Solution to Test Paper 3
From Mechanics
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Revision as of 09:20, 8 April 2012
| 1 (a) | \displaystyle \begin{align}
& 0=24.5-9.8t \\ & t=\frac{24.5}{9.8}=2.5\text{ seconds} \\ \end{align} | (3 marks)
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| 1 (b) | \displaystyle \begin{align}
& {{0}^{2}}={{24.5}^{2}}+2\times (-9.8)\times s \\ & s=\frac{{{24.5}^{2}}}{2\times 9.8}=30.625\text{ m} \\ \end{align}
\displaystyle s=\frac{1}{2}(24.5+0)\times 2.5=30.625\text{ m} | (3 marks) |
| 1 (c) | 61.25 m
| (1 marks) |
| (7 marks)
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| 2 (a) | 
| (1 marks)
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| 2 (b) | T = 4900 N | (1 marks)
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| 2 (c) | \displaystyle \begin{align}
& T-4900=500\times 0.6 \\ & T=300+4900=5200\text{ N} \\ \end{align} | (3 mark) |
| 3 (d) | \displaystyle \begin{align}
& 4800-4900=500a \\ & a=\frac{-100}{500}=-0.2\text{ m}{{\text{s}}^{\text{-2}}} \\ & 0.\text{2 m}{{\text{s}}^{-\text{2}}}\quad \text{downwards} \\ \end{align} | (4 mark) |
| (9 marks)
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| 3 (a) | \displaystyle R=5\times 9.8\cos 42{}^\circ =36.4\text{ N} | (2 mark)
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| 3 (b) | \displaystyle F=0.4R=14.6\text{ N} | (2 marks)
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| 3 (c) | \displaystyle \begin{align}
& 5\times 9.8\sin 42{}^\circ -14.6=5a \\ & a=\frac{18.22}{5}=3.64\text{ m}{{\text{s}}^{\text{-2}}} \\ \end{align} | (3 marks)
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| 3 (d) | \displaystyle \begin{align}
& 2=\frac{1}{2}\times 3.64{{t}^{2}} \\ & t=\sqrt{\frac{4}{3.64}}=1.05\text{ seconds} \\ \end{align}
| (3 marks)
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| (10 marks)
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| 4 (a) | \displaystyle \begin{align}
& Q\cos 40{}^\circ =8\times 9.8\sin 40{}^\circ \\ & Q=\frac{8\times 9.8\sin 40{}^\circ }{\cos 40{}^\circ }=65.8\text{ N} \\ \end{align} | (3 mark)
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| 4 (b) | \displaystyle \begin{align}
& Q\cos 40{}^\circ +F=8\times 9.8\sin 40{}^\circ \\ & Q\sin 40{}^\circ +8\times 9.8\cos 40{}^\circ =R \\ & Q\cos 40{}^\circ =8\times 9.8\sin 40{}^\circ \\ & \text{ }-0.3(Q\sin 40{}^\circ +8\times 9.8\cos 40{}^\circ ) \\ & Q=\frac{8\times 9.8\sin 40{}^\circ -0.3\times 8\times 9.8\cos 40{}^\circ }{\cos 40{}^\circ +0.3\times \sin 40{}^\circ } \\ & \text{ }=33.8\text{ N} \\ \end{align} | (5 marks)
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| (8 marks)
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| 5 (a) | \displaystyle \mathbf{v}=(\mathbf{i}+\mathbf{j})+(0.05\mathbf{i}-0.1\mathbf{j})t
| (2 marks)
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| 5 (b) | \displaystyle \begin{align}
& -(1+0.05t)=1-0.1t \\ & 0.05t=2 \\ & t=40 \\ \end{align} | (4 marks)
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| (6 marks)
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