CAIE FP2 2011 June — Question 5 12 marks

Exam BoardCAIE
ModuleFP2 (Further Pure Mathematics 2)
Year2011
SessionJune
Marks12
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicCircular Motion 2
TypeParticle on outer surface of sphere
DifficultyChallenging +1.2 This is a standard Further Maths mechanics problem on motion on a sphere with projectile motion afterward. It requires energy conservation, circular motion equations (N=0 at loss of contact), and projectile trajectory analysis, but follows a well-established template with straightforward algebra. The multi-part structure and specific numerical verification elevate it slightly above average difficulty.
Spec6.05e Radial/tangential acceleration

5 A particle \(P\) of mass \(m\) is placed at the point \(Q\) on the outer surface of a fixed smooth sphere with centre \(O\) and radius \(a\). The acute angle between \(O Q\) and the upward vertical is \(\alpha\), where \(\cos \alpha = \frac { 9 } { 10 }\). The particle is released from rest and begins to move in a vertical circle on the surface of the sphere. Show that \(P\) loses contact with the sphere when \(O P\) makes an angle \(\theta\) with the upward vertical, where \(\cos \theta = \frac { 3 } { 5 }\), and find the speed of \(P\) at this instant. Show that, in the subsequent motion, when \(P\) is at a distance \(\frac { 7 } { 5 } a\) from the vertical diameter through \(O\), its distance below the horizontal through \(O\) is \(\frac { 31 } { 30 } a\).

Question 5:
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\(\frac{1}{2}mv^2 = mga(\cos\alpha - \cos\theta)\)M1 A1 Use conservation of energy
\(mv^2/a + R = mg\cos\theta\)M1 Equate radial forces [may imply \(R=0\)]
\(mv^2/a = mg\cos\theta\)A1 Take \(R=0\) when contact lost
\(2(\cos\alpha - \cos\theta) = \cos\theta\) Eliminate \(v^2\)
\(\cos\theta = \frac{2}{3}\cos\alpha = 3/5\) A.G.B1 [or can show \(R=0\) when \(\cos\theta = 3/5\)]
\(v = \sqrt{(3ag/5)}\) or \(\sqrt{(6a)}\)B1 Find \(v\)
\(x = 7a/5 - 4a/5\ [= 3a/5]\)B1 Find horizontal distance moved
\(t = x/v\cos\theta\)M1 *EITHER:* Find time of subsequent motion
\([= a/v = \sqrt{(5a/3g)} = \sqrt{(a/6)}]\)
\(h = (v\sin\theta)t + \frac{1}{2}gt^2\)M1 Find distance \(h\) fallen in time \(t\)
\(h = x\tan\theta + \frac{1}{2}g(x/v\cos\theta)^2\)(M2) *OR:* Use trajectory eqn to find \(h\)
\(h = 4a/5 + 5a/6 = 49a/30\)M1 A1 Substitute and simplify \(h\)
\(h - 3a/5 = 31a/30\) A.G.B1 Find distance below horizontal through \(O\)
Total: [12]
## Question 5:

| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{1}{2}mv^2 = mga(\cos\alpha - \cos\theta)$ | M1 A1 | Use conservation of energy |
| $mv^2/a + R = mg\cos\theta$ | M1 | Equate radial forces [may imply $R=0$] |
| $mv^2/a = mg\cos\theta$ | A1 | Take $R=0$ when contact lost |
| $2(\cos\alpha - \cos\theta) = \cos\theta$ | | Eliminate $v^2$ |
| $\cos\theta = \frac{2}{3}\cos\alpha = 3/5$ **A.G.** | B1 | [or can show $R=0$ when $\cos\theta = 3/5$] |
| $v = \sqrt{(3ag/5)}$ or $\sqrt{(6a)}$ | B1 | Find $v$ |
| $x = 7a/5 - 4a/5\ [= 3a/5]$ | B1 | Find horizontal distance moved |
| $t = x/v\cos\theta$ | M1 | *EITHER:* Find time of subsequent motion |
| $[= a/v = \sqrt{(5a/3g)} = \sqrt{(a/6)}]$ | | |
| $h = (v\sin\theta)t + \frac{1}{2}gt^2$ | M1 | Find distance $h$ fallen in time $t$ |
| $h = x\tan\theta + \frac{1}{2}g(x/v\cos\theta)^2$ | (M2) | *OR:* Use trajectory eqn to find $h$ |
| $h = 4a/5 + 5a/6 = 49a/30$ | M1 A1 | Substitute and simplify $h$ |
| $h - 3a/5 = 31a/30$ **A.G.** | B1 | Find distance below horizontal through $O$ |

**Total: [12]**

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5 A particle $P$ of mass $m$ is placed at the point $Q$ on the outer surface of a fixed smooth sphere with centre $O$ and radius $a$. The acute angle between $O Q$ and the upward vertical is $\alpha$, where $\cos \alpha = \frac { 9 } { 10 }$. The particle is released from rest and begins to move in a vertical circle on the surface of the sphere. Show that $P$ loses contact with the sphere when $O P$ makes an angle $\theta$ with the upward vertical, where $\cos \theta = \frac { 3 } { 5 }$, and find the speed of $P$ at this instant.

Show that, in the subsequent motion, when $P$ is at a distance $\frac { 7 } { 5 } a$ from the vertical diameter through $O$, its distance below the horizontal through $O$ is $\frac { 31 } { 30 } a$.

\hfill \mbox{\textit{CAIE FP2 2011 Q5 [12]}}