| Exam Board | Edexcel |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2023 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circular Motion 1 |
| Type | Banked track – with friction (find maximum/minimum speed or friction coefficient) |
| Difficulty | Standard +0.3 This is a standard M3 banked track problem with two parts: (a) showing a given result using circular motion equations with no friction (routine application of resolving forces and F=mrω²), and (b) finding maximum speed with friction (standard extension requiring friction force in the limiting case). The mathematics is straightforward once the force diagram is set up correctly, making this slightly easier than average for M3 content. |
| Spec | 6.05c Horizontal circles: conical pendulum, banked tracks |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(R\cos\alpha = mg\) | M1 A1 | Resolve vertically; correct no. of terms; condone sign errors and sin/cos confusion |
| \(R\sin\alpha = \dfrac{m\!\left(\frac{1}{4}gr\right)}{r}\) | M1A1 | Equation of motion horizontally; correct no. of terms; \(V\) does not need to be substituted; allow \(r\omega^2\) for acceleration but not \(a\) |
| OR: \(mg\sin\alpha = \dfrac{m\!\left(\frac{1}{4}gr\right)}{r}\cos\alpha\) | M2 A2 | Equation of motion down the plane; correct no. of terms; condone sign errors and sin/cos confusion |
| \(\tan\alpha = \dfrac{1}{4}\) * | A1* (5) | Correctly obtain given answer, written exactly; note: use of \(\theta\) instead of \(\alpha\) penalises only last mark; maximum M1A1M1A1A0 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Vert equil: \(S\cos\alpha - F\sin\alpha = mg\) | M1A1 | Resolve vertically or equation of motion perpendicular; correct no. of terms; condone sign errors and sin/cos confusion; M0 if \(R\) from (a) used |
| Perp N2L: \(S - mg\cos\alpha = \dfrac{mV^2}{r}\sin\alpha\) | ||
| N2L towards \(O\): \(S\sin\alpha + F\cos\alpha = \dfrac{mV^2}{r}\) | M1A1 | Equation of motion towards centre or parallel to plane; correct no. of terms; condone sign errors and sin/cos confusion |
| Parallel N2L: \(F + mg\sin\alpha = \dfrac{mV^2}{r}\cos\alpha\) | ||
| \(F = \mu S\) | B1 | |
| Eliminate \(F\), substitute trig, solve for \(V\) in terms of \(\mu\), \(r\) and \(g\) | dM1 | Dependent on previous M marks |
| \(V = \sqrt{rg\dfrac{(1+4\mu)}{(4-\mu)}}\) oe | A1 (7) |
# Question 4(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $R\cos\alpha = mg$ | M1 A1 | Resolve vertically; correct no. of terms; condone sign errors and sin/cos confusion |
| $R\sin\alpha = \dfrac{m\!\left(\frac{1}{4}gr\right)}{r}$ | M1A1 | Equation of motion horizontally; correct no. of terms; $V$ does not need to be substituted; allow $r\omega^2$ for acceleration but not $a$ |
| **OR:** $mg\sin\alpha = \dfrac{m\!\left(\frac{1}{4}gr\right)}{r}\cos\alpha$ | M2 A2 | Equation of motion down the plane; correct no. of terms; condone sign errors and sin/cos confusion |
| $\tan\alpha = \dfrac{1}{4}$ * | A1* (5) | Correctly obtain given answer, written exactly; note: use of $\theta$ instead of $\alpha$ penalises only last mark; maximum M1A1M1A1A0 |
---
# Question 4(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Vert equil: $S\cos\alpha - F\sin\alpha = mg$ | M1A1 | Resolve vertically or equation of motion perpendicular; correct no. of terms; condone sign errors and sin/cos confusion; M0 if $R$ from (a) used |
| Perp N2L: $S - mg\cos\alpha = \dfrac{mV^2}{r}\sin\alpha$ | | |
| N2L towards $O$: $S\sin\alpha + F\cos\alpha = \dfrac{mV^2}{r}$ | M1A1 | Equation of motion towards centre or parallel to plane; correct no. of terms; condone sign errors and sin/cos confusion |
| Parallel N2L: $F + mg\sin\alpha = \dfrac{mV^2}{r}\cos\alpha$ | | |
| $F = \mu S$ | B1 | |
| Eliminate $F$, substitute trig, solve for $V$ in terms of $\mu$, $r$ and $g$ | dM1 | Dependent on previous M marks |
| $V = \sqrt{rg\dfrac{(1+4\mu)}{(4-\mu)}}$ oe | A1 (7) | |4.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{dceb2432-117c-40fe-bf3d-782beeb42e41-12_360_1004_246_534}
\captionsetup{labelformat=empty}
\caption{Figure 4}
\end{center}
\end{figure}
A car is travelling round a circular track. The track is banked at an angle $\alpha$ to the horizontal, as shown in Figure 4.
The car and driver are modelled as a particle.\\
The car moves round the track with constant speed in a horizontal circle of radius $r$.\\
When the car is moving with speed $\frac { 1 } { 2 } \sqrt { g r }$ round the circle, there is no sideways friction between the tyres of the car and the track.
\begin{enumerate}[label=(\alph*)]
\item Show that $\tan \alpha = \frac { 1 } { 4 }$
The sideways friction between the tyres of the car and the track has coefficient of friction $\mu$, where $\mu < 4$
The maximum speed at which the car can move round the circle without slipping sideways is $V$.
\item Find $V$ in terms of $\mu , r$ and $g$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M3 2023 Q4 [12]}}