Solution to Test Paper 3 - Revision history

Ian at 09:36, 8 April 2012

Ian — Sun, 08 Apr 2012 09:36:59 GMT

←Older revision		Revision as of 09:36, 8 April 2012
Line 15:		Line 15:
	& {{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} \\
		+	& \\
		+	& \text{OR} \\
		+	& \\
		+	& s=\frac{1}{2}(24\textrm{.}5+0)\times 2\textrm{.}5=30\textrm{.}625\text{ m} \\
	\end{align}</math>		\end{align}</math>

-
-	OR
-
-	<p>
-
-	<math>s=\frac{1}{2}(24\textrm{.}5+0)\times 2.5=30\textrm{.}625\text{ m}</math>
-
-	</p>
	\| (3 marks)		\| (3 marks)

Ian at 09:30, 8 April 2012

Ian — Sun, 08 Apr 2012 09:30:04 GMT

←Older revision		Revision as of 09:30, 8 April 2012
Line 3:		Line 3:
	\| 1 (a)		\| 1 (a)
	\|<math>\begin{align}		\|<math>\begin{align}
-	& 0=24.5-9.8t \\	+	& 0=24\textrm{.}5-9\textrm{.}8t \\
-	& t=\frac{24.5}{9.8}=2.5\text{ seconds} \\	+	& t=\frac{24\textrm{.}5}{9\textrm{.}8}=2\textrm{.}5\text{ seconds} \\
	\end{align}</math>		\end{align}</math>

Line 13:		Line 13:
	\| 1 (b)		\| 1 (b)
	\| <math>\begin{align}		\| <math>\begin{align}
-	& {{0}^{2}}={{24.5}^{2}}+2\times (-9.8)\times s \\	+	& {{0}^{2}}={{24\textrm{.}5}^{2}}+2\times (-9\textrm{.}8)\times s \\
-	& s=\frac{{{24.5}^{2}}}{2\times 9.8}=30.625\text{ m} \\	+	& s=\frac{{{24\textrm{.}5}^{2}}}{2\times 9\textrm{.}8}=30\textrm{.}625\text{ m} \\
	\end{align}</math>		\end{align}</math>

Line 22:		Line 22:
	<p>		<p>

-	<math>s=\frac{1}{2}(24.5+0)\times 2.5=30.625\text{ m}</math>	+	<math>s=\frac{1}{2}(24\textrm{.}5+0)\times 2.5=30\textrm{.}625\text{ m}</math>

	</p>		</p>
Line 57:		Line 57:
	\| 2 (c)		\| 2 (c)
	\| <math>\begin{align}		\| <math>\begin{align}
-	& T-4900=500\times 0.6 \\	+	& T-4900=500\times 0\textrm{.}6 \\
	& T=300+4900=5200\text{ N} \\		& T=300+4900=5200\text{ N} \\
	\end{align}</math>		\end{align}</math>
Line 64:		Line 64:

	\|-		\|-
-	\| 3 (d)	+	\| 2 (d)
	\| <math>\begin{align}		\| <math>\begin{align}
	& 4800-4900=500a \\		& 4800-4900=500a \\
-	& a=\frac{-100}{500}=-0.2\text{ m}{{\text{s}}^{\text{-2}}} \\	+	& a=\frac{-100}{500}=-0\textrm{.}2\text{ m}{{\text{s}}^{\text{-2}}} \\
-	& 0.\text{2 m}{{\text{s}}^{-\text{2}}}\quad \text{downwards} \\	+	& 0\textrm{.}\text{2 m}{{\text{s}}^{-\text{2}}}\quad \text{downwards} \\
	\end{align}</math>		\end{align}</math>

Line 81:		Line 81:
	\|-		\|-
	\| 3 (a)		\| 3 (a)
-	\|<math>R=5\times 9.8\cos 42{}^\circ =36.4\text{ N}</math>	+	\|<math>R=5\times 9\textrm{.}8\cos 42{}^\circ =36\textrm{.}4\text{ N}</math>
	\| (2 mark)		\| (2 mark)

