| Exam Board | OCR |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2009 |
| Session | January |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circular Motion 2 |
| Type | Energy considerations in circular motion |
| Difficulty | Standard +0.8 This is a substantial M3 question combining elastic strings, equilibrium, SHM derivation, and vertical circular motion with energy methods. Part (i) is routine verification, but parts (ii)-(iii) require careful algebraic manipulation to show SHM, and part (iv) involves energy considerations. The vertical circle component adds complexity. This is above-average difficulty for A-level but standard for M3 module work. |
| Spec | 4.10f Simple harmonic motion: x'' = -omega^2 x6.02h Elastic PE: 1/2 k x^26.02i Conservation of energy: mechanical energy principle6.05e Radial/tangential acceleration |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(T_A = (24 \times 0.45)/0.6\), \(T_B = (24 \times 0.15)/0.6\) | M1 | For using \(T = \lambda x/L\) for PA or PB |
| \(T_A - T_B = 18 - 6 = 12 = W \Rightarrow P\) in equilibrium | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Extensions are \(0.45 + x\) and \(0.15 - x\) | B1 | |
| Tensions are \(18 + 40x\) and \(6 - 40x\) | B1 | AG From \(T = \lambda x/L\) for PA and PB |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \([12 + (6-40x) - (18+40x) = 12\ddot{x}/g]\) | M1 | For using Newton's second law (4 terms required) |
| \(\ddot{x} = -80gx/12 \Rightarrow\) SHM | A1 | |
| Period is \(0.777\)s | A1 | AG From Period \(= 2\pi\sqrt{12/(80g)}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(v_{\max} = 0.15\sqrt{80g/12}\) | M1 | For using \(v_{\max} = A\omega\) or \(v_{\max} = 2\pi A/T\) or conservation of energy (5 terms needed) |
| or \(v_{\max} = 2\pi \times 0.15/0.777\) | ||
| or \(\frac{1}{2}(12/g)v_{\max}^2 + mg(0.15) + 24\{0.45^2 + 0.15^2 - 0.6^2\}/(2\times0.6) = 0\) | A1 | |
| Speed is \(1.21\ \text{ms}^{-1}\) |
## Question 3:
**Part (i)**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $T_A = (24 \times 0.45)/0.6$, $T_B = (24 \times 0.15)/0.6$ | M1 | For using $T = \lambda x/L$ for PA or PB |
| $T_A - T_B = 18 - 6 = 12 = W \Rightarrow P$ in equilibrium | A1 | |
**Part (ii)**
| Answer/Working | Mark | Guidance |
|---|---|---|
| Extensions are $0.45 + x$ and $0.15 - x$ | B1 | |
| Tensions are $18 + 40x$ and $6 - 40x$ | B1 | AG From $T = \lambda x/L$ for PA and PB |
**Part (iii)**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $[12 + (6-40x) - (18+40x) = 12\ddot{x}/g]$ | M1 | For using Newton's second law (4 terms required) |
| $\ddot{x} = -80gx/12 \Rightarrow$ SHM | A1 | |
| Period is $0.777$s | A1 | AG From Period $= 2\pi\sqrt{12/(80g)}$ |
**Part (iv)**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $v_{\max} = 0.15\sqrt{80g/12}$ | M1 | For using $v_{\max} = A\omega$ or $v_{\max} = 2\pi A/T$ or conservation of energy (5 terms needed) |
| or $v_{\max} = 2\pi \times 0.15/0.777$ | | |
| or $\frac{1}{2}(12/g)v_{\max}^2 + mg(0.15) + 24\{0.45^2 + 0.15^2 - 0.6^2\}/(2\times0.6) = 0$ | A1 | |
| Speed is $1.21\ \text{ms}^{-1}$ | | |
---3\\
\includegraphics[max width=\textwidth, alt={}, center]{14403602-94a6-4441-a673-65f9b98180e5-3_387_181_274_982}\\
$A$ and $B$ are fixed points with $B$ at a distance of 1.8 m vertically below $A$. One end of a light elastic string of natural length 0.6 m and modulus of elasticity 24 N is attached to $A$, and one end of an identical elastic string is attached to $B$. A particle $P$ of weight 12 N is attached to the other ends of the strings (see diagram).\\
(i) Verify that $P$ is in equilibrium when it is at a distance of 1.05 m vertically below $A$.\\
$P$ is released from rest at the point 1.2 m vertically below $A$ and begins to move.\\
(ii) Show that, when $P$ is $x \mathrm {~m}$ below its equilibrium position, the tensions in $P A$ and $P B$ are $( 18 + 40 x ) \mathrm { N }$ and $( 6 - 40 x ) \mathrm { N }$ respectively.\\
(iii) Show that $P$ moves with simple harmonic motion of period 0.777 s , correct to 3 significant figures.\\
(iv) Find the speed with which $P$ passes through the equilibrium position.\\
\includegraphics[max width=\textwidth, alt={}, center]{14403602-94a6-4441-a673-65f9b98180e5-3_540_655_1564_744}
One end of a light inextensible string of length 0.5 m is attached to a fixed point $O$. A particle $P$ of mass 0.2 kg is attached to the other end of the string. With the string taut and horizontal, $P$ is projected with a velocity of $3 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ vertically downward. $P$ begins to move in a vertical circle with centre $O$. While the string remains taut the angular displacement of $O P$ is $\theta$ radians from its initial position, and the speed of $P$ is $v \mathrm {~m} \mathrm {~s} ^ { - 1 }$ (see diagram).\\
\hfill \mbox{\textit{OCR M3 2009 Q3 [9]}}