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Author Topic: 2010 VCAA question 16 -17  (Read 423 times)  Share 

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Wingtips

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2010 VCAA question 16 -17
« on: June 08, 2012, 11:43:18 pm »
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Just wanted to know for question 16 - 17, whether my explanation would be correct:

Q16 Total momentum is always conserved in an interaction between objects in an isolated system. (Law of Conservation of Mass) [Same for question 17]

I didn't mention "before, during and after a collision." Thought it was kind of self-explanatory with the word "always conserved."

Edit: Of the motion section. >< Sorry
« Last Edit: June 09, 2012, 12:05:32 am by Wingtips »

StumbleBum

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Re: 2010 VCAA question 16 -17
« Reply #1 on: June 09, 2012, 12:08:26 am »
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Seen as they refer to the momentum before, during and after in the question i would recommend also referring to it in your answers, remember that it can't necessarily hurt to put a little bit more than required (provided that its not wrong or completely irrelevant).
2011: Mathematical Methods (CAS) [36]

2012: English [35+] | Specialist Mathematics [35+] | Further Mathematics [45+] | Physics [40+] | Accounting [38+] |

Wingtips

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Re: 2010 VCAA question 16 -17
« Reply #2 on: June 09, 2012, 12:19:28 am »
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Yep, fair enough. It's one of the moments where it's like "Maybe I should it put it in. Hmm, nah should be alright." Haha

Wingtips

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Re: 2010 VCAA question 16 -17
« Reply #3 on: June 09, 2012, 11:00:19 am »
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Could someone also please explain question 14 of motion, same exam?

Mr. Study

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Re: 2010 VCAA question 16 -17
« Reply #4 on: June 09, 2012, 11:08:41 am »
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Hey,

We can work out the gravitational potential energy of the spring, due to Figure B giving us the Change in Height.

Us=2x10x0.4=8J

Now in figure B, The spring has stretched by 0.4 m. However, The due it it being stretched, when compared to Figure A, it will now have elastic potential energy within the spring.

E=0.5x50x0.42=4J

So, We can say that From Figure A to B, the spring has lost 8 J, as it 'doesn't have any gravitational potential energy' in Figure B. However, it has gained 4J, within the spring itself, stored as elastic potential energy.

Does that help? :S
« Last Edit: June 09, 2012, 11:10:19 am by Mr. Study »
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Wingtips

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Re: 2010 VCAA question 16 -17
« Reply #5 on: June 09, 2012, 11:15:30 am »
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Yep, makes sense. So Ug = m*g*change in h. Not simply height? I've been going by the Heinemann textbook and all it says is height. Or is it just because you assume the 'zero point' is when the spring is not extended and then from A to B the height increases to 0.4m?
« Last Edit: June 09, 2012, 11:17:57 am by Wingtips »

StumbleBum

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Re: 2010 VCAA question 16 -17
« Reply #6 on: June 09, 2012, 11:52:48 am »
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∆Ug=mg∆h
Ug=mgh
So you would use the first one when there has been a change in height and you want to know how much Ug it has lost (or gained). And you would use the second one when you know the height relative to the ground and want to know the Ug the particle possesses. 
Does that clear things up?
2011: Mathematical Methods (CAS) [36]

2012: English [35+] | Specialist Mathematics [35+] | Further Mathematics [45+] | Physics [40+] | Accounting [38+] |