Line 87:		Line 87:
	\|-		\|-
	\| 3 (b)		\| 3 (b)
-	\| <math>F=0.4R=14.6\text{ N}</math>	+	\| <math>F=0\textrm{.}4R=14\textrm{.}6\text{ N}</math>

	\| (2 marks)		\| (2 marks)
Line 95:		Line 95:
	\| 3 (c)		\| 3 (c)
	\| <math>\begin{align}		\| <math>\begin{align}
-	& 5\times 9.8\sin 42{}^\circ -14.6=5a \\	+	& 5\times 9\textrm{.}8\sin 42{}^\circ -14\textrm{.}6=5a \\
-	& a=\frac{18.22}{5}=3.64\text{ m}{{\text{s}}^{\text{-2}}} \\	+	& a=\frac{18\textrm{.}22}{5}=3\textrm{.}64\text{ m}{{\text{s}}^{\text{-2}}} \\
	\end{align}</math>		\end{align}</math>

Line 105:		Line 105:
	\| 3 (d)		\| 3 (d)
	\| <math>\begin{align}		\| <math>\begin{align}
-	& 2=\frac{1}{2}\times 3.64{{t}^{2}} \\	+	& 2=\frac{1}{2}\times 3\textrm{.}64{{t}^{2}} \\
-	& t=\sqrt{\frac{4}{3.64}}=1.05\text{ seconds} \\	+	& t=\sqrt{\frac{4}{3\textrm{.}64}}=1\textrm{.}05\text{ seconds} \\
	\end{align}</math>		\end{align}</math>

Line 124:		Line 124:
	\| 4 (a)		\| 4 (a)
	\| <math>\begin{align}		\| <math>\begin{align}
-	& Q\cos 40{}^\circ =8\times 9.8\sin 40{}^\circ \\	+	& Q\cos 40{}^\circ =8\times 9\textrm{.}8\sin 40{}^\circ \\
-	& Q=\frac{8\times 9.8\sin 40{}^\circ }{\cos 40{}^\circ }=65.8\text{ N} \\	+	& Q=\frac{8\times 9\textrm{.}8\sin 40{}^\circ }{\cos 40{}^\circ }=65\textrm{.}8\text{ N} \\
	\end{align}</math>		\end{align}</math>

Line 134:		Line 134:
	\| 4 (b)		\| 4 (b)
	\|<math>\begin{align}		\|<math>\begin{align}
-	& Q\cos 40{}^\circ +F=8\times 9.8\sin 40{}^\circ \\	+	& Q\cos 40{}^\circ +F=8\times 9\textrm{.}8\sin 40{}^\circ \\
-	& Q\sin 40{}^\circ +8\times 9.8\cos 40{}^\circ =R \\	+	& Q\sin 40{}^\circ +8\times 9\textrm{.}8\cos 40{}^\circ =R \\
-	& Q\cos 40{}^\circ =8\times 9.8\sin 40{}^\circ \\	+	& Q\cos 40{}^\circ =8\times 9\textrm{.}8\sin 40{}^\circ \\
-	& \text{ }-0.3(Q\sin 40{}^\circ +8\times 9.8\cos 40{}^\circ ) \\	+	& \text{ }-0\textrm{.}3(Q\sin 40{}^\circ +8\times 9\textrm{.}8\cos 40{}^\circ ) \\
-	& Q=\frac{8\times 9.8\sin 40{}^\circ -0.3\times 8\times 9.8\cos 40{}^\circ }{\cos 40{}^\circ +0.3\times \sin 40{}^\circ } \\	+	& Q=\frac{8\times 9\textrm{.}8\sin 40{}^\circ -0.\textrm{.}\times 8\times 9\textrm{.}8\cos 40{}^\circ }{\cos 40{}^\circ +0.\times \sin 40{}^\circ } \\
-	& \text{ }=33.8\text{ N} \\	+	& \text{ }=33\textrm{.}8\text{ N} \\
	\end{align}</math>		\end{align}</math>

