50 replies to this topic

[englisher]

i got 3 as D as it says that "its speed increases at a constant speed" which seems to me to meen that the lift is accelerating downwards

Hi.

The answer was B because:

It was accelerating downwards at a constant rate (speed increasing at a constant rate)
a=Fun/m . "a" is constant, m is constant, so Fun is constant
Weight is greater than "S" as it is accelerating downwards, where S is the force against the man, and hence the man's force against the scales
Fun = weight - "S"
S= weight - Fun

Therefore, S is less than weight.

(Now why did I think the reading on the scales was W in the exam?)

[BuckminsterFullerene]

i got 3 as D as it says that "its speed increases at a constant speed" which seems to me to meen that the lift is accelerating downwards

Hi.

The answer was B because:

It was accelerating downwards at a constant rate (speed increasing at a constant rate)
a=Fun/m so Fun is constant
Weight is greater than "S" as it is accelerating downwards
Fun = weight - "S"
S= weight - Fun

Therefore, S is less than weight.

(Now why did I think the reading on the scales was W in the exam?)

Are you sure that's correct, because thats what I got!

[englisher]

Hi mate,

Yeah that's the right answer. Just sat and worked it out.

Cheers

[mikey4020]

I've dropped a few silly marks in the "multiple guess" questions but hopefully I've got about 12 right at least because I never do well in multiple choice! I was really glad there was no questions in the multi-choice that give you three statements and then ask you which is true!!

I felt the written went OK though I messed up Q21 though forgot direction at part (d) and (a) physics teacher nearly hit me lol

Also can someone tell me what the voltage on the graph should have been in Q25 [c]

[englisher]

12V dropping to 8V

[mikey4020]

How do you get the 8V? I ended up with 4.8 or something!

[Kiltox]

How do you get the 8V? I ended up with 4.8 or something!

The current increases to 2A because the resistance is 4 ohms instead of 8, therefore the "lost volts" is 4v and the terminal potential difference is 8v

[englisher]

How do you get the 8V? I ended up with 4.8 or something!

And the 4.8V answer arises from not including the big resitor in your initial calculation, I think.
You may well get the mark if they allow continuation from question b).

[davm88]

i got 3 as D as it says that "its speed increases at a constant speed" which seems to me to meen that the lift is accelerating downwards

hey i got the same as u! i dnt understand the solutions up here, if its "speed increases at a constant rate" then its gettin faster and faster. meaning its a constant acceleration, but not a constant speed. if its gettin faster and faster, eventually it'll b freefalling, meanin the dude(ette) will b weightless. meanin the weight "continually decreases from 700N", evetually to nothin. am i not rite!

[weerydo]

thts wot i thot m8, and i am still sure i am. maybe many ppl didnt reed it correctly and they just red constant and thot constand speed so a set amount less than 700n. so any1 else got any views on this? its question 3 of multi choice

[anon]

Higher Physics 2007 Solution

[/size]








1 D

2 C

3 B

4 D

5 A

6 C

7 A

8 E

9 A

10 A

11 E

12 C

13 D

14 A

15 E

16 C

17 E

18 D

19 B

20 B








[size="3"]21 a) Scale diagram or use cosine rule (c² = a² + b² -2abcosC) MARKS

Resultant = 350m 1

For direction use sine rule a/sinA = b/sinB

Direction = 38.2 degrees 1



b) v_X = s/t =350/66 = 5.3 m/s 2



c) t_Y = d/v = 400/6.5 = 61.5s so Y arrives first. 2



d) 350m @ 218 degrees. 1





22 a) F = mgsin Ө = 60 x 9.8 x sin 22 = 220N 1



b) Unbalanced force = 220 - 180 = 40N

a = F/m = 40/60 = 0.67 m/s² 2



c) v² = u² + 2as

= 0² + 2 x 0.67 x 50 = 67

so v = 8.19 m/s 2



d) If the mass is less then the force down the slope will be less, and since the friction force is the same, the unbalanced force, and hence the acceleration and speed at the foot of the slope will all be less. 2



