| Exam Board | OCR MEI |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2013 |
| Session | January |
| Marks | 18 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circular Motion 2 |
| Type | Particle on outer surface of sphere |
| Difficulty | Standard +0.3 This is a standard M3/Further Mechanics circular motion question with routine application of energy conservation and Newton's second law in radial/tangential components. Part (a) requires standard energy methods and resolving forces at leaving point; part (b) is a straightforward conical pendulum with two strings requiring basic trigonometry and force resolution. All techniques are textbook exercises with no novel insight required, making it slightly easier than average. |
| Spec | 3.03d Newton's second law: 2D vectors6.02i Conservation of energy: mechanical energy principle6.02j Conservation with elastics: springs and strings6.05a Angular velocity: definitions6.05c Horizontal circles: conical pendulum, banked tracks6.05e Radial/tangential acceleration |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{1}{2}mv^2 - \frac{1}{2}m(1.2)^2 = mg(0.8 - 0.8\cos\frac{1}{6}\pi)\) | M1, A1 | Equation involving initial KE, final KE and attempt at PE |
| \(v^2 = 3.5407\) | A1 | |
| Radial component \(\left(\frac{v^2}{0.8}\right)\) is 4.43 ms\(^{-2}\) (3 sf) | A1 | |
| \((\pm)mg\sin\frac{1}{6}\pi = ma_T\) | M1 | Allow M1 for \(\cos\frac{1}{6}\pi\) used instead of \(\sin\frac{1}{6}\pi\); but M0 for \(a_T = mg\sin\frac{1}{6}\pi\) |
| Tangential component is 4.9 ms\(^{-2}\) | A1 | Allow \(\frac{1}{2}g\) |
| [5] |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{1}{2}mv^2 - \frac{1}{2}m(1.2)^2 = mg(0.8 - 0.8\cos\theta)\) | M1 | Equation involving initial KE, final KE and attempt at PE in general position. Between OP and upward vertical. Allow \(mgh\) for PE if \(h\) is linked to \(\theta\) in later work |
| M1 | Equation involving resolved component of weight and \(v^2/r\) | |
| \(mg\cos\theta - R = \frac{mv^2}{0.8}\) | A1 | \(R\) may be omitted |
| Leaves surface when \(R = 0\) | M1 | May be implied. E.g. Implied by \(mg\cos\theta = \frac{mv^2}{0.8}\) |
| \(v^2 - 1.44 = 2 \times 9.8 \times 0.8\left(1 - \frac{v^2}{7.84}\right)\) | M1 | Obtaining equation in \(v\) or \(\theta\). Dependent on previous M1M1 |
| Speed is 2.39 ms\(^{-1}\) | A1 | |
| [6] |
| Answer | Marks | Guidance |
|---|---|---|
| \(T_R \sin\alpha + T_S \sin\beta = mg\) | M1 | Resolving vertically (three terms). \(\alpha = RQC = 53.1°, \beta = SQC = 16.3°\) |
| \(0.8T_R + 0.287T_S = 0.9 \times 9.8 \, (= 8.82)\) | A1 | Allow \(\sin 53.1°\), etc |
| \(T_R \cos\alpha + T_S \cos\beta = m\frac{v^2}{r}\) | M1 | Horizontal equation of motion. Three terms, and \(v^2/r\) |
| \(0.6T_R + 0.96T_S = 0.9 \times \frac{5^2}{2.4} \, (= 9.375)\) | A1 | |
| Tension in string RQ is 9.737 N | A1 | |
| Tension in string SQ is 3.68 N | A1 | |
| [7] | Dependent on previous M1M1 |
### Part (a)(i)
| $\frac{1}{2}mv^2 - \frac{1}{2}m(1.2)^2 = mg(0.8 - 0.8\cos\frac{1}{6}\pi)$ | M1, A1 | Equation involving initial KE, final KE and attempt at PE |
|---|---|---|
| $v^2 = 3.5407$ | A1 | |
|---|---|---|
