| Exam Board | CAIE |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2017 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Hooke's law and elastic energy |
| Type | Particle attached to two separate elastic strings |
| Difficulty | Standard +0.3 This is a straightforward statics problem requiring resolution of forces and application of Hooke's law. Students must find the extension using Pythagoras, apply T = λx/l for the elastic string, then resolve forces horizontally and vertically. While it involves multiple steps (geometry, Hooke's law, two-directional resolution), these are standard M2 techniques with no novel insight required, making it slightly easier than average. |
| Spec | 3.03m Equilibrium: sum of resolved forces = 06.02h Elastic PE: 1/2 k x^2 |
| Answer | Marks | Guidance |
|---|---|---|
| 2(i) | 7 = 0.35λ / 0.25 | M1 |
| λ = 5 | A1 | |
| Total: | 2 |
| Answer | Marks | Guidance |
|---|---|---|
| 2(ii) | EE =0.352 × 5 / (2 × 0.25) or 0.052 × 5 / (2 × 0.05) | B1 |
| PE = mg × 0.3sin30 | B1 | |
| mg × 0.3sin30 = 0.352 × 5 / (2 × 0.25) − 0.052 × 5 / (2 × 0.25) | M1 | Sets up a 3 term energy equation involving EE, KE and PE |
| m = 0.8 | A1 | |
| Total: | 4 | |
| Question | Answer | Marks |
Question 2:
--- 2(i) ---
2(i) | 7 = 0.35λ / 0.25 | M1 | Uses T = λx / L
λ = 5 | A1
Total: | 2
--- 2(ii) ---
2(ii) | EE =0.352 × 5 / (2 × 0.25) or 0.052 × 5 / (2 × 0.05) | B1 | Uses EE = λx2 / 2L
PE = mg × 0.3sin30 | B1
mg × 0.3sin30 = 0.352 × 5 / (2 × 0.25) − 0.052 × 5 / (2 × 0.25) | M1 | Sets up a 3 term energy equation involving EE, KE and PE
m = 0.8 | A1
Total: | 4
Question | Answer | Marks | Guidance\includegraphics{figure_1}
One end of a light inextensible string is attached to a fixed point $A$. The other end of the string is attached to a particle $P$ of mass $m$ kg which hangs vertically below $A$. The particle is also attached to one end of a light elastic string of natural length $0.25$ m. The other end of this string is attached to a point $B$ which is $0.6$ m from $P$ and on the same horizontal level as $P$. Equilibrium is maintained by a horizontal force of magnitude $7$ N applied to $P$ (see Fig. 1).
\begin{enumerate}[label=(\roman*)]
\item Calculate the modulus of elasticity of the elastic string. [2]
\item Find the value of $m$. [4]
\end{enumerate}
\hfill \mbox{\textit{CAIE M2 2017 Q2 [6]}}