.6 kg is released from rest and slides down a line of greatest slope of a rough plane. The plane is inclined at $30 ^ { \circ }$ to the horizontal. When P has moved 12 m , its speed is $4 \mathrm {~m} \mathrm {~s} ^ { - 1 }$. Given that friction is the only non-gravitational resistive force acting on P , find
\begin{enumerate}[label=(\alph*)]
\item the work done against friction as the speed of $P$ increases from $0 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ to $4 \mathrm {~m} \mathrm {~s} ^ { - 1 }$,
\item the coefficient of friction between the particle and the plane.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M2 Q2 [8]}}