| Exam Board | OCR |
|---|---|
| Module | M3 (Mechanics 3) |
| Session | Specimen |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Simple Harmonic Motion |
| Type | Small oscillations: rigid body compound pendulum |
| Difficulty | Standard +0.8 This is a multi-part M3 question combining circular motion with energy conservation and tension analysis. Part (i) requires standard radial/transverse acceleration components. Parts (ii)-(iv) involve energy conservation, resolving forces in circular motion, and finding when the string becomes slack—all requiring careful application of multiple mechanics principles across several steps. This is more demanding than typical A-level questions but uses standard M3 techniques without requiring novel insight. |
| Spec | 3.04a Calculate moments: about a point3.04b Equilibrium: zero resultant moment and force6.02d Mechanical energy: KE and PE concepts6.02e Calculate KE and PE: using formulae6.04e Rigid body equilibrium: coplanar forces6.05d Variable speed circles: energy methods6.05e Radial/tangential acceleration |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| (i) Radial acc is \(\frac{6.5^2}{1.5} = 28.2 \text{ m s}^{-2}\) | B1 | For correct value \(28.2\) |
| Transverse acc is \(g\sin40° = 6.30 \text{ m s}^{-2}\) | B1 [2] | For correct value \(6.30\) |
| (ii) \(\frac{1}{2}\times0.3\times(6.5^2-v^2) = 0.3\times9.8\times1.5(\cos40°-\cos\theta°)\) | M1 | For equating PE gain to KE loss, or equiv |
| B1 | For correct expression for PE gain | |
| B1 | For correct expression for KE loss | |
| Hence \(42.25 - v^2 = 29.4(\cos40° - \cos\theta°)\) | ||
| i.e. \(v^2 \approx 19.7 + 29.4\cos\theta°\) | A1 [4] | For showing given answer correctly |
| (iii) \(T - 0.3g\cos\theta° = 0.3\times\frac{v^2}{1.5}\) | M1 | For use of Newton II, including \(\frac{mv^2}{r}\) term |
| A1 | For correct (unsimplified) equation | |
| Hence \(T = 2.94\cos\theta° + 0.2(19.7 + 29.4\cos\theta°)\) | M1 | For substitution, to obtain expression for \(T\) |
| \(= 3.95 + 8.82\cos\theta°\) | A1 [4] | For correct answer |
| (iv) \(T=0\) when \(3.95 + 8.82\cos\theta° = 0\) | M1 | For equating \(T\) to zero to find \(\cos\theta\) |
| Hence \(v^2 = 19.7 + 29.4\times\left(-\frac{3.95}{8.82}\right) \Rightarrow v \approx 2.56\) | M1 | For using this \(\cos\theta\) to find \(v\) |
| A1 [3] | For correct answer \(2.56\) |
# Question 5:
| Answer/Working | Marks | Guidance |
|---|---|---|
| **(i)** Radial acc is $\frac{6.5^2}{1.5} = 28.2 \text{ m s}^{-2}$ | B1 | For correct value $28.2$ |
| Transverse acc is $g\sin40° = 6.30 \text{ m s}^{-2}$ | B1 **[2]** | For correct value $6.30$ |
| **(ii)** $\frac{1}{2}\times0.3\times(6.5^2-v^2) = 0.3\times9.8\times1.5(\cos40°-\cos\theta°)$ | M1 | For equating PE gain to KE loss, or equiv |
| | B1 | For correct expression for PE gain |
| | B1 | For correct expression for KE loss |
| Hence $42.25 - v^2 = 29.4(\cos40° - \cos\theta°)$ | | |
| i.e. $v^2 \approx 19.7 + 29.4\cos\theta°$ | A1 **[4]** | For showing given answer correctly |
| **(iii)** $T - 0.3g\cos\theta° = 0.3\times\frac{v^2}{1.5}$ | M1 | For use of Newton II, including $\frac{mv^2}{r}$ term |
| | A1 | For correct (unsimplified) equation |
| Hence $T = 2.94\cos\theta° + 0.2(19.7 + 29.4\cos\theta°)$ | M1 | For substitution, to obtain expression for $T$ |
