| Exam Board | Edexcel |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2021 |
| Session | January |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circular Motion 1 |
| Type | Circular motion with rod |
| Difficulty | Standard +0.3 This is a standard M3 circular motion problem requiring resolution of forces and application of F=mrω². The geometry is straightforward (3-4-5 triangle), and the method is routine: resolve horizontally for centripetal force and vertically for equilibrium. Part (b) requires understanding that increased radius increases required centripetal force. Slightly above average due to the two-rod setup and multi-step calculation, but follows a well-practiced template. |
| Spec | 3.03d Newton's second law: 2D vectors3.03m Equilibrium: sum of resolved forces = 06.05c Horizontal circles: conical pendulum, banked tracks |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\cos\theta = \frac{4}{5}, \sin\theta = \frac{3}{5}\) | B1 | Correct trig used anywhere |
| \(\omega = \pi\) | B1 | Angular speed seen |
| \(T_A\cos\theta - T_B\cos\theta = 600g\) | M1A1 | Attempt at vertical resolution; correct equation in \(\theta\), \(m=600\) needs to be used now or later |
| \((T_A - T_B = 750g)\) | ||
| \(T_A\sin\theta + T_B\sin\theta = 600 \times \omega^2 \times (5\sin\theta)\) | M1A1 | Attempt at horizontal resolution, acceleration in either form; correct equation in \(\omega\) and \(\theta\), \(m\) and \(R\) must be substituted |
| \((T_A + T_B = 3000\pi^2)\) | ||
| Solve simultaneously | dM1 | Dependent on both previous M marks |
| \(T_A = 1500\pi^2 + 375g = 18000\text{N},\ 18500\text{N},\ 18\text{kN},\ 18.5\text{kN}\) | A1 | Correct \(T_A\) (must be 2/3 s.f.) |
| \(T_B = 1500\pi^2 - 375g = 11000\text{N},\ 11100\text{N},\ 11\text{kN},\ 11.1\text{kN}\) | A1 | Correct \(T_B\) (must be 2/3 s.f.); only penalise over-accuracy on \(T_A\) if in both |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| If the length of the arms increased, then the radius of the circle would increase. | B1 | Correct statement about the effect on the radius of the motion |
| Therefore the total tension would increase. | dB1 | Conclusion that total tension would be greater; must reference total tension; dependent on correct radius statement |
## Question 3:
### Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\cos\theta = \frac{4}{5}, \sin\theta = \frac{3}{5}$ | B1 | Correct trig used anywhere |
| $\omega = \pi$ | B1 | Angular speed seen |
| $T_A\cos\theta - T_B\cos\theta = 600g$ | M1A1 | Attempt at vertical resolution; correct equation in $\theta$, $m=600$ needs to be used now or later |
| $(T_A - T_B = 750g)$ | | |
| $T_A\sin\theta + T_B\sin\theta = 600 \times \omega^2 \times (5\sin\theta)$ | M1A1 | Attempt at horizontal resolution, acceleration in either form; correct equation in $\omega$ and $\theta$, $m$ and $R$ must be substituted |
| $(T_A + T_B = 3000\pi^2)$ | | |
| Solve simultaneously | dM1 | Dependent on both previous M marks |
| $T_A = 1500\pi^2 + 375g = 18000\text{N},\ 18500\text{N},\ 18\text{kN},\ 18.5\text{kN}$ | A1 | Correct $T_A$ (must be 2/3 s.f.) |
| $T_B = 1500\pi^2 - 375g = 11000\text{N},\ 11100\text{N},\ 11\text{kN},\ 11.1\text{kN}$ | A1 | Correct $T_B$ (must be 2/3 s.f.); only penalise over-accuracy on $T_A$ if in both |
### Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| If the length of the arms increased, then the radius of the circle would increase. | B1 | Correct statement about the effect on the radius of the motion |
| Therefore the total tension would increase. | dB1 | Conclusion that **total tension** would be greater; must reference total tension; dependent on correct radius statement |
---3.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{8a687d17-ec7e-463f-84dd-605f5c230db1-08_506_527_251_712}
\captionsetup{labelformat=empty}
\caption{Figure 2}
\end{center}
\end{figure}
A fairground ride consists of a cabin $C$ that travels in a horizontal circle with a constant angular speed about a fixed vertical central axis. The cabin is attached to one end of each of two rigid arms, each of length 5 m . The other end of the top arm is attached to the fixed point $A$ at the top of the central axis of the ride. The other end of the lower arm is attached to the fixed point $B$ on the central axis, where $A B$ is 8 m , as shown in Figure 2.
Both arms are free to rotate about the central axis.
The arms are modelled as light inextensible rods.
The cabin, together with the people inside, is modelled as a particle.
The cabin completes one revolution every 2 seconds.
Given that the combined mass of the cabin and the people is 600 kg ,
\begin{enumerate}[label=(\alph*)]
\item find
\begin{enumerate}[label=(\roman*)]
\item the tension in the upper arm of the ride,
\item the tension in the lower arm of the ride.
In a refined model, it is assumed that both arms stretch to a length of 5.1 m .
\end{enumerate}\item State how this would affect the sum of the tensions in the two arms, justifying your answer.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M3 2021 Q3 [11]}}