| Exam Board | Edexcel |
|---|---|
| Module | M3 (Mechanics 3) |
| Marks | 20 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circular Motion 2 |
| Type | Vertical circle: complete revolution conditions |
| Difficulty | Standard +0.8 This is a multi-part vertical circle problem requiring energy conservation, tension conditions for complete circles, and projectile motion after the string slackens. Part (a) is a standard derivation (showing u ≥ √(3g)), but parts (b) and (c) require solving for when tension becomes zero with a specific initial speed, then analyzing subsequent projectile motion—this combination of techniques and the need to work with a case where complete circles don't occur makes it moderately challenging. |
| Spec | 3.02h Motion under gravity: vector form6.02i Conservation of energy: mechanical energy principle6.05d Variable speed circles: energy methods6.05e Radial/tangential acceleration |
| Answer | Marks |
|---|---|
| (a) full circles if \(T \geq 0\) at max height | B1 |
| resolve \(\downarrow\): \(T + mg = \frac{mv^2}{r}\) \(\therefore \frac{mv^2}{0.6} \geq mg\) | M1 A1 |
| con. of ME: \(\frac{1}{2}m(u^2 - v^2) = mg \times 1.2\) | M1 A1 |
| \(\therefore v^2 = u^2 - 2.4g\) | M1 |
| combining, \(\frac{u^2 - 2.4g}{0.6} \geq g\) | M1 |
| giving \(u^2 \geq 3g \therefore u \geq \sqrt{3g}\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| resolve \(\leftarrow\): \(T - mg \cos \theta = \frac{mv^2}{r}\) | M1 | |
| when slack, \(T = 0 \therefore v^2 = "0.6g \cos \theta\) | A1 | |
| con. of ME: \(\frac{1}{2}m(u^2 - v^2) = 0.6mg(1 - \cos \theta)\) | M1 A1 | |
| \(\therefore v^2 = 25 - 1.2g(1 - \cos \theta)\) | M1 | |
| combining, \("0.6g \cos \theta = 25 - 1.2g(1 - \cos \theta)\) | M1 | |
| giving \(\cos \theta = \frac{1.2g - 25}{1.8g} \therefore \theta = 139°\) (nearest deg) | A1 | |
| (c) \(v^2 = "0.6g(\frac{1.2g - 25}{1.8g})\) giving \(v^2 = 4.413\) | M1 | |
| vertically: \(u = \sqrt{4.413} \sin 41°\), \(a = -g\), \(v = 0\) (for greatest height) | M1 | |
| using \(v^2 = u^2 + 2as\), \(0 = 4.413 \sin^2 41° - 2gs\) | M1 A1 | |
| giving \(s = 0.098 \text{ m} = 10 \text{ cm (nearest cm)}\) | A1 | (20) |
(a) full circles if $T \geq 0$ at max height | B1 |
resolve $\downarrow$: $T + mg = \frac{mv^2}{r}$ $\therefore \frac{mv^2}{0.6} \geq mg$ | M1 A1 |
con. of ME: $\frac{1}{2}m(u^2 - v^2) = mg \times 1.2$ | M1 A1 |
$\therefore v^2 = u^2 - 2.4g$ | M1 |
combining, $\frac{u^2 - 2.4g}{0.6} \geq g$ | M1 |
giving $u^2 \geq 3g \therefore u \geq \sqrt{3g}$ | A1 |
(b) [Diagram showing circular motion with angle $\theta$, tension T, velocity v, and weight mg with horizontal velocity u]
resolve $\leftarrow$: $T - mg \cos \theta = \frac{mv^2}{r}$ | M1 |
when slack, $T = 0 \therefore v^2 = "0.6g \cos \theta$ | A1 |
con. of ME: $\frac{1}{2}m(u^2 - v^2) = 0.6mg(1 - \cos \theta)$ | M1 A1 |
$\therefore v^2 = 25 - 1.2g(1 - \cos \theta)$ | M1 |
combining, $"0.6g \cos \theta = 25 - 1.2g(1 - \cos \theta)$ | M1 |
giving $\cos \theta = \frac{1.2g - 25}{1.8g} \therefore \theta = 139°$ (nearest deg) | A1 |
(c) $v^2 = "0.6g(\frac{1.2g - 25}{1.8g})$ giving $v^2 = 4.413$ | M1 |
vertically: $u = \sqrt{4.413} \sin 41°$, $a = -g$, $v = 0$ (for greatest height) | M1 |
using $v^2 = u^2 + 2as$, $0 = 4.413 \sin^2 41° - 2gs$ | M1 A1 |
giving $s = 0.098 \text{ m} = 10 \text{ cm (nearest cm)}$ | A1 | (20)7. A particle of mass 0.5 kg is hanging vertically at one end of a light inextensible string of length 0.6 m . The other end of the string is attached to a fixed point.
The particle is given an initial horizontal speed of $u \mathrm {~ms} ^ { - 1 }$.
\begin{enumerate}[label=(\alph*)]
\item Show that the particle will perform complete circles if $u \geq \sqrt { 3 g }$.
Given that $u = 5$,
\item find, correct to the nearest degree, the angle through which the string turns before it becomes slack,
\item find, correct to the nearest centimetre, the greatest height the particle reaches above its position when the string becomes slack.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M3 Q7 [20]}}