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CfE Higher Physics

No I'm not stupid just forget and lack a lot of confidence that's all I put a lot of pressure on myself unnecessarily in some cases if that makes sense I hope your revision is going well

@Labrador99 @JP298

Had a few questions. I tried a past paper and got these questions wrong but don't understand how. Here's the link and would appreciate it if you could walk me through the answers real quick(only multi choice):
http://mrmackenzie.wikispaces.com/file/view/1992_HigherI.pdf

Questions: 6,15,16,20,22,23,25

Thanks so much if you could :smile:

Original post by Jeff548

@Labrador99 @JP298

Had a few questions. I tried a past paper and got these questions wrong but don't understand how. Here's the link and would appreciate it if you could walk me through the answers real quick(only multi choice):
http://mrmackenzie.wikispaces.com/file/view/1992_HigherI.pdf

Questions: 6,15,16,20,22,23,25

Thanks so much if you could :smile:

Here is Q6- I will post the other ones as I do them :smile:

Q6. Total mass of balloon + people = 300 + 250 = 550kg
The balloon floats at a steady height, therefore, the forces acting on the balloon vertically are balanced, therefore, upwards thrust = weight.
Note that 10 is used for g, but the datasheet for CfE higher says 9.8…I think the really old exams maybe used 10?
W = mg
W = 550 x 10
W = 5500N
Therefore upwards thrust = 5500N
Therefore answer E

Original post by Jeff548

@Labrador99 @JP298

Had a few questions. I tried a past paper and got these questions wrong but don't understand how. Here's the link and would appreciate it if you could walk me through the answers real quick(only multi choice):
http://mrmackenzie.wikispaces.com/file/view/1992_HigherI.pdf

Questions: 6,15,16,20,22,23,25

Thanks so much if you could :smile:

Q15. This isn’t a great explanation, so if anyone else can explain it better,please do!...
The thermistor is placed in a beaker of melting ice, therefore temperaturedecreases. This causes the resistance ofthe thermistor to increase. When theWheatstone bridge is balanced, R_Q/R_P = R_R/R_t…Iwould say pick random numbers for each of the resistances and substitute theminto that equation. Then increase R_t. Try doing each of the possible answers inturn until one of them makes the equation make sense like it did in thebeginning…By doing this, you will find that when R_tincreases, thisequation can only be kept true by increasing R_R.Therefore, answer A.

Original post by Jeff548

@Labrador99 @JP298

Had a few questions. I tried a past paper and got these questions wrong but don't understand how. Here's the link and would appreciate it if you could walk me through the answers real quick(only multi choice):
http://mrmackenzie.wikispaces.com/file/view/1992_HigherI.pdf

Questions: 6,15,16,20,22,23,25

Thanks so much if you could :smile:

Q16. I’m going to describe the first 2 resistors as ‘line 1’ (i.e the 6Ω resistor and the 6Ω resistor) and the second set of 2 resistors (2Ω and 4Ω) as line 2.
In both line 1 and line 2, the total voltage will be 12V as the voltage in parallel parts of the circuit equals the supply voltage.
Line 1- The voltage will be shared by proportion as the 2 resistors are in series… The ratio of the 2 resistors in line 1 is 6:6 (1:1), therefore, the voltage will split in a ratio of 1:1, so the voltage through each of the 6Ω resistors will be 6V.
Line 2- Same sort of idea- The voltage willbe shared by proportion as the 2 resistors are in series … The ratio of the 2 resistors in line 1 is 2:4 (1:2), therefore, the voltage will split in a ratioof 1:2. There are therefore 3 ’parts’. Each part is 4V since 12/3 = 4. The 2Ω resistor gets 1 part and the 4Ω resistor gets 2 parts, so the voltage through the 2Ω resistoris 4V and the voltage through the 4Ω resistor is 8V.
You then need to compare either the voltages through the top 2 or the bottom 2 resistors (either combination will work)…I’ll do the top 2… The voltage through the 6Ω resistor is 6V and the voltage through the 2Ω resistor is 4V. The potential difference is therefore 6 - 4 = 2V. (Try this with the bottom set of resistors and you'll see that it works too!) Therefore, answer B. Hope this makes sense- if it doesn't, let me know :smile:

(edited 8 years ago)

Original post by Jeff548

@Labrador99 @JP298

Had a few questions. I tried a past paper and got these questions wrong but don't understand how. Here's the link and would appreciate it if you could walk me through the answers real quick(only multi choice):
http://mrmackenzie.wikispaces.com/file/view/1992_HigherI.pdf

Questions: 6,15,16,20,22,23,25

Thanks so much if you could :smile:

I'm not sure about 20, but I'll have a look at my notes later and try and figure it out if no one else can help. Here's Q22...
R is a minimum, so the equation Path difference = (m + 1/2)x λ is used. R is the first order minimum, so m = 1 and the question states that λ = 3cm. If you substitute these in and simplify, you get Path difference = 4.5cm, therefore, answer C.

