Solution to Test Paper 2

From Mechanics

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| '''(8 marks)'''
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| 5 (a)
| 5 (a)
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| <math>\mathbf{v}=(6\mathbf{i}+4\mathbf{j})+(0\textrm{.}2\mathbf{i}-0\textrm{.}4\mathbf{j})t</math>
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| M1
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A1
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| (2 marks)
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| M1: Use of <math>\mathbf{v}=\mathbf{u}+\mathbf{a}t</math>
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A1: Correct expression
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| 5 (b)
| 5 (b)
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| <math>\begin{align}
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& 4-0\textrm{.}4t=0 \\
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& t=10 \ \text{s}\\
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\end{align}</math>
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| M1
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A1
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A1
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| (3 marks)
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| M1: Using <math>\mathbf{j}</math> component equal to zero.
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A1: Correct equation.
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A1: Correct time.
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| 5 (c)
| 5 (c)
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| <math>\begin{align}
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& \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}} \\
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& =270\mathbf{i}-60\mathbf{j} \\
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& r=\sqrt{{{270}^{2}}+{{60}^{2}}}=277\ \text{m} \\
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\end{align}</math>
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| M1
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A1
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M1
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A1
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| (4 marks)
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| M1: Using <math>\mathbf{r}=\mathbf{u}t+\frac{1}{2}\mathbf{a}{{t}^{2}}</math>
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A1: Correct position vector.
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M1: Finding distance.
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A1: Correct distance
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| '''(8 marks)'''
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KEY
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M1: Method Mark
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A1: Accuracy Mark following a method mark
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B1: Accuracy Mark not following a method mark
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AG: Answer Given in Question – Working must justify answer.
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TOTAL: '''40 Marks'''

Revision as of 16:43, 19 January 2011

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} \\ \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.
(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}} \\ \end{align}

Car is slowing down


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

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