| Exam Board | WJEC |
|---|---|
| Module | Further Unit 3 (Further Unit 3) |
| Year | 2019 |
| Session | June |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circular Motion 2 |
| Type | Vertical circle: tension at specific point |
| Difficulty | Standard +0.8 This is a multi-part vertical circle problem requiring energy conservation, circular motion dynamics, and work-energy with friction. Parts (a)-(c) are standard Further Maths mechanics (energy equation, resolving forces radially, interpreting R≥0), but part (d) adds a non-trivial extension with friction over unknown distance. The question requires multiple techniques and careful reasoning but follows established patterns for FM vertical circle problems. |
| Spec | 6.02d Mechanical energy: KE and PE concepts6.02e Calculate KE and PE: using formulae6.02i Conservation of energy: mechanical energy principle6.05d Variable speed circles: energy methods6.05e Radial/tangential acceleration6.05f Vertical circle: motion including free fall |
| Answer | Marks | Guidance |
|---|---|---|
| (a) Conservation of energy: | M1 | KE and PE in dim. correct equation |
| \(\frac{1}{2}mu^2 = \frac{1}{2}mv^2 + mgr(1 - \cos \theta)\) | A1, A1 | KE; PE |
| \(v^2 = u^2 - 2gr(1 - \cos \theta)\) | ||
| \(v^2 = 60g - 20g(1 - \cos \theta)\) | A1 | |
| or \(v^2 = \begin{cases} 40g + 20g \cos \theta \\ 20g(2 + \cos \theta) \end{cases}\) |
| Answer | Marks | Guidance |
|---|---|---|
| (b) N2L towards centre: | M1 | Dim. correct equation, \(R\) and \(mg \cos \theta\) opposing |
| \(R - mg \cos \theta = \frac{mv^2}{10}\) | A1 | |
| \(R = \frac{m}{10}(40g + 20g \cos \theta) + mg \cos \theta\) | m1 | Substitute their \(v^2\) |
| \(R = 4mg + 2mg \cos \theta + mg \cos \theta\) | ||
| \(R = 4mg + 3mg \cos \theta\) | ||
| \(R = mg(4 + 3 \cos \theta)\) | A1 | Convincing |
| Answer | Marks | Guidance |
|---|---|---|
| (c) Test for \(R = 0\): | M1 | si |
| \(mg(4 + 3 \cos \theta) = 0\) | ||
| \(\cos \theta = -\frac{4}{3}\) | ||
| which is not possible (i.e. car will perform loop) | A1 | Convincing |
| Answer | Marks | Guidance |
|---|---|---|
| Consider \(R\) when \(\theta = 180°\): | (M1), (A1) | si, Convincing |
| \(R = mg(4 + 3(-1)) = mg > 0\) | ||
| (i.e. car will perform loop) | [(2)] | |
| (d) Loss in PE \(= mg(30 - 28)\) | B1 | |
| \(= 2mg\) | ||
| Work-energy principle: | M1 | Used, \(F \times d = E\) |
| \(\frac{mg}{32} \times d = 2mg\) | ||
| \(d = 2 \times 32\) | ||
| \(d = 64\) (m) | A1 | cao |
**(a)** Conservation of energy: | M1 | KE and PE in dim. correct equation
$\frac{1}{2}mu^2 = \frac{1}{2}mv^2 + mgr(1 - \cos \theta)$ | A1, A1 | KE; PE
$v^2 = u^2 - 2gr(1 - \cos \theta)$ | |
$v^2 = 60g - 20g(1 - \cos \theta)$ | A1 |
or $v^2 = \begin{cases} 40g + 20g \cos \theta \\ 20g(2 + \cos \theta) \end{cases}$ | |
**Subtotal: [4]**
**(b)** N2L towards centre: | M1 | Dim. correct equation, $R$ and $mg \cos \theta$ opposing
$R - mg \cos \theta = \frac{mv^2}{10}$ | A1 |
$R = \frac{m}{10}(40g + 20g \cos \theta) + mg \cos \theta$ | m1 | Substitute their $v^2$
$R = 4mg + 2mg \cos \theta + mg \cos \theta$ | |
$R = 4mg + 3mg \cos \theta$ | |
$R = mg(4 + 3 \cos \theta)$ | A1 | Convincing
**Subtotal: [4]**
**(c)** Test for $R = 0$: | M1 | si
$mg(4 + 3 \cos \theta) = 0$ | |
$\cos \theta = -\frac{4}{3}$ | |
which is not possible (i.e. car will perform loop) | A1 | Convincing
**Subtotal: [2]**
**Alternative solution**
Consider $R$ when $\theta = 180°$: | (M1), (A1) | si, Convincing
$R = mg(4 + 3(-1)) = mg > 0$ | |
(i.e. car will perform loop) | [(2)] |
**(d)** Loss in PE $= mg(30 - 28)$ | B1 |
$= 2mg$ | |
Work-energy principle: | M1 | Used, $F \times d = E$
$\frac{mg}{32} \times d = 2mg$ | |
$d = 2 \times 32$ | |
$d = 64$ (m) | A1 | cao
**Subtotal: [3]**
**Total for Question 6: 13**
---6. The diagram shows a rollercoaster at an amusement park where a car is projected from a launch point $O$ so that it performs a loop before instantaneously coming to rest at point $C$. The car then performs the same journey in reverse.\\
\includegraphics[max width=\textwidth, alt={}, center]{b430aa50-27e3-46f7-afef-7b8e75d46e1f-5_677_1733_552_166}
The loop section is modelled by considering the track to be a vertical circle of radius 10 m and the car as a particle of mass $m$ kg moving on the inside surface of the circular loop. You may assume that the track is smooth.
At point $A$, which is the lowest point of the circle, the car has velocity $u \mathrm {~ms} ^ { - 1 }$ such that $u ^ { 2 } = 60 g$. When the car is at point $B$ the radius makes an angle $\theta$ with the downward vertical.
\begin{enumerate}[label=(\alph*)]
\item Find, in terms of $\theta$ and $g$, an expression for $v ^ { 2 }$, where $v \mathrm {~ms} ^ { - 1 }$ is the speed of the car at $B$.
\item Show that $R \mathrm {~N}$, the reaction of the track on the car at $B$, is given by
$$R = m g ( 4 + 3 \cos \theta ) .$$
\item Explain why the expression for $R$ in part (b) shows that the car will perform a complete loop.
\item This model predicts that the car will stop at $C$ at a vertical height of 30 m above $A$. However, after the car has completed the loop, the track becomes rough and the car only reaches a point $D$ at a vertical height of 28 m above $A$. The resistance to motion of the car beyond the loop is of constant magnitude $\frac { m g } { 32 } \mathrm {~N}$. Calculate the length of the rough track between $A$ and $D$.
\end{enumerate}
\hfill \mbox{\textit{WJEC Further Unit 3 2019 Q6 [13]}}