VCE Specialist 3/4 Question Thread!

[Sine]

Is a force required to push up an object up an inclined plane?

My teacher told us that this was a misconception??

Well no a force isn't required to push the object up an inclined plane it could also be pulled up the plane.

You just need a net upwards force (along inclined plane) that defies any frictional force (if existent)

[Rieko Ioane]

Hi,

For these 2 2008 VCAA Qs https://imgur.com/a/Babm6
I don't understand why the worked solutions do what they did.

Q15) Why can't we multiply the magnitudes to the vectors, then sum these 2 vectors and then take their magnitudes? I don't understand what VCAA did here,

Part f) Not sure how to do this. Apparently you use U1/2 geometry but I did shit in that so...

[Eric11267]

Hi,

For these 2 2008 VCAA Qs https://imgur.com/a/Babm6
I don't understand why the worked solutions do what they did.

Q15) Why can't we multiply the magnitudes to the vectors, then sum these 2 vectors and then take their magnitudes? I don't understand what VCAA did here,

Part f) Not sure how to do this. Apparently you use U1/2 geometry but I did shit in that so...

For the first question you need to scale up the vectors so that they have the magnitudes provided. Since the second one has a magnitude of 4 it would be 2i+2root(3)j
After you've added the actual vectors together, then you can take magnitude

As for the second question its circle geometry from last year. To find the area of the shaded region you need to first find the area of that non shaded segment. To find the area of the sector you need to take the segment between the two points of intersections and subtract the isoscoles triangle formed by the points of intersection and the centre of the circle. I've attached an image to make this more clear because I know what I just said is probably really confusing.

[Shadowxo]

Hi,

For these 2 2008 VCAA Qs https://imgur.com/a/Babm6
I don't understand why the worked solutions do what they did.

Q15) Why can't we multiply the magnitudes to the vectors, then sum these 2 vectors and then take their magnitudes? I don't understand what VCAA did here,

Part f) Not sure how to do this. Apparently you use U1/2 geometry but I did shit in that so...

Just to elaborate on Eric's answer,
15 - The reason you can't do what you did is because these forces aren't perpendicular. That formula uses pythag, c=√(a²+b²). As these forces aren't perpendicular, you can't use pythag to find the overall magnitude. You instead need to find the magnitude of the combined (added) forces.
Also, to find the force for Q, you know the magnitude of i+√3j is √(1+3)=2, so as the magnitude of Q is 4, you can multiply i+√3j by 2 to double the magnitude. So, you end up with Q= 2i+2√3j

f) uses unit 1/2 work which I believe isn't in the 3/4 course or going to be in the exam. So, don't worry about it too much

[Rieko Ioane]

Thanks both of you.

Looking back at Q15, why couldn't you multiply Q by four, then sum Q and P and then take the sum's magnitude?

[Eric11267]

Thanks both of you.

Looking back at Q15, why couldn't you multiply Q by four, then sum Q and P and then take the sum's magnitude?

If you multiplied it by four, your vector for Q would have a magnitude of 8 rather than the 4 specified in the question. You would have to see that i+root(3)j already has a magnitude of two, so would need to be multiplied by two to get a magnitude of 4.
So you would multiply by two then sum the vectors and then take the magnitude

[Rieko Ioane]

If you multiplied it by four, your vector for Q would have a magnitude of 8 rather than the 4 specified in the question. You would have to see that i+root(3)j already has a magnitude of two, so would need to be multiplied by two to get a magnitude of 4.
So you would multiply by two then sum the vectors and then take the magnitude

Thanks.

--------
For this https://m.imgur.com/a/UshsZ
On the RHS of the page where I have | 1/2ln(2) - ln(2) |, can I not evaluate this by making everything positive like I have shown. Do I have to leave the modulus signs in there unless I multiply the whole inside by -1?

I'm having trouble understanding how to evaluate modulus signs with constants inside of it.

Thanks.

[Sine]

Thanks.

--------
For this https://m.imgur.com/a/UshsZ
On the RHS of the page where I have | 1/2ln(2) - ln(2) |, can I not evaluate this by making everything positive like I have shown. Do I have to leave the modulus signs in there unless I multiply the whole inside by -1?

I'm having trouble understanding how to evaluate modulus signs with constants inside of it.

Thanks.

modulus sign is basically taking the magnitude of the result so if you have a positive number inside a modolus sign nothing changes e.g. |3| = 3
but if it's negative |-3| = 3

[joejoh]

Hey,
Could someone please explain how to do part bii for this question (comes from Cambridge chapter review Qn 4). Answer is (12.648,13.772). The solutions don't really make much sense to me... a worded explanation would be very helpful if possible.

[S200]

Pssst...
I think we'll need an attachment or something b4 we can help...

[Shadowxo]

Hey,
Could someone please explain how to do part bii for this question (comes from Cambridge chapter review Qn 4). Answer is (12.648,13.772). The solutions don't really make much sense to me... a worded explanation would be very helpful if possible.

Hi, is it bii you want help with or biii? As the answer you provided didn't make much sense, I checked my old textbook and the answer you've provided corresponds with biii. Happy to help with either

[joejoh]

Hi, is it bii you want help with or biii? As the answer you provided didn't make much sense, I checked my old textbook and the answer you've provided corresponds with biii. Happy to help with either

Apologies, its question 4biii

[Eric11267]

Apologies, its question 4biii

The question asks for an interval using the combined data. But since the samples are different sizes, the sample mean of the combined data isn't just the average of the two numbers. Lets just try to find the sample mean of the 70 total people.
For 20 people the mean was 12.5 minutes giving us 20x12.5=250 total headache time
For 50 people the mean was 13.5 minutes giving us 50x13.5=675 total headache time
Adding those together I get 250+675=925 minutes of headache time for 70 people.
925/70= sample mean of 13.2143
Then you just use the standard confidence interval formula. Since you know the standard deviation is 2.4 for both groups, you can just change the sample size to 70.
So your interval becomes
 (13.2143-1.96x2.4/root(70),13.2143+1.96x2.4/root(70))

[humblepie]

Would really appreciate some help with questions G) and L)

I've attached the solutions (which are kinda confusing) - it would be great if someone could explain them to me

Thanks in advance!

[RuiAce]

Would really appreciate some help with questions G) and L)

I've attached the solutions (which are kinda confusing) - it would be great if someone could explain them to me

Thanks in advance!

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