| Exam Board | Edexcel |
|---|---|
| Module | M3 (Mechanics 3) |
| Session | Specimen |
| Marks | 15 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circular Motion 1 |
| Type | Vertical circle – string/rod (tension and energy) |
| Difficulty | Challenging +1.2 This is a standard M3 vertical circular motion problem requiring energy conservation and force analysis to find when the string goes slack, followed by projectile motion. While it involves multiple steps and two distinct phases of motion, the techniques are routine for this module with no novel insight required—students following the standard method (tension=0 condition, energy equation, then projectile motion) will succeed systematically. |
| Spec | 6.02i Conservation of energy: mechanical energy principle6.05e Radial/tangential acceleration |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Notes |
| \(R(\searrow)\; T + mg\cos\theta = \dfrac{mv^2}{a}\) | M1 A1 | |
| Energy: \(\dfrac{1}{2}mu^2 - \dfrac{1}{2}mv^2 = mga(1+\cos\theta)\) | M1 A1 A1 | |
| \(u^2 = 3ga \Rightarrow v^2 = ga(1-2\cos\theta)\) | ||
| \(T = -mg\cos\theta + \dfrac{mv^2}{a} = mg(-3\cos\theta+1)\) | M1 A1 | |
| \(T=0 \Rightarrow \cos\theta = \dfrac{1}{3}\) | M1 A1 | (9) |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Notes |
| \(v^2 = \dfrac{ga}{3}\) | B1 | |
| \(\sin^2\theta = 1 - \left(-\dfrac{1}{3}\right)^2 = \dfrac{8}{9}\) | M1 A1 | |
| \(H = \dfrac{v^2\sin^2\theta}{2g} = \dfrac{ga}{3}\cdot\dfrac{8}{9}\cdot\dfrac{1}{2g} = \dfrac{4a}{27}\) | M1 M1 A1 | (6) |
# Question 6:
## Part (a):
| Working/Answer | Marks | Notes |
|---|---|---|
| $R(\searrow)\; T + mg\cos\theta = \dfrac{mv^2}{a}$ | M1 A1 | |
| Energy: $\dfrac{1}{2}mu^2 - \dfrac{1}{2}mv^2 = mga(1+\cos\theta)$ | M1 A1 A1 | |
| $u^2 = 3ga \Rightarrow v^2 = ga(1-2\cos\theta)$ | | |
| $T = -mg\cos\theta + \dfrac{mv^2}{a} = mg(-3\cos\theta+1)$ | M1 A1 | |
| $T=0 \Rightarrow \cos\theta = \dfrac{1}{3}$ | M1 A1 | (9) |
## Part (b):
| Working/Answer | Marks | Notes |
|---|---|---|
| $v^2 = \dfrac{ga}{3}$ | B1 | |
| $\sin^2\theta = 1 - \left(-\dfrac{1}{3}\right)^2 = \dfrac{8}{9}$ | M1 A1 | |
| $H = \dfrac{v^2\sin^2\theta}{2g} = \dfrac{ga}{3}\cdot\dfrac{8}{9}\cdot\dfrac{1}{2g} = \dfrac{4a}{27}$ | M1 M1 A1 | (6) |
---6. A particle $P$ is attached to one end of a light inextensible string of length $a$. The other end of the string is attached to a fixed point $O$. The particle is hanging in equilibrium below $O$ when it receives a horizontal impulse giving it a speed $u$, where $u ^ { 2 } = 3 g a$. The string becomes slack when $P$ is at the point $B$. The line $O B$ makes an angle $\theta$ with the upward vertical.
\begin{enumerate}[label=(\alph*)]
\item Show that $\cos \theta = \frac { 1 } { 3 }$.\\
(9)
\item Show that the greatest height of $P$ above $B$ in the subsequent motion is $\frac { 4 a } { 27 }$.\\
(6)
\end{enumerate}
\hfill \mbox{\textit{Edexcel M3 Q6 [15]}}