| Exam Board | CAIE |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2016 |
| Session | March |
| Marks | 9 |
| 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 +0.3 This is a standard A-level mechanics question on elastic strings requiring equilibrium analysis (Hooke's law), then energy conservation for two different scenarios. While it involves multiple parts and careful bookkeeping of energy terms (KE, GPE, EPE), the techniques are routine for M2 students with no novel problem-solving required. Slightly above average due to the multi-step nature and potential for sign errors. |
| Spec | 6.02g Hooke's law: T = k*x or T = lambda*x/l6.02i Conservation of energy: mechanical energy principle |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(24e/0.8 = 0.2g\) | M1 | |
| \(e = 0.2\) | A1 | Total: 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(24 \times 0.2^2/(2 \times 0.8)\ (= 0.6)\) | B1\(\checkmark\) | ft(cv0.2) Initial EE |
| \(0.6 \times 4.5^2/2 + 0.6gd + 24 \times 0.2^2/(2 \times 0.8) = 0.6 \times 3.5^2/2 + 24 \times (0.2+d)^2/(2 \times 0.8)\) | M1 | PE/EE/KE balance attempt |
| A1 | \(d =\) distance particle falls | |
| \(d = 0.4\) so \(AP\ (= 0.8 + 0.2 + 0.4) = 1.4\text{m}\) | A1 | Total: 4 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(24 \times 0.2^2/(2 \times 0.8) + 0.6 \times 4.5^2/2 = 0.6v^2/2 + 0.6g \times 0.5\) | M1, A1 | PE/EE/KE balance, 4 terms. Award B1ft for initial KE if not already seen in part (ii) |
| \(v = 3.5\ \text{m s}^{-1}\) | A1 | Total: 3 |
## Question 5:
### Part (i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $24e/0.8 = 0.2g$ | M1 | |
| $e = 0.2$ | A1 | **Total: 2** |
### Part (ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $24 \times 0.2^2/(2 \times 0.8)\ (= 0.6)$ | B1$\checkmark$ | ft(cv0.2) Initial EE |
| $0.6 \times 4.5^2/2 + 0.6gd + 24 \times 0.2^2/(2 \times 0.8) = 0.6 \times 3.5^2/2 + 24 \times (0.2+d)^2/(2 \times 0.8)$ | M1 | PE/EE/KE balance attempt |
| | A1 | $d =$ distance particle falls |
| $d = 0.4$ so $AP\ (= 0.8 + 0.2 + 0.4) = 1.4\text{m}$ | A1 | **Total: 4** |
### Part (iii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $24 \times 0.2^2/(2 \times 0.8) + 0.6 \times 4.5^2/2 = 0.6v^2/2 + 0.6g \times 0.5$ | M1, A1 | PE/EE/KE balance, 4 terms. Award B1ft for initial KE if not already seen in part (ii) |
| $v = 3.5\ \text{m s}^{-1}$ | A1 | **Total: 3** |
---5 A particle $P$ of mass 0.6 kg is attached to one end of a light elastic string of natural length 0.8 m and modulus of elasticity 24 N . The other end of the string is attached to a fixed point $A$, and $P$ hangs in equilibrium.\\
(i) Calculate the extension of the string.\\
$P$ is projected vertically downwards from the equilibrium position with speed $4.5 \mathrm {~m} \mathrm {~s} ^ { - 1 }$.\\
(ii) Find the distance $A P$ when the speed of $P$ is $3.5 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ and $P$ is below the equilibrium position.\\
(iii) Calculate the speed of $P$ when it is 0.5 m above the equilibrium position.
\hfill \mbox{\textit{CAIE M2 2016 Q5 [9]}}