Solution to Test Paper 3
From Mechanics
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| [[Image:test3ans1.gif]] | | [[Image:test3ans1.gif]] | ||
| - | | ( | + | | (1 marks) |
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| T = 4900 N | | T = 4900 N | ||
| - | | ( | + | | (1 marks) |
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\end{align}</math> | \end{align}</math> | ||
| - | | ( | + | | (3 mark) |
| + | |- | ||
| + | | 3 (d) | ||
| + | | <math>\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}</math> | ||
| + | |||
| + | | (4 mark) | ||
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| - | | '''( | + | | '''(9 marks)''' |
Revision as of 09:09, 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)
|
| 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)
| ||
| 2 (a) | 
| (1 marks)
|
| 2 (b) | T = 4900 N | (1 marks)
|
| 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)
| ||
| 3 (a) | 
| (1 mark)
|
| 3 (b) | \displaystyle F=2\times 1\textrm{.}8=3\textrm{.}6\text{ N} | (2 marks)
|
| 3 (c) | \displaystyle R=2\times 9\textrm{.}8=19\textrm{.}6\text{ N} | (2 marks)
|
| 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}
| (3 marks)
|
| (8 marks)
| ||
| 4 (a) | 
| (1 mark)
|
| 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} | (3 marks)
|
| 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)
|
| (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} AG | (3 marks)
|
| 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)
|
| 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)
|
| 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)
|
| (11 marks)
|
AG: Answer Given in Question – Working must justify answer.
