| Exam Board | CAIE |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2009 |
| Session | November |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Hooke's law and elastic energy |
| Type | Vertical elastic string: general symbolic/proof questions |
| Difficulty | Standard +0.3 This is a standard elastic string energy problem requiring conservation of energy (GPE + EPE = KE) to derive the given velocity equation, then differentiation or force analysis at the lowest point. The algebra is straightforward and the method is a textbook application of energy principles in mechanics, making it slightly easier than average for A-level Further Maths M2. |
| Spec | 3.02f Non-uniform acceleration: using differentiation and integration6.02i Conservation of energy: mechanical energy principle6.02j Conservation with elastics: springs and strings |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(EE = \frac{1}{2}(15\,m)x^2/3\) | B1 | |
| M1 | For using Loss of \(PE =\) Gain in \(KE + EE\) | |
| \(\frac{1}{2}mv^2 + \frac{1}{2}(15\,m)x^2/3 = mg(3 + x^2)\) | A1ft | Ft error in EE |
| \(v^2 = 5(12 + 4x - x)\) | A1 | AG Subtotal: 4 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \([a = -2.5(2x - 4)\) or \(a = g - 15x/3]\) | M1 | For using \(a = v(dv/dx)\) or \(a = (mg - \lambda x/L)/m\) |
| \([v = 0 \rightarrow x = 6 \rightarrow a = -20]\) | M1 | For finding \(x\) at the lowest point and substituting |
| Magnitude is \(20\text{ ms}^{-2}\); direction is upwards | A1 | Subtotal: 3 |
## Question 4:
### Part (i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $EE = \frac{1}{2}(15\,m)x^2/3$ | B1 | |
| | M1 | For using Loss of $PE =$ Gain in $KE + EE$ |
| $\frac{1}{2}mv^2 + \frac{1}{2}(15\,m)x^2/3 = mg(3 + x^2)$ | A1ft | Ft error in EE |
| $v^2 = 5(12 + 4x - x)$ | A1 | AG **Subtotal: 4** |
### Part (ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $[a = -2.5(2x - 4)$ or $a = g - 15x/3]$ | M1 | For using $a = v(dv/dx)$ or $a = (mg - \lambda x/L)/m$ |
| $[v = 0 \rightarrow x = 6 \rightarrow a = -20]$ | M1 | For finding $x$ at the lowest point and substituting |
| Magnitude is $20\text{ ms}^{-2}$; direction is upwards | A1 | **Subtotal: 3** | **Total: 7** |
---4 One end of a light elastic string of natural length 3 m and modulus of elasticity 15 mN is attached to a fixed point $O$. A particle $P$ of mass $m \mathrm {~kg}$ is attached to the other end of the string. $P$ is released from rest at $O$ and moves vertically downwards. When the extension of the string is $x \mathrm {~m}$ the velocity of $P$ is $v \mathrm {~m} \mathrm {~s} ^ { - 1 }$.\\
(i) Show that $v ^ { 2 } = 5 \left( 12 + 4 x - x ^ { 2 } \right)$.\\
(ii) Find the magnitude of the acceleration of $P$ when it is at its lowest point, and state the direction of this acceleration.
\hfill \mbox{\textit{CAIE M2 2009 Q4 [7]}}