Line 155:		Line 155:
	\|-		\|-
	\| 5 (a)		\| 5 (a)
-	\| <math>\mathbf{v}=(\mathbf{i}+\mathbf{j})+(0.05\mathbf{i}-0.1\mathbf{j})t</math>	+	\| <math>\mathbf{v}=(\mathbf{i}+\mathbf{j})+(0\textrm{.}05\mathbf{i}-0\textrm{.}1\mathbf{j})t</math>


Line 164:		Line 164:
	\| 5 (b)		\| 5 (b)
	\| <math>\begin{align}		\| <math>\begin{align}
-	& -(1+0.05t)=1-0.1t \\	+	& -(1+0\textrm{.}05t)=1-0\textrm{.}1t \\
-	& 0.05t=2 \\	+	& 0\textrm{.}05t=2 \\
	& t=40 \\		& t=40 \\
	\end{align}</math>		\end{align}</math>

Ian at 09:20, 8 April 2012

Ian — Sun, 08 Apr 2012 09:20:48 GMT

←Older revision		Revision as of 09:20, 8 April 2012
Line 181:		Line 181:

	\|}		\|}
-
-	AG: Answer Given in Question – Working must justify answer.

Ian at 09:19, 8 April 2012

Ian — Sun, 08 Apr 2012 09:19:43 GMT

←Older revision		Revision as of 09:19, 8 April 2012
Line 123:		Line 123:
	\|-		\|-
	\| 4 (a)		\| 4 (a)
-	\| ~~[[Image:test1ans2~~.~~gif]]~~	+	\| <math>\begin{align}
-	\| (1 mark)	+	& Q\cos 40{}^\circ =8\times 9.8\sin 40{}^\circ \\
		+	& Q=\frac{8\times 9.8\sin 40{}^\circ }{\cos 40{}^\circ }=65.8\text{ N} \\
		+	\end{align}</math>
		+
		+	\| (3 mark)


	\|-		\|-
	\| 4 (b)		\| 4 (b)
-	\| <math>\begin{align}	+	\|<math>\begin{align}
-	& ~~R+T~~\~~sin 30~~{}^\circ =~~200~~\times 9 \~~textrm~~{.}8 \\	+	& Q\cos 40{}^\circ +F=8\times 9.8\sin 40{}^\circ \\
-	& R+0 \~~textrm~~{.}~~5T=1960~~ \\	+	& Q\sin 40{}^\circ +8\times 9.8\cos 40{}^\circ =R \\
-	& R=~~1960~~-0 \~~textrm~~{.}5T \\	+	& Q\cos 40{}^\circ =8\times 9.8\sin 40{}^\circ \\
		+	& \text{ }-0.3(Q\sin 40{}^\circ +8\times 9.8\cos 40{}^\circ ) \\
		+	& Q=\frac{8\times 9.8\sin 40{}^\circ -0.3\times 8\times 9.8\cos 40{}^\circ }{\cos 40{}^\circ +0.3\times \sin 40{}^\circ } \\
		+	& \text{ }=33.8\text{ N} \\
	\end{align}</math>		\end{align}</math>
-	\| (3 marks)	+
		+	\| (5 marks)


-	\|-
-	\| 4 (c)
-	\| <math></math>
-	<math>\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}</math>
-	\| (4 marks)


Line 157:		Line 155:
	\|-		\|-
	\| 5 (a)		\| 5 (a)
-	\| <math>~~\begin{align}~~	+	\| <math>\mathbf{v}=(\mathbf{i}+\mathbf{j})+(0.05\mathbf{i}-0.1\mathbf{j})t</math>
-	~~& 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}~~</math>	+

