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 with algebraic manipulation and the constraint R_B = 10R_A, the techniques are routine for Further Maths students: setting up force equations at two points, using conservation of energy, and finding the loss-of-contact condition. The problem is slightly above average difficulty due to the multi-step nature and algebraic complexity, but follows a well-established template for this topic.
Spec6.02i Conservation of energy: mechanical energy principle6.05d Variable speed circles: energy methods
3 \includegraphics[max width=\textwidth, alt={}, center]{2c6b6722-ebba-4ade-9a9d-dd70e61cf52b-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
Working/AnswerMark 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 \(v_B^2\)
\(\left[v_B^2 = \frac{17ag}{5}\right]\)
\(u^2 = \left(\frac{5ag}{3}\right),\quad \cos\alpha = ag\ \mathbf{A.G.}\)M1 A1
Working/AnswerMark Guidance
\(\frac{1}{2}mv^2 = \frac{1}{2}mu^2 + mga(\cos\alpha - \cos\theta)\) Use conservation of energy for loss of contact
\(\underline{or}\quad \frac{1}{2}mv_B^2 - mga(\cos\alpha + \cos\theta)\)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:

| Working/Answer | 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 $v_B^2$ |
| $\left[v_B^2 = \frac{17ag}{5}\right]$ | | |
| $u^2 = \left(\frac{5ag}{3}\right),\quad \cos\alpha = ag\ \mathbf{A.G.}$ | M1 A1 | |

| Working/Answer | Mark | Guidance |
|---|---|---|
| $\frac{1}{2}mv^2 = \frac{1}{2}mu^2 + mga(\cos\alpha - \cos\theta)$ | | Use conservation of energy for loss of contact |
| $\underline{or}\quad \frac{1}{2}mv_B^2 - mga(\cos\alpha + \cos\theta)$ | 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]{2c6b6722-ebba-4ade-9a9d-dd70e61cf52b-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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