| Exam Board | WJEC |
|---|---|
| Module | Further Unit 3 (Further Unit 3) |
| Year | 2018 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circular Motion 1 |
| Type | Banked track – no friction (find speed or radius) |
| Difficulty | Challenging +1.2 This is a standard circular motion problem on a banked track with no friction. Part (a) requires resolving forces perpendicular to motion (routine). Part (b) involves resolving horizontally to find centripetal force and applying v²/r, which is textbook application. Part (c) tests understanding of the no-friction assumption. While it's Further Maths content (making it inherently harder than core), the problem follows a standard template with straightforward resolution of forces and direct application of circular motion formulas. The multi-part structure and 11 marks indicate moderate length, but no novel insight is required. |
| Spec | 6.05c Horizontal circles: conical pendulum, banked tracks |
| Answer | Marks | Guidance |
|---|---|---|
| 6(a) | Resolve vertically | M1 |
| \(R\cos 60° = 1200g\) | A1 | |
| \(R = 2400g = 23520\) (N) | A1 | cao |
| 6(b)(i) | N2L towards centre | M1 |
| \(R\sin 60° = 1200a\) | ||
| \(R\sin 60° = 1200 \times \frac{v^2}{r}\) | m1 | |
| \(23520 \times \frac{\sqrt{3}}{2} = 1200 \times \frac{40^2}{r}\) | A1 | |
| \(r = 94.26\) (m) | A1 | |
| 6(b)(ii) | \(\omega = \frac{v}{r} = 0.424\) rad s\(^{-1}\) | B1 |
| B1 | units | |
| 6(c) | The assumption was made that there are no external forces acting on the vehicle. If there is an external force with component acting horizontally towards the centre of motion, then the LHS of the equation in (b)(i) would be larger resulting in a smaller radius \(r\). Similarly, if the component of force is acting away from the centre, the radius would be larger. | B1E1 |
6(a) | Resolve vertically | M1 | dim correct equation |
| $R\cos 60° = 1200g$ | A1 | |
| $R = 2400g = 23520$ (N) | A1 | cao |
6(b)(i) | N2L towards centre | M1 | dim correct equ. |
| $R\sin 60° = 1200a$ | | |
| $R\sin 60° = 1200 \times \frac{v^2}{r}$ | m1 | |
| $23520 \times \frac{\sqrt{3}}{2} = 1200 \times \frac{40^2}{r}$ | A1 | |
| $r = 94.26$ (m) | A1 | |
6(b)(ii) | $\omega = \frac{v}{r} = 0.424$ rad s$^{-1}$ | B1 | |
| | B1 | units |
6(c) | The assumption was made that there are no external forces acting on the vehicle. If there is an external force with component acting horizontally towards the centre of motion, then the LHS of the equation in (b)(i) would be larger resulting in a smaller radius $r$. Similarly, if the component of force is acting away from the centre, the radius would be larger. | B1E1 | |A vehicle of mass 1200 kg is moving with a constant speed of $40\text{ ms}^{-1}$ around a horizontal circular path which is on a test track banked at an angle of 60° to the horizontal. There is no tendency to sideslip at this speed. The vehicle is modelled as a particle.
\begin{enumerate}[label=(\alph*)]
\item Calculate the normal reaction of the track on the vehicle. [3]
\item Determine
\begin{enumerate}[label=(\roman*)]
\item the radius of the circular path,
\item the angular speed of the vehicle and clearly state its units. [6]
\end{enumerate}
\item What further assumption have you made in your solution to (b)? Briefly explain what effect this assumption has on the radius of the circular path. [2]
\end{enumerate}
\hfill \mbox{\textit{WJEC Further Unit 3 2018 Q6 [11]}}