| Exam Board | Edexcel |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2021 |
| Session | October |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Advanced work-energy problems |
| Type | Rod or pendulum circular motion energy |
| Difficulty | Standard +0.8 This M3 circular motion problem requires energy conservation with a non-standard starting position (angle α), converting tan α = 3/4 to cos α = 4/5, then applying Newton's second law for circular motion to find when tension/thrust is zero. It combines multiple techniques (energy methods, circular motion dynamics, trigonometry) with careful algebraic manipulation, making it moderately challenging but within standard M3 scope. |
| Spec | 6.02i Conservation of energy: mechanical energy principle6.05e Radial/tangential acceleration |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\frac{1}{2}mv^2 - \frac{1}{2}mu^2 = mga(\cos\alpha - \cos\theta)\) | M1A2,1,0 | M1 for energy equation with 2 KE terms and 2 PE terms; \(\cos\alpha\) must be seen. A2 for correct equation, A1A0 for at most one error |
| \(v^2 = u^2 + \frac{2ag}{5}(4 - 5\cos\theta)\) * | A1* (4) | A1* for given answer correctly obtained |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(T + mg\cos\theta = \frac{mv^2}{a}\) (\(T\) may be omitted) | M1A2,1,0 | M1 for equation of motion towards \(O\) with all necessary terms; condone sign errors and sin/cos confusion; \(mg\) must be resolved. A2 for correct equation (allow \(-T\)), A1A0 for at most one error |
| Use of \(T=0\) and substitute for \(v^2\) and \(u^2\) | DM1 | Depends on first M mark; substitute to obtain equation in \(\cos\beta\) |
| \(mg\cos\beta = \frac{m}{a}\left(\frac{6ag}{5} + \frac{2ag}{5}(4-5\cos\beta)\right)\) | A1 | Correct unsimplified equation following substitution |
| \(\cos\beta = \frac{14}{15}\) (0.93 or better) | A1 (6) | cao |
## Question 6:
### Part 6(a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{1}{2}mv^2 - \frac{1}{2}mu^2 = mga(\cos\alpha - \cos\theta)$ | M1A2,1,0 | M1 for energy equation with 2 KE terms and 2 PE terms; $\cos\alpha$ must be seen. A2 for correct equation, A1A0 for at most one error |
| $v^2 = u^2 + \frac{2ag}{5}(4 - 5\cos\theta)$ * | A1* (4) | A1* for given answer correctly obtained |
### Part 6(b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $T + mg\cos\theta = \frac{mv^2}{a}$ ($T$ may be omitted) | M1A2,1,0 | M1 for equation of motion towards $O$ with all necessary terms; condone sign errors and sin/cos confusion; $mg$ must be resolved. A2 for correct equation (allow $-T$), A1A0 for at most one error |
| Use of $T=0$ and substitute for $v^2$ and $u^2$ | DM1 | Depends on first M mark; substitute to obtain equation in $\cos\beta$ |
| $mg\cos\beta = \frac{m}{a}\left(\frac{6ag}{5} + \frac{2ag}{5}(4-5\cos\beta)\right)$ | A1 | Correct unsimplified equation following substitution |
| $\cos\beta = \frac{14}{15}$ (0.93 or better) | A1 (6) | cao |
---6.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{9777abb8-a564-40d5-8d96-d5649913737b-20_534_551_248_699}
\captionsetup{labelformat=empty}
\caption{Figure 3}
\end{center}
\end{figure}
A light rod of length $a$ is free to rotate in a vertical plane about a horizontal axis through one end $O$. A particle $P$ of mass $m$ is attached to the other end of the rod. The particle $P$ is held at rest with the rod making an angle $\alpha$ with the upward vertical through $O$, where $\tan \alpha = \frac { 3 } { 4 }$
The particle $P$ is then projected with speed $u$ in a direction which is perpendicular to the rod. At the instant when the rod makes an angle $\theta$ with the upward vertical through $O$, the speed of $P$ is $v$, as shown in Figure 3.
Air resistance is assumed to be negligible.
\begin{enumerate}[label=(\alph*)]
\item Show that $v ^ { 2 } = u ^ { 2 } + \frac { 2 a g } { 5 } ( 4 - 5 \cos \theta )$
It is given that $u ^ { 2 } = \frac { 6 a g } { 5 }$ and $P$ moves in complete vertical circles.
When $\theta = \beta$, the force exerted on $P$ by the rod is zero.
\item Find the value of $\cos \beta$
\end{enumerate}
\hfill \mbox{\textit{Edexcel M3 2021 Q6 [10]}}