VCE Physics Question Thread!

[knightrider]

Just wanted to clarify something

lets take the situation of a projectile motion.

Say you use a shortcut formula and then get the answer wrong , is it true that you will not get method marks too?

Also say if you do things from first principles and then get the answer wrong, is it true that you will still have a chance to get method marks ?

[JI2015]

Hi knightrider,

In physics, the examiners don't clearly tell us how marks are awarded and teachers often judge answers based on their own expectations. The exam reports are not in-depth and usually only present one interpretation especially for worded questions.

My advice would be to show as much working out as possible so that the assessor can see if you made a careless error, and therefore be more inclined to award you marks for demonstration of the correct physics principles.

So for projectile motion, I would write the 'shortcut' formula, then write it again with the known information substituted in and then solve and write the answer. Assessors are familiar with these formulas, and if you use them in the correct situation but just enter something wrong in the calculator, you may only lose 1 mark as they are aware that you knew what you were doing.
 
However, if you use the 'shortcut' formulas when they are not applicable and as a result get the wrong answer, you will not get marks.

Lastly, to address your point on the traditional method of dealing with projectile motion: It may take too long in an exam to set all your equations up and will not be advantageous in dealing with a question that the 'shortcut' formulas can be used for. Some questions, you have to set it out the long way, and method marks will be awarded if you demonstrate correct physics principles.

As long as you demonstrate an understanding of the situation presented to you, and apply relevant and correct physics ideas, then you will most likely gain some marks with or without a correct answer!

[knightrider]

Hi knightrider,

In physics, the examiners don't clearly tell us how marks are awarded and teachers often judge answers based on their own expectations. The exam reports are not in-depth and usually only present one interpretation especially for worded questions.

My advice would be to show as much working out as possible so that the assessor can see if you made a careless error, and therefore be more inclined to award you marks for demonstration of the correct physics principles.

So for projectile motion, I would write the 'shortcut' formula, then write it again with the known information substituted in and then solve and write the answer. Assessors are familiar with these formulas, and if you use them in the correct situation but just enter something wrong in the calculator, you may only lose 1 mark as they are aware that you knew what you were doing.
 
However, if you use the 'shortcut' formulas when they are not applicable and as a result get the wrong answer, you will not get marks.

Lastly, to address your point on the traditional method of dealing with projectile motion: It may take too long in an exam to set all your equations up and will not be advantageous in dealing with a question that the 'shortcut' formulas can be used for. Some questions, you have to set it out the long way, and method marks will be awarded if you demonstrate correct physics principles.

As long as you demonstrate an understanding of the situation presented to you, and apply relevant and correct physics ideas, then you will most likely gain some marks with or without a correct answer!

Thanks so much for clarifying JI2015 

[JI2015]

 I'm pretty sure Method 1 is correct, although you would have to be quite pedantic to take away marks for the second working.

Great job learning data analysis principles as they are often neglected and not taken seriously!

EDIT:

The percentage error for (10.5 +/- 0.5) is supposed to have 2 sig. figs. and therefore it become 4.8%

4.8%+0.9%=5.7% total percentage error

[Syndicate]

Hi, I was wondering which of these working outs are correct:
(21.4 +/- 0.2) * (10.5 +/- 0.5)

1)
0.2 / 21.4 = 0.0093457943925234
= 0.009 (1 s.f)
= 0.9%
0.5 / 10.5 = 0.0476190476190476
= 0.05 (1 s.f)
= 5%
0.9% + 5% = 5.9%
21.4 * 10.5 = 224.7
= 225 (3 s.f)

Final answer = 225 +/- 5.9%

2)
0.2 / 21.4 = 0.0093457943925234
0.5 / 10.5 = 0.0476190476190476
0.0093457943925234 + 0.0476190476190476 = 0.05696484201
= 5.7%
21.4 * 10.5 = 224.7
= 225 (3 s.f)

Final answer = 225 +/- 5.7%

the second seems more accurate, as you have added all the decimals points together, while in the first one, you rounded 0.0476190.... to 0.05 (see the difference).

[JI2015]

the second seems more accurate, as you have added all the decimals points together, while in the first one, you rounded 0.476190.... to 0.05 (see the difference).

In my notes I had written down that you add the individual percentage errors together and those are supposed to be to 1 or 2 sig. figs. and therefore the top calculation seems right. I'm not completely certain though.

