CAIE FP2 2016 June — Question 4 10 marks

Exam BoardCAIE
ModuleFP2 (Further Pure Mathematics 2)
Year2016
SessionJune
Marks10
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicCircular Motion 2
TypeParticle on inner surface of sphere/bowl
DifficultyChallenging +1.2 This is a standard two-part circular motion problem requiring energy conservation and the condition for losing contact (N=0 giving v²=g·r·cosθ). While it involves multiple steps (finding θ, then u, then projectile motion), each step follows well-established methods taught in Further Maths mechanics. The algebraic manipulation is moderate but routine for this level.
Spec6.02d Mechanical energy: KE and PE concepts6.02e Calculate KE and PE: using formulae6.02i Conservation of energy: mechanical energy principle6.05c Horizontal circles: conical pendulum, banked tracks6.05e Radial/tangential acceleration

4 A particle \(P\) is at rest at the lowest point on the smooth inner surface of a hollow sphere with centre \(O\) and radius \(a\). The particle is projected horizontally with speed \(u\) and begins to move in a vertical circle on the inner surface of the sphere. The particle loses contact with the sphere at the point \(A\), where \(O A\) makes an angle \(\theta\) with the upward vertical through \(O\). Given that the speed of \(P\) at \(A\) is \(\sqrt { } \left( \frac { 3 } { 5 } a g \right)\), find \(u\) in terms of \(a\) and \(g\). Find, in terms of \(a\), the greatest height above the level of \(O\) achieved by \(P\) in its subsequent motion. (You may assume that \(P\) achieves its greatest height before it makes any further contact with the sphere.)

Question 4:
Part 1 (finding \(u\)):
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\(\frac{1}{2}mv^2 = \frac{1}{2}mu^2 - mga(1 + \cos\theta)\)M1 A1 Find \(v^2\) at \(A\) from conservation of energy
\(mv^2/a = mg\cos\theta\)B1 Use \(F = ma\) radially at \(A\) with \(R = 0\)
\(\cos\theta = 3/5\) Use \(v^2 = 3ag/5\) and eliminate \(\cos\theta\) to find \(u\)
\(u^2 = v^2 + 2ag(1 + \cos\theta) = 19ag/5\)
\(u = \sqrt{(19ag/5)}\) or \(1.95\sqrt{(ag)}\)M1 A1
or \(\sqrt{(38ag)}\) or \(6.16\sqrt{(a)}\) allow numerical value of \(g\); Total: [5]
Part 2 (finding \(h_O\)):
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\(V = v\sin\theta [= 4v/5]\)M1 Find or imply vertical component of speed at \(A\)
\(h_A = V^2/2g = 24a/125\) or \(0.192a\)M1 A1 Find greatest height \(h_A\) reached above \(A\)
\(h_O = h_A + a\cos\theta\)M1 A1 Find greatest height \(h_O\) reached above \(O\)
\(= 99a/125\) or \(0.792a\) Total: [5]
## Question 4:

### Part 1 (finding $u$):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{1}{2}mv^2 = \frac{1}{2}mu^2 - mga(1 + \cos\theta)$ | M1 A1 | Find $v^2$ at $A$ from conservation of energy |
| $mv^2/a = mg\cos\theta$ | B1 | Use $F = ma$ radially at $A$ with $R = 0$ |
| $\cos\theta = 3/5$ | | Use $v^2 = 3ag/5$ and eliminate $\cos\theta$ to find $u$ |
| $u^2 = v^2 + 2ag(1 + \cos\theta) = 19ag/5$ | | |
| $u = \sqrt{(19ag/5)}$ or $1.95\sqrt{(ag)}$ | M1 A1 | |
| or $\sqrt{(38ag)}$ or $6.16\sqrt{(a)}$ | | allow numerical value of $g$; Total: [5] |

### Part 2 (finding $h_O$):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $V = v\sin\theta [= 4v/5]$ | M1 | Find or imply vertical component of speed at $A$ |
| $h_A = V^2/2g = 24a/125$ or $0.192a$ | M1 A1 | Find greatest height $h_A$ reached above $A$ |
| $h_O = h_A + a\cos\theta$ | M1 A1 | Find greatest height $h_O$ reached above $O$ |
| $= 99a/125$ or $0.792a$ | | Total: [5] |

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4 A particle $P$ is at rest at the lowest point on the smooth inner surface of a hollow sphere with centre $O$ and radius $a$. The particle is projected horizontally with speed $u$ and begins to move in a vertical circle on the inner surface of the sphere. The particle loses contact with the sphere at the point $A$, where $O A$ makes an angle $\theta$ with the upward vertical through $O$. Given that the speed of $P$ at $A$ is $\sqrt { } \left( \frac { 3 } { 5 } a g \right)$, find $u$ in terms of $a$ and $g$.

Find, in terms of $a$, the greatest height above the level of $O$ achieved by $P$ in its subsequent motion. (You may assume that $P$ achieves its greatest height before it makes any further contact with the sphere.)

\hfill \mbox{\textit{CAIE FP2 2016 Q4 [10]}}