Solution to Test Paper 2

From Mechanics

(Difference between revisions)
Jump to: navigation, search
Line 6: Line 6:
| 1 (a)
| 1 (a)
| <math>\begin{align}
| <math>\begin{align}
-
& 44.1=\frac{1}{2}\times 9.8{{t}^{2}} \\
+
& 44 \ \textrm{.}1=\frac{1}{2}\times 9 \ \ 8{{t}^{2}} \\
-
& t=\sqrt{\frac{44.1}{4.9}}=3\text{ s} \\
+
& t=\sqrt{\frac{44 \textrm{.}1}{4 \textrm{.}9}}=3\text{ s} \\
\end{align}</math>
\end{align}</math>
Line 13: Line 13:
<math>\begin{align}
<math>\begin{align}
-
& s=\frac{1}{2}\times 9.8\times {{3}^{2}}=44.1 \\
+
& s=\frac{1}{2}\times 9 \textrm{.}8\times {{3}^{2}}=44 \textrm{.}1 \\
& \text{AG} \\
& \text{AG} \\
& \therefore \text{Hits ground after 3 seconds} \\
& \therefore \text{Hits ground after 3 seconds} \\
Line 48: Line 48:
|-
|-
| 1 (b)
| 1 (b)
-
| 2
+
| <math>\begin{align}
-
| 3
+
& {{v}^{2}}={{0}^{2}}+2\times 9 \textrm{.}8\times 44 \textrm{.}1 \\
-
| 4
+
& v=\sqrt{864 \textrm{.}36}=29 \ \textrm{.}4\text{ m}{{\text{s}}^{-1}} \\
 +
\end{align}</math>
 +
 
 +
OR
 +
 
 +
<math>\begin{align}
 +
& v=0+9 \textrm{.}8\times 3 \\
 +
& v=29 \textrm{.}4 \text{ m}{{\text{s}}^{-1}} \\
 +
\end{align}</math>
 +
 
 +
 
 +
|M1
 +
 
 +
A1
 +
 
 +
A1
 +
 
 +
 
 +
| (3 marks)
 +
 
 +
 
 +
| M1: Use of constant acceleration equation with <math>v=0</math>
 +
 
 +
A1: Correct equation.
 +
 
 +
A1: Correct <math>v</math>.
 +
 
|-
|-
| 1 (c)
| 1 (c)
-
| 2
+
| Air resistance would slow the ball down.
-
| 3
+
| B1
-
| 4
+
| (1 mark)
 +
| B1: Sensible statement about air resistance.
|-
|-
Line 63: Line 90:
| 3
| 3
| 4
| 4
 +
| 5
|-
|-
Line 69: Line 97:
| 3
| 3
| 4
| 4
 +
| 5
|-
|-
Line 75: Line 104:
| 3
| 3
| 4
| 4
 +
| 5
|-
|-
Line 81: Line 111:
| 3
| 3
| 4
| 4
 +
| 5
|-
|-
Line 87: Line 118:
| 3
| 3
| 4
| 4
 +
| 5
|-
|-
Line 92: Line 124:
| 2
| 2
| 3
| 3
-
| 4
+
| 4| 5
-
 
+
|-
|-
| 3 (a)
| 3 (a)
Line 99: Line 130:
| 3
| 3
| 4
| 4
 +
| 5
|-
|-
Line 105: Line 137:
| 3
| 3
| 4
| 4
 +
| 5
|-
|-
Line 111: Line 144:
| 3
| 3
| 4
| 4
 +
| 5
|-
|-
Line 117: Line 151:
| 3
| 3
| 4
| 4
 +
| 5
|-
|-
Line 123: Line 158:
| 3
| 3
| 4
| 4
 +
| 5
|-
|-
Line 129: Line 165:
| 3
| 3
| 4
| 4
 +
| 5
|-
|-
Line 135: Line 172:
| 3
| 3
| 4
| 4
 +
| 5
|-
|-
Line 141: Line 179:
| 3
| 3
| 4
| 4
 +
| 5
|-
|-
Line 147: Line 186:
| 3
| 3
| 4
| 4
 +
| 5
|-
|-
Line 153: Line 193:
| 3
| 3
| 4
| 4
-
 
+
| 5
|-
|-
| 5 (c)
| 5 (c)
Line 159: Line 199:
| 3
| 3
| 4
| 4
-
 
+
| 5
|}
|}

Revision as of 14:13, 19 January 2011

Solutions
1 (a) \displaystyle \begin{align}

& 44 \ \textrm{.}1=\frac{1}{2}\times 9 \ \ 8{{t}^{2}} \\ & t=\sqrt{\frac{44 \textrm{.}1}{4 \textrm{.}9}}=3\text{ s} \\ \end{align}

OR

\displaystyle \begin{align} & s=\frac{1}{2}\times 9 \textrm{.}8\times {{3}^{2}}=44 \textrm{.}1 \\ & \text{AG} \\ & \therefore \text{Hits ground after 3 seconds} \\ \end{align}



M1

A1

A1


(M1)

(A1)

(A1)


(3 marks) M1: Use of constant acceleration

equation with \displaystyle v=0

A1: Correct equation

A1: Correct \displaystyle s


1 (b) \displaystyle \begin{align}

& {{v}^{2}}={{0}^{2}}+2\times 9 \textrm{.}8\times 44 \textrm{.}1 \\ & v=\sqrt{864 \textrm{.}36}=29 \ \textrm{.}4\text{ m}{{\text{s}}^{-1}} \\ \end{align}

OR

\displaystyle \begin{align} & v=0+9 \textrm{.}8\times 3 \\ & v=29 \textrm{.}4 \text{ m}{{\text{s}}^{-1}} \\ \end{align}


M1

A1

A1


(3 marks)


M1: Use of constant acceleration equation with \displaystyle v=0

A1: Correct equation.

A1: Correct \displaystyle v.


1 (c) Air resistance would slow the ball down. B1 (1 mark) B1: Sensible statement about air resistance.
1 2 3 4 5
2 (a) 2 3 4 5
2 (b) 2 3 4 5
2 (c) 2 3 4 5
2 (d) 2 3 4 5
1 2 3 5
3 (a) 2 3 4 5
3 (b) 2 3 4 5
3 (c) 2 3 4 5
1 2 3 4 5
4 (a) 2 3 4 5
4 (b) 2 3 4 5
4 (c) 2 3 4 5
1 2 3 4 5
5 (a) 2 3 4 5
5 (b) 2 3 4 5
5 (c) 2 3 4 5


Retrieved from "http://wiki.sommarmatte.se/wikis/2009/bridgecoursemechanics/index.php/Solution_to_Test_Paper_2"