CAIE FP2 2014 November — Question 3 10 marks

Exam BoardCAIE
ModuleFP2 (Further Pure Mathematics 2)
Year2014
SessionNovember
Marks10
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicCircular Motion 2
TypeParticle on outer surface of cylinder
DifficultyChallenging +1.2 This is a standard circular motion problem on the inner surface of a cylinder requiring application of Newton's second law in the radial direction and energy conservation. While it involves two parts and algebraic manipulation with the constraint R_B = 10R_A, the setup is conventional and the techniques are well-practiced in Further Maths. The loss of contact condition (R=0) is a standard concept. More routine than average FM problems but requires careful bookkeeping.
Spec6.02i Conservation of energy: mechanical energy principle6.05d Variable speed circles: energy methods6.05e Radial/tangential acceleration
3 \includegraphics[max width=\textwidth, alt={}, center]{5d40f5b4-e3d4-482c-8d8d-05a01bd3b43f-2_413_414_1155_863} A smooth cylinder of radius \(a\) is fixed with its axis horizontal. The point \(O\) is the centre of a circular cross-section of the cylinder. The line \(A O B\) is a diameter of this circular cross-section and the radius \(O A\) makes an angle \(\alpha\) with the upward vertical (see diagram). It is given that \(\cos \alpha = \frac { 3 } { 5 }\). A particle \(P\) of mass \(m\) moves on the inner surface of the cylinder in the plane of the cross-section. The particle passes through \(A\) with speed \(u\) along the surface in the downwards direction. The magnitude of the reaction between \(P\) and the inner surface of the sphere is \(R _ { A }\) when \(P\) is at \(A\), and is \(R _ { B }\) when \(P\) is at \(B\). It is given that \(R _ { B } = 10 R _ { A }\). Show that \(u ^ { 2 } = a g\). The particle loses contact with the surface of the cylinder when \(O P\) makes an angle \(\theta\) with the upward vertical. Find the value of \(\cos \theta\).
Question 3:
AnswerMarks Guidance
Answer/WorkingMark Guidance
\(\frac{1}{2}mv_B^2 = \frac{1}{2}mu^2 + 2mga\cos\alpha\)B1 Conservation of energy
\(\left[v_B^2 = u^2 + \frac{12ag}{5}\right]\)
\(R_A = \frac{mu^2}{a} - mg\cos\alpha\)B1 Use \(F=ma\) radially at \(A\) and \(B\) (B1 for either)
\(R_B = \frac{mv_B^2}{a} + mg\cos\alpha\)
\(\frac{mv_B^2}{a} + mg\cos\alpha = 10\left(\frac{mu^2}{a} - mg\cos\alpha\right)\)M1 A1 Equate \(R_B\) to \(10R_A\)
\(u^2 + 4ag\cos\alpha = 10u^2 - 11ag\cos\alpha\) Eliminate \(v_B^2\)
\(\left[v_B^2 = \frac{17ag}{5}\right]\)
\(u^2 = \left(\frac{5ag}{3}\right)\ \cos\alpha = ag\) A.G.M1 A1
\(\frac{1}{2}mv^2 = \frac{1}{2}mu^2 + mga(\cos\alpha - \cos\theta)\) Conservation of energy for loss of contact
\(\underline{or}\ \frac{1}{2}mv_B^2 - mga(\cos\alpha + \cos\theta)\)B1
\(\frac{mv^2}{a} = mg\cos\theta\)B1 Use \(F=ma\) radially with \(R=0\)
\(ga + 2ga(\cos\alpha - \cos\theta) = ga\cos\theta\)M1 A1 Eliminate \(v^2\) with \(u^2 = ag\) to find \(\cos\theta\)
\(\cos\theta = \frac{1}{3}(2\cos\alpha + 1) = \frac{11}{15}\)
Total: [10]
## Question 3:

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{1}{2}mv_B^2 = \frac{1}{2}mu^2 + 2mga\cos\alpha$ | B1 | Conservation of energy |
| $\left[v_B^2 = u^2 + \frac{12ag}{5}\right]$ | | |
| $R_A = \frac{mu^2}{a} - mg\cos\alpha$ | B1 | Use $F=ma$ radially at $A$ and $B$ (B1 for either) |
| $R_B = \frac{mv_B^2}{a} + mg\cos\alpha$ | | |
| $\frac{mv_B^2}{a} + mg\cos\alpha = 10\left(\frac{mu^2}{a} - mg\cos\alpha\right)$ | M1 A1 | Equate $R_B$ to $10R_A$ |
| $u^2 + 4ag\cos\alpha = 10u^2 - 11ag\cos\alpha$ | | Eliminate $v_B^2$ |
| $\left[v_B^2 = \frac{17ag}{5}\right]$ | | |
| $u^2 = \left(\frac{5ag}{3}\right)\ \cos\alpha = ag$ **A.G.** | M1 A1 | |
| $\frac{1}{2}mv^2 = \frac{1}{2}mu^2 + mga(\cos\alpha - \cos\theta)$ | | Conservation of energy for loss of contact |
| $\underline{or}\ \frac{1}{2}mv_B^2 - mga(\cos\alpha + \cos\theta)$ | B1 | |
| $\frac{mv^2}{a} = mg\cos\theta$ | B1 | Use $F=ma$ radially with $R=0$ |
| $ga + 2ga(\cos\alpha - \cos\theta) = ga\cos\theta$ | M1 A1 | Eliminate $v^2$ with $u^2 = ag$ to find $\cos\theta$ |
| $\cos\theta = \frac{1}{3}(2\cos\alpha + 1) = \frac{11}{15}$ | | |

**Total: [10]**

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\includegraphics[max width=\textwidth, alt={}, center]{5d40f5b4-e3d4-482c-8d8d-05a01bd3b43f-2_413_414_1155_863}

A smooth cylinder of radius $a$ is fixed with its axis horizontal. The point $O$ is the centre of a circular cross-section of the cylinder. The line $A O B$ is a diameter of this circular cross-section and the radius $O A$ makes an angle $\alpha$ with the upward vertical (see diagram). It is given that $\cos \alpha = \frac { 3 } { 5 }$. A particle $P$ of mass $m$ moves on the inner surface of the cylinder in the plane of the cross-section. The particle passes through $A$ with speed $u$ along the surface in the downwards direction. The magnitude of the reaction between $P$ and the inner surface of the sphere is $R _ { A }$ when $P$ is at $A$, and is $R _ { B }$ when $P$ is at $B$. It is given that $R _ { B } = 10 R _ { A }$. Show that $u ^ { 2 } = a g$.

The particle loses contact with the surface of the cylinder when $O P$ makes an angle $\theta$ with the upward vertical. Find the value of $\cos \theta$.

\hfill \mbox{\textit{CAIE FP2 2014 Q3 [10]}}
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