| Exam Board | Edexcel |
|---|---|
| Module | M3 (Mechanics 3) |
| Marks | 15 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circular Motion 2 |
| Type | Projectile motion after leaving circle |
| Difficulty | Challenging +1.2 This is a multi-part circular motion problem requiring application of energy conservation and Newton's second law for circular motion. While it involves several steps (finding reaction force change, leaving speed, and final speed), the techniques are standard M3 material with straightforward application of familiar formulas. The geometry is clearly defined and the problem follows a predictable structure, making it moderately above average difficulty but not requiring novel insight. |
| Spec | 3.02h Motion under gravity: vector form6.02i Conservation of energy: mechanical energy principle6.05d Variable speed circles: energy methods |
| Answer | Marks | Guidance |
|---|---|---|
| (a) just before B, resolve \(\downarrow\): \(60g - R_1 = 0\) \(\therefore R_1 = 60g\) | M1 | |
| just after B, resolve \(\downarrow\): \(60g - R_2 = \frac{m v^2}{30} = \frac{60 \times 4^2}{30} = 288\) | M1 A1 | |
| \(\therefore R_2 = 60g - 288\) so loss of reaction = 288 N | A1 | |
| (b) resolve \(\leftarrow\): \(mg\cos\theta - R = \frac{m v^2}{30}\) | M1 A1 | |
| at P, \(R = 0\) \(\therefore v^2 = 30g\cos\theta\) | M1 | |
| con. of ME: \(\frac{1}{2}m(v^2 - 12^2) = mg \times 30(1 - \cos\theta)\) | M1 A1 | |
| \(\therefore v^2 = 144 + 60g(1 - \cos\theta)\) | A1 | |
| combining, \(v^2 = 144 + 60g(1 - \frac{v^2}{30g})\) | M1 | |
| giving \(v^2 = 144 + 60g - 2v^2\) \(\therefore 3v^2 = 144 + 60g\) so, \(v^2 = 48 + 20g\) \(\therefore v = 15.6\) ms\(^{-1}\) (3sf) | A1 | |
| (c) con. of ME: \(\frac{1}{2}m(v^2 - 12^2) = mg \times 30\) | M1 A1 | |
| giving \(v^2 = 144 + 60g\) \(\therefore v = 27.1\) ms\(^{-1}\) (3sf) | A1 | (15) |
| Answer | Marks |
|---|---|
| Total | (75) |
(a) just before B, resolve $\downarrow$: $60g - R_1 = 0$ $\therefore R_1 = 60g$ | M1 |
just after B, resolve $\downarrow$: $60g - R_2 = \frac{m v^2}{30} = \frac{60 \times 4^2}{30} = 288$ | M1 A1 |
$\therefore R_2 = 60g - 288$ so loss of reaction = 288 N | A1 |
(b) resolve $\leftarrow$: $mg\cos\theta - R = \frac{m v^2}{30}$ | M1 A1 |
at P, $R = 0$ $\therefore v^2 = 30g\cos\theta$ | M1 |
con. of ME: $\frac{1}{2}m(v^2 - 12^2) = mg \times 30(1 - \cos\theta)$ | M1 A1 |
$\therefore v^2 = 144 + 60g(1 - \cos\theta)$ | A1 |
combining, $v^2 = 144 + 60g(1 - \frac{v^2}{30g})$ | M1 |
giving $v^2 = 144 + 60g - 2v^2$ $\therefore 3v^2 = 144 + 60g$ so, $v^2 = 48 + 20g$ $\therefore v = 15.6$ ms$^{-1}$ (3sf) | A1 |
(c) con. of ME: $\frac{1}{2}m(v^2 - 12^2) = mg \times 30$ | M1 A1 |
giving $v^2 = 144 + 60g$ $\therefore v = 27.1$ ms$^{-1}$ (3sf) | A1 | (15)
---
**Total** | (75)7.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{cab238c9-f4e2-4637-a079-f74779548f49-4_300_952_1201_497}
\captionsetup{labelformat=empty}
\caption{Fig. 3}
\end{center}
\end{figure}
Figure 3 shows a vertical cross-section through part of a ski slope consisting of a horizontal section $A B$ followed by a downhill section $B C$. The point $O$ is on the same horizontal level as $C$ and $B C$ is a circular arc of radius 30 m and centre $O$, such that $\angle B O C = 90 ^ { \circ }$.
A skier of mass 60 kg is skiing at $12 \mathrm {~ms} ^ { - 1 }$ along $A B$.
\begin{enumerate}[label=(\alph*)]
\item Assuming that friction and air resistance may be neglected, find the magnitude of the loss in reaction between the skier and the surface at $B$.\\
(4 marks)\\
The skier subsequently leaves the slope at the point $P$.
\item Find, correct to 3 significant figures, the speed at which the skier leaves the slope.
\item Find, correct to 3 significant figures, the speed of the skier immediately before hitting the ground again at the point $D$ which is on the same horizontal level as $C$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M3 Q7 [15]}}