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Solution to Test Paper 2

From Mechanics

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| 4 (a)
| 4 (a)
| [[Image:test2ans4.gif]]
| [[Image:test2ans4.gif]]
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| 3
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| B1
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| 4
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| (1 mark)
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| 5
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| B1: Correct force diagram
|-
|-
| 4 (b)
| 4 (b)
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| 2
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| <math>\begin{align}
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| 3
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& 100a=200-980\sin 5{}^\circ \\
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| 4
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& a=\frac{200-980\sin 5{}^\circ }{100}=1\textrm{.}15\ \text{m}{{\text{s}}^{-2}} \\
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| 5
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\end{align}</math>
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| M1A1
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 +
M1
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 +
A1
 +
 
 +
| (4 marks)
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| M1: Three term equation of motion
 +
 
 +
A1: Correct equation
 +
 
 +
M1: Rearranging equation.
 +
 
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A1: Correct <math>a</math>.
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|-
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| 4 (c)
| 4 (c)
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| 2
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|
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| 3
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<math>\begin{align}
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| 4
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& s=0\times 5+\frac{1}{2}\times 1\textrm{.}15\times {{5}^{2}} \\
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| 5
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& =14\textrm{.}4\ \text{m} \\
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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 a constant acceleration equation
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A1: Correct equation
 +
 
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A1: Correct distance.
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|-
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Revision as of 16:22, 19 January 2011

Solutions
1 (a) 44 .1=219.8t2t=4.944.1=3 sORs=219.832=44.1AGHits ground after 3 seconds M1

A1

A1


(M1)

(A1)

(A1)


(3 marks) M1: Use of constant acceleration

equation with v=0

A1: Correct equation

A1: Correct s


1 (b) v2=02+29.844.1v=864.36=29 .4 ms−1ORv=0+9.83v=29.4 ms−1M1

A1

A1


(3 marks)


M1: Use of constant acceleration equation with v=0

A1: Correct equation.

A1: Correct 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) P=900N B1 (1 mark) B1: Correct value for P.
2 (c) P−900=20001.2P=2400+900=3300N M1

A1

A1

(1 mark) M1: Three term equation of motion

A1: Correct equation

A1: Correct P.


2 (d) 800−900=2000aa=2000−100=−0.05 ms−2Carisslowingdown M1

A1

A1

A1

(4 marks) M1: Three term equation of motion

A1: Correct equation

A1: Correct a

A1: Correct statement


(9 marks)
3 (a) R=209.8=196 N M1

A1

(2 marks) M1: Use of R=mg

A1: Correct R.


3 (b) F=0.4196=78.4 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.


1 2 3 4 5
5 (a) 2 3 4 5
5 (b) 2 3 4 5
5 (c) 2 3 4 5


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