| 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 standard two-part mechanics problem involving Hooke's law and energy conservation. Part (i) requires straightforward resolution of forces at equilibrium to find the modulus of elasticity. Part (ii) uses energy conservation with elastic potential energy, which is a routine application of standard formulas. The geometry is clearly specified, and the solution path follows standard textbook methods without requiring novel insight. |
| Spec | 6.02h Elastic PE: 1/2 k x^26.02j Conservation with elastics: springs and strings |
| 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 $P$ is released from rest by removing the $7$ N force. In its subsequent motion $P$ first comes to instantaneous rest at a point where $BP = 0.3$ m and the elastic string makes an angle of $30°$ with the horizontal (see Fig. 2).
\includegraphics{figure_2}
Find the value of $m$. [4]
\end{enumerate}
\hfill \mbox{\textit{CAIE M2 2017 Q2 [6]}}