| Exam Board | CAIE |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2012 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Hooke's law and elastic energy |
| Type | Particle at midpoint of string between two horizontal fixed points: vertical motion |
| Difficulty | Challenging +1.2 This is a multi-step energy conservation problem requiring understanding of elastic strings, equilibrium positions, and energy methods. While it involves several concepts (elastic potential energy, gravitational potential energy, finding equilibrium), the solution follows a standard framework: energy conservation between release and maximum speed point, recognizing that maximum speed occurs at equilibrium where net force is zero. The calculations are straightforward once the setup is understood, making it moderately above average difficulty for A-level mechanics. |
| Spec | 6.02h Elastic PE: 1/2 k x^26.02i Conservation of energy: mechanical energy principle |
| Answer | Marks | Guidance |
|---|---|---|
| Length \(= \sqrt{1.2^2 + 0.5^2} = 1.3\) | B1 | Pythagoras on \(\frac{1}{4}\) string |
| \(2 \times [14.3 \times (1.3 - 1.1)/1.1] \times [0.5/1.3] = mg\) | M1* D* M1 | Uses \(T = \lambda x/L\); Component(s) \(T\) equated to weight |
| \(m = 0.2\) | A1 | [4] |
| Answer | Marks | Guidance |
|---|---|---|
| \(0.2v^2/2 = 0.2g \times 0.5 - [14.3 \times 0.2^2/(2 \times 1.1) - 14.3 \times 0.1^2/(2 \times 1.1)] \times 2\) | M1 | KE/EE/PE balance (4 terms) |
| \(v = 2.47\,\mathrm{ms}^{-1}\) | A1 | \(\unicode{x2713}\) candidate's value of \(m\) from (i) |
**(i)**
Length $= \sqrt{1.2^2 + 0.5^2} = 1.3$ | B1 | Pythagoras on $\frac{1}{4}$ string
$2 \times [14.3 \times (1.3 - 1.1)/1.1] \times [0.5/1.3] = mg$ | M1* D* M1 | Uses $T = \lambda x/L$; Component(s) $T$ equated to weight
$m = 0.2$ | A1 | [4]
**(ii)**
$0.2v^2/2 = 0.2g \times 0.5 - [14.3 \times 0.2^2/(2 \times 1.1) - 14.3 \times 0.1^2/(2 \times 1.1)] \times 2$ | M1 | KE/EE/PE balance (4 terms)
$v = 2.47\,\mathrm{ms}^{-1}$ | A1 | $\unicode{x2713}$ candidate's value of $m$ from (i) | A1 | [3] | [7]3 A light elastic string has natural length 2.2 m and modulus of elasticity 14.3 N . A particle $P$ of mass $m \mathrm {~kg}$ is attached to the mid-point of the string. The ends of the string are attached to fixed points $A$ and $B$ which are 2.4 m apart at the same horizontal level. $P$ is released from rest at the mid-point of $A B$. In the subsequent motion $P$ has its greatest speed at a point 0.5 m below $A B$.\\
(i) Find $m$.\\
(ii) Calculate the greatest speed of $P$.
\hfill \mbox{\textit{CAIE M2 2012 Q3 [7]}}