| Exam Board | OCR MEI |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2012 |
| Session | January |
| Marks | 18 |
| 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 M3 circular motion question covering routine applications of Newton's second law in circular motion and energy conservation. Parts (i)-(iii) are textbook exercises requiring direct formula application (T = mv²/r ± mg), while part (iv) involves resolving forces with two strings in a conical pendulum setup—slightly more complex but still a standard M3 problem type with clear geometric setup provided. |
| Spec | 6.05c Horizontal circles: conical pendulum, banked tracks6.05d Variable speed circles: energy methods6.05e Radial/tangential acceleration |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\frac{1}{2}m(15^2-10^2) = m\times9.8\times(5-5\cos\theta)\) | M1 | Equation involving KE and PE; \(\theta\) is angle with upward vertical |
| \(\cos\theta = -\frac{27}{98}\quad [\theta=106°]\) | A1 | |
| \(T+(0.72)(9.8)\cos\theta = 0.72\times\frac{15^2}{5}\) | M1 | Using acceleration \(v^2/r\) |
| Tension is 34.3 N (3 sf) | A1, A1 | If evaluated, FT their \(\cos\theta\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\cos\alpha = 0.96\quad [\alpha=16.26°]\) | B1 | \(\alpha\) is angle between AP and vertical |
| \(T\cos\alpha = 0.72\times9.8\) | M1 | Resolve vertically |
| Tension is 7.35 N | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(T\sin\alpha = (0.72)(1.4\omega^2)\) | M1, A1 | Horizontal equation of motion; Or \(T\sin\alpha=(0.72)(v^2/1.4)\), \((v=2.00)\) |
| \(\omega = 1.429\) | ||
| Period \(\frac{2\pi}{\omega}\) with numerical \(\omega\) | M1 | Or \(\frac{2\pi\times1.4}{v}\) with numerical \(v\) |
| Time for one revolution is \(\frac{2\pi}{\omega} = 4.40\) s (3 sf) | A1 | FT is \(\frac{11.92}{\sqrt{T}}\) (only); Accept \(1.4\pi\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(T_{PA}\cos\alpha = T_{PB}\cos\alpha + 0.72\times9.8\) | M1, A1 | Resolving vertically, 3 terms required |
| \(T_{PA} - T_{PB} = 7.35\) | ||
| \(T_{PA}\sin\alpha + T_{PB}\sin\alpha = 0.72\times\frac{7^2}{1.4}\) | M1, A1 | Horizontal equation of motion, 3 terms required |
| \(T_{PA} + T_{PB} = 90\) | ||
| Tension in PA is 48.7 N (3 sf) | M1 | Obtaining tension in at least one string; Dependent on previous M1M1 |
| Tension in PB is 41.3 N (3 sf) | A1 | CAO |
# Question 2:
## Part (i)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{1}{2}m(15^2-10^2) = m\times9.8\times(5-5\cos\theta)$ | M1 | Equation involving KE and PE; $\theta$ is angle with upward vertical |
| $\cos\theta = -\frac{27}{98}\quad [\theta=106°]$ | A1 | |
| $T+(0.72)(9.8)\cos\theta = 0.72\times\frac{15^2}{5}$ | M1 | Using acceleration $v^2/r$ |
| Tension is 34.3 N (3 sf) | A1, A1 | If evaluated, FT their $\cos\theta$ |
## Part (ii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\cos\alpha = 0.96\quad [\alpha=16.26°]$ | B1 | $\alpha$ is angle between AP and vertical |
| $T\cos\alpha = 0.72\times9.8$ | M1 | Resolve vertically |
| Tension is 7.35 N | A1 | |
## Part (iii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $T\sin\alpha = (0.72)(1.4\omega^2)$ | M1, A1 | Horizontal equation of motion; Or $T\sin\alpha=(0.72)(v^2/1.4)$, $(v=2.00)$ |
| $\omega = 1.429$ | | |
| Period $\frac{2\pi}{\omega}$ with numerical $\omega$ | M1 | Or $\frac{2\pi\times1.4}{v}$ with numerical $v$ |
| Time for one revolution is $\frac{2\pi}{\omega} = 4.40$ s (3 sf) | A1 | FT is $\frac{11.92}{\sqrt{T}}$ (only); Accept $1.4\pi$ |
## Part (iv)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $T_{PA}\cos\alpha = T_{PB}\cos\alpha + 0.72\times9.8$ | M1, A1 | Resolving vertically, 3 terms required |
| $T_{PA} - T_{PB} = 7.35$ | | |
| $T_{PA}\sin\alpha + T_{PB}\sin\alpha = 0.72\times\frac{7^2}{1.4}$ | M1, A1 | Horizontal equation of motion, 3 terms required |
| $T_{PA} + T_{PB} = 90$ | | |
| Tension in PA is 48.7 N (3 sf) | M1 | Obtaining tension in at least one string; Dependent on previous M1M1 |
| Tension in PB is 41.3 N (3 sf) | A1 | CAO |
---2 A light inextensible string of length 5 m has one end attached to a fixed point A and the other end attached to a particle P of mass 0.72 kg .
At first, P is moving in a vertical circle with centre A and radius 5 m . When P is at the highest point of the circle it has speed $10 \mathrm {~ms} ^ { - 1 }$.\\
(i) Find the tension in the string when the speed of P is $15 \mathrm {~ms} ^ { - 1 }$.
The particle P now moves at constant speed in a horizontal circle with radius 1.4 m and centre at the point C which is 4.8 m vertically below A .\\
(ii) Find the tension in the string.\\
(iii) Find the time taken for P to make one complete revolution.
Another light inextensible string, also of length 5 m , now has one end attached to P and the other end attached to the fixed point B which is 9.6 m vertically below A . The particle P then moves with constant speed $7 \mathrm {~ms} ^ { - 1 }$ in the circle with centre C and radius 1.4 m , as shown in Fig. 2.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{86d79489-aec1-4c94-bef6-45b007f818a0-3_693_465_1078_817}
\captionsetup{labelformat=empty}
\caption{Fig. 2}
\end{center}
\end{figure}
(iv) Find the tension in the string PA and the tension in the string PB .
\hfill \mbox{\textit{OCR MEI M3 2012 Q2 [18]}}