| Exam Board | Edexcel |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2012 |
| Session | January |
| Marks | 15 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circular Motion 2 |
| Type | Vertical circle: string becomes slack |
| Difficulty | Standard +0.8 This is a standard M3 vertical circle problem requiring energy conservation and circular motion dynamics, but the multi-part structure (deriving tension formula, finding speed when string slackens, then projectile motion) and the need to recognize that T=0 defines the slack condition elevates it above routine exercises. The calculations are straightforward once the method is identified, placing it moderately above average difficulty. |
| Spec | 3.02h Motion under gravity: vector form6.02i Conservation of energy: mechanical energy principle |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| GPE gained \(= mgl(1-\cos\theta)\) | ||
| \(\frac{1}{2}m\frac{11gl}{4} = mgl(1-\cos\theta) + \frac{1}{2}mv^2\) | M1 A1 A1 | |
| \(v^2 = gl\left(\frac{11}{4} - 2 + 2\cos\theta\right) = gl\left(\frac{3}{4} + 2\cos\theta\right)\) | ||
| \(T - mg\cos\theta = \frac{mv^2}{l}\) | M1, A1 A1 | |
| \(T - mg\cos\theta = mg\left(\frac{3}{4} + 2\cos\theta\right)\) | M1 | |
| \(T = mg\left(\frac{3}{4} + 3\cos\theta\right) = 3mg\left(\cos\theta + \frac{1}{4}\right)\) | A1 | (8) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(T = 0 \Rightarrow \cos\theta = -\frac{1}{4}\) | M1 | |
| \(v^2 = gl\left(\frac{3}{4} + 2\cos\theta\right) = \frac{gl}{4}\), \(v = \sqrt{\frac{gl}{4}}\) | M1, A1 | (3) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Horizontal component of velocity at \(B = \sqrt{\frac{gl}{4}}\times\cos(180°-\theta) = \frac{1}{4}\sqrt{\frac{gl}{4}}\) | B1ft | |
| Extra height \(h\): \(mgh + \frac{1}{2}m\frac{gl}{64} = \frac{1}{2}m\frac{gl}{4}\) | M1 A1 | |
| \(h = \left(\frac{1}{8} - \frac{1}{128}\right)l = \frac{15l}{128}\) (0.117\(l\)) | A1 | (4) Total: 15 |
## Question 6:
### Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| GPE gained $= mgl(1-\cos\theta)$ | | |
| $\frac{1}{2}m\frac{11gl}{4} = mgl(1-\cos\theta) + \frac{1}{2}mv^2$ | M1 A1 A1 | |
| $v^2 = gl\left(\frac{11}{4} - 2 + 2\cos\theta\right) = gl\left(\frac{3}{4} + 2\cos\theta\right)$ | | |
| $T - mg\cos\theta = \frac{mv^2}{l}$ | M1, A1 A1 | |
| $T - mg\cos\theta = mg\left(\frac{3}{4} + 2\cos\theta\right)$ | M1 | |
| $T = mg\left(\frac{3}{4} + 3\cos\theta\right) = 3mg\left(\cos\theta + \frac{1}{4}\right)$ | A1 | (8) |
### Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $T = 0 \Rightarrow \cos\theta = -\frac{1}{4}$ | M1 | |
| $v^2 = gl\left(\frac{3}{4} + 2\cos\theta\right) = \frac{gl}{4}$, $v = \sqrt{\frac{gl}{4}}$ | M1, A1 | (3) |
### Part (c):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Horizontal component of velocity at $B = \sqrt{\frac{gl}{4}}\times\cos(180°-\theta) = \frac{1}{4}\sqrt{\frac{gl}{4}}$ | B1ft | |
| Extra height $h$: $mgh + \frac{1}{2}m\frac{gl}{64} = \frac{1}{2}m\frac{gl}{4}$ | M1 A1 | |
| $h = \left(\frac{1}{8} - \frac{1}{128}\right)l = \frac{15l}{128}$ (0.117$l$) | A1 | (4) Total: 15 |\begin{enumerate}
\item A particle $P$ of mass $m$ is attached to one end of a light inextensible string of length $l$. The other end of the string is attached to a fixed point $O$. The particle is hanging in equilibrium at the point $A$, vertically below $O$, when it is set in motion with a horizontal speed $\frac { 1 } { 2 } \sqrt { } ( 11 g l )$. When the string has turned through an angle $\theta$ and the string is still taut, the tension in the string is $T$.\\
(a) Show that $T = 3 m g \left( \cos \theta + \frac { 1 } { 4 } \right)$.
\end{enumerate}
At the instant when $P$ reaches the point $B$, the string becomes slack. Find\\
(b) the speed of $P$ at $B$,\\
(c) the maximum height above $B$ reached by $P$ before it starts to fall.\\
\hfill \mbox{\textit{Edexcel M3 2012 Q6 [15]}}