| Exam Board | CAIE |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2017 |
| Session | November |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable Force |
| Type | Elastic string with variable force |
| Difficulty | Standard +0.8 This question requires understanding of elastic strings, energy methods, and kinematics with variable forces. Part (i) involves finding where tension equals weight (zero acceleration condition), then applying energy conservation with elastic potential energy. Part (ii) requires identifying the region where the string is slack and applying constant acceleration equations. The multi-step reasoning and integration of elastic string theory with energy methods makes this moderately challenging, though the techniques are standard for M2 level. |
| Spec | 6.02e Calculate KE and PE: using formulae6.02i Conservation of energy: mechanical energy principle |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(0.3g = 24e\) | M1 | Use \(T = \lambda x/L\) |
| \(e = 0.1\) | A1 | |
| \(EE = 24 \times (1.2-0.8)^2/(2 \times 0.8)\) or \(24 \times 0.1^2/(2 \times 0.8)\) | B1 | Use \(EE = \lambda x^2/(2L)\) |
| \(0.3v^2/2 = 0.3 \times 4^2/2 + 24 \times (1.2-0.8)^2/(2 \times 0.8) - 24 \times 0.1^2/(2 \times 0.8) - 0.3g(1.2-0.8)\) | M1 | Sets up a 5 term energy equation involving \(EE\), \(KE\) and \(PE\) |
| \(v = 5 \text{ m s}^{-1}\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(0.5 \times 5^2/2 + 24 \times 0.1^2/(2 \times 0.8) = 0.3(x + 0.9) \times 10\) | M1 | Sets up a 3 term energy equation where \(x\) is the distance above 0 when \(v = 0\) |
| \(x = 0.4\) | A1 | |
| Distance moved \(= 0.8 + 0.4 = 1.2 \text{ m}\) | A1 | AG |
## Question 5(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $0.3g = 24e$ | M1 | Use $T = \lambda x/L$ |
| $e = 0.1$ | A1 | |
| $EE = 24 \times (1.2-0.8)^2/(2 \times 0.8)$ or $24 \times 0.1^2/(2 \times 0.8)$ | B1 | Use $EE = \lambda x^2/(2L)$ |
| $0.3v^2/2 = 0.3 \times 4^2/2 + 24 \times (1.2-0.8)^2/(2 \times 0.8) - 24 \times 0.1^2/(2 \times 0.8) - 0.3g(1.2-0.8)$ | M1 | Sets up a 5 term energy equation involving $EE$, $KE$ and $PE$ |
| $v = 5 \text{ m s}^{-1}$ | A1 | |
## Question 5(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $0.5 \times 5^2/2 + 24 \times 0.1^2/(2 \times 0.8) = 0.3(x + 0.9) \times 10$ | M1 | Sets up a 3 term energy equation where $x$ is the distance above 0 when $v = 0$ |
| $x = 0.4$ | A1 | |
| Distance moved $= 0.8 + 0.4 = 1.2 \text{ m}$ | A1 | AG |
---5 One end of a light elastic string of natural length 0.8 m and modulus of elasticity 24 N is attached to a fixed point $O$. The other end of the string is attached to a particle $P$ of mass $0.3 \mathrm {~kg} . P$ is projected vertically upwards with speed $4 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ from a position 1.2 m vertically below $O$.\\
(i) Calculate the speed of the particle at the position where it is moving with zero acceleration. [5\\
(ii) Show that the particle moves 1.2 m while moving upwards with constant deceleration.\\
\hfill \mbox{\textit{CAIE M2 2017 Q5 [8]}}