(edited 8 years ago)

Original post by Jeff548

@Labrador99 @JP298

Had a few questions. I tried a past paper and got these questions wrong but don't understand how. Here's the link and would appreciate it if you could walk me through the answers real quick(only multi choice):
http://mrmackenzie.wikispaces.com/file/view/1992_HigherI.pdf

Questions: 6,15,16,20,22,23,25

Thanks so much if you could :smile:

Here's Q23 :smile:

23. For questions like this, I suggest that you do a drawing of the situation, without all of the unnecessary information that the SQA like to provide. When you do this, you should realise that the angles given are not the angle of incidence or the angle of refraction (if you can’t see why they are not, look at your definitions of angleof incidence and angle of refraction from N5). You will then find that the angle of incidence is 40 and the angle of refraction is 60. You need to use the equation n = Sinθ₁/Sin θ₂, where n is the refractive index, θ₁ angle made with the normal in the air and θ₂ is the angle made with the normal in the material (in this case the water). When you do this, you get n = 1.35, therefore answer C.

(edited 8 years ago)

Original post by Jeff548

@Labrador99 @JP298

Had a few questions. I tried a past paper and got these questions wrong but don't understand how. Here's the link and would appreciate it if you could walk me through the answers real quick(only multi choice):
http://mrmackenzie.wikispaces.com/file/view/1992_HigherI.pdf

Questions: 6,15,16,20,22,23,25

Thanks so much if you could :smile:

25. When a ray of light enters a medium at 90 degrees to the normal, it does not change direction, so B and E can be ruled out immediately. I would advise drawing a normal line at every glass/air boundary for this one. When the light goes from a medium with a higher refractive index to a medium with a lower refractive index (glass to air) it bends away from the normal line. And when the light goes from a medium with a lower refractive index to a medium with a higher refractive index (air to glass) it bends towards the normal line. Using this, the answer is A.
Hope these answers have helped and that your studying is going well…If any of them don’t make sense let me know and if anyone has a better way of explaining, feel free :smile:

Can anyone help Jeff548 with question 20?

Reply 48

Asklepios

19

Original post by Jeff548

@Labrador99 @JP298

Had a few questions. I tried a past paper and got these questions wrong but don't understand how. Here's the link and would appreciate it if you could walk me through the answers real quick(only multi choice):
http://mrmackenzie.wikispaces.com/file/view/1992_HigherI.pdf

Questions: 6,15,16,20,22,23,25

Thanks so much if you could :smile:

Original post by Labrador99

25. When a ray of light enters a medium at 90 degrees to the normal, it does not change direction, so B and E can be ruled out immediately. I would advise drawing a normal line at every glass/air boundary for this one. When the light goes from a medium with a higher refractive index to a medium with a lower refractive index (glass to air) it bends away from the normal line. And when the light goes from a medium with a lower refractive index to a medium with a higher refractive index (air to glass) it bends towards the normal line. Using this, the answer is A.
Hope these answers have helped and that your studying is going well…If any of them don’t make sense let me know and if anyone has a better way of explaining, feel free :smile:

Can anyone help Jeff548 with question 20?

For Q20, the emf is still 10V, so the maximum voltage across the capacitor after charging will still be 10V. However, with increased capacitance and increased circuit resistance, what will be the effect on charging time?

Original post by Asklepios

For Q20, the emf is still 10V, so the maximum voltage across the capacitor after charging will still be 10V. However, with increased capacitance and increased circuit resistance, what will be the effect on charging time?

I agree that the maximum voltage across the capacitor will still be 10V. Therefore, the answer cannot be a graph that show the voltage increasing beyond 10V, ruling out answer A. Any graph that shows the maximum voltage to be less than 10V (i.e the graph levels off, without reaching 10V) can also be ruled out, eliminating answer B. Increased capacitance and increased resistance both result in the capacitor taking longer to charge (see below for explanation from BBC bitesize). The original set up took over 4 seconds to fully charge, so any graph that shows charging to take the same amount of time or less should be eliminated- so both C and D can be ruled out.
This leaves only E- The graph is still increasing at just over 4 seconds, so it is still charging and could go up to 10V and it also shows that it takes longer for the graph to reach 10V.
I looked at the answers on another part of the site(http://mrmackenzie.org/webtribe/Physics1992p1solns.html) and E is correct, though I'm not 100% sure about my reasoning- Asklepios what do you think?

