| Exam Board | OCR |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2011 |
| Session | June |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circular Motion 2 |
| Type | Vertical circle: string becomes slack |
| Difficulty | Standard +0.3 This is a standard M3 vertical circle problem with projectile motion after the string slackens. Part (i) uses energy conservation (routine for M3), part (ii) applies the slack condition T=0 (standard technique), and part (iii) requires projectile motion with energy methods. While multi-step, all techniques are textbook applications with no novel insight required, making it slightly easier than average. |
| Spec | 6.05d Variable speed circles: energy methods6.05e Radial/tangential acceleration |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{1}{2}mv^2 = \frac{1}{2}m5.6^2 - mg0.8(1 - \cos\theta)\) \(v^2 = 15.68(1 + \cos\theta)\) | M1, A1, A1, M1 | For using the principle of conservation of energy. Allow sign error, \(\sin/\cos\); need 3 terms. For using Newton's second law. Allow sign error and/or \(\sin/\cos\) and/or \(m\) omitted |
| \([T - 0.3g\cos\theta = 0.3 \times 15.68(1 + \cos\theta)/0.8]\) \(\text{Tension is } 2.94(3\cos\theta + 2) \text{ N oe}\) | A1, M1, A1 [7] | For substituting for \(v^2\) |
| Answer | Marks | Guidance |
|---|---|---|
| \(\theta \text{ is } 131.8° \text{ (or } 2.3 \text{ rads) Accept } 132° \text{ (exact)}\) \(v \text{ is } 2.29\) | M1, A1, B1 [3] | For putting \(T = 0\) and attempting to solve accept \(\theta = \cos^{-1}(-2/3)\) \(\sqrt{15.68/3}\) exact |
| Answer | Marks | Guidance |
|---|---|---|
| \([\text{speed} = | v\cos(180° - \theta) | =\) \(\sqrt{15.68/3 \times (2/3)}]\) \(\text{Speed at greatest height is } 1.52 \text{ ms}^{-1}\) |
| \(0.3g H = \frac{1}{2}0.3(5.6^2 - 1.52^2)\) \(\text{Greatest height is } 1.48 \text{ m}\) | M1, A1 |
| Answer | Marks | Guidance |
|---|---|---|
| \([0 = 2.286_... ^2 \times (1-4/9) -19.6y,\) \(H = 0.8(1 + 2/3) + y]\) \(H = 1.3333... + 0.1481... (4/3 + 4/27)\) \(\text{Greatest height is } 1.48 \text{ m } (40/27)\) \([\frac{1}{2}m(2.286...^2 - \text{speed}^2) = mg \times 0.1481...]\) speed\(^2 = 2.286_...^2 - 19.6 \times 0.1481...\) ] or \([\frac{1}{2}m(5.6^2 - \text{speed}^2) = mg \times 1.481...]\) speed\(^2 = 5.6^2 - 19.6 \times 1.481...]\) \(\text{Speed at greatest height is } 1.52 \text{ ms}^{-1}\) | M1, A1, M1, A1 | For using \(0^2 = \dot{y}^2 - 2gy\) and \(H = 0.8(1 + \cos(180° - \theta)) + y\). For using the principle of conservation of energy |
**Part (i):**
$\frac{1}{2}mv^2 = \frac{1}{2}m5.6^2 - mg0.8(1 - \cos\theta)$ $v^2 = 15.68(1 + \cos\theta)$ | M1, A1, A1, M1 | For using the principle of conservation of energy. Allow sign error, $\sin/\cos$; need 3 terms. For using Newton's second law. Allow sign error and/or $\sin/\cos$ and/or $m$ omitted |
$[T - 0.3g\cos\theta = 0.3 \times 15.68(1 + \cos\theta)/0.8]$ $\text{Tension is } 2.94(3\cos\theta + 2) \text{ N oe}$ | A1, M1, A1 [7] | For substituting for $v^2$ |
**Part (ii):**
$\theta \text{ is } 131.8° \text{ (or } 2.3 \text{ rads) Accept } 132° \text{ (exact)}$ $v \text{ is } 2.29$ | M1, A1, B1 [3] | For putting $T = 0$ and attempting to solve accept $\theta = \cos^{-1}(-2/3)$ $\sqrt{15.68/3}$ exact |
**Part (iii):**
$[\text{speed} = |v\cos(180° - \theta)| =$ $\sqrt{15.68/3 \times (2/3)}]$ $\text{Speed at greatest height is } 1.52 \text{ ms}^{-1}$ | M1, A1 [4] | For using 'speed at max. height = horiz. comp. of vel. when string becomes slack' |
$0.3g H = \frac{1}{2}0.3(5.6^2 - 1.52^2)$ $\text{Greatest height is } 1.48 \text{ m}$ | M1, A1 | |
**ALTERNATIVE for (iii):**
$[0 = 2.286_... ^2 \times (1-4/9) -19.6y,$ $H = 0.8(1 + 2/3) + y]$ $H = 1.3333... + 0.1481... (4/3 + 4/27)$ $\text{Greatest height is } 1.48 \text{ m } (40/27)$ $[\frac{1}{2}m(2.286...^2 - \text{speed}^2) = mg \times 0.1481...]$ speed$^2 = 2.286_...^2 - 19.6 \times 0.1481...$ ] or $[\frac{1}{2}m(5.6^2 - \text{speed}^2) = mg \times 1.481...]$ speed$^2 = 5.6^2 - 19.6 \times 1.481...]$ $\text{Speed at greatest height is } 1.52 \text{ ms}^{-1}$ | M1, A1, M1, A1 | For using $0^2 = \dot{y}^2 - 2gy$ and $H = 0.8(1 + \cos(180° - \theta)) + y$. For using the principle of conservation of energy |One end of a light inextensible string of length $0.8$ m is attached to a fixed point $O$. A particle $P$ of mass $0.5$ kg is attached to the other end of the string. $P$ is projected horizontally from the point $0.8$ m vertically below $O$ with speed $5.6$ m s$^{-1}$. $P$ starts to move in a vertical circle with centre $O$. The speed of $P$ is $v$ m s$^{-1}$ when the string makes an angle $\theta$ with the downward vertical.
\begin{enumerate}[label=(\roman*)]
\item While the string remains taut, show that $v^2 = 15.68(1 + \cos \theta)$, and find the tension in the string in terms of $\theta$. [7]
\item For the instant when the string becomes slack, find the value of $\theta$ and the value of $v$. [3]
\item Find, in either order, the speed of $P$ when it is at its greatest height after the string becomes slack, and the greatest height reached by $P$ above its point of projection. [4]
\end{enumerate}
\hfill \mbox{\textit{OCR M3 2011 Q7 [14]}}