Solution to Test Paper 2

From Mechanics

(Difference between revisions)
Jump to: navigation, search
(New page: {| class="wikitable" style="text-align: center; width: 200px; height: 200px;" |- | 1 || 2 || 3 |- | 2 || 4 || 6 |- | 3 || 6 || 9 |- | 4 || 8 || 12 |- | 5 || 10 || 15 |})
Current revision (10:22, 8 April 2012) (edit) (undo)
 
(12 intermediate revisions not shown.)
Line 1: Line 1:
-
{| class="wikitable" style="text-align: center; width: 200px; height: 200px;"
+
{| border="1"
 +
|+ Solutions
 +
| 1 (a)
 +
| <math>\begin{align}
 +
& 44\textrm{.}1=\frac{1}{2}\times 9\textrm{.}8{{t}^{2}} \\
 +
& t=\sqrt{\frac{44\textrm{.}1}{4\textrm{.}9}}=3\text{ s} \\
 +
& \\
 +
& \text{OR} \\
 +
& \\
 +
& s=\frac{1}{2}\times 9\textrm{.}8\times {{3}^{2}}=44\textrm{.}1 \\
 +
& \therefore \text{Hits ground after 3 seconds} \\
 +
\end{align}</math>
 +
 
 +
AG
 +
 
 +
 
 +
|M1
 +
 
 +
A1
 +
 
 +
A1
 +
 
 +
 
 +
(M1)
 +
 
 +
(A1)
 +
 
 +
(A1)
 +
 
 +
 
 +
 
 +
| (3 marks)
 +
 
 +
| M1: Use of constant acceleration
 +
equation with <math>v=0</math>
 +
 
 +
A1: Correct equation
 +
 
 +
A1: Correct <math>s</math>
|-
|-
 +
| 1 (b)
 +
| <math>\begin{align}
 +
& {{v}^{2}}={{0}^{2}}+2\times 9\textrm{.}8\times 44\textrm{.}1 \\
 +
& v=\sqrt{864\textrm{.}36}=29\textrm{.}4m \\
 +
& \\
 +
& \text{OR} \\
 +
& \\
 +
& v=0+9\textrm{.}8\times 3 \\
 +
& v=29\textrm{.}4 \\
 +
\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 || 2 || 3
 
|-
|-
 +
| 1 (c)
 +
| Air resistance would slow the ball down.
 +
| B1
 +
| (1 mark)
 +
| B1: Sensible statement about air resistance.
-
| 2 || 4 || 6
 
|-
|-
 +
|
 +
|
 +
|
 +
| '''(7 marks)'''
 +
|
-
| 3 || 6 || 9
 
|-
|-
 +
| 2 (a)
 +
| [[Image:test2ans2.gif]]
 +
| B1
 +
| (1 mark)
 +
| B1: Correct horizontal forces.
 +
Ignore any vertical forces.
 +
-
| 4 || 8 || 12
 
