Solution to Test Paper 2

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{| border="1"
{| border="1"
|+ Solutions
|+ Solutions
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| 1 (a)
| 1 (a)
| <math>\begin{align}
| <math>\begin{align}
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& 44 \ \textrm{.}1=\frac{1}{2}\times 9 \textrm{.} 8{{t}^{2}} \\
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& 44\textrm{.}1=\frac{1}{2}\times 9\textrm{.}8{{t}^{2}} \\
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& t=\sqrt{\frac{44 \textrm{.}1}{4 \textrm{.}9}}=3\text{ s} \\
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& t=\sqrt{\frac{44\textrm{.}1}{4\textrm{.}9}}=3\text{ s} \\
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\end{align}</math>
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& \\
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& \text{OR} \\
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OR
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& \\
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& s=\frac{1}{2}\times 9\textrm{.}8\times {{3}^{2}}=44\textrm{.}1 \\
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<math>\begin{align}
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& s=\frac{1}{2}\times 9 \textrm{.}8\times {{3}^{2}}=44 \textrm{.}1 \\
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& \text{AG} \\
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& \therefore \text{Hits ground after 3 seconds} \\
& \therefore \text{Hits ground after 3 seconds} \\
\end{align}</math>
\end{align}</math>
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+
AG
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| 1 (b)
| 1 (b)
| <math>\begin{align}
| <math>\begin{align}
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& {{v}^{2}}={{0}^{2}}+2\times 9 \textrm{.}8\times 44 \textrm{.}1 \\
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& {{v}^{2}}={{0}^{2}}+2\times 9\textrm{.}8\times 44\textrm{.}1 \\
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& v=\sqrt{864 \textrm{.}36}=29 \ \textrm{.}4\text{ m}{{\text{s}}^{-1}} \\
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& v=\sqrt{864\textrm{.}36}=29\textrm{.}4m \\
 +
& \\
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& \text{OR} \\
 +
& \\
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& v=0+9\textrm{.}8\times 3 \\
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& v=29\textrm{.}4 \\
\end{align}</math>
\end{align}</math>
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OR
 
- 
-
<math>\begin{align}
 
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& v=0+9 \textrm{.}8\times 3 \\
 
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& v=29 \textrm{.}4 \text{ m}{{\text{s}}^{-1}} \\
 
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\end{align}</math>
 
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& 800-900=2000a \\
& 800-900=2000a \\
& a=\frac{-100}{2000}=-0\textrm{.}05 \text{ m}{{\text{s}}^{-2}} \\
& a=\frac{-100}{2000}=-0\textrm{.}05 \text{ m}{{\text{s}}^{-2}} \\
 +
& \text{Car is slowing down} \\
\end{align}</math>
\end{align}</math>
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Car is slowing down
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|-
|-
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| 1
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| 2
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|
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| 3
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|
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| 4
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| '''(8 marks)'''
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| 5
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|
|-
|-
| 5 (a)
| 5 (a)
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| 2
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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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| 3
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| M1
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| 4
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| 5
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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
 +
 
|-
|-
| 5 (b)
| 5 (b)
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| 2
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| <math>\begin{align}
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| 3
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& 4-0\textrm{.}4t=0 \\
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| 4
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& t=10 \ \text{s}\\
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| 5
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\end{align}</math>
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 +
| M1
 +
 
 +
A1
 +
 
 +
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.
 +
 
|-
|-
| 5 (c)
| 5 (c)
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| 2
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| <math>\begin{align}
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| 3
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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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| 4
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& =270\mathbf{i}-60\mathbf{j} \\
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| 5
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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
 +
 
 +
M1
 +
 
 +
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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|-
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|
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|
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|
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| '''(8 marks)'''
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|}
|}
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 +
KEY
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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

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