23 a) p₁V₁ = p₂V₂ so 750 x 8 = 125 V₂

so V₂ = 0.48m³ 2



b) No. of balloons = 0.48/0.02 = 24 2



c) Since r = m/V, and the same mass of gas occupies 6 times the volume, then the density of the helium will be less in the balloons than in the cylinder. 2

24 a) E_k at A = ¹/₂mv² = 2.24 x 10^-14 J

so gain in E_k = (3.05 – 2.24) x 10^-14 J = 8.1 x 10^-15 J 2



b) QV = Gain in E_k so V = 8.1 x 10^-15 /3.2 x 10^-19 = 25.3 kV 2



c) Since the charge on the electron is half of the charge on the alpha particle, then the work done and hence the kinetic energy gained by the electron will be half that gained by the alpha particle. 2



25 a) Emf = 12 V 1



b) (i) I = lost volts/r = 2.4/2 = 1.2A 2



(ii) R = V/I = 9.6/1.2 = 8 ohms 1



c) Total resistance in circuit is now 6 ohms,

hence I = 12/6 = 2A and V = IR = 2 x 4 = 8V,

so graph is same except pd drops from 12V to 8V

when the switch is closed. 2





26 a) I = V/R = 12/480,000 = 2.5 x 10 ^-5 A 2



b) Q = VC = 8.2 x 2200 x 10 ^-6 = 0.018C 3



c) E = ¹/₂ CV² = ½ x 2200 x 10 ^-6 x 12² = 0.158J 2





27 a) 2200/5000 = R/750 so R = 330 ohms 2



b) (i) Differential mode 1



(ii) As the temperature falls the resistance of the thermistor increases and so the voltage at Q falls below the voltage level at P. This results in a rise in the output voltage of the op-amp and hence an increase in the gate voltage. When this rises above 2V, the MOSFET will switch ON and the LED will light. 2



(iii) V_o = (V₂ - V₁)R_f /R₁

(V₂ - V₁) = V_o x R₁ /R_f

= 3 x 40/100 = 1.2V 2









28 a) For a maximum, the waves arrive in phase or one crest meets another crest or constructive interference occurs 1



28 (cont) b) (i) A) Mean = 6.6/6 = 1.1m 1



B) (1.13 – 1.07)/6 = 0.01m 1



(ii) % error in AB = 0.01/1.1 x 100 = 0.91%

% error in BC = 10/270 x 100 = 3.7%

So % error in BC is larger 2



(iii) d sin Ө = n l

so l = ½ x 4 x 10^-6 x 0.27/1.1 = 491nm

3.7% of 491nm = 18nm

so l = 491 +/- 18nm 3





29 a) n = sinq_air/sinq_medium so sinq_medium = sinq_air/n = sin 50/1.5

so q_medium = 33.3 degrees



b) n = l _air /l _medium

so l _air = 1.5 x 420 = 630nm 2



c) Red light is refracted less than blue light, as it travels faster than blue light in glass and hence the refracted angle for blue light will be less than for red light. 1



30 a) f₀ = v/l = 3 x 10⁸/605 x 10^-9

= 4.96 x 10¹⁴ Hz

E = hf₀ = 6.63 x 10^-34 x 4.96 x 10¹⁴ = 3.29 x 10^-19 J 2



b) (i) E_k = hf - hf₀ = 5.12 – 3.29 = 1.83 x 10 ^-19 J 1



(ii) I = Nhf so if I is reduced the no. of photons striking the metal, and hence the no. of electrons released will be reduced, resulting in a lower ammeter reading. 2







31 a) Mass before = 398.26 x 10 ^-27 kg

Mass after = 391.170 x 10 ^-27 kg + 6.645 x 10 ^-27 kg

= 398.615 x 10 ^-27 kg



so lost mass = 1.1 x 10 ^-29 kg



Energy released = mc² = 1.1 x 10 ^-29 x (3 x 10⁸ )²

= 9.9 x 10 ^-13 J 3



b) Absorbed dose rate = D/t so D = Absorbed dose rate x time

= 4 x 2 = 8 micro Grays

H = D w_R = 8 x 3 = 24 micro Sieverts 3

[davm88]

hey, are these solutions accurate or just sumbodies answers? im still sticking to my guns about 3 being D, but want to know other peoples views on this.