From BBC Bitesize:
"If a larger value of capacitance were used with the same value of resistance in the above circuit it would be able to store more charge. As a result, it would take longer to charge up to the supply voltage during charging and longer to lose all its charge when discharging.
If a larger value of resistance were used with the same value of capacitance in the above circuit, then a smaller current would flow, therefore it would take longer for the capacitor to charge up and longer for it to discharge."(http://www.bbc.co.uk/bitesize/higher/physics/elect/capacitance/revision/2/)

Reply 50

molly221

3

Can someone please explain to me how to do this. I tried it and got the answer of 9 ohms (D). The correct answer of the marking scheme is B. Why is this?
Thanks :smile:

Original post by molly221

Can someone please explain to me how to do this. I tried it and got the answer of 9 ohms (D). The correct answer of the marking scheme is B. Why is this?
Thanks :smile:

The worked solution is attached :smile:

Reply 52

molly221

3

Original post by Labrador99

The worked solution is attached :smile:

Thank you so much @Labrador99! Really helped! You are so good at physics :smile:

Reply 53

molly221

3

Attachment not found

Can anyone please explain to me how to do this. Thanks

It is only Q5! i know how to do Q15. It will not let me delete though. Thanks

(edited 8 years ago)

Screen Shot 2016-04-13 at 16.19.41.png247.7KB

Screen Shot 2016-04-13 at 16.36.12.png280.5KB

Can anyone help???

Original post by Labrador99

I agree that the maximum voltage across the capacitor will still be 10V. Therefore, the answer cannot be a graph that show the voltage increasing beyond 10V, ruling out answer A. Any graph that shows the maximum voltage to be less than 10V (i.e the graph levels off, without reaching 10V) can also be ruled out, eliminating answer B. Increased capacitance and increased resistance both result in the capacitor taking longer to charge (see below for explanation from BBC bitesize). The original set up took over 4 seconds to fully charge, so any graph that shows charging to take the same amount of time or less should be eliminated- so both C and D can be ruled out.
This leaves only E- The graph is still increasing at just over 4 seconds, so it is still charging and could go up to 10V and it also shows that it takes longer for the graph to reach 10V.
I looked at the answers on another part of the site(http://mrmackenzie.org/webtribe/Physics1992p1solns.html) and E is correct, though I'm not 100% sure about my reasoning- Asklepios what do you think?

From BBC Bitesize:
"If a larger value of capacitance were used with the same value of resistance in the above circuit it would be able to store more charge. As a result, it would take longer to charge up to the supply voltage during charging and longer to lose all its charge when discharging.
If a larger value of resistance were used with the same value of capacitance in the above circuit, then a smaller current would flow, therefore it would take longer for the capacitor to charge up and longer for it to discharge."(http://www.bbc.co.uk/bitesize/higher/physics/elect/capacitance/revision/2/)

Yeah that's correct

Original post by molly221

Thank you so much @Labrador99! Really helped! You are so good at physics :smile:

You're welcome- I'm really bad at the explaining questions though!

Reply 57

Zrehman24

2

Very worried as one of my friends told me that there physics teacher said that the old higher was easier than the current cfe higher!! I have bought a specimen paper book for physics for cfe and working through that if I work through them will that be the right style and level of questions in the cfe higher physics exam?

Original post by molly221

Can anyone help???

First, you need to find the current when the switch is open:
E = 12V
V = 12V
Lost volts = 0V
R = 3Ω
r = 3Ω
I = ?

E = V + Ir
V = IR, therefore,
E = IR + Ir
E = I(R + r)
12 = I(3 + 3)
12 = 6I
I = 2A

Then, find the current when the switch isclosed:
E = 12V
V = 12V
Lost volts = 0V
R = ?
r = 3Ω
I = ?
To find R:The resistors are in parallel, so
1/R_T = 1/R₁ + 1/R₂
1/ R_T = 1/3 + 1/6
1/ R_T = 2/6 + 1/6
1/ R_T = 3/6
R_T = 6/3
R_T = 2Ω
Then, do similar to before:
E = V + Ir
V = IR, therefore,
E = IR + Ir
E = I(R + r)
12 = I(2 + 3)
12 = 5I
I = 2.4A
Therefore, answer B
Hope this helps :smile:

Reply 59

molly221

3

Original post by Labrador99

First, you need to find the current when the switch is open:
E = 12V
V = 12V
Lost volts = 0V
R = 3Ω
r = 3Ω
I = ?

E = V + Ir
V = IR, therefore,
E = IR + Ir
E = I(R + r)
12 = I(3 + 3)
12 = 6I
I = 2A

Then, find the current when the switch isclosed:
E = 12V
V = 12V
Lost volts = 0V
R = ?
r = 3Ω
I = ?
To find R:The resistors are in parallel, so
1/R_T = 1/R₁ + 1/R₂
1/ R_T = 1/3 + 1/6
1/ R_T = 2/6 + 1/6
1/ R_T = 3/6
R_T = 6/3
R_T = 2Ω
Then, do similar to before:
E = V + Ir
V = IR, therefore,
E = IR + Ir
E = I(R + r)
12 = I(2 + 3)
12 = 5I
I = 2.4A
Therefore, answer B
Hope this helps :smile:

Thank you, again really helped.

CfE Higher Physics

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