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Solution to Test Paper 3

From Mechanics

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Current revision (09:36, 8 April 2012) (edit) (undo)
 
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& {{0}^{2}}={{24\textrm{.}5}^{2}}+2\times (-9\textrm{.}8)\times s \\
& {{0}^{2}}={{24\textrm{.}5}^{2}}+2\times (-9\textrm{.}8)\times s \\
& s=\frac{{{24\textrm{.}5}^{2}}}{2\times 9\textrm{.}8}=30\textrm{.}625\text{ m} \\
& s=\frac{{{24\textrm{.}5}^{2}}}{2\times 9\textrm{.}8}=30\textrm{.}625\text{ m} \\
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& \\
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& \text{OR} \\
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& \\
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& s=\frac{1}{2}(24\textrm{.}5+0)\times 2\textrm{.}5=30\textrm{.}625\text{ m} \\
\end{align}</math>
\end{align}</math>
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OR
 
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<p>
 
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<math>s=\frac{1}{2}(24\textrm{.}5+0)\times 2.5=30\textrm{.}625\text{ m}</math>
 
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</p>
 
| (3 marks)
| (3 marks)

Current revision

Solutions
1 (a) 0=24.5−9.8tt=9.824.5=2.5 seconds (3 marks)


1 (b) 02=24.52+2(−9.8)ss=24.5229.8=30.625 mORs=21(24.5+0)2.5=30.625 m (3 marks)
1 (c) 61.25 m


(1 marks)
(7 marks)


2 (a) Image:test3ans1.gif (1 marks)


2 (b) T = 4900 N (1 marks)


2 (c) T−4900=5000.6T=300+4900=5200 N (3 mark)
2 (d) 4800−4900=500aa=500−100=−0.2 ms-20.2 ms−2downwards (4 mark)
(9 marks)


3 (a) R=59.8cos42=36.4 N (2 mark)


3 (b) F=0.4R=14.6 N (2 marks)


3 (c) 59.8sin42−14.6=5aa=518.22=3.64 ms-2 (3 marks)


3 (d) 2=213.64t2t=43.64=1.05 seconds  (3 marks)


(10 marks)


4 (a) Qcos40=89.8sin40Q=cos4089.8sin40=65.8 N (3 mark)


4 (b) Qcos40+F=89.8sin40Qsin40+89.8cos40=RQcos40=89.8sin40 −0.3(Qsin40+89.8cos40)Q=cos40+0sin4089.8sin40−0.89.8cos40 =33.8 N (5 marks)



(8 marks)


5 (a) v=(i+j)+(0.05i−0.1j)t


(2 marks)


5 (b) −(1+0.05t)=1−0.1t0.05t=2t=40 (4 marks)



(6 marks)


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