Physics Space Question Watch

OOOllie
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#1
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I was looking at the past paper on the back of the AQA Physics paper and have spent countless hours on the question. I have absolutely no idea how to work it out, I took a look at the mark scheme and it told me the answer but not how to work out it out.

The question I am looking at is 3.4 and all the details can be found from this screenshot:
http://prntscr.com/juy6mk

Thank you in advance
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kunoichinin
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Think of units. Acceleration (ms^-2) is 1/200000 as earth. It took 7 hours (25200 secs)

Velocity is m/s and (m/s^2) / (s) = m/s

so 9.8 * 1/200000 * 25200 is average velocity.

Not 100% sure though.
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OOOllie
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Thank you for your response, I have tried your calculation and it gave me the answer 1.23. The mark scheme says 0.89 so I don't know.
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June_.
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Oo that is tricky.

I got 2.5 m/s (rounded to 1dp)
Started off with supposing there was no air resistance, and as Ek = Egp when falling, I re-arranged it to:

V^2 = 2 x h x gfs

Then plugged in some numbers and rearranged to find h then V:

h^2 / 25200^2 = 2 x h x (9.8 / 200000)
h = 25200^2 x 2 x (9.8/200000) = 62233.92 metres

V = 62233.92 / 25200 = 2.5 m/s

I probably made a rediculous amount of mistakes seeing as I didn't get 0.89 but eh, it was a fun question to tackle anyway! Thank you for posting!!
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Joinedup
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used s=1/2 a t2 to get s=15.55x103 m

then
15.55x103/25200 to get 0.617 m/s

I think people getting 1.23 (which is 2x 0.617) are calculating final velocity... it'd be another good method providing you remember to divide by 2

Fwiw philae was actually separated from rosetta at about 22.5 km from the comet's centre of mass according to ESA https://www.esa.int/Our_Activities/S...for_separation

The comet is 4.3 x 4.1km size and irregular in shape (somewhat like a rubber duck) if you said philae had 20x103m to go in 25200 s the average of that would be 0.79 m/s
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OOOllie
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See, I understand that since I do additional maths at GCSE and we are taught mechanics but in Physics we are not so we wouldn't be expected to use that and it is only a 3 mark question.
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RogerOxon
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(Original post by OOOllie)
The question I am looking at is 3.4 and all the details can be found from this screenshot:
http://prntscr.com/juy6mk
This looks like quite a poorly written question to me.

I have to assume that the initial speed of descent is 0, and that gravity is approximately constant from where it is dropped (which I doubt):

s=ut+\frac{at^2}{2}=\frac{at^2}{  2}

v_{average}=\frac{s}{t}=\frac{at  }{2}

\therefore v_{average}=\frac{at}{2}=\frac{2  5200g}{400000}=0.618 ms^{-1}

From what you said, that's not the mark scheme answer. I could do a full calculus answer, but I doubt that's the level expected by this question. I'm not seeing a simple way, unless you were given equations to memorise.
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OOOllie
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Alright, my physics GCSE is over now anyway. Thank you so much though and I assume it was the question then,
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