| Exam Board | CAIE |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2017 |
| Session | March |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Hooke's law and elastic energy |
| Type | Vertical elastic string: projected from equilibrium or other point |
| Difficulty | Standard +0.8 This is a multi-part elastic string problem requiring equilibrium analysis, energy conservation with elastic potential energy, and then a collision followed by further energy analysis. Part (i) is routine, but parts (ii) and (iii) require careful application of energy methods and understanding of the collision's effect on the system, making this moderately challenging for A-level mechanics. |
| Spec | 6.02g Hooke's law: T = k*x or T = lambda*x/l6.02h Elastic PE: 1/2 k x^26.02i Conservation of energy: mechanical energy principle6.03i Coefficient of restitution: e |
| Answer | Marks | Guidance |
|---|---|---|
| 7(i) | 0.4g = 24e/0.6 | M1 |
| e = 0.1 m | A1 | |
| Total: | 2 |
| Answer | Marks | Guidance |
|---|---|---|
| 7(ii) | Initial EE = 24 x 0.12/(2 x 0.6) (= 0.2 J) | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| x 0.12/(2 x 0.6) | M1 A1 | Set up a 4 term energy equation involving |
| Answer | Marks | Guidance |
|---|---|---|
| d = 0.5 m | A1 | |
| Total: | 4 | |
| Question | Answer | Marks |
| Answer | Marks | Guidance |
|---|---|---|
| 7(iii) | e = 0.2 | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| 0.6) – 0.8g x 0.4 | M1 A1 | Set up a 4 term energy equation in EE, PE |
| Answer | Marks |
|---|---|
| v = 2 2 = 2.83 ms–1 | A1 |
| Total: | 4 |
Question 7:
--- 7(i) ---
7(i) | 0.4g = 24e/0.6 | M1 | Uses T = λx/L
e = 0.1 m | A1
Total: | 2
--- 7(ii) ---
7(ii) | Initial EE = 24 x 0.12/(2 x 0.6) (= 0.2 J) | B1 | Uses EE = λx2/2L
0.4 x 52/2 + 0.4gd=24(0.1 + d)2/(2 x 0.6) –24
x 0.12/(2 x 0.6) | M1 A1 | Set up a 4 term energy equation involving
EE, PE and KE
d = 0.5 m | A1
Total: | 4
Question | Answer | Marks | Guidance
--- 7(iii) ---
7(iii) | e = 0.2 | B1
0.8v2/2=24 x 0.62/(2 x 0.6)– 24 x 0.22/(2 x
0.6) – 0.8g x 0.4 | M1 A1 | Set up a 4 term energy equation in EE, PE
and KE
v = 2 2 = 2.83 ms–1 | A1
Total: | 4One end of a light elastic string of natural length $0.6 \text{ m}$ and modulus of elasticity $24 \text{ N}$ is attached to a fixed point $O$. The other end of the string is attached to a particle $P$ of mass $0.4 \text{ kg}$ which hangs in equilibrium vertically below $O$.
\begin{enumerate}[label=(\roman*)]
\item Calculate the extension of the string. [2]
\end{enumerate}
$P$ is projected vertically downwards from the equilibrium position with speed $5 \text{ m s}^{-1}$.
\begin{enumerate}[label=(\roman*)]
\setcounter{enumi}{1}
\item Calculate the distance $P$ travels before it is first at instantaneous rest. [4]
\end{enumerate}
When $P$ is first at instantaneous rest a stationary particle of mass $0.4 \text{ kg}$ becomes attached to $P$.
\begin{enumerate}[label=(\roman*)]
\setcounter{enumi}{2}
\item Find the greatest speed of the combined particle in the subsequent motion. [4]
\end{enumerate}
\hfill \mbox{\textit{CAIE M2 2017 Q7 [10]}}