| Exam Board | CAIE |
|---|---|
| Module | Further Paper 3 (Further Paper 3) |
| Year | 2022 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circular Motion 2 |
| Type | Vertical circle: tension at specific point |
| Difficulty | Challenging +1.8 This is a challenging vertical circle problem requiring energy conservation and Newton's second law at two different positions, with careful angle geometry. The given speed relationship and tension ratio create a system requiring algebraic manipulation, but the provided value of u simplifies calculations. More demanding than standard circular motion but follows established methods. |
| Spec | 3.03d Newton's second law: 2D vectors6.02i Conservation of energy: mechanical energy principle6.05f Vertical circle: motion including free fall |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\frac{1}{2}mv^2 - \frac{1}{2}mu^2 = mga(\cos\theta + \cos\alpha)\) | M1 | Energy equation with all necessary terms, GPE terms must be resolved, allow sin/cos mix, allow sign error |
| \(\frac{1}{2}m(2u)^2 - \frac{1}{2}mu^2 = mga(\cos\theta + \cos\alpha)\) | A1 | \(2u\) may be substituted later. Implied by \(\frac{3}{2} \times \frac{2}{3}ag = ga(\cos\theta + \cos\alpha)\) |
| At \(A\): \(T + mg\cos\theta = \frac{m}{a}u^2\) | B1 | N2L |
| Also: \(10T - mg\cos\alpha = \frac{m}{a}4u^2\) | B1 | N2L and use of tension \(10T\) |
| Use all three equations to find \(T\) in terms of \(m\) and \(g\) only | M1 | Might see: \(9T - mg(\cos\theta + \cos\alpha) = \frac{3m}{a} \times \frac{2}{3}ga\); \((\cos\theta + \cos\alpha) = 1\); \((10\cos\theta + \cos\alpha) = 4\) |
| \(T = \frac{1}{3}mg\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Substitute back: \(10 \times \frac{1}{3}mg - mg\cos\alpha = \frac{4m}{a} \times \frac{2}{3}ga\) | M1 | Any appropriate method to obtain \(\cos\alpha\) |
| \(\cos\alpha = \frac{2}{3}\) | A1 |
## Question 4(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{1}{2}mv^2 - \frac{1}{2}mu^2 = mga(\cos\theta + \cos\alpha)$ | M1 | Energy equation with all necessary terms, GPE terms must be resolved, allow sin/cos mix, allow sign error |
| $\frac{1}{2}m(2u)^2 - \frac{1}{2}mu^2 = mga(\cos\theta + \cos\alpha)$ | A1 | $2u$ may be substituted later. Implied by $\frac{3}{2} \times \frac{2}{3}ag = ga(\cos\theta + \cos\alpha)$ |
| At $A$: $T + mg\cos\theta = \frac{m}{a}u^2$ | B1 | N2L |
| Also: $10T - mg\cos\alpha = \frac{m}{a}4u^2$ | B1 | N2L and use of tension $10T$ |
| Use all three equations to find $T$ in terms of $m$ and $g$ only | M1 | Might see: $9T - mg(\cos\theta + \cos\alpha) = \frac{3m}{a} \times \frac{2}{3}ga$; $(\cos\theta + \cos\alpha) = 1$; $(10\cos\theta + \cos\alpha) = 4$ |
| $T = \frac{1}{3}mg$ | A1 | |
---
## Question 4(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Substitute back: $10 \times \frac{1}{3}mg - mg\cos\alpha = \frac{4m}{a} \times \frac{2}{3}ga$ | M1 | Any appropriate method to obtain $\cos\alpha$ |
| $\cos\alpha = \frac{2}{3}$ | A1 | |
---4 One end of a light inextensible string of length $a$ is attached to a fixed point $O$. A particle of mass $m$ is attached to the other end of the string and is held with the string taut at the point $A$. At $A$ the string makes an angle $\theta$ with the upward vertical through $O$. The particle is projected perpendicular to the string in a downward direction from $A$ with a speed $u$. It moves along a circular path in the vertical plane.
When the string makes an angle $\alpha$ with the downward vertical through $O$, the speed of the particle is $2 u$ and the magnitude of the tension in the string is 10 times its magnitude at $A$.
It is given that $\mathrm { u } = \sqrt { \frac { 2 } { 3 } \mathrm { ga } }$.
\begin{enumerate}[label=(\alph*)]
\item Find, in terms of $m$ and $g$, the magnitude of the tension in the string at $A$.
\item Find the value of $\cos \alpha$.
\end{enumerate}
\hfill \mbox{\textit{CAIE Further Paper 3 2022 Q4 [8]}}