hungry_soul
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The test was really hard 
Ill be lucky to get a credit.
Ill be lucky to get a credit.
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Physics Olympiad (2013) (1 Viewer)
- Thread starter Thushan
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hungry_soul
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the last question wasn't that hard imo, it was so random though
hungry_soul
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can someone please explain the mechanics question, i'm in Year 10 and haven't started physics so i had no idea what i was doing. just point me in the direction of the relevant required physics
Guys what did you get for the MCQ question with the 8kg mass hanging down from the pulley and the 5kg mass on the smooth horizontal surface?
For those who didn't do the olympiad today, i'll try to regurgitate the question (with words):
A mass of 8kg hangs from a pulley. It is connected to a 5kg mass that lies on a smooth horizontal table with a string.
Find the acceleration of the 8kg mass (which is hanging beneath the table)
a. 15ms2
b. 9.8ms2
c. 6ms2
d. 3.8ms2
e. forgot (something like 0.5ms2)
For those who didn't do the olympiad today, i'll try to regurgitate the question (with words):
A mass of 8kg hangs from a pulley. It is connected to a 5kg mass that lies on a smooth horizontal table with a string.
Find the acceleration of the 8kg mass (which is hanging beneath the table)
a. 15ms2
b. 9.8ms2
c. 6ms2
d. 3.8ms2
e. forgot (something like 0.5ms2)
The answer for that is c) 6ms2. That is what I got with my calculations. I can try and explain it to you if you want.
Both masses are accelerated, one up and one across, but both have an acceleration of a. Since it is a pulley system, they both have the same acceleration. Let their acceleration be a. Two equations can be set up for this, one for the 5kg block and one for the 8kg block.
So let m1 (the mass of the first block) =5kg
and let m2 = 8kg
The net force is equal to the mass x the acceleration or F=ma
1. T=m1 x a (this is the equation for the forces on the first mass)
2. m2 x g - T = m2 x a (this is the equation for the net forces on the second mass)
Using these two equations, T and a can be found.
m2 x g - m1 x a = m2 x a
a = (m2 x g) / (m1 + m2)
substituting these values
m1 = 5kg
m2 = 8kg
a = 9.8
the answer is close to 6ms2.
So let m1 (the mass of the first block) =5kg
and let m2 = 8kg
The net force is equal to the mass x the acceleration or F=ma
1. T=m1 x a (this is the equation for the forces on the first mass)
2. m2 x g - T = m2 x a (this is the equation for the net forces on the second mass)
Using these two equations, T and a can be found.
m2 x g - m1 x a = m2 x a
a = (m2 x g) / (m1 + m2)
substituting these values
m1 = 5kg
m2 = 8kg
a = 9.8
the answer is close to 6ms2.
how did you design your experiment?
I didn't even bother to include a verb at the start of each method since I got distracted. That test was so weird, it tells you not to hurt your rulers, your pens and it tells you not to eat your test paper - so I was like wtf? and did I forget to mention that pesky rhyme at the end of the test paper?
Andrew Pan
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ahahaha they always have a good sense of humour, at the end of last year's paper it was 'that is all your honour'
They seem to be pretty funny
Kurosaki
True Fail Kid
Nice, I got a bronze for chem as well ^___^. Was pretty surprising to hear since I thought I messed up real bad. My school got another bronze in physics and one gold for Chem and another gold for physics. Pretty good since we didn't get trained and just self taught ourselves..
RealiseNothing
what is that?It is Cowpea
Beast!
Kurosaki
True Fail Kid
Haha not really, there was this really easy section (which I still screwed a bit in ). It had a lot of VERY simple calculations, based on how much volume of water gets displaced when you throw some metal into it. And they gave you the density of the metal and everything so it was fairly easy to do.
). It had a lot of VERY simple calculations, based on how much volume of water gets displaced when you throw some metal into it. And they gave you the density of the metal and everything so it was fairly easy to do.
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