| Exam Board | CAIE |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2011 |
| Session | June |
| Marks | 9 |
| 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 force. Part (i) involves finding when tension equals weight (zero acceleration condition), then applying energy conservation. Part (ii) requires identifying the region where string is slack (constant acceleration under gravity only) and using kinematic equations. The multi-step reasoning across different force regimes and the need to recognize when the string becomes slack elevates this above routine mechanics problems. |
| Spec | 6.02h Elastic PE: 1/2 k x^26.02i Conservation of energy: mechanical energy principle6.02j Conservation with elastics: springs and strings |
| Answer | Marks | Guidance |
|---|---|---|
| (i) \(0.24g = 12(x) 0.5\); \(x = 0.1\) OR \(\frac{1}{2} \times 0.24 \times 3^2 + 12 \times (0.8 - 0.5)^2/(2 \times 0.5) = 0.24v^2/2 + 12 \times 0.1^2/(2 \times 0.5) + 0.24g(0.8 - 0.5 - 0.1)\); \(v = 3.61\) ms\(^{-1}\) OR \(0.24vdv/dx = mg - 12v/0.5\); \(0.24v^2/2 = 2.4x - 12x^2\) (+ c); \(v = 3, x = 0.3, c = 1.44\); \(x = 0.1, v = 3.61\) ms\(^{-1}\) | M1, A1 [5] | Finds position for equilibrium; Energy balance, initial to equilibrium positions; Using Newton's Second Law; Or uses limits |
| (ii) \(0.24 \times 3^2/2 + 12 \times (0.8 - 0.5)^2/(2 \times 0.5) = 0.24g(0.8 + x)\); \(x = 0.1\)m; \(s = (0.5 + 0.1) = 0.6\) m OR \(\frac{1}{2} \times 12 \times 0.3^2/0.5 + \frac{1}{2} \times 0.24 \times 3^2 = \frac{1}{2} \times 0.24v^2 + 0.24 \times 10s\); \(v = \sqrt{12}\); Either \(0 = 12 - 2 \times 10s\); \(s = 0.6\) Or \(\frac{1}{2} \times 0.24 \times 12 = 0.24 \times 10s\); \(s = 0.6\) OR \(\frac{1}{2} \times 12 \times 0.3^2/0.5 + \frac{1}{2} \times 0.24 \times 3^2 = 0.24 \times 10y\); \(y = 0.9\); \(s = 0.9 - 0.3 = 0.6\) | M1, A1, A1, A1 [4] | Initial KE + initial EE = Final PE; Initial EE + Initial KE = (KE + PE) at equilibrium position; Using \(v^2 = u^2 + 2as\); Using KE at equilibrium position = Final PE; Initial EE + Initial KE = Final PE where y is the distance above the start |
(i) $0.24g = 12(x) 0.5$; $x = 0.1$ OR $\frac{1}{2} \times 0.24 \times 3^2 + 12 \times (0.8 - 0.5)^2/(2 \times 0.5) = 0.24v^2/2 + 12 \times 0.1^2/(2 \times 0.5) + 0.24g(0.8 - 0.5 - 0.1)$; $v = 3.61$ ms$^{-1}$ OR $0.24vdv/dx = mg - 12v/0.5$; $0.24v^2/2 = 2.4x - 12x^2$ (+ c); $v = 3, x = 0.3, c = 1.44$; $x = 0.1, v = 3.61$ ms$^{-1}$ | M1, A1 [5] | Finds position for equilibrium; Energy balance, initial to equilibrium positions; Using Newton's Second Law; Or uses limits
(ii) $0.24 \times 3^2/2 + 12 \times (0.8 - 0.5)^2/(2 \times 0.5) = 0.24g(0.8 + x)$; $x = 0.1$m; $s = (0.5 + 0.1) = 0.6$ m OR $\frac{1}{2} \times 12 \times 0.3^2/0.5 + \frac{1}{2} \times 0.24 \times 3^2 = \frac{1}{2} \times 0.24v^2 + 0.24 \times 10s$; $v = \sqrt{12}$; Either $0 = 12 - 2 \times 10s$; $s = 0.6$ Or $\frac{1}{2} \times 0.24 \times 12 = 0.24 \times 10s$; $s = 0.6$ OR $\frac{1}{2} \times 12 \times 0.3^2/0.5 + \frac{1}{2} \times 0.24 \times 3^2 = 0.24 \times 10y$; $y = 0.9$; $s = 0.9 - 0.3 = 0.6$ | M1, A1, A1, A1 [4] | Initial KE + initial EE = Final PE; Initial EE + Initial KE = (KE + PE) at equilibrium position; Using $v^2 = u^2 + 2as$; Using KE at equilibrium position = Final PE; Initial EE + Initial KE = Final PE where y is the distance above the start4 One end of a light elastic string of natural length 0.5 m and modulus of elasticity 12 N is attached to a fixed point $O$. The other end of the string is attached to a particle $P$ of mass $0.24 \mathrm {~kg} . P$ is projected vertically upwards with speed $3 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ from a position 0.8 m vertically below $O$.\\
(i) Calculate the speed of the particle when it is moving upwards with zero acceleration.\\
(ii) Show that the particle moves 0.6 m while it is moving upwards with constant acceleration.
\hfill \mbox{\textit{CAIE M2 2011 Q4 [9]}}