| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2010 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Work done and energy |
| Type | Energy method - smooth inclined plane (no resistance) |
| Difficulty | Moderate -0.3 This is a straightforward energy conservation problem with standard mechanics. Part (i) requires simple application of KE = PE with no friction. Part (ii) adds work done against friction but still follows a direct energy accounting approach (initial KE = PE gained + work against friction). The calculations are routine and the problem structure is typical for M1 level, making it slightly easier than average A-level difficulty. |
| Spec | 6.02a Work done: concept and definition6.02e Calculate KE and PE: using formulae |
| Answer | Marks | Guidance |
|---|---|---|
| (i) | M1 | For using KE loss = PE gain or \(0^2 = u^2 - 2(g\sin\alpha)(0.45/\sin\alpha)\) |
| \(\frac{1}{2}(m)u^2 = (m)g(0.45)\) | A1 | |
| Speed is \(3 \text{ ms}^{-1}\) | A1 | |
| [3] | ||
| (ii) \([PE \text{ gain} = \frac{1}{2}0.3 \times 3^2 = 0.39]\) | M1 | For using PE gain = KE lost − WD |
| PE gain is \(0.96 \text{ J}\) | A1ft | ft incorrect \(u\) |
| \([0.3gh = 0.96]\) | DM1 | For using PE = \(mgh\); dependent on the given WD being reflected in the value for PE used |
| R is \(0.32 \text{ m}\) higher than the level of P | A1 | |
| [4] |
**(i)** | M1 | For using KE loss = PE gain or $0^2 = u^2 - 2(g\sin\alpha)(0.45/\sin\alpha)$
$\frac{1}{2}(m)u^2 = (m)g(0.45)$ | A1 |
Speed is $3 \text{ ms}^{-1}$ | A1 |
| | | **[3]**
**(ii)** $[PE \text{ gain} = \frac{1}{2}0.3 \times 3^2 = 0.39]$ | M1 | For using PE gain = KE lost − WD
PE gain is $0.96 \text{ J}$ | A1ft | ft incorrect $u$
$[0.3gh = 0.96]$ | DM1 | For using PE = $mgh$; dependent on the given WD being reflected in the value for PE used
R is $0.32 \text{ m}$ higher than the level of P | A1 |
| | | **[4]**
---$P$ and $Q$ are fixed points on a line of greatest slope of an inclined plane. The point $Q$ is at a height of 0.45 m above the level of $P$. A particle of mass 0.3 kg moves upwards along the line $PQ$.
\begin{enumerate}[label=(\roman*)]
\item Given that the plane is smooth and that the particle just reaches $Q$, find the speed with which it passes through $P$. [3]
\item It is given instead that the plane is rough. The particle passes through $P$ with the same speed as that found in part (i), and just reaches a point $R$ which is between $P$ and $Q$. The work done against the frictional force in moving from $P$ to $R$ is 0.39 J. Find the potential energy gained by the particle in moving from $P$ to $R$ and hence find the height of $R$ above the level of $P$. [4]
\end{enumerate}
\hfill \mbox{\textit{CAIE M1 2010 Q5 [7]}}