Solution to Test Paper 1
From Mechanics
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| 1 (a) | | 1 (a) | ||
|<math>\begin{align} | |<math>\begin{align} | ||
| - | & {{0}^{2}}={{7}^{2}}+2(-\textrm{.}8)s \\ | + | & {{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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& R=1960-0 \textrm{.}5T \\ | & R=1960-0 \textrm{.}5T \\ | ||
\end{align}</math> | \end{align}</math> | ||
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| + | AG | ||
| (3 marks) | | (3 marks) | ||
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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}}} \\ | & \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}}} \\ | ||
\end{align}</math> | \end{align}</math> | ||
| - | | (3 marks) | + | |
| + | | (3 marks) | ||
Current revision
| 1 (a) | \displaystyle \begin{align}
& {{0}^{2}}={{7}^{2}}+2(-9\textrm{.}8)s \\ & s=\frac{49}{19\textrm{.}6}=2\textrm{.}5 \\ & \text{Max Height }=\text{ 5}+\text{2}\text{.5}=\text{7}\text{.5 m} \\ \end{align} | (3 marks)
|
| 1 (b) | \displaystyle \begin{align}
& 0=7-9\textrm{.}8t \\ & t=\frac{7}{9\textrm{.}8}=0\textrm{.}714\text{ s} \\ \end{align} | (3 marks) |
| (6 marks)
| ||
| 2 (a) | \displaystyle \begin{align}
& T-800\times 9\textrm{.}8=800\times 0\textrm{.}2 \\ & T=\text{7840}+\text{160}=\text{8000 N} \\ \end{align} | (3 marks)
|
| 2 (b) | \displaystyle \begin{align}
& T-800\times 9\textrm{.}8=800\times (-0\textrm{.}2) \\ & T=\text{7840}-\text{160}=\text{7680 N} \\ \end{align} | (3 marks)
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| 2 (c) | \displaystyle T=800\times 9\textrm{.}8=7840\text{ N} | (1 mark)
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| (7 marks)
| ||
| 3 (a) | 
| (1 mark)
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| 3 (b) | \displaystyle F=2\times 1\textrm{.}8=3\textrm{.}6\text{ N} | (2 marks)
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| 3 (c) | \displaystyle R=2\times 9\textrm{.}8=19\textrm{.}6\text{ N} | (2 marks)
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| 3 (d) | \displaystyle \begin{align}
& 3 \textrm{.}6=19 \textrm{.}6\mu \\ & \mu =\frac{3 \textrm{.}6}{19 \textrm{.}6}=0 \textrm{.}184 \\ \end{align} | (3 marks)
|
| (8 marks)
| ||
| 4 (a) | 
| (1 mark)
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| 4 (b) | \displaystyle \begin{align}
& R+T\sin 30{}^\circ =200\times 9 \textrm{.}8 \\ & R+0 \textrm{.}5T=1960 \\ & R=1960-0 \textrm{.}5T \\ \end{align} AG | (3 marks)
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| 4 (c) | \displaystyle
\displaystyle \begin{align} & F=T\cos 30{}^\circ \\ & T\cos 30{}^\circ =0\textrm{.}6(1960-0\textrm{.}5T) \\ & T(\cos 30{}^\circ +0\textrm{.}3)=1176 \\ & T=\frac{1176}{(\cos 30{}^\circ +0\textrm{.}3)}=1010\text{ N} \\ \end{align} | (4 marks)
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| (8 marks)
| ||
| 5 (a) | \displaystyle \begin{align}
& 5\mathbf{i}-2\mathbf{j}=4\mathbf{i}+3\mathbf{j}+10\mathbf{a} \\ & \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}}} \\ \end{align} | (3 marks)
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| 5 (b) | \displaystyle \mathbf{r}=(4\mathbf{i}+3\mathbf{j})t+0\textrm{.}5(0\textrm{.}1\mathbf{i}-0\textrm{.}5\mathbf{j}){{t}^{2}} | (2 marks)
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| 5 (c) | \displaystyle \begin{align}
& \mathbf{r}=(4t+0\textrm{.}05{{t}^{2}})\mathbf{i}+(3t-0\textrm{.}25{{t}^{2}})\mathbf{j} \\ & 3t-0\textrm{.}25{{t}^{2}}=0 \\ & t(3-0\textrm{.}25t)=0 \\ & t=0\text{ or }t=\frac{3}{0\textrm{.}25}=12\text{ s} \\ & t=12\text{ s} \\ \end{align} | (3 marks)
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| 5 (d) | \displaystyle \begin{align}
& \mathbf{v}=(4+0\textrm{.}1t)\mathbf{i}+(3-0\textrm{.}5t)\mathbf{j} \\ & 3-0\textrm{.}5t=0 \\ & t=6 \text{ s} \\ \end{align} | (3 marks)
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| (11 marks)
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AG: Answer Given in Question – Working must justify answer.
