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Two particles A and B are released from rest from different starting points above a horizontal surface - AQA - A-Level Maths Pure - Question 16 - 2020 - Paper 2

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Two particles A and B are released from rest from different starting points above a horizontal surface. A is released from a height of h metres. B is released at a... show full transcript

Worked Solution & Example Answer:Two particles A and B are released from rest from different starting points above a horizontal surface - AQA - A-Level Maths Pure - Question 16 - 2020 - Paper 2

Step 1

Find h in terms of g

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Answer

For particle A, using the equation of motion:

s=ut+12at2s = ut + \frac{1}{2} a t^2

where u=0u = 0, a=ga = g, and t=5t = 5. Thus,

h=12gt2h = \frac{1}{2} g t^2

Substituting the value of tt:

h=12g(52)=25g2h = \frac{1}{2} g (5^2) = \frac{25g}{2}

Step 2

Use kh = \frac{1}{2} gt^2 for particle B

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Answer

For particle B, it is released at time tt after A:

Using the formula:

kh=12g(5t)2kh = \frac{1}{2} g (5 - t)^2

Substituting the expression for hh from particle A:

k(25g2)=12g(5t)2k \left( \frac{25g}{2} \right) = \frac{1}{2} g (5 - t)^2

This simplifies to:

25k=(5t)225k = (5 - t)^2

Step 3

Deduce time for B + t = 5

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Taking the square root of both sides:

5t=5k5 - t = 5\sqrt{k}

Rearranging gives:

t=55kt = 5 - 5\sqrt{k}

Thus:

t=5(1k)t = 5(1 - \sqrt{k})

Step 4

Final Result

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Answer

This shows that the time for particle B to reach the surface can be expressed as:

t=5(1k)t = 5(1 - \sqrt{k})

which completes the proof.

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