| $= 3.95 + 8.82\cos\theta°$ | A1 **[4]** | For correct answer |
| **(iv)** $T=0$ when $3.95 + 8.82\cos\theta° = 0$ | M1 | For equating $T$ to zero to find $\cos\theta$ |
| Hence $v^2 = 19.7 + 29.4\times\left(-\frac{3.95}{8.82}\right) \Rightarrow v \approx 2.56$ | M1 | For using this $\cos\theta$ to find $v$ |
| | A1 **[3]** | For correct answer $2.56$ |
---5\\
\includegraphics[max width=\textwidth, alt={}, center]{bfa6d51d-0992-4f43-adab-77ce893c1ca9-3_576_535_258_804}
A particle $P$ of mass 0.3 kg is moving in a vertical circle. It is attached to the fixed point $O$ at the centre of the circle by a light inextensible string of length 1.5 m . When the string makes an angle of $40 ^ { \circ }$ with the downward vertical, the speed of $P$ is $6.5 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ (see diagram). Air resistance may be neglected.\\
(i) Find the radial and transverse components of the acceleration of $P$ at this instant.
In the subsequent motion, with the string still taut and making an angle $\theta ^ { \circ }$ with the downward vertical, the speed of $P$ is $v \mathrm {~m} \mathrm {~s} ^ { - 1 }$\\
(ii) Use conservation of energy to show that $v ^ { 2 } \approx 19.7 + 29.4 \cos \theta ^ { \circ }$.\\
(iii) Find the tension in the string in terms of $\theta$.\\
(iv) Find the value of $v$ at the instant when the string becomes slack.\\
\includegraphics[max width=\textwidth, alt={}, center]{bfa6d51d-0992-4f43-adab-77ce893c1ca9-3_574_842_1640_664}
A step-ladder is modelled as two uniform rods $A B$ and $A C$, freely jointed at $A$. The rods are in equilibrium in a vertical plane with $B$ and $C$ in contact with a rough horizontal surface. The rods have equal lengths; $A B$ has weight 150 N and $A C$ has weight 270 N . The point $A$ is 2.5 m vertically above the surface, and $B C = 1.6 \mathrm {~m}$ (see diagram).\\
(i) Find the horizontal and vertical components of the force acting on $A C$ at $A$.\\
(ii) The coefficient of friction has the same value $\mu$ at $B$ and at $C$, and the step-ladder is on the point of slipping. Giving a reason, state whether the equilibrium is limiting at $B$ or at $C$, and find $\mu$.\\
\includegraphics[max width=\textwidth, alt={}, center]{bfa6d51d-0992-4f43-adab-77ce893c1ca9-4_648_227_269_982}
Two points $A$ and $B$ lie on a vertical line with $A$ at a distance 2.6 m above $B$. A particle $P$ of mass 10 kg is joined to $A$ by an elastic string and to $B$ by another elastic string (see diagram). Each string has natural length 0.8 m and modulus of elasticity 196 N . The strings are light and air resistance may be neglected.\\
(i) Verify that $P$ is in equilibrium when $P$ is vertically below $A$ and the length of the string $P A$ is 1.5 m .
The particle is set in motion along the line $A B$ with both strings remaining taut. The displacement of $P$ below the equilibrium position is denoted by $x$ metres.\\
(ii) Show that the tension in the string $P A$ is $245 ( 0.7 + x )$ newtons, and the tension in the string $P B$ is $245 ( 0.3 - x )$ newtons.\\
(iii) Show that the motion of $P$ is simple harmonic.\\
(iv) Given that the amplitude of the motion is 0.25 m , find the proportion of time for which $P$ is above the mid-point of $A B$.
\hfill \mbox{\textit{OCR M3 Q5 [13]}}