VCE Physics Exam - 13/11/19 - Discussion/Questions/Solutions

[3086]

Anyone have an estimate at what the A+ and A cutoffs will be this year?

It felt harder than 2018 for me but idk if that’s just cuz I was under more pressure due to exam conditions.

probably be around 88% since it was "easy" according to many people. Tho I guess q19 threw many people off so it could be around 86%.

If it is this high, SAC results are going to play a more crucial role in determining your study score according to my teacher.

[sadia2089]

I was wondering, for the last question, if you drew the graph wrong (i even put g=9.8 ), but i do the rest of the working out correctly according to MY graph, will i get consequential marks for those parts?

[monopoly]

I was wondering, for the last question, if you drew the graph wrong (i even put g=9.8 ), but i do the rest of the working out correctly according to MY graph, will i get consequential marks for those parts?

Ye u will cze the nxt qns ans depended on that graph so ye

[wanigara]

Thats what I got so hopefully others are wrong 

Edit: Actually I think I got 15, 35 oops

Q19. Most part of the question 19 is ok. (total 18 marks)

Plotting the graph is straight forward (just to time mass in kg x 10 to convert to force, The compression in mm in x axis.
When you plot the graph you'll get two linear sections. Gradient of first linear graph is the K of spring A (which is 150 N/m). The gradient of the second liner section is the sum of K values of A and B which was 500 N/m. Hence the K of the spring B = 350 N/m.  The scales need to be selected to spread your data more than 50% of the area. Uncertainty in compression was +/- 2.0mm . If you have selected scales appropriately, the error bars are like half a box left and right.
(c) (i) Area under the graph of spring A = 1/2*150*0.08^2 = 0.48J
C(ii) Spring Pot energy stored in the spring when the system is compressed by 80 mm. Need to calculate the area under both  straight lines= 1/2*0.06*9 + (1/2(9+19)*0.02 =0.55 J
(iii) Work done only by spring B to compress to 80 mm = 1/2 * 350*0.02^2  (because Spring B only compresses from 60 mm to 80 mm). = 0.07 J

(b) How these type spring system can be used in small bumps and severe bumps.

Low K spring for small bumps and combination of Both springs for sever bumps to absorbs the shock.

[Matthew_Whelan]

Q19. Most part of the question 19 is ok. (total 18 marks)

Plotting the graph is straight forward (just to time mass in kg x 10 to convert to force, The compression in mm in x axis.
When you plot the graph you'll get two linear sections. Gradient of first linear graph is the K of spring A (which is 150 N/m). The gradient of the second liner section is the sum of K values of A and B which was 500 N/m. Hence the K of the spring B = 350 N/m.  The scales need to be selected to spread your data more than 50% of the area. Uncertainty in compression was +/- 2.0mm . If you have selected scales appropriately, the error bars are like half a box left and right.
(c) (i) Area under the graph of spring A = 1/2*150*0.08^2 = 0.48J
C(ii) Spring Pot energy stored in the spring when the system is compressed by 80 mm. Need to calculate the area under both  straight lines= 1/2*0.06*9 + (1/2(9+19)*0.02 =0.55 J
(iii) Work done only by spring B to compress to 80 mm = 1/2 * 350*0.02^2  (because Spring B only compresses from 60 mm to 80 mm). = 0.07 J

(b) How these type spring system can be used in small bumps and severe bumps.

Low K spring for small bumps and combination of Both springs for sever bumps to absorbs the shock.

I realised afterwards I converted mm to m wrong like a dumbass, hopefully I'll get consequential marks since I did everything else right. Also was it really 18 marks?

[monopoly]

It was 16 marks

[Ryan_DS]

I have some answers here. Please let me know if you agree/disagree with anything.

https://filebin.net/8tdp5nycdwy46qol

For 6c, I thought the answer was to move the step down transformer to the end of the circuit, after running the extention cord for 200m at 240v without changing anything, cause the question specifically said 'with existing equipment'

[JeromeTT]

For 6c, I thought the answer was to move the step down transformer to the end of the circuit, after running the extention cord for 200m at 240v without changing anything, cause the question specifically said 'with existing equipment'

This is what i wrote as well

[mada]

Both of them would be linear

Do you know exactly why it should be linear and not curvy for the second spring?

[Erutepa]

Do you know exactly why it should be linear and not curvy for the second spring?

When you plot out the values and draw error bars and what not, the data points seemed at first look very linear to me and you can definitely draw a straight line that passes within the error bars of each point. I can't remember what the points looked like specifically, so perhaps you could have drawn a curved line that suited the plotted points well, however the fact that they wanted you to calculate the K value for the line indicated that the line had a constant gradient and thus must be a linear line.
Hopefully this makes sense, but don't stress if you drew a curvy line - i don't think you'd lose many marks for it and there's nothing you can do about it now. Enjoy your break!

[Ryan_DS]

Do you know exactly why it should be linear and not curvy for the second spring?

Think of it from a physics perspective. You know Hooke's law is linear since F=kx, and if you imagine the system initially, there is only one spring active. Then, once the second one comes it should still be linear as both springs are now applying a force that still scales linearly. It doesn't make sense for it to be curved since Hooke's law is only ever linear, no matter the number of springs or their arrangement.