| Exam Board | OCR |
|---|---|
| Module | M4 (Mechanics 4) |
| Year | 2007 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Moments |
| Type | Rod or object resting on curved surface |
| Difficulty | Challenging +1.8 This M4 rotation question requires finding angular acceleration using moment of inertia about a non-center point, applying energy conservation for angular speed, resolving forces in a rotating reference frame, and deriving a slipping condition. The multi-step nature, use of parallel axis theorem, and coordination of rotational dynamics with friction makes this substantially harder than average, though the techniques are standard for Further Maths M4. |
| Spec | 1.10d Vector operations: addition and scalar multiplication1.10e Position vectors: and displacement1.10f Distance between points: using position vectors6.05f Vertical circle: motion including free fall |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Relative velocity on bearing 050 | M1 A1 | Correct velocity diagram; or \(\binom{u\sin 50}{u\cos 50} = \binom{16\sin\alpha}{16\cos\alpha} - \binom{12\sin 345}{12\cos 345}\) |
| \(\frac{\sin\theta}{12} = \frac{\sin 115}{16}\), \(\theta = 42.8°\) | M1 | or eliminating \(u\) (or \(\alpha\)) |
| Bearing of \(\mathbf{v}_B\) is \(007.2°\) | A1 | |
| \(\frac{u}{\sin 22.2} = \frac{16}{\sin 115}\), \(u = 6.66\) | M1 A1 | or obtaining equation for \(u\) (or \(\alpha\)) |
| Time taken is \(\frac{2400}{6.664} = 360\) s | M1*A1 ft | For equations in \(\alpha\) and \(t\): M1*M1A1 for equations; M1 for eliminating \(t\) (or \(\alpha\)); A1 for \(\alpha=7.2\); M1A1 ft for equation for \(t\); A1 cao for \(t=360\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Relative velocity perpendicular to \(\mathbf{v}_B\) | M1 A1 | Correct velocity diagram |
| \(\cos\phi = \frac{10}{12}\), \(\phi = 33.6°\) | M1 | For alternative methods: M2 for completely correct method |
| Bearing of \(\mathbf{v}_B\) is \(018.6°\) | A1 | A2 for 018.6 (give A1 for a correct relevant angle) |
# Question 5:
## Part (i):
| Answer | Mark | Guidance |
|--------|------|----------|
| Relative velocity on bearing 050 | M1 A1 | Correct velocity diagram; or $\binom{u\sin 50}{u\cos 50} = \binom{16\sin\alpha}{16\cos\alpha} - \binom{12\sin 345}{12\cos 345}$ |
| $\frac{\sin\theta}{12} = \frac{\sin 115}{16}$, $\theta = 42.8°$ | M1 | or eliminating $u$ (or $\alpha$) |
| Bearing of $\mathbf{v}_B$ is $007.2°$ | A1 | |
| $\frac{u}{\sin 22.2} = \frac{16}{\sin 115}$, $u = 6.66$ | M1 A1 | or obtaining equation for $u$ (or $\alpha$) |
| Time taken is $\frac{2400}{6.664} = 360$ s | M1*A1 ft | For equations in $\alpha$ and $t$: M1*M1A1 for equations; M1 for eliminating $t$ (or $\alpha$); A1 for $\alpha=7.2$; M1A1 ft for equation for $t$; A1 cao for $t=360$ |
## Part (ii):
| Answer | Mark | Guidance |
|--------|------|----------|
| Relative velocity perpendicular to $\mathbf{v}_B$ | M1 A1 | Correct velocity diagram |
| $\cos\phi = \frac{10}{12}$, $\phi = 33.6°$ | M1 | For alternative methods: M2 for completely correct method |
| Bearing of $\mathbf{v}_B$ is $018.6°$ | A1 | A2 for 018.6 (give A1 for a correct relevant angle) |
---5 A ship $S$ is travelling with constant speed $12 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ on a course with bearing $345 ^ { \circ }$. A patrol boat $B$ spots the ship $S$ when $S$ is 2400 m from $B$ on a bearing of $050 ^ { \circ }$. The boat $B$ sets off in pursuit, travelling with constant speed $v \mathrm {~m} \mathrm {~s} ^ { - 1 }$ in a straight line.\\
(i) Given that $v = 16$, find the bearing of the course which $B$ should take in order to intercept $S$, and the time taken to make the interception.\\
(ii) Given instead that $v = 10$, find the bearing of the course which $B$ should take in order to get as close as possible to $S$.\\
\includegraphics[max width=\textwidth, alt={}, center]{181fad74-6e60-4435-a176-3edff5062c32-4_337_954_278_544}
A uniform rod $A B$ has mass $m$ and length $2 a$. The point $P$ on the rod is such that $A P = \frac { 2 } { 3 } a$. The rod is placed in a horizontal position perpendicular to the edge of a rough horizontal table, with $A P$ in contact with the table and $P B$ overhanging the edge. The rod is released from rest in this position. When it has rotated through an angle $\theta$, and no slipping has occurred at $P$, the normal reaction acting on the rod at $P$ is $R$ and the frictional force is $F$ (see diagram).\\
(i) Show that the angular acceleration of the rod is $\frac { 3 g \cos \theta } { 4 a }$.\\
(ii) Find the angular speed of the rod, in terms of $a , g$ and $\theta$.\\
(iii) Find $F$ and $R$ in terms of $m , g$ and $\theta$.\\
(iv) Given that the coefficient of friction between the rod and the edge of the table is $\mu$, show that the rod is on the point of slipping at $P$ when $\tan \theta = \frac { 1 } { 2 } \mu$.\\
\includegraphics[max width=\textwidth, alt={}, center]{181fad74-6e60-4435-a176-3edff5062c32-5_677_624_269_753}
A smooth circular wire, with centre $O$ and radius $a$, is fixed in a vertical plane. The highest point on the wire is $A$ and the lowest point on the wire is $B$. A small ring $R$ of mass $m$ moves freely along the wire. A light elastic string, with natural length $a$ and modulus of elasticity $\frac { 1 } { 2 } m g$, has one end attached to $A$ and the other end attached to $R$. The string $A R$ makes an angle $\theta$ (measured anticlockwise) with the downward vertical, so that $O R$ makes an angle $2 \theta$ with the downward vertical (see diagram). You may assume that the string does not become slack.\\
(i) Taking $A$ as the level for zero gravitational potential energy, show that the total potential energy $V$ of the system is given by
$$V = m g a \left( \frac { 1 } { 4 } - \cos \theta - \cos ^ { 2 } \theta \right) .$$
(ii) Show that $\theta = 0$ is the only position of equilibrium.\\
(iii) By differentiating the energy equation with respect to time $t$, show that
$$\frac { \mathrm { d } ^ { 2 } \theta } { \mathrm {~d} t ^ { 2 } } = - \frac { g } { 4 a } \sin \theta ( 1 + 2 \cos \theta ) .$$
(iv) Deduce the approximate period of small oscillations about the equilibrium position $\theta = 0$.
\hfill \mbox{\textit{OCR M4 2007 Q5 [12]}}