| Exam Board | OCR MEI |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2010 |
| Session | January |
| Marks | 18 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circular Motion 2 |
| Type | Vertical circle: string becomes slack |
| Difficulty | Standard +0.3 This is a standard M3 vertical circle problem with predictable parts: energy conservation (shown result), tension formula, and finding slack condition. Part (iv) adds a straightforward horizontal circle with two strings requiring force resolution. All techniques are routine for this module with no novel insights required, making it slightly easier than average. |
| Spec | 6.02d Mechanical energy: KE and PE concepts6.05c Horizontal circles: conical pendulum, banked tracks6.05d Variable speed circles: energy methods6.05e Radial/tangential acceleration |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| By conservation of energy: \(\frac{1}{2}\times0.6\times6^2 - \frac{1}{2}\times0.6v^2 = 0.6\times9.8(1.25 - 1.25\cos\theta)\) | M1, A1 | Equation involving KE and PE |
| \(36 - v^2 = 24.5 - 24.5\cos\theta\) | ||
| \(v^2 = 11.5 + 24.5\cos\theta\) | E1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(T - 0.6\times9.8\cos\theta = 0.6\times\dfrac{v^2}{1.25}\) | M1, A1 | For acceleration \(\dfrac{v^2}{r}\) |
| \(T - 5.88\cos\theta = 0.48(11.5 + 24.5\cos\theta)\) | M1 | Substituting for \(v^2\) |
| \(T = 5.52 + 17.64\cos\theta\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| String becomes slack when \(T = 0\) | M1 | |
| \(\cos\theta = -\dfrac{5.52}{17.64}\) \((\theta = 108.2°\) or \(1.889\) rad\()\) | A1 | May be implied |
| \(v^2 = 11.5 - 24.5\times\dfrac{5.52}{17.64}\) | M1 | or \(0.6\times9.8\times\dfrac{5.52}{17.64} = 0.6\times\dfrac{v^2}{1.25}\); or \(-0.6\times9.8\times\dfrac{v^2-11.5}{24.5} = 0.6\times\dfrac{v^2}{1.25}\) |
| Speed is \(1.96 \text{ ms}^{-1}\) (3 sf) | A1 cao |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(T_1\cos\theta = mg\); \(T_1\times\dfrac{1.2}{1.25} = 0.6\times9.8\) (where \(\theta\) is angle COP) | M1, A1 | Resolving vertically |
| Tension in OP is \(6.125\) N | A1 | |
| \(T_1\sin\theta + T_2 = \dfrac{mv^2}{0.35}\) | M1 | Horizontal equation (three terms) |
| \(6.125\times\dfrac{0.35}{1.25} + T_2 = \dfrac{0.6\times1.4^2}{0.35}\) | F1B1 | For LHS and RHS |
| Tension in CP is \(1.645\) N | A1 |
# Question 3:
## Part (i)
| Answer/Working | Mark | Guidance |
|---|---|---|
| By conservation of energy: $\frac{1}{2}\times0.6\times6^2 - \frac{1}{2}\times0.6v^2 = 0.6\times9.8(1.25 - 1.25\cos\theta)$ | M1, A1 | Equation involving KE and PE |
| $36 - v^2 = 24.5 - 24.5\cos\theta$ | | |
| $v^2 = 11.5 + 24.5\cos\theta$ | E1 | |
## Part (ii)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $T - 0.6\times9.8\cos\theta = 0.6\times\dfrac{v^2}{1.25}$ | M1, A1 | For acceleration $\dfrac{v^2}{r}$ |
| $T - 5.88\cos\theta = 0.48(11.5 + 24.5\cos\theta)$ | M1 | Substituting for $v^2$ |
| $T = 5.52 + 17.64\cos\theta$ | A1 | |
## Part (iii)
| Answer/Working | Mark | Guidance |
|---|---|---|
| String becomes slack when $T = 0$ | M1 | |
| $\cos\theta = -\dfrac{5.52}{17.64}$ $(\theta = 108.2°$ or $1.889$ rad$)$ | A1 | May be implied |
| $v^2 = 11.5 - 24.5\times\dfrac{5.52}{17.64}$ | M1 | or $0.6\times9.8\times\dfrac{5.52}{17.64} = 0.6\times\dfrac{v^2}{1.25}$; or $-0.6\times9.8\times\dfrac{v^2-11.5}{24.5} = 0.6\times\dfrac{v^2}{1.25}$ |
| Speed is $1.96 \text{ ms}^{-1}$ (3 sf) | A1 cao | |
## Part (iv)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $T_1\cos\theta = mg$; $T_1\times\dfrac{1.2}{1.25} = 0.6\times9.8$ (where $\theta$ is angle COP) | M1, A1 | Resolving vertically |
| Tension in OP is $6.125$ N | A1 | |
| $T_1\sin\theta + T_2 = \dfrac{mv^2}{0.35}$ | M1 | Horizontal equation (three terms) |
| $6.125\times\dfrac{0.35}{1.25} + T_2 = \dfrac{0.6\times1.4^2}{0.35}$ | F1B1 | For LHS and RHS |
| Tension in CP is $1.645$ N | A1 | |
---3 A particle P of mass 0.6 kg is connected to a fixed point O by a light inextensible string of length 1.25 m . When it is 1.25 m vertically below $\mathrm { O } , \mathrm { P }$ is set in motion with horizontal velocity $6 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ and then moves in part of a vertical circle with centre O and radius 1.25 m . When OP makes an angle $\theta$ with the downward vertical, the speed of P is $v \mathrm {~m} \mathrm {~s} ^ { - 1 }$, as shown in Fig. 3.1.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{023afdfb-21b6-40fe-9a09-e6769667ee7b-3_602_627_484_758}
\captionsetup{labelformat=empty}
\caption{Fig. 3.1}
\end{center}
\end{figure}
(i) Show that $v ^ { 2 } = 11.5 + 24.5 \cos \theta$.\\
(ii) Find the tension in the string in terms of $\theta$.\\
(iii) Find the speed of P at the instant when the string becomes slack.
A second light inextensible string, of length 0.35 m , is attached to P , and the other end of this string is attached to a point C which is 1.2 m vertically below O . The particle P now moves in a horizontal circle with centre C and radius 0.35 m , as shown in Fig. 3.2. The speed of P is $1.4 \mathrm {~m} \mathrm {~s} ^ { - 1 }$.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{023afdfb-21b6-40fe-9a09-e6769667ee7b-3_518_488_1701_826}
\captionsetup{labelformat=empty}
\caption{Fig. 3.2}
\end{center}
\end{figure}
(iv) Find the tension in the string OP and the tension in the string CP.
\hfill \mbox{\textit{OCR MEI M3 2010 Q3 [18]}}