| Exam Board | Edexcel |
|---|---|
| Module | M3 (Mechanics 3) |
| Marks | 13 |
| 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 +0.8 This M3 circular motion problem requires resolving forces in two directions on a banked track, applying friction at limiting equilibrium, and manipulating trigonometric expressions to find maximum speed—significantly more complex than standard horizontal circular motion questions. The 'show that' part (b) requires careful algebraic manipulation and percentage calculation, making this harder than average but within reach of well-prepared M3 students. |
| Spec | 3.03t Coefficient of friction: F <= mu*R model3.03v Motion on rough surface: including inclined planes6.05c Horizontal circles: conical pendulum, banked tracks |
| Answer | Marks | Guidance |
|---|---|---|
| resolve \(\uparrow\): \(R - mg = 0, R = mg\) | M1 | |
| resolve \(\leftarrow\): \(\mu R = \frac{mv^2}{r}\) \(\therefore \frac{2}{5}R = \frac{mv^2}{40}\) | M1 A1 | |
| combining, \(\frac{2}{5}mg = \frac{mv^2}{40}\) | M1 | |
| \(\therefore v^2 = \frac{2}{5}g \times 40\) giving \(v = 12.5\) ms\(^{-1}\) (3sf) | A1 | |
| resolve \(\uparrow\): \(R\cos 25 - \mu R\sin 25 - mg = 0\) | M1 A1 | |
| \(R = \frac{mg}{\cos 25 - \frac{1}{2}\sin 25}\) | ||
| resolve \(\leftarrow\): \(R\sin 25 + \mu R\cos 25 = \frac{mv^2}{r}\) | M1 A1 | |
| combining, \(\frac{mg(\sin 25 + \frac{1}{2}\cos 25)}{\cos 25 - \frac{1}{2}\sin 25} = \frac{mv^2}{40}\) | M1 | |
| giving \(v^2 = \frac{40g(\sin 25 + \frac{1}{2}\cos 25)}{\cos 25 - \frac{1}{2}\sin 25} \therefore v = 20.43\) | A1 | |
| % increase \(= \frac{20.43 - 12.52}{12.52} \times 100\% = 63\%\) (nearest WN) | M1 A1 | (13) |
| resolve $\uparrow$: $R - mg = 0, R = mg$ | M1 | |
| resolve $\leftarrow$: $\mu R = \frac{mv^2}{r}$ $\therefore \frac{2}{5}R = \frac{mv^2}{40}$ | M1 A1 | |
| combining, $\frac{2}{5}mg = \frac{mv^2}{40}$ | M1 | |
| $\therefore v^2 = \frac{2}{5}g \times 40$ giving $v = 12.5$ ms$^{-1}$ (3sf) | A1 | |
| resolve $\uparrow$: $R\cos 25 - \mu R\sin 25 - mg = 0$ | M1 A1 | |
| $R = \frac{mg}{\cos 25 - \frac{1}{2}\sin 25}$ | | |
| resolve $\leftarrow$: $R\sin 25 + \mu R\cos 25 = \frac{mv^2}{r}$ | M1 A1 | |
| combining, $\frac{mg(\sin 25 + \frac{1}{2}\cos 25)}{\cos 25 - \frac{1}{2}\sin 25} = \frac{mv^2}{40}$ | M1 | |
| giving $v^2 = \frac{40g(\sin 25 + \frac{1}{2}\cos 25)}{\cos 25 - \frac{1}{2}\sin 25} \therefore v = 20.43$ | A1 | |
| % increase $= \frac{20.43 - 12.52}{12.52} \times 100\% = 63\%$ (nearest WN) | M1 A1 | (13) |6. A car is travelling on a horizontal racetrack round a circular bend of radius 40 m . The coefficient of friction between the car and the road is $\frac { 2 } { 5 }$.
\begin{enumerate}[label=(\alph*)]
\item Find the maximum speed at which the car can travel round the bend without slipping, giving your answer correct to 3 significant figures.\\
(5 marks)\\
The owner of the track decides to bank the corner at an angle of $25 ^ { \circ }$ in order to enable the cars to travel more quickly.
\item Show that this increases the maximum speed at which the car can travel round the bend without slipping by 63\%, correct to the nearest whole number.\\
(8 marks)
\end{enumerate}
\hfill \mbox{\textit{Edexcel M3 Q6 [13]}}