| Exam Board | OCR |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2012 |
| Session | June |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circular Motion 1 |
| Type | Two strings, two fixed points |
| Difficulty | Standard +0.3 This is a standard M2 circular motion problem with two strings requiring resolution of forces and application of F=mrω². Part (i) involves straightforward geometry to find angles, then resolving vertically and horizontally. Part (ii) requires recognizing that minimum speed occurs when one string goes slack. Slightly above average difficulty due to the two-string setup and the conceptual understanding needed for part (ii), but follows standard M2 patterns. |
| Spec | 6.05c Horizontal circles: conical pendulum, banked tracks |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\sin\theta = 0.8\) or \(\cos\theta = 0.6\) or \(\tan\theta = 4/3\) or \(\theta = 53.1°\) | B1 | \(\theta\) is angle AP makes with horizontal |
| *M1 | Attempt to resolve horizontally and use N2L with a version of acceleration, not just \(a\). Allow \(T_A = T_B\) for M1 only | |
| \(T_A\cos\theta + T_B\cos\theta = 2 \times 1.2 \times 4^2\) | A1 | Use their \(\theta\) |
| *M1 | Attempt to resolve vertically | |
| \(T_A\sin\theta = T_B\sin\theta + 2g\) | A1 | Use their \(\theta\) |
| Solve simultaneously to get at least \(T_A\) or \(T_B\) | Dep*M1 | |
| \(T_A = 44.25\) and \(T_B = 19.75\) | A1 | For both. Allow \(44.2\), \(44.3\), \(19.7\), \(19.8\) |
| [7] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(T_B = 0\) | B1 | May be implied |
| *M1 | Attempt to resolve horizontally and use N2L with a version of acceleration, not just \(a\) | |
| \(T_A\cos\theta = 2v^2/1.2\) | A1 | Use their \(\theta\) |
| \(T_A\sin\theta = 2g\) | B1 | Use their \(\theta\) |
| Solve for \(v\) or \(\omega\) | Dep*M1 | |
| \(v = 2.97\) | A1 | |
| [6] |
## Question 5:
### Part (i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\sin\theta = 0.8$ or $\cos\theta = 0.6$ or $\tan\theta = 4/3$ or $\theta = 53.1°$ | B1 | $\theta$ is angle AP makes with horizontal |
| | *M1 | Attempt to resolve horizontally and use N2L with a version of acceleration, not just $a$. Allow $T_A = T_B$ for M1 only |
| $T_A\cos\theta + T_B\cos\theta = 2 \times 1.2 \times 4^2$ | A1 | Use their $\theta$ |
| | *M1 | Attempt to resolve vertically |
| $T_A\sin\theta = T_B\sin\theta + 2g$ | A1 | Use their $\theta$ |
| Solve simultaneously to get at least $T_A$ or $T_B$ | Dep*M1 | |
| $T_A = 44.25$ and $T_B = 19.75$ | A1 | For both. Allow $44.2$, $44.3$, $19.7$, $19.8$ |
| **[7]** | | |
### Part (ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $T_B = 0$ | B1 | May be implied |
| | *M1 | Attempt to resolve horizontally and use N2L with a version of acceleration, not just $a$ |
| $T_A\cos\theta = 2v^2/1.2$ | A1 | Use their $\theta$ |
| $T_A\sin\theta = 2g$ | B1 | Use their $\theta$ |
| Solve for $v$ or $\omega$ | Dep*M1 | |
| $v = 2.97$ | A1 | |
| **[6]** | | |
---5 A particle $P$, of mass 2 kg , is attached to fixed points $A$ and $B$ by light inextensible strings, each of length 2 m . $A$ and $B$ are 3.2 m apart with $A$ vertically above $B$. The particle $P$ moves in a horizontal circle with centre at the mid-point of $A B$.\\
(i) Find the tension in each string when the angular speed of $P$ is $4 \mathrm { rads } ^ { - 1 }$.\\
(ii) Find the least possible speed of $P$.
\hfill \mbox{\textit{OCR M2 2012 Q5 [13]}}