Solution to Test Paper 1

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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 \\
+
& {{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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|-
|-
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| cells in the next row go here
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| 1 (b)
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| more cells in the same row here
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| <math>\begin{align}
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| even more cells in the same row here
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& 0=7-9\textrm{.}8t \\
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& t=\frac{7}{9\textrm{.}8}=0\textrm{.}714\text{ s} \\
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\end{align}</math>
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| (3 marks)
|-
|-
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| Row heading 1
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|
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| Cell 2 || Cell 3
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|
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| '''(6 marks)'''
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|-
|-
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| Row heading A
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| 2 (a)
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| Cell B
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| <math>\begin{align}
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| Cell C
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& T-800\times 9\textrm{.}8=800\times 0\textrm{.}2 \\
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& T=\text{7840}+\text{160}=\text{8000 N} \\
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\end{align}</math>
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| (3 marks)
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|-
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| 2 (b)
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| <math>\begin{align}
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& T-800\times 9\textrm{.}8=800\times (-0\textrm{.}2) \\
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& T=\text{7840}-\text{160}=\text{7680 N} \\
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\end{align}</math>
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| (3 marks)
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|-
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| 2 (c)
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| <math>T=800\times 9\textrm{.}8=7840\text{ N}</math>
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| (1 mark)
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|-
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|
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|
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| '''(7 marks)'''
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|-
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| 3 (a)
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|[[Image:test1ans1.gif]]
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| (1 mark)
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|-
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| 3 (b)
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| <math>F=2\times 1\textrm{.}8=3\textrm{.}6\text{ N}</math>
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| (2 marks)
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|-
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| 3 (c)
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| <math>R=2\times 9\textrm{.}8=19\textrm{.}6\text{ N}</math>
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| (2 marks)
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|-
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| 3 (d)
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| <math>\begin{align}
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& 3 \textrm{.}6=19 \textrm{.}6\mu \\
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& \mu =\frac{3 \textrm{.}6}{19 \textrm{.}6}=0 \textrm{.}184 \\
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\end{align}</math>
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| (3 marks)
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|-
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|
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|
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| '''(8 marks)'''
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|-
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| 4 (a)
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| [[Image:test1ans2.gif]]
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| (1 mark)
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|-
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| 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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| (3 marks)
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|-
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| 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} \\
 +
\end{align}</math>
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| (4 marks)
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|-
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|
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|
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| '''(8 marks)'''
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|-
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| 5 (a)
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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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| (3 marks)
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|-
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| 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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| (2 marks)
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|-
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| 5 (c)
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| <math>\begin{align}
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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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| (3 marks)
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|-
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| 5 (d)
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| <math>\begin{align}
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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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| (3 marks)
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|-
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|
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|
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| '''(11 marks)'''
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|}
|}
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 +
AG: Answer Given in Question – Working must justify answer.

Current revision

Solutions
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)


2 (c) \displaystyle T=800\times 9\textrm{.}8=7840\text{ N} (1 mark)


(7 marks)


3 (a) Image:test1ans1.gif (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}

& 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) Image:test1ans2.gif (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}

AG

(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}

(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.

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