| Exam Board | OCR |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2008 |
| Session | June |
| Marks | 12 |
| 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 at equal angles. Part (i) requires resolving forces vertically and horizontally with centripetal force equation - straightforward application of standard techniques. Part (ii) involves finding when lower string becomes slack (tension = 0), requiring similar force resolution. Slightly above average difficulty due to two-string setup and two parts, but follows predictable M2 patterns with no novel insight required. |
| Spec | 6.05c Horizontal circles: conical pendulum, banked tracks |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(T\cos60° = S\cos60° + 4.9\) | M1 | Resolving vertically, nb for M1: must be components – all 4 cases |
| A1 | ||
| \(T\sin60° + S\sin60° = 0.5 \times 3^2/0.4\) | M1 | Res. Horiz. \(mr\omega^2\) ok if \(\omega \neq 3\) |
| A1 | If equal tensions \(2T = 45/4\) M1 only | |
| \((S+9.8)\sin60° + S\sin60° = 45/4\) | M1 | |
| \(S = 1.60\) N | A1 | |
| \(T = 11.4\) N | A1 | 7 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(T\cos60° = 4.9\), \(T = 9.8\) | M1, A1 | Resolving vertically (component) |
| \(T\sin60° = 0.5 \times 0.4\omega^2\) | M1, A1 | Resolving horiz. (component) |
| \(\omega = 6.51\) rad s\(^{-1}\) | A1 | 5 |
# Question 6:
## Part (i)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $T\cos60° = S\cos60° + 4.9$ | M1 | Resolving vertically, nb for M1: must be components – all 4 cases |
| | A1 | |
| $T\sin60° + S\sin60° = 0.5 \times 3^2/0.4$ | M1 | Res. Horiz. $mr\omega^2$ ok if $\omega \neq 3$ |
| | A1 | If equal tensions $2T = 45/4$ M1 only |
| $(S+9.8)\sin60° + S\sin60° = 45/4$ | M1 | |
| $S = 1.60$ N | A1 | |
| $T = 11.4$ N | A1 | **7** |
## Part (ii)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $T\cos60° = 4.9$, $T = 9.8$ | M1, A1 | Resolving vertically (component) |
| $T\sin60° = 0.5 \times 0.4\omega^2$ | M1, A1 | Resolving horiz. (component) |
| $\omega = 6.51$ rad s$^{-1}$ | A1 | **5** | or 6.5 |
---6\\
\includegraphics[max width=\textwidth, alt={}, center]{6ae57fe9-3b6f-46c2-95b8-d48903ed796b-4_794_735_264_705}
A particle $P$ of mass 0.5 kg is attached to points $A$ and $B$ on a fixed vertical axis by two light inextensible strings of equal length. Both strings are taut and each is inclined at $60 ^ { \circ }$ to the vertical (see diagram). The particle moves with constant speed $3 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ in a horizontal circle of radius 0.4 m .\\
(i) Calculate the tensions in the two strings.
The particle now moves with constant angular speed $\omega$ rad s $^ { - 1 }$ and the string $B P$ is on the point of becoming slack.\\
(ii) Calculate $\omega$.
\hfill \mbox{\textit{OCR M2 2008 Q6 [12]}}