-	AG	+	\| (2 marks)
-	\| (3 marks)	+


	\|-		\|-
	\| 5 (b)		\| 5 (b)
-	\| <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>
-	\| (2 marks)
-
-
-	\|-
-	\| 5 (c)
	\| <math>\begin{align}		\| <math>\begin{align}
-	& ~~\mathbf{r}=~~(4t+0~~\textrm{~~.~~}05{{t}^{2}}~~)~~\mathbf{i}+(3t-0\textrm{.}25{{t}^{2}})\mathbf{j} \\~~	+	& -(1+0.05t)=1-0.1t \\
-	~~& 3t-0\textrm{.}25{{t}^{2}}~~=~~0 \\~~	+	& 0.05t=2 \\
-	~~& t(3~~-0~~\textrm{~~.~~}25t)=0~~ \\	+	& t=40 \\
-	& t=0~~\text{ or }t=\frac{3}{0\textrm{~~.~~}25}~~=~~12\text{ s}~~ \\	+
-	& t=~~12\text{ s}~~ \\	+
	\end{align}</math>		\end{align}</math>
-	\| (3 marks)	+
		+	\| (4 marks)


-	\|-
-	\| 5 (d)
-	\| <math>\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}</math>
-	\| (3 marks)


Line 197:		Line 177:
	\|		\|
	\|		\|
-	\| '''(11 marks)'''	+	\| '''(6 marks)'''

Ian at 09:13, 8 April 2012

Ian — Sun, 08 Apr 2012 09:13:37 GMT

←Older revision		Revision as of 09:13, 8 April 2012
Line 81:		Line 81:
	\|-		\|-
	\| 3 (a)		\| 3 (a)
-	\|~~[[Image:test1ans1~~.~~gif]]~~	+	\|<math>R=5\times 9.8\cos 42{}^\circ =36.4\text{ N}</math>
-	\| (1 mark)	+	\| (2 mark)


	\|-		\|-
	\| 3 (b)		\| 3 (b)
-	\| <math>F=~~2\times 1\textrm{~~.}8=~~3\textrm{~~.}6\text{ N}</math>	+	\| <math>F=0.4R=14.6\text{ N}</math>
		+
	\| (2 marks)		\| (2 marks)

Line 93:		Line 94:
	\|-		\|-
	\| 3 (c)		\| 3 (c)
-	\| <math>~~R=2~~\times 9\~~textrm~~{.}8=19\~~textrm~~{.}6\text{ N}</math>	+	\| <math>\begin{align}
-	\| (2 marks)	+	& 5\times 9.8\sin 42{}^\circ -14.6=5a \\
		+	& a=\frac{18.22}{5}=3.64\text{ m}{{\text{s}}^{\text{-2}}} \\
		+	\end{align}</math>
		+
		+	\| (3 marks)


Line 100:		Line 105:
	\| 3 (d)		\| 3 (d)
	\| <math>\begin{align}		\| <math>\begin{align}
-	& ~~4800-4900=500a \\~~	+	& 2=\frac{1}{2}\times 3.64{{t}^{2}} \\
-	~~& a~~=\frac{~~-100~~}{~~500~~}~~=-0~~.~~2\text~~{ ~~m}{{\text~~{s}}^{~~\text{-~~2}}} \\	+	& t=\sqrt{\frac{4}{3.64}}=1.05\text{ seconds} \\
-	& 0.\~~text{2 m}{~~{\~~text~~{s}}^{~~-\text{2~~}}~~}\quad~~ \text{~~downwards~~} \\	+
	\end{align}</math>		\end{align}</math>
		+


Line 113:		Line 118:
	\|		\|
	\|		\|
-	\| '''(8 marks)'''	+	\| '''(10 marks)'''

Ian at 09:09, 8 April 2012

Ian — Sun, 08 Apr 2012 09:09:14 GMT

←Older revision		Revision as of 09:09, 8 April 2012
Line 44:		Line 44:
	\| [[Image:test3ans1.gif]]		\| [[Image:test3ans1.gif]]

