| Exam Board | OCR |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2012 |
| Session | January |
| Marks | 15 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Advanced work-energy problems |
| Type | Elastic string ring or sphere surface |
| Difficulty | Challenging +1.3 This M3 circular motion problem requires resolving forces with elastic strings, applying energy conservation, and analyzing equilibrium conditions including a numerical equation to solve. While it involves multiple concepts (elasticity, circular motion, energy), the steps are fairly standard for M3 level with clear guidance through parts (i)-(iii), making it moderately above average difficulty but not requiring exceptional insight. |
| Spec | 1.09a Sign change methods: locate roots6.02h Elastic PE: 1/2 k x^26.02i Conservation of energy: mechanical energy principle6.05e Radial/tangential acceleration |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Tension in string \(T = mg\sin\alpha\) | M1, B1 | For using \(T = \lambda x/L\) |
| For using \(e = R\alpha - 2R/3\) | B1 | \(mg\sin\alpha = 1.2mg\left(R\alpha - \dfrac{2R}{3}\right) \div \dfrac{2R}{3}\) |
| \(1.8\alpha - \sin\alpha - 1.2 = 0\) | A1 | AG; establish result. By iteration: \(\alpha = (1.2 + \sin\alpha)/1.8\) M1 |
| Finding \(f(1.175)\) and \(f(1.185)\) correctly | M1 | \(\approx -0.008\) and \(\approx +0.0065\); start \([1, 2]\) and 1 iteration A1 |
| Correct conclusion | A1 | AG \(\alpha = 1.18\) correct to 3 significant figures; at least 1 more iteration, conclusion \(1.18(0427)\) A1 |
| [7] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Direction is towards \(O\) | B1 | |
| [1] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Gain in EE \(= 1.2mg(1.18R - 2R/3)^2 \div (2 \times 2R/3)\) | M1* | For using EE \(= \lambda e^2 \div (2L)\) and PE \(= mgh\) |
| PE loss \(= mgR(\cos 2/3 - \cos 1.18)\) | A1, A1 | Ignore signs; allow \(\alpha\) for \(1.18\) for A1A1; allow sign errors |
| \(\frac{1}{2}mv^2 =\) PE loss \(-\) EE gain | M1 | Need \(1.18\) here |
| \(v^2 = 2gR[\cos 2/3 - \cos 1.18 - 0.9(1.18 - 2/3)^2]\) | A1 | If candidates use \(mR\theta\) use equivalent scheme |
| Acceleration is \(3.29\text{ ms}^{-2}\) | *M1, A1 | For using acceleration \(= v^2/R\) |
| [7] |
## Question 7:
### Part (i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Tension in string $T = mg\sin\alpha$ | M1, B1 | For using $T = \lambda x/L$ |
| For using $e = R\alpha - 2R/3$ | B1 | $mg\sin\alpha = 1.2mg\left(R\alpha - \dfrac{2R}{3}\right) \div \dfrac{2R}{3}$ |
| $1.8\alpha - \sin\alpha - 1.2 = 0$ | A1 | AG; establish result. By iteration: $\alpha = (1.2 + \sin\alpha)/1.8$ M1 |
| Finding $f(1.175)$ and $f(1.185)$ correctly | M1 | $\approx -0.008$ and $\approx +0.0065$; start $[1, 2]$ and 1 iteration A1 |
| Correct conclusion | A1 | AG $\alpha = 1.18$ correct to 3 significant figures; at least 1 more iteration, conclusion $1.18(0427)$ A1 |
| **[7]** | | |
### Part (ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Direction is towards $O$ | B1 | |
| **[1]** | | |
### Part (iii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Gain in EE $= 1.2mg(1.18R - 2R/3)^2 \div (2 \times 2R/3)$ | M1* | For using EE $= \lambda e^2 \div (2L)$ and PE $= mgh$ |
| PE loss $= mgR(\cos 2/3 - \cos 1.18)$ | A1, A1 | Ignore signs; allow $\alpha$ for $1.18$ for A1A1; allow sign errors |
| $\frac{1}{2}mv^2 =$ PE loss $-$ EE gain | M1 | Need $1.18$ here |
| $v^2 = 2gR[\cos 2/3 - \cos 1.18 - 0.9(1.18 - 2/3)^2]$ | A1 | If candidates use $mR\theta$ use equivalent scheme |
| Acceleration is $3.29\text{ ms}^{-2}$ | *M1, A1 | For using acceleration $= v^2/R$ |
| **[7]** | | |7\\
\includegraphics[max width=\textwidth, alt={}, center]{43ed8ec7-67f1-418a-8d4e-ee96448647fd-4_351_314_255_861}
One end of a light elastic string, of natural length $\frac { 2 } { 3 } R \mathrm {~m}$ and with modulus of elasticity 1.2 mgN , is attached to the highest point $A$ of a smooth fixed sphere with centre $O$ and radius $R \mathrm {~m}$. A particle $P$ of mass $m \mathrm {~kg}$ is attached to the other end of the string and is in contact with the surface of the sphere, where the angle $A O P$ is equal to $\theta$ radians (see diagram).\\
(i) Given that $P$ is in equilibrium at the point where $\theta = \alpha$, show that $1.8 \alpha - \sin \alpha - 1.2 = 0$. Hence show that $\alpha = 1.18$ correct to 3 significant figures.\\
$P$ is now released from rest at the point of the surface of the sphere where $\theta = \frac { 2 } { 3 }$, and starts to move downwards on the surface. For an instant when $\theta = \alpha$,\\
(ii) state the direction of the acceleration of $P$,\\
(iii) find the magnitude of the acceleration of $P$.
\hfill \mbox{\textit{OCR M3 2012 Q7 [15]}}