| Radial component $\left(\frac{v^2}{0.8}\right)$ is 4.43 ms$^{-2}$ (3 sf) | A1 | |
|---|---|---|
| $(\pm)mg\sin\frac{1}{6}\pi = ma_T$ | M1 | Allow M1 for $\cos\frac{1}{6}\pi$ used instead of $\sin\frac{1}{6}\pi$; but M0 for $a_T = mg\sin\frac{1}{6}\pi$ |
|---|---|---|
| Tangential component is 4.9 ms$^{-2}$ | A1 | Allow $\frac{1}{2}g$ |
|---|---|---|
| | [5] | |
|---|---|---|
### Part (a)(ii)
| $\frac{1}{2}mv^2 - \frac{1}{2}m(1.2)^2 = mg(0.8 - 0.8\cos\theta)$ | M1 | Equation involving initial KE, final KE and attempt at PE in general position. Between OP and upward vertical. Allow $mgh$ for PE if $h$ is linked to $\theta$ in later work |
|---|---|---|
| | M1 | Equation involving resolved component of weight and $v^2/r$ |
|---|---|---|
| $mg\cos\theta - R = \frac{mv^2}{0.8}$ | A1 | $R$ may be omitted |
|---|---|---|
| Leaves surface when $R = 0$ | M1 | May be implied. E.g. Implied by $mg\cos\theta = \frac{mv^2}{0.8}$ |
|---|---|---|
| $v^2 - 1.44 = 2 \times 9.8 \times 0.8\left(1 - \frac{v^2}{7.84}\right)$ | M1 | Obtaining equation in $v$ or $\theta$. Dependent on previous M1M1 |
|---|---|---|
| Speed is 2.39 ms$^{-1}$ | A1 | |
|---|---|---|
| | [6] | |
|---|---|---|
### Part (b)
| $T_R \sin\alpha + T_S \sin\beta = mg$ | M1 | Resolving vertically (three terms). $\alpha = RQC = 53.1°, \beta = SQC = 16.3°$ |
|---|---|---|
| $0.8T_R + 0.287T_S = 0.9 \times 9.8 \, (= 8.82)$ | A1 | Allow $\sin 53.1°$, etc |
| $T_R \cos\alpha + T_S \cos\beta = m\frac{v^2}{r}$ | M1 | Horizontal equation of motion. Three terms, and $v^2/r$ |
|---|---|---|
| $0.6T_R + 0.96T_S = 0.9 \times \frac{5^2}{2.4} \, (= 9.375)$ | A1 | |
| Tension in string RQ is 9.737 N | A1 | |
|---|---|---|
| Tension in string SQ is 3.68 N | A1 | |
| | [7] | Dependent on previous M1M1 |
|---|---|---|2
\begin{enumerate}[label=(\alph*)]
\item A fixed solid sphere with a smooth surface has centre O and radius 0.8 m . A particle P is given a horizontal velocity of $1.2 \mathrm {~ms} ^ { - 1 }$ at the highest point on the sphere, and it moves on the surface of the sphere in part of a vertical circle of radius 0.8 m .
\begin{enumerate}[label=(\roman*)]
\item Find the radial and tangential components of the acceleration of P at the instant when OP makes an angle $\frac { 1 } { 6 } \pi$ radians with the upward vertical. (You may assume that P is still in contact with the sphere.)
\item Find the speed of P at the instant when it leaves the surface of the sphere.
\end{enumerate}\item Two fixed points R and S are 2.5 m apart with S vertically below R . A particle Q of mass 0.9 kg is connected to R and to S by two light inextensible strings; Q is moving in a horizontal circle at a constant speed of $5 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ with both strings taut. The radius of the circle is 2.4 m and the centre C of the circle is 0.7 m vertically below S, as shown in Fig. 2.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{3f674569-7e99-4ba8-84f1-a1eb438e30ed-2_547_720_1946_644}
\captionsetup{labelformat=empty}
\caption{Fig. 2}
\end{center}
\end{figure}
Find the tension in the string RQ and the tension in the string $S Q$.
\end{enumerate}
\hfill \mbox{\textit{OCR MEI M3 2013 Q2 [18]}}