|-
|-
 +
| 2 (b)
 +
| <math>P = 900 \text{N}</math>
 +
| B1
 +
| (1 mark)
 +
| B1: Correct value for <math>P</math>.
 +
 +
|-
 +
| 2 (c)
 +
| <math>\begin{align}
 +
& P-900=2000\times 1\textrm{.}2 \\
 +
& P=2400+900=3300 \text{N} \\
 +
\end{align}</math>
 +
 +
| M1
 +
 +
A1
 +
 +
A1
 +
 +
| (1 mark)
 +
| M1: Three term equation of motion
 +
 +
A1: Correct equation
 +
 +
A1: Correct <math>P</math>.
 +
 +
 +
|-
 +
| 2 (d)
 +
| <math>\begin{align}
 +
& 800-900=2000a \\
 +
& a=\frac{-100}{2000}=-0\textrm{.}05 \text{ m}{{\text{s}}^{-2}} \\
 +
& \text{Car is slowing down} \\
 +
\end{align}</math>
 +
 +
 +
 +
 +
| M1
 +
 +
A1
 +
 +
A1
 +
 +
A1
 +
 +
| (4 marks)
 +
| M1: Three term equation of motion
 +
A1: Correct equation
 +
 +
A1: Correct <math>a</math>
 +
 +
A1: Correct statement
 +
 +
 +
|-
 +
|
 +
|
 +
|
 +
| '''(9 marks)'''
 +
|
 +
 +
|-
 +
| 3 (a)
 +
| <math>R=20\times 9\textrm{.}8=196</math> N
 +
|M1
 +
 +
A1
 +
 +
| (2 marks)
 +
| M1: Use of <math>R=mg</math>
 +
 +
A1: Correct <math>R</math>.
 +
 +
 +
|-
 +
| 3 (b)
 +
| <math>F=0\textrm{.}4\times 196=78\textrm{.}4</math> N
 +
| M1
 +
 +
A1
 +
| (2 Marks)
 +
| M1: Use of <math>F=\mu R</math>
 +
 +
A1: Correct <math>F</math>.
 +
 +
 +
|-
 +
| 3 (c)
 +
| <math>\begin{align}
 +
& 100-78 \textrm{.}4=20a \\
 +
& a=\frac{100-78 \textrm{.}4}{20}=1 \textrm{.}08 \text{ m}{{\text{s}}^{-2}} \\
 +
\end{align}</math>
 +
 +
| M1
 +
 +
A1
 +
 +
A1
 +
 +
| (3 marks)
 +
| M1: Three term equation of motion
 +
 +
A1: Correct equation
 +
 +
A1: Correct <math>a</math>.
 +
 +
 +
|-
 +
|
 +
|
 +
|
 +
| '''(8 marks)'''
 +
|
 +
 +
|-
 +
| 4 (a)
 +
| [[Image:test2ans4.gif]]
 +
| B1
 +
| (1 mark)
 +
| B1: Correct force diagram
 +
 +
|-
 +
| 4 (b)
 +
| <math>\begin{align}
 +
& 100a=200-980\sin 5{}^\circ \\
 +
& a=\frac{200-980\sin 5{}^\circ }{100}=1\textrm{.}15\ \text{m}{{\text{s}}^{-2}} \\
 +
\end{align}</math>
 +
 +
| M1A1
 +
 +
M1
 +
 +
A1
 +
 +
| (4 marks)
 +
| M1: Three term equation of motion
 +
 +
A1: Correct equation
 +
 +
M1: Rearranging equation.
 +
 +
A1: Correct <math>a</math>.
 +
 +
 +
|-
 +
| 4 (c)
 +
|
 +
<math>\begin{align}
 +
& s=0\times 5+\frac{1}{2}\times 1\textrm{.}15\times {{5}^{2}} \\
 +
& =14\textrm{.}4\ \text{m} \\
 +
\end{align}</math>
 +
 +
| M1
 +
 +
A1
 +
 +
A1
 +
 +
| (3 marks)
 +
| M1: Using a constant acceleration equation
 +
 +
A1: Correct equation
 +
 +
A1: Correct distance.
 +
 +
 +
|-
 +
|
 +
|
 +
|
 +
| '''(8 marks)'''
 +
|
 +
 +
|-
 +
| 5 (a)
 +
| <math>\mathbf{v}=(6\mathbf{i}+4\mathbf{j})+(0\textrm{.}2\mathbf{i}-0\textrm{.}4\mathbf{j})t</math>
 +
| M1
 +
 +
A1
 +
| (2 marks)
 +
| M1: Use of <math>\mathbf{v}=\mathbf{u}+\mathbf{a}t</math>
 +
 +
A1: Correct expression
 +
 +
 +
|-
 +
| 5 (b)
 +
| <math>\begin{align}
 +
& 4-0\textrm{.}4t=0 \\
 +
& t=10 \ \text{s}\\
 +
\end{align}</math>
 +
 +
| M1
 +
 +
A1
 +
 +
A1
 +
 +
| (3 marks)
 +
| M1: Using <math>\mathbf{j}</math> component equal to zero.
 +
 +
A1: Correct equation.
 +
 +
A1: Correct time.
 +
 +
|-
 +
| 5 (c)
 +
| <math>\begin{align}
 +
& \mathbf{r}=(6\mathbf{i}+4\mathbf{j})\times 30+\frac{1}{2}(0\textrm{.}2\mathbf{i}-0\textrm{.}4\mathbf{j})\times {{30}^{2}} \\
 +
& =270\mathbf{i}-60\mathbf{j} \\
 +
& r=\sqrt{{{270}^{2}}+{{60}^{2}}}=277\ \text{m} \\
 +
\end{align}</math>
 +
 +
| M1
 +
 +
A1
 +
 +
M1
 +
 +
A1
 +
 +
| (4 marks)
 +
| M1: Using <math>\mathbf{r}=\mathbf{u}t+\frac{1}{2}\mathbf{a}{{t}^{2}}</math>
 +
 +
A1: Correct position vector.
 +
 +
M1: Finding distance.
 +
 +
A1: Correct distance
 +
 +
|-
 +
|
 +
|
 +
|
 +
| '''(8 marks)'''
 +
 +
-
| 5 || 10 || 15
 
|}
|}
 +
 +
KEY
 +
M1: Method Mark
 +
 +
A1: Accuracy Mark following a method mark
 +
 +
B1: Accuracy Mark not following a method mark
 +
 +
AG: Answer Given in Question – Working must justify answer.
 +
 +
TOTAL: '''40 Marks'''

Current revision

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

& 44\textrm{.}1=\frac{1}{2}\times 9\textrm{.}8{{t}^{2}} \\ & t=\sqrt{\frac{44\textrm{.}1}{4\textrm{.}9}}=3\text{ s} \\ & \\ & \text{OR} \\ & \\ & s=\frac{1}{2}\times 9\textrm{.}8\times {{3}^{2}}=44\textrm{.}1 \\ & \therefore \text{Hits ground after 3 seconds} \\ \end{align}

AG


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{.}4m \\ & \\ & \text{OR} \\ & \\ & v=0+9\textrm{.}8\times 3 \\ & v=29\textrm{.}4 \\ \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.
(7 marks)
2 (a) Image:test2ans2.gif B1 (1 mark) B1: Correct horizontal forces.

