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Solution to Test Paper 1

From Mechanics

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Current revision (09:55, 8 April 2012) (edit) (undo)
 
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| 1 (a)
| 1 (a)
|<math>\begin{align}
|<math>\begin{align}
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& {{0}^{2}}={{7}^{2}}+2(-\textrm{.}8)s \\
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& {{0}^{2}}={{7}^{2}}+2(-9\textrm{.}8)s \\
& s=\frac{49}{19\textrm{.}6}=2\textrm{.}5 \\
& s=\frac{49}{19\textrm{.}6}=2\textrm{.}5 \\
& \text{Max Height }=\text{ 5}+\text{2}\text{.5}=\text{7}\text{.5 m} \\
& \text{Max Height }=\text{ 5}+\text{2}\text{.5}=\text{7}\text{.5 m} \\
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& \mu =\frac{3 \textrm{.}6}{19 \textrm{.}6}=0 \textrm{.}184 \\
& \mu =\frac{3 \textrm{.}6}{19 \textrm{.}6}=0 \textrm{.}184 \\
\end{align}</math>
\end{align}</math>
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| '''(8 marks)'''
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| (3 marks)
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| 4 (a)
| 4 (a)
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| [[Image:test1ans2.gif]]
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| (1 mark)
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| 4 (b)
| 4 (b)
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| <math>\begin{align}
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& R+T\sin 30{}^\circ =200\times 9 \textrm{.}8 \\
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& R+0 \textrm{.}5T=1960 \\
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& R=1960-0 \textrm{.}5T \\
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\end{align}</math>
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AG
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| 4 (c)
| 4 (c)
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| <math></math>
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<math>\begin{align}
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& F=T\cos 30{}^\circ \\
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& T\cos 30{}^\circ =0\textrm{.}6(1960-0\textrm{.}5T) \\
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& T(\cos 30{}^\circ +0\textrm{.}3)=1176 \\
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& T=\frac{1176}{(\cos 30{}^\circ +0\textrm{.}3)}=1010\text{ N} \\
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\end{align}</math>
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| '''(8 marks)'''
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| 5 (a) go here
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| <math>\begin{align}
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& 5\mathbf{i}-2\mathbf{j}=4\mathbf{i}+3\mathbf{j}+10\mathbf{a} \\
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& \mathbf{a}=\frac{1}{10}(\mathbf{i}-5\mathbf{j})=\left( 0\textrm{.}1\mathbf{i}-0\textrm{.}5\mathbf{j} \right)\text{ m}{{\text{s}}^{\text{-2}}} \\
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\end{align}</math>
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| 5 (b)
| 5 (b)
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| <math>\mathbf{r}=(4\mathbf{i}+3\mathbf{j})t+0\textrm{.}5(0\textrm{.}1\mathbf{i}-0\textrm{.}5\mathbf{j}){{t}^{2}}</math>
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| 5 (c)
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& \mathbf{r}=(4t+0\textrm{.}05{{t}^{2}})\mathbf{i}+(3t-0\textrm{.}25{{t}^{2}})\mathbf{j} \\
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& 3t-0\textrm{.}25{{t}^{2}}=0 \\
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& t(3-0\textrm{.}25t)=0 \\
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& t=0\text{ or }t=\frac{3}{0\textrm{.}25}=12\text{ s} \\
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& t=12\text{ s} \\
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\end{align}</math>
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| 5 (d)
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& \mathbf{v}=(4+0\textrm{.}1t)\mathbf{i}+(3-0\textrm{.}5t)\mathbf{j} \\
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& 3-0\textrm{.}5t=0 \\
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& t=6 \text{ s} \\
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\end{align}</math>
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| '''(11 marks)'''
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|}
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AG: Answer Given in Question – Working must justify answer.
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Current revision

Solutions
1 (a) 02=72+2(−9.8)ss=4919.6=2.5Max Height = 5+2.5=7.5 m (3 marks)


1 (b) 0=7−9.8tt=79.8=0.714 s (3 marks)
(6 marks)


2 (a) T−8009.8=8000.2T=7840+160=8000 N (3 marks)


2 (b) T−8009.8=800(−0.2)T=7840−160=7680 N (3 marks)


2 (c) T=8009.8=7840 N (1 mark)


(7 marks)


3 (a) Image:test1ans1.gif (1 mark)


3 (b) F=21.8=3.6 N (2 marks)


3 (c) R=29.8=19.6 N (2 marks)


3 (d) 3.6=19.6=3.619.6=0.184 (3 marks)


(8 marks)


4 (a) Image:test1ans2.gif (1 mark)


4 (b) R+Tsin30=2009.8R+0.5T=1960R=1960−0.5TAG (3 marks)


4 (c)

F=Tcos30Tcos30=0.6(1960−0.5T)T(cos30+0.3)=1176T=1176(cos30+0.3)=1010 N

(4 marks)


(8 marks)


5 (a) 5i−2j=4i+3j+10aa=110(i−5j)=0.1i−0.5j ms-2  (3 marks)


5 (b) r=(4i+3j)t+0.5(0.1i−0.5j)t2 (2 marks)


5 (c) r=(4t+0.05t2)i+(3t−0.25t2)j3t−0.25t2=0t(3−0.25t)=0t=0 or t=30.25=12 st=12 s (3 marks)


5 (d) v=(4+0.1t)i+(3−0.5t)j3−0.5t=0t=6 s (3 marks)


(11 marks)


AG: Answer Given in Question – Working must justify answer.

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