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Solution to Test Paper 2

From Mechanics

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Line 5: Line 5:
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| 1 (a)
| 1 (a)
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| <math>\begin{align}
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& 44.1=\frac{1}{2}\times 9.8{{t}^{2}} \\
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| 4
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& t=\sqrt{\frac{44.1}{4.9}}=3\text{ s} \\
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\end{align}</math>
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OR
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<math>\begin{align}
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& s=\frac{1}{2}\times 9.8\times {{3}^{2}}=44.1 \\
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& \text{AG} \\
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& \therefore \text{Hits ground after 3 seconds} \\
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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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(M1)
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(A1)
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(A1)
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| (3 marks)
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| M1: Use of constant acceleration
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equation with <math>v=0</math>
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A1: Correct equation
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A1: Correct <math>s</math>
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Revision as of 13:12, 19 January 2011

Solutions
1 (a) 441=2198t2t=49441=3 sORs=219832=441AGHits 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) 2 3 4
1 (c) 2 3 4
1 2 3 4
2 (a) 2 3 4
2 (b) 2 3 4
2 (c) 2 3 4
2 (d) 2 3 4
1 2 3 4
3 (a) 2 3 4
3 (b) 2 3 4
3 (c) 2 3 4
1 2 3 4
4 (a) 2 3 4
4 (b) 2 3 4
4 (c) 2 3 4
1 2 3 4
5 (a) 2 3 4
5 (b) 2 3 4
5 (c) 2 3 4