Ignore any vertical forces.


2 (b) \displaystyle P = 900 \text{N} B1 (1 mark) B1: Correct value for \displaystyle P.
2 (c) \displaystyle \begin{align}

& P-900=2000\times 1\textrm{.}2 \\ & P=2400+900=3300 \text{N} \\ \end{align}

M1

A1

A1

(1 mark) M1: Three term equation of motion

A1: Correct equation

A1: Correct \displaystyle P.


2 (d) \displaystyle \begin{align}

& 800-900=2000a \\ & a=\frac{-100}{2000}=-0\textrm{.}05 \text{ m}{{\text{s}}^{-2}} \\ & \text{Car is slowing down} \\ \end{align}



M1

A1

A1

A1

(4 marks) M1: Three term equation of motion

A1: Correct equation

A1: Correct \displaystyle a

A1: Correct statement


(9 marks)
3 (a) \displaystyle R=20\times 9\textrm{.}8=196 N M1

A1

(2 marks) M1: Use of \displaystyle R=mg

A1: Correct \displaystyle R.


3 (b) \displaystyle F=0\textrm{.}4\times 196=78\textrm{.}4 N M1

A1

(2 Marks) M1: Use of \displaystyle F=\mu R

A1: Correct \displaystyle F.


3 (c) \displaystyle \begin{align}

& 100-78 \textrm{.}4=20a \\ & a=\frac{100-78 \textrm{.}4}{20}=1 \textrm{.}08 \text{ m}{{\text{s}}^{-2}} \\ \end{align}

M1

A1

A1

(3 marks) M1: Three term equation of motion

A1: Correct equation

A1: Correct \displaystyle a.


(8 marks)
4 (a) Image:test2ans4.gif B1 (1 mark) B1: Correct force diagram
4 (b) \displaystyle \begin{align}

& 100a=200-980\sin 5{}^\circ \\ & a=\frac{200-980\sin 5{}^\circ }{100}=1\textrm{.}15\ \text{m}{{\text{s}}^{-2}} \\ \end{align}

M1A1

M1

A1

(4 marks) M1: Three term equation of motion

A1: Correct equation

M1: Rearranging equation.

A1: Correct \displaystyle a.


4 (c)

\displaystyle \begin{align} & s=0\times 5+\frac{1}{2}\times 1\textrm{.}15\times {{5}^{2}} \\ & =14\textrm{.}4\ \text{m} \\ \end{align}

M1

A1

A1

(3 marks) M1: Using a constant acceleration equation

A1: Correct equation

A1: Correct distance.


(8 marks)
5 (a) \displaystyle \mathbf{v}=(6\mathbf{i}+4\mathbf{j})+(0\textrm{.}2\mathbf{i}-0\textrm{.}4\mathbf{j})t M1

A1

(2 marks) M1: Use of \displaystyle \mathbf{v}=\mathbf{u}+\mathbf{a}t

A1: Correct expression


5 (b) \displaystyle \begin{align}

& 4-0\textrm{.}4t=0 \\ & t=10 \ \text{s}\\ \end{align}

M1

A1

A1

(3 marks) M1: Using \displaystyle \mathbf{j} component equal to zero.

A1: Correct equation.

A1: Correct time.

5 (c) \displaystyle \begin{align}

& \mathbf{r}=(6\mathbf{i}+4\mathbf{j})\times 30+\frac{1}{2}(0\textrm{.}2\mathbf{i}-0\textrm{.}4\mathbf{j})\times {{30}^{2}} \\ & =270\mathbf{i}-60\mathbf{j} \\ & r=\sqrt{{{270}^{2}}+{{60}^{2}}}=277\ \text{m} \\ \end{align}

M1

A1

M1

A1

(4 marks) M1: Using \displaystyle \mathbf{r}=\mathbf{u}t+\frac{1}{2}\mathbf{a}{{t}^{2}}

A1: Correct position vector.

M1: Finding distance.

A1: Correct distance

(8 marks)


KEY M1: Method Mark

A1: Accuracy Mark following a method mark

B1: Accuracy Mark not following a method mark

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

TOTAL: 40 Marks

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