Explanation

 (a) 

If d = deceleration, then retarding force = \(md\).

This is also the force down the plane \(mg \sin \theta\)

\(\therefore md = mg\sin \theta \implies d = g \sin \theta\)

= \(10 \sin 30° = 10 \times \frac{1}{2} = 5 ms^{-2}\)

(i) \(v^{2} = u^{2} + 2as\)

At the highest point, v = 0 ms\(^{-1}\).

\(0^{2} = u^{2} - 2(5)(15) \)

\(u^{2} = 150 \implies u = \sqrt{150} = 12.247 ms^{-1}\)

(ii) \(v = u - dt\)

\(0 = 12.247 - 5t \implies 5t = 12.247\)

\(t = \frac{12.247}{5} = 2.4494 \approxeq 2.45 s\)

(b)(i) Resultant force 

\(F = \begin{pmatrix} 10 \cos 0° \\ 10 \sin 0° \end{pmatrix} + \begin{pmatrix} 5\sqrt{3} \cos 30° \\ 5\sqrt{3} \sin 30° \end{pmatrix} + \begin{pmatrix} 5 \cos 30° \\ 5 \sin 30° \end{pmatrix} + \begin{pmatrix} 5 \cos 300° \\ 5 \sin 300° \end{pmatrix} + \begin{pmatrix} 5\sqrt{3} \cos 330° \\ 5 \sqrt{3} \sin 330° \end{pmatrix}\)

= \(\begin{pmatrix} 10 \times 1 \\ 10 \times 0 \end{pmatrix} + \begin{pmatrix} 5\sqrt{3} \times \frac{\sqrt{3}}{2} \\ 5 \sqrt{3} \times \frac{1}{2} \end{pmatrix} + \begin{pmatrix} 5 \times \frac{1}{2} \\ 5 \times \frac{\sqrt{3}}{2} \end{pmatrix} + \begin{pmatrix} 5 \times \frac{1}{2} \\ -5 \times \frac{\sqrt{3}}{2} \end{pmatrix} + \begin{pmatrix} 5\sqrt{3} \times \frac{\sqrt{3}}{2} \\ -5\sqrt{3} \times \frac{1}{2} \end{pmatrix}\)

= \(\begin{pmatrix} 10 \\ 0 \end{pmatrix} + \begin{pmatrix} 7.5 \\ 2.5 \sqrt{3} \end{pmatrix} + \begin{pmatrix} 2.5 \\ 2.5\sqrt{3} \end{pmatrix} + \begin{pmatrix} 2.5 \\ -2.5 \sqrt{3} \end{pmatrix} + \begin{pmatrix} 7.5 \\ -2.5 \sqrt{3} \end{pmatrix}\)

= \(\begin{pmatrix} 30 \\ 0 \end{pmatrix}\)

Magnitude of resultant = \(\sqrt{30^{2}} = 30N\)

(ii) \(F = ma \implies a = \frac{F}{m}\)

= \(\frac{30}{5} = 6 ms^{-2}\)

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