CAIE FP2 2014 November — Question 2 5 marks

Exam BoardCAIE
ModuleFP2 (Further Pure Mathematics 2)
Year2014
SessionNovember
Marks5
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TopicOblique and successive collisions
TypeSphere rebounds off fixed wall obliquely
DifficultyStandard +0.3 This is a standard 2D collision problem requiring resolution of velocity components, application of Newton's experimental law (coefficient of restitution), and solving a trigonometric equation. While it involves multiple steps, the techniques are routine for Further Maths students and the problem follows a well-established template with no novel insight required.
Spec6.03k Newton's experimental law: direct impact6.03l Newton's law: oblique impacts
2 \includegraphics[max width=\textwidth, alt={}, center]{5d40f5b4-e3d4-482c-8d8d-05a01bd3b43f-2_312_409_525_868} A small smooth ball \(P\) is moving on a smooth horizontal plane with speed \(4 \mathrm {~m} \mathrm {~s} ^ { - 1 }\). It strikes a smooth vertical barrier at an angle \(\alpha\) (see diagram). The coefficient of restitution between \(P\) and the barrier is 0.4 . Given that the speed of \(P\) is halved as a result of the collision, find the value of \(\alpha\).
Question 2:
AnswerMarks Guidance
Answer/WorkingMark Guidance
\(V\cos\beta = 4\cos\alpha\)B1 Speed component along barrier
\(V\sin\beta = 0.4 \times 4\sin\alpha\)B1 Speed component normal to barrier
\(V^2 = 2^2 = 1.6^2\sin^2\alpha + 16\cos^2\alpha\)M1 Find \(\beta\) by eliminating \(\alpha\) with \(V=2\)
\(1 - \sin^2\alpha + 0.16\sin^2\alpha = 0.25\)
\(\sin^2\alpha = \frac{0.75}{0.84} = \frac{25}{28} = 0.8929\)
\(\underline{or}\quad \cos^2\alpha = \frac{3}{28} = 0.1071\)
\(\alpha = 1.24\) rad \(or\ 70.9°\)M1 A1
Total: [5]
## Question 2:

| Answer/Working | Mark | Guidance |
|---|---|---|
| $V\cos\beta = 4\cos\alpha$ | B1 | Speed component along barrier |
| $V\sin\beta = 0.4 \times 4\sin\alpha$ | B1 | Speed component normal to barrier |
| $V^2 = 2^2 = 1.6^2\sin^2\alpha + 16\cos^2\alpha$ | M1 | Find $\beta$ by eliminating $\alpha$ with $V=2$ |
| $1 - \sin^2\alpha + 0.16\sin^2\alpha = 0.25$ | | |
| $\sin^2\alpha = \frac{0.75}{0.84} = \frac{25}{28} = 0.8929$ | | |
| $\underline{or}\quad \cos^2\alpha = \frac{3}{28} = 0.1071$ | | |
| $\alpha = 1.24$ rad $or\ 70.9°$ | M1 A1 | |

**Total: [5]**

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\includegraphics[max width=\textwidth, alt={}, center]{5d40f5b4-e3d4-482c-8d8d-05a01bd3b43f-2_312_409_525_868}

A small smooth ball $P$ is moving on a smooth horizontal plane with speed $4 \mathrm {~m} \mathrm {~s} ^ { - 1 }$. It strikes a smooth vertical barrier at an angle $\alpha$ (see diagram). The coefficient of restitution between $P$ and the barrier is 0.4 . Given that the speed of $P$ is halved as a result of the collision, find the value of $\alpha$.

\hfill \mbox{\textit{CAIE FP2 2014 Q2 [5]}}
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