CAIE FP2 2015 June — Question 3 9 marks

Exam BoardCAIE
ModuleFP2 (Further Pure Mathematics 2)
Year2015
SessionJune
Marks9
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicCircular Motion 1
TypeVertical circle – surface contact (sphere/track, leaving surface)
DifficultyStandard +0.8 This is a classic vertical circular motion problem requiring energy conservation and Newton's second law in the radial direction. While the derivation of the reaction force formula is standard Further Maths fare, the second part requires students to combine the loss-of-contact condition (R=0) with the constraint v=2u and the energy equation, involving algebraic manipulation of multiple equations. This is moderately challenging but follows established patterns for FP2/FM mechanics.
Spec1.02f Solve quadratic equations: including in a function of unknown3.03d Newton's second law: 2D vectors6.05d Variable speed circles: energy methods

3 A particle \(P\), of mass \(m\), is placed at the highest point of a fixed solid smooth sphere with centre \(O\) and radius \(a\). The particle \(P\) is given a horizontal speed \(u\) and it moves in part of a vertical circle, with centre \(O\), on the surface of the sphere. When \(O P\) makes an angle \(\theta\) with the upward vertical, and \(P\) is still in contact with the surface of the sphere, the speed of \(P\) is \(v\) and the reaction of the sphere on \(P\) has magnitude \(R\). Show that \(R = m g ( 3 \cos \theta - 2 ) - \frac { m u ^ { 2 } } { a }\). The particle loses contact with the sphere at the instant when \(v = 2 u\). Find \(u\) in terms of \(a\) and \(g\).

Question 3:
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\(\frac{1}{2}mv^2 = \frac{1}{2}mu^2 + mga(1-\cos\theta)\)M1 A1 Find \(v^2\) from conservation of energy
\(R = mg\cos\theta - mv^2/a\)B1 Find \(R\) by using \(F = ma\) radially
\(R = mg(3\cos\theta - 2) - mu^2/a\)M1 A1 Eliminate \(v^2\) to find \(R\): AG
\(u^2 = ag(3\cos\theta - 2)\) *or* \(v^2 = ag\cos\theta\)B1 Find \(u^2\) or \(v^2\) in terms of \(\cos\theta\) when \(R=0\)
*EITHER:* \(4u^2 = u^2 + 2ag - \frac{2}{3}(u^2 + 2ag)\)M1 A1 Replace \(\cos\theta\) in energy eqn with \(v = 2u\)
*or* \(u^2 + 2ag - 8u^2\)
*OR:* \([v^2/ag =]\ 4(3\cos\theta - 2) = \cos\theta\) Find \(\cos\theta\) and substitute in energy eqn
\(\cos\theta = 8/11\)
\(4u^2 = u^2 + 2ag(1 - 8/11)\)(M1 A1)
\(u = \sqrt{(2ag/11)}\) *or* \(0.426\sqrt{(ag)}\)A1
Total: 9 marks
## Question 3:
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{1}{2}mv^2 = \frac{1}{2}mu^2 + mga(1-\cos\theta)$ | M1 A1 | Find $v^2$ from conservation of energy |
| $R = mg\cos\theta - mv^2/a$ | B1 | Find $R$ by using $F = ma$ radially |
| $R = mg(3\cos\theta - 2) - mu^2/a$ | M1 A1 | Eliminate $v^2$ to find $R$: AG |
| $u^2 = ag(3\cos\theta - 2)$ *or* $v^2 = ag\cos\theta$ | B1 | Find $u^2$ or $v^2$ in terms of $\cos\theta$ when $R=0$ |
| *EITHER:* $4u^2 = u^2 + 2ag - \frac{2}{3}(u^2 + 2ag)$ | M1 A1 | Replace $\cos\theta$ in energy eqn with $v = 2u$ |
| *or* $u^2 + 2ag - 8u^2$ | | |
| *OR:* $[v^2/ag =]\ 4(3\cos\theta - 2) = \cos\theta$ | | Find $\cos\theta$ and substitute in energy eqn |
| $\cos\theta = 8/11$ | | |
| $4u^2 = u^2 + 2ag(1 - 8/11)$ | (M1 A1) | |
| $u = \sqrt{(2ag/11)}$ *or* $0.426\sqrt{(ag)}$ | A1 | |
| **Total: 9 marks** | | |
3 A particle $P$, of mass $m$, is placed at the highest point of a fixed solid smooth sphere with centre $O$ and radius $a$. The particle $P$ is given a horizontal speed $u$ and it moves in part of a vertical circle, with centre $O$, on the surface of the sphere. When $O P$ makes an angle $\theta$ with the upward vertical, and $P$ is still in contact with the surface of the sphere, the speed of $P$ is $v$ and the reaction of the sphere on $P$ has magnitude $R$. Show that $R = m g ( 3 \cos \theta - 2 ) - \frac { m u ^ { 2 } } { a }$.

The particle loses contact with the sphere at the instant when $v = 2 u$. Find $u$ in terms of $a$ and $g$.

\hfill \mbox{\textit{CAIE FP2 2015 Q3 [9]}}