-	\| (3 marks)	+	\| (1 marks)


Line 51:		Line 51:
	\| T = 4900 N		\| T = 4900 N

-	\| (3 marks)	+	\| (1 marks)


Line 61:		Line 61:
	\end{align}</math>		\end{align}</math>

-	\| (1 mark)	+	\| (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)

	\|-		\|-
	\|		\|
	\|		\|
-	\| '''(7 marks)'''	+	\| '''(9 marks)'''

Ian at 08:54, 8 April 2012

Ian — Sun, 08 Apr 2012 08:54:53 GMT

←Older revision		Revision as of 08:54, 8 April 2012
Line 42:		Line 42:
	\|-		\|-
	\| 2 (a)		\| 2 (a)
-	\| ~~<math>\begin{align}~~	+	\| [[Image:test3ans1.gif]]
-	~~& T-800\times 9\textrm{~~.~~}8=800\times 0\textrm{.}2 \\~~	+
-	~~& T=\text{7840}+\text{160}=\text{8000 N} \\~~	+
-	~~\end{align}</math>~~	+

	\| (3 marks)		\| (3 marks)
Line 52:		Line 49:
	\|-		\|-
	\| 2 (b)		\| 2 (b)
-	\| ~~<math>\begin{align}~~	+	\| T = 4900 N
-	& T~~-800\times 9\textrm{.}8~~=~~800\times (-0\textrm{.}2) \\~~	+
-	~~& T=\text{7840}-\text{160}=\text{7680~~ N~~} \\~~	+
-	~~\end{align}</math>~~	+

	\| (3 marks)		\| (3 marks)
Line 62:		Line 56:
	\|-		\|-
	\| 2 (c)		\| 2 (c)
-	\| <math>T=~~800~~\times ~~9\textrm{~~.}8=~~7840~~\text{ N}</math>	+	\| <math>\begin{align}
		+	& T-4900=500\times 0.6 \\
		+	& T=300+4900=5200\text{ N} \\
		+	\end{align}</math>
		+
	\| (1 mark)		\| (1 mark)

Line 93:		Line 91:
	\| 3 (d)		\| 3 (d)
	\| <math>\begin{align}		\| <math>\begin{align}
-	& 3 \~~textrm~~{.}6=19 \~~textrm~~{.}6\mu \\	+	& 4800-4900=500a \\
-	& \~~mu =\frac~~{3 \~~textrm~~{.}6}{19 \~~textrm~~{.}6}=0 \~~textrm~~{.}~~184~~ \\	+	& 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>		\end{align}</math>
		+
		+
		+
	\| (3 marks)		\| (3 marks)

Ian at 16:05, 4 April 2012

Ian — Wed, 04 Apr 2012 16:05:58 GMT

←Older revision		Revision as of 16:05, 4 April 2012
Line 3:		Line 3:
	\| 1 (a)		\| 1 (a)
	\|<math>\begin{align}		\|<math>\begin{align}
-	& 0=24.5-9.8t \\	+	& 0=24.5-9.8t \\
-	& t=\frac{24.5}{9.8}=2.5\text{ seconds} \\	+	& t=\frac{24.5}{9.8}=2.5\text{ seconds} \\
	\end{align}</math>		\end{align}</math>

Line 13:		Line 13:
	\| 1 (b)		\| 1 (b)
	\| <math>\begin{align}		\| <math>\begin{align}
-	& {{0}^{2}}={{24.5}^{2}}+2\times (-9.8)\times s \\	+	& {{0}^{2}}={{24.5}^{2}}+2\times (-9.8)\times s \\
-	& s=\frac{{{24.5}^{2}}}{2\times 9.8}=30.625\text{ m} \\	+	& s=\frac{{{24.5}^{2}}}{2\times 9.8}=30.625\text{ m} \\
	\end{align}</math>		\end{align}</math>
		+

