CAIE FP2 2014 November — Question 2 5 marks

Exam BoardCAIE
ModuleFP2 (Further Pure Mathematics 2)
Year2014
SessionNovember
Marks5
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Mark schemeDownload PDF ↗
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 the Pythagorean theorem. 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 impact
2 \includegraphics[max width=\textwidth, alt={}, center]{2c6b6722-ebba-4ade-9a9d-dd70e61cf52b-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
Working/AnswerMark 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 \(V=2\)
\(1 - \sin^2\alpha + 0.16\sin^2\alpha = 0.25\)
\(\sin^2\alpha = \frac{0.75}{0.84} = \frac{25}{28} = 0.8929\)
\(\underline{or}\quad \cos^2\alpha = \frac{3}{28} = 0.1071\)
\(\alpha = 1.24\ \text{rad}\ or\ 70.9°\)M1 A1
Total: [5]
## Question 2:

| Working/Answer | 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 $V=2$ |
| $1 - \sin^2\alpha + 0.16\sin^2\alpha = 0.25$ | | |
| $\sin^2\alpha = \frac{0.75}{0.84} = \frac{25}{28} = 0.8929$ | | |
| $\underline{or}\quad \cos^2\alpha = \frac{3}{28} = 0.1071$ | | |
| $\alpha = 1.24\ \text{rad}\ or\ 70.9°$ | M1 A1 | |

**Total: [5]**

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\includegraphics[max width=\textwidth, alt={}, center]{2c6b6722-ebba-4ade-9a9d-dd70e61cf52b-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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