| Exam Board | CAIE |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2014 |
| Session | November |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Hooke's law and elastic energy |
| Type | Particle attached to two separate elastic strings |
| Difficulty | Challenging +1.2 This is a two-part mechanics problem involving elastic strings in equilibrium and energy methods. Part (i) requires setting up simultaneous equations using Hooke's law and equilibrium conditions—straightforward but requiring careful algebra. Part (ii) applies conservation of energy with elastic potential energy, which is a standard M2 technique but requires tracking multiple energy terms. The problem is more involved than basic mechanics questions but follows well-established methods without requiring novel insight. |
| 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 principle |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(T = \frac{12(d-0.6)}{0.6} = \frac{21(d-0.7)}{0.7}\) | M1 | Uses \(T = \frac{\lambda\text{ ext}}{l}\) either string; must be total length, not extension |
| \(d = 0.9\) | A1 | |
| \(T = \frac{12(0.2-0.6)}{0.6} = \frac{21(0.2-0.7)}{0.7}\) | M1 | Equation with both sides correct |
| \(W = 12\) | A1 | [4] Finds either Tension \((=6)\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\text{PE} = 0.2\times12\ (=2.4)\) | B1\(\checkmark\) | ft their value of \(d\) and their value of \(W\); PE/KE/EE balance |
| \(\frac{12\times0.1^2}{2\times0.6}+12\times0.2 = \frac{\frac{12}{g}v^2}{2}+\frac{12\times0.3^2}{2\times0.6}+\frac{21\times0.2^2}{2\times0.7}\) | M1, A1 | |
| \(v = 1.29\text{ ms}^{-1}\) | A1 | [4] |
## Question 5:
### Part (i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $T = \frac{12(d-0.6)}{0.6} = \frac{21(d-0.7)}{0.7}$ | M1 | Uses $T = \frac{\lambda\text{ ext}}{l}$ either string; must be total length, not extension |
| $d = 0.9$ | A1 | |
| $T = \frac{12(0.2-0.6)}{0.6} = \frac{21(0.2-0.7)}{0.7}$ | M1 | Equation with both sides correct |
| $W = 12$ | A1 | [4] Finds either Tension $(=6)$ |
### Part (ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\text{PE} = 0.2\times12\ (=2.4)$ | B1$\checkmark$ | ft their value of $d$ and their value of $W$; PE/KE/EE balance |
| $\frac{12\times0.1^2}{2\times0.6}+12\times0.2 = \frac{\frac{12}{g}v^2}{2}+\frac{12\times0.3^2}{2\times0.6}+\frac{21\times0.2^2}{2\times0.7}$ | M1, A1 | |
| $v = 1.29\text{ ms}^{-1}$ | A1 | [4] |
---5\\
\includegraphics[max width=\textwidth, alt={}, center]{81be887c-ab01-4327-a5df-f25c68a6fdb6-2_337_517_1749_813}
Two light elastic strings each have one end attached to a fixed horizontal beam. One string has natural length 0.6 m and modulus of elasticity 12 N ; the other string has natural length 0.7 m and modulus of elasticity 21 N . The other ends of the strings are attached to a small block $B$ of weight $W \mathrm {~N}$. The block hangs in equilibrium $d \mathrm {~m}$ below the beam, with both strings vertical (see diagram).\\
(i) Given that the tensions in the strings are equal, find $d$ and $W$.
The small block is now raised vertically to the point 0.7 m below the beam, and then released from rest.\\
(ii) Find the greatest speed of the block in its subsequent motion.
\hfill \mbox{\textit{CAIE M2 2014 Q5 [8]}}