	OR		OR
		+
		+	<p>

	<math>s=\frac{1}{2}(24.5+0)\times 2.5=30.625\text{ m}</math>		<math>s=\frac{1}{2}(24.5+0)\times 2.5=30.625\text{ m}</math>

-		+	</p>
	\| (3 marks)		\| (3 marks)
		+
		+	\|-
		+	\| 1 (c)
		+	\|61.25 m
		+
		+
		+	\| (1 marks)

	\|-		\|-
	\|		\|
	\|		\|
-	\| '''(6 marks)'''	+	\| '''(7 marks)'''

Ian: New page: {| border="1" |+ Solutions | 1 (a) | $\begin{align} & 0=24.5-9.8t \\ & t=\frac{24.5}{9.8}=2.5\text{ seconds} \\ \end{align}$ | (3 marks) |- | 1 (b) | $\begin{align} ...$

Ian — Wed, 04 Apr 2012 15:45:38 GMT

New page: {| border="1" |+ Solutions | 1 (a) |<math>\begin{align} & 0=24.5-9.8t \\ & t=\frac{24.5}{9.8}=2.5\text{ seconds} \\ \end{align}</math> | (3 marks) |- | 1 (b) | <math>\begin{align} ...

New page

{| border="1"
|+ Solutions
| 1 (a)
|<math>\begin{align}
& 0=24.5-9.8t \\
& t=\frac{24.5}{9.8}=2.5\text{ seconds} \\
\end{align}</math>

| (3 marks)

|-
| 1 (b)
| <math>\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}</math>

OR

<math>s=\frac{1}{2}(24.5+0)\times 2.5=30.625\text{ m}</math>

| (3 marks)

|-
|
|
| '''(6 marks)'''

|-
| 2 (a)
| <math>\begin{align}
& T-800\times 9\textrm{.}8=800\times 0\textrm{.}2 \\
& T=\text{7840}+\text{160}=\text{8000 N} \\
\end{align}</math>

| (3 marks)

|-
| 2 (b)
| <math>\begin{align}
& T-800\times 9\textrm{.}8=800\times (-0\textrm{.}2) \\
& T=\text{7840}-\text{160}=\text{7680 N} \\
\end{align}</math>

| (3 marks)

|-
| 2 (c)
| <math>T=800\times 9\textrm{.}8=7840\text{ N}</math>
| (1 mark)

|-
|
|
| '''(7 marks)'''

|-
| 3 (a)
|[[Image:test1ans1.gif]]
| (1 mark)

|-
| 3 (b)
| <math>F=2\times 1\textrm{.}8=3\textrm{.}6\text{ N}</math>
| (2 marks)

|-
| 3 (c)
| <math>R=2\times 9\textrm{.}8=19\textrm{.}6\text{ N}</math>
| (2 marks)

|-
| 3 (d)
| <math>\begin{align}
& 3 \textrm{.}6=19 \textrm{.}6\mu \\
& \mu =\frac{3 \textrm{.}6}{19 \textrm{.}6}=0 \textrm{.}184 \\
\end{align}</math>
| (3 marks)

|-
|
|
| '''(8 marks)'''

|-
| 4 (a)
| [[Image:test1ans2.gif]]
| (1 mark)

|-
| 4 (b)
| <math>\begin{align}
& R+T\sin 30{}^\circ =200\times 9 \textrm{.}8 \\
& R+0 \textrm{.}5T=1960 \\
& R=1960-0 \textrm{.}5T \\
\end{align}</math>
| (3 marks)

|-
| 4 (c)
| <math></math>
<math>\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}</math>
| (4 marks)

|-
|
|
| '''(8 marks)'''

|-
| 5 (a)
| <math>\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}</math>

AG
| (3 marks)

|-
| 5 (b)
| <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>
| (2 marks)

|-
| 5 (c)
| <math>\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}</math>
| (3 marks)

|-
| 5 (d)
| <math>\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}</math>
| (3 marks)

|-
|
|
| '''(11 marks)'''

|}

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