Processing Math: Done
To print higher-resolution math symbols, click the
Hi-Res Fonts for Printing button on the jsMath control panel.
jsMath

Solution to Test Paper 3

From Mechanics

(Difference between revisions)
Jump to: navigation, search
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)'''

Revision as of 09:13, 8 April 2012

Solutions
1 (a) 0=245−98tt=98245=25 seconds (3 marks)


1 (b) 02=2452+2(−98)ss=2452298=30625 mORs=21(245+0)25=30625 m (3 marks)
1 (c) 61.25 m


(1 marks)
(7 marks)


2 (a) Image:test3ans1.gif (1 marks)


2 (b) T = 4900 N (1 marks)


2 (c) T−4900=50006T=300+4900=5200 N (3 mark)
3 (d) 4800−4900=500aa=500−100=−02 ms-202 ms−2downwards (4 mark)
(9 marks)


3 (a) R=598cos42=364 N (2 mark)


3 (b) F=04R=146 N (2 marks)


3 (c) 598sin42−146=5aa=51822=364 ms-2 (3 marks)


3 (d) 2=21364t2t=4364=105 seconds  (3 marks)


(10 marks)


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


4 (b) R+Tsin30=2009.8R+0.5T=1960R=1960−0.5T (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-2AG  (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.

Retrieved from "http://wiki.math.se/wikis/2009/bridgecoursemechanics/index.php/Solution_to_Test_Paper_3"