[englisher]

thts wot i thot m8, and i am still sure i am. maybe many ppl didnt reed it correctly and they just red constant and thot constand speed so a set amount less than 700n. so any1 else got any views on this? its question 3 of multi choice

The first post on this page I went through it very clearly!!

hey, are these solutions accurate or just sumbodies answers? im still sticking to my guns about 3 being D, but want to know other peoples views on this.

See the first post on the top of this page!!!

[englisher]

Double post.

[englisher]

snip solutions

Lost 3 marks

[babeasc]

i cant even remember what i put dont want to look at the paper again just depresses me. everyone found it easy i thought it was okayish but not too great o well nothing i can do now

I thought the paper was easy, but looking through the answers i put there are actually BILLIONS (now exxageration ) of mistakes

Oh well, its all done now!

[sunnyho]

Higher Physics 2007 Solution

[/size]





[size=3]


1 D

2 C

3 B

4 D

5 A

6 C

7 A

8 E

9 A

10 .....................

Thank you very much for the solution!!
I think I lost 14 marks, so total will be 84%...
Hope it will be a A pass...

[davm88]

thts wot i thot m8, and i am still sure i am. maybe many ppl didnt reed it correctly and they just red constant and thot constand speed so a set amount less than 700n. so any1 else got any views on this? its question 3 of multi choice

The first post on this page I went through it very clearly!!

hey, are these solutions accurate or just sumbodies answers? im still sticking to my guns about 3 being D, but want to know other peoples views on this.

See the first post on the top of this page!!!

i know, but i stil dnt understand. can u see where im comin from tho? tht this lift IS gettin faster and faster, and that the guy inside will EVENTUALLY get weightless?

[Yan]

Thank you very much for the solution!!
I think I lost 14 marks, so total will be 84%...
Hope it will be a A pass...

I think that'll be like a Band 1 A pass... 84% is really really high! Well done

I haven't bothered to count up all the marks I've lost.. can't be bothered plus it'll just depress me

[englisher]

thts wot i thot m8, and i am still sure i am. maybe many ppl didnt reed it correctly and they just red constant and thot constand speed so a set amount less than 700n. so any1 else got any views on this? its question 3 of multi choice

The first post on this page I went through it very clearly!!

hey, are these solutions accurate or just sumbodies answers? im still sticking to my guns about 3 being D, but want to know other peoples views on this.

See the first post on the top of this page!!!

i know, but i stil dnt understand. can u see where im comin from tho? tht this lift IS gettin faster and faster, and that the guy inside will EVENTUALLY get weightless?

Just think of the equations. If he was going infinitely fast, at a constant speed, it would be precisely the same for him as if he was going at zero velocity. The upward force would equal the downward force so there would be no unbalanced force. Therefore, he wouldn't ever get weightless travelling at a constant speed. Surely then at every speed up to that point he would not become "weightless"? That doesn't quite answer your question so I'll go through the first post again.

W = 700N

Force upwards from floor of lift (balanced by force on the scales pushing up) = "S"

He is accelerating downwards, there must be an unbalanced force, so Weight is greater than S (down greater than up). Accelerating at a constant rate, so "a" is constant. For the purposes of illustration, constant1 = a constant 2 =m
a = F(un)/m
constant1 = F(un)/constant2
Therefore, F(un) = constant1 x constant2
Unbalanced force is constant

Do you follow up to this point?

Unbalanced force = Weight - "S" (Remember how weight was greater?)
S = Weight - unbalanced force
= 700 - constant
= a constant less than 700N

Hope you get it now.