[Syndicate]

In my notes I had written down that you add the individual percentage errors together and those are supposed to be to 1 or 2 sig. figs. and therefore the top calculation seems right. I'm not completely certain though.

hey Jl2015

What I did was, that I added those two answers together (0.09... + 0.47...), and get an answer closer to 0.56 , which seems much closer to 5.7% than 5.9%

[JI2015]

hey Jl2015

What I did was, that if you added those two answers together (0.09... + 0.47...), i get a answer higher to 0.56 , which seems much closer to 5.7%

Take a look at my notes screenshot, that is the correct way to deal with these calculations. I modified my initial response as I found an error in the working out.

[JI2015]

Sorry to be pedantic, but I thought when we divide 10.5 by 0.5 we take our answer to the least amount of sig figs, which is 1. Or is there a special exception with percentage errors?

dividing 0.5 by 10.5 yields 0.0476, to convert it into a percentage we multiply it by 100 and we get 4.76%

Since 4.76% is > 1% then it will have two sig figs and it will become 4.8%

This is just how it's done.

[JI2015]

Here is a resource with heaps of data analysis examples and explanations that my school provided me. Make sure to check it out and read some of the explanations.

Spoiler

http://pegsnet.pegs.vic.edu.au/studentdownloads/Physics/Students/UNIT%201/Data%20analysis/DataAnalysisBooklet.pdf

Spoiler

Heaps of physics booklets are also available on my school's website so just take what you need, but let's agree that I did not send you, haha: http://pegsnet.pegs.vic.edu.au/studentdownloads/Physics/Students /

[YellowTongue]

Can someone please help me with this question. I just can't figure it out...

A ball with a mass of 100g is thrown at a wall horizontally with a speed of 4.0m/s. Calculate the force that the wall exerts on the ball during impact.

Thanks 

[JI2015]

Can someone please help me with this question. I just can't figure it out...

A ball with a mass of 100g is thrown at a wall horizontally with a speed of 4.0m/s. Calculate the force that the wall exerts on the ball during impact.

Thanks 

Where did you find this question? You need more information.

If you are given a time of contact with the wall and a rebound velocity then you can use the change of momentum/ impulse formulas.

Otherwise there isn't much you can do with the information you have provided.

[YellowTongue]

Sorry, I looked up the worked solution on the internet and found that I needed to use Hooke's Law and the elastic potential energy formula from a graph, which I didn't think I'd have to do because that was all covered in the next section of the chapter. I understand it now though 

[Muchos Help]

http://m.imgur.com/ZjeJkM2 - In the attached for Ex15,Q7: I initially got this question wrong but got it correct in the end by using a different equation of motion equation.

My variables were x=?, u=5, *v=5, a=0 and t=0.6. For v, is 5 incorrect? I thought that horizontal velocity was constant throughout the entire flight, but I tried using v2 = u2 +2ax but it was wrong, then used x = ut + 0.5at2 which I got the correct answer from.

http://imgur.com/yr7WRUb, similarly for this question. Why can't you use the horizontal direction's variables then use v=u+at?
The correct answer for this Q is considering the vertical direction and I used x=vt-0.5at2 but I encountered a problem, why does x= -5 if I take down to be positive? Whereas when you use x=ut+0.5at2, x needs to = 5 for the answer to be correct (1second).

Sorry if my writing is confusing, I'll clarify if needed.

[Syndicate]

http://m.imgur.com/ZjeJkM2 - In the attached for Ex15,Q7: I initially got this question wrong but got it correct in the end by using a different equation of motion equation.

My variables were x=?, u=5, *v=5, a=0 and t=0.6. For v, is 5 incorrect? I thought that horizontal velocity was constant throughout the entire flight, but I tried using v2 = u2 +2ax but it was wrong, then used x = ut + 0.5at2 which I got the correct answer from.

http://imgur.com/yr7WRUb, similarly for this question. Why can't you use the horizontal direction's variables then use v=u+at?
The correct answer for this Q is considering the vertical direction and I used x=vt-0.5at2 but I encountered a problem, why does x= -5 if I take down to be positive? Whereas when you use x=ut+0.5at2, x needs to = 5 for the answer to be correct (1second).

Sorry if my writing is confusing, I'll clarify if needed.

The second rule you have used is right, as it is to find the horizontal displacement. If you re-read the question, it says that your intial horizontal velocity is 5 m/s , therefore, you are required to use this formula to find the horizontal displacemnt. It can also be used to find the vertical displacement, by simply subbing in your intial vertical veloctiy, instead of intial horizontal velocity.

As for your first formula. It used to find the final vertical/ horizontal velocity{squared} (velocity = speed in given direction, think of vectors), while, you had to find the displacment (distanced covered from the beginning and concluding points)

Displacement formula : d =

where
d = displacment
v = intial vertical/horizontal velocity
t = time
a = acceleration

Do note: whenever you are working out with any of the vertical formulas, acceleration will always = 9.8 m/s/s

Hope this helps