| Exam Board | OCR |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2011 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Hooke's law and elastic energy |
| Type | Particle in circular motion with string/rod |
| Difficulty | Challenging +1.8 This M3 question requires energy methods with elastic strings on a curved surface, resolving forces in polar coordinates to find transverse acceleration, and applying conservation principles. The multi-step nature, combination of gravitational and elastic PE changes, and the non-trivial verification that transverse acceleration is zero at a specific angle make this significantly harder than average, though the calculations follow standard M3 techniques once the approach is identified. |
| Spec | 6.02h Elastic PE: 1/2 k x^26.02i Conservation of energy: mechanical energy principle6.05e Radial/tangential acceleration |
| Answer | Marks | Guidance |
|---|---|---|
| \(PE \text{ gain } = 6x0.8(\sqrt{3}/2 - 1/\sqrt{2})\) \(= 2.4(\sqrt{3} - \sqrt{2})\) | M1, A1, M1 | For using \(PE \text{ gain } = W(h_B - h_X)\) Shown fully, with no slips. For using \(EE \text{ loss } = \lambda(e^2_2 - e^2_1)/2l\). Allow slips for M1. |
| \(EE \text{ loss } = \frac{9}{2(\pi/10)}[(0.8\pi/4 - \pi/10)^2 - (0.8\pi/6 - \pi/10)^2]\) | A1 | |
| \(EE \text{ loss } = 45\pi[(0.2 - 0.1)^2 - (0.4 - 0.3)^2 \div 9]\) \(= 5\pi(9x0.01 - 0.01) = 40\pi/100 = 0.4\pi J\) | A1, A1 [5] | Fully correct. No slips in simplification. |
| Answer | Marks | Guidance |
|---|---|---|
| \(T = 9(0.8\pi/6 - \pi/10) \cdot (\pi/10)\) | B1 | For attempting to show that \(W\sin\theta - T = 0\) at \(Y\) by subst \(\theta = \pi/6\) |
| \(W\sin\theta - T = 6 \times \sin(\pi/6) - 90 \times (0.2 - 6) = 0\) \(\Rightarrow\) transverse acceleration is zero | A1, M1 | No slips. For using \(KE \text{ gain } = EE \text{ loss } - PE \text{ gain}\) at \(Y\). Need 3 terms, allow sign errors and/or \(g\) omitted. |
| \(\frac{1}{2}(6/9.8)v^2 = 0.4\pi - 2.4(\sqrt{3} - \sqrt{2})\) \(\text{Maximum speed is } 1.27 \text{ ms}^{-1}\) | A1, A1 [6] |
**Part (i):**
$PE \text{ gain } = 6x0.8(\sqrt{3}/2 - 1/\sqrt{2})$ $= 2.4(\sqrt{3} - \sqrt{2})$ | M1, A1, M1 | For using $PE \text{ gain } = W(h_B - h_X)$ Shown fully, with no slips. For using $EE \text{ loss } = \lambda(e^2_2 - e^2_1)/2l$. Allow slips for M1. |
$EE \text{ loss } = \frac{9}{2(\pi/10)}[(0.8\pi/4 - \pi/10)^2 - (0.8\pi/6 - \pi/10)^2]$ | A1 |
$EE \text{ loss } = 45\pi[(0.2 - 0.1)^2 - (0.4 - 0.3)^2 \div 9]$ $= 5\pi(9x0.01 - 0.01) = 40\pi/100 = 0.4\pi J$ | A1, A1 [5] | Fully correct. No slips in simplification. |
**Part (ii):**
$T = 9(0.8\pi/6 - \pi/10) \cdot (\pi/10)$ | B1 | For attempting to show that $W\sin\theta - T = 0$ at $Y$ by subst $\theta = \pi/6$ |
$W\sin\theta - T = 6 \times \sin(\pi/6) - 90 \times (0.2 - 6) = 0$ $\Rightarrow$ transverse acceleration is zero | A1, M1 | No slips. For using $KE \text{ gain } = EE \text{ loss } - PE \text{ gain}$ at $Y$. Need 3 terms, allow sign errors and/or $g$ omitted. |
$\frac{1}{2}(6/9.8)v^2 = 0.4\pi - 2.4(\sqrt{3} - \sqrt{2})$ $\text{Maximum speed is } 1.27 \text{ ms}^{-1}$ | A1, A1 [6] |\includegraphics{figure_6}
A particle $P$ of weight $6$ N is attached to the highest point $A$ of a fixed smooth sphere by a light elastic string. The sphere has centre $O$ and radius $0.8$ m. The string has natural length $\frac{1}{10}\pi$ m and modulus of elasticity $9$ N. $P$ is released from rest at a point $X$ on the sphere where $OX$ makes an angle of $\frac{1}{3}\pi$ radians with the upwards vertical. $P$ remains in contact with the sphere as it moves upwards to $A$. At time $t$ seconds after the release, $OP$ makes an angle of $\theta$ radians with the upwards vertical (see diagram). When $\theta = \frac{1}{4}\pi$, $P$ passes through the point $Y$.
\begin{enumerate}[label=(\roman*)]
\item Show that as $P$ moves from $X$ to $Y$ its gravitational potential energy increases by $2.4(\sqrt{3} - \sqrt{2})$ J and the elastic potential energy in the string decreases by $0.4\pi$ J. [5]
\item Verify that the transverse acceleration of $P$ is zero when $\theta = \frac{1}{4}\pi$, and hence find the maximum speed of $P$. [6]
\end{enumerate}
\hfill \mbox{\textit{OCR M3 2011 Q6 [11]}}