| Exam Board | CAIE |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2006 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Hooke's law and elastic energy |
| Type | Elastic string on rough inclined plane |
| Difficulty | Standard +0.8 This is a multi-step energy problem requiring careful bookkeeping of gravitational PE, elastic PE, and friction work on an inclined plane. Part (i) tests standard energy expressions, while part (ii) requires setting up an energy equation at the turning point (v=0) and solving a quadratic. The combination of elastic strings, inclined planes, and friction makes this moderately challenging, but the structure is guided and the techniques are standard for M2 level. |
| Spec | 3.03r Friction: concept and vector form3.03v Motion on rough surface: including inclined planes6.02h Elastic PE: 1/2 k x^26.02i Conservation of energy: mechanical energy principle |
| Answer | Marks | Guidance |
|---|---|---|
| PE loss is 0.48d | M1, A1 | 2 |
| Answer | Marks | Guidance |
|---|---|---|
| EE gain is \(\lambda x^2/2L\) | M1, A1 | 2 |
| Answer | Marks | Guidance |
|---|---|---|
| WD \(= 0.08d\) | M1, A1 | 2 |
| Answer | Marks |
|---|---|
| \(d^2 - 6d + 4 = 0\) | M1, A1 |
| Answer | Marks | Guidance |
|---|---|---|
| For obtaining and selecting relevant correct root | M1, A1 | |
| \(d = 5.24\) (or \(d = 3 + \sqrt{5}\)) | B1 | 3 |
**(i)(a)** $0.052 \times 10(12/13)$
PE loss is 0.48d | M1, A1 | 2 | For using PE loss $= mgo\sin \alpha$
**(i)(b)** $0.8 \times (+1-2)^2$
EE gain is $\lambda x^2/2L$ | M1, A1 | 2 |
**(i)(c)** $F = 0.4 \times 0.052 \times 10 \times (5/13)$
WD $= 0.08d$ | M1, A1 | 2 | For using $F = \mu mg \cos \alpha$
**(ii)** $0.48d - 0.2(d-2)^2 + 0.08d \rightarrow$
$d^2 - 6d + 4 = 0$ | M1, A1 |
For using PE loss $=$ EE gain + WD
From correct simplification
For obtaining and selecting relevant correct root | M1, A1
$d = 5.24$ (or $d = 3 + \sqrt{5}$) | B1 | 3 |6 A light elastic string has natural length 2 m and modulus of elasticity 0.8 N . One end of the string is attached to a fixed point $O$ of a rough plane which is inclined at an angle $\alpha$ to the horizontal, where $\sin \alpha = \frac { 12 } { 13 }$. A particle $P$ of mass 0.052 kg is attached to the other end of the string. The coefficient of friction between the particle and the plane is 0.4 . $P$ is released from rest at $O$.\\
(i) When $P$ has moved $d$ metres down the plane from $O$, where $d > 2$, find expressions in terms of $d$ for
\begin{enumerate}[label=(\alph*)]
\item the loss in gravitational potential energy of $P$,
\item the gain in elastic potential energy of the string,
\item the work done by the frictional force acting on $P$.\\
(ii) Show that $d ^ { 2 } - 6 d + 4 = 0$ when $P$ is at its lowest point, and hence find the value of $d$ in this case.
\end{enumerate}
\hfill \mbox{\textit{CAIE M2 2006 Q6 [9]}}