• We need your support!

    We are currently struggling to cover the operational costs of Xtremepapers, as a result we might have to shut this website down. Please donate if we have helped you and help make a difference in other students' lives!
    Click here to Donate Now (View Announcement)

AS Physics (9702) Paper 22 Answers and Discussion

Messages
185
Reaction score
104
Points
53
For the moment: perpendicular distance = 0 m...so moment = 0 Nm
You are as stupid as they come.
For the momentum graphs, must we show all calculations?

Yes, hence they gave 5 marks for it.

Grade thresholds will be AROUND 35 to 40(max) Your-Blood because I see a LOT of stupid people commenting here. Those of you that know you got more than 40, can stop worrying.
 
Messages
375
Reaction score
226
Points
53
hahah i made such a silly mistake . in the first Q to find the wind speed instead of taking cube root i took square root :p Man I couldnt believe what i did :p
 
Messages
4
Reaction score
0
Points
11
The charge you calculated in the previous part was the total charge in the given amount of time (Something in Megaseconds)
The next question asked you to calculate the number of electrons per second, so you had to divide the charge obtained in the previous part by the elementary charge of an electron (1.6*10^-19 C)
After that you were supposed to divide the calculated value by the time.
The charge you calculated in the previous part was the total charge in the given amount of time (Something in Megaseconds)
The next question asked you to calculate the number of electrons per second, so you had to divide the charge obtained in the previous part by the elementary charge of an electron (1.6*10^-19 C)
After that you were supposed to divide the calculated value by the time.

The answer to current is 13.3A. Do me a favor and divide this value by 1.6x10^-19 and tell me what answer you get :) It's the same as the answer given in the first post. My method was correct. I don't see why you guys have to involve time here. It is simple the charge passing through a circuit in ONE second is EQUAL to the value of current as Q=It where t=1. How do you get the number of electrons? You divide the charge by elementary charge and that's how I did it. Gives the same value :)
 
Messages
99
Reaction score
65
Points
18
You are as stupid as they come.


Yes, hence they gave 5 marks for it.

Grade thresholds will be AROUND 35 to 40(max) Your-Blood because I see a LOT of stupid people commenting here. Those of you that know you got more than 40, can stop worrying.

Dude don't hurt anyone by using such harsh language.
And I don't think showing calculations were important. Your graphs itself were giving your calculations.
Well who knows
 
Messages
306
Reaction score
278
Points
38
the line goes flat for the resultant displacement of the two waves...but there they dint ask us to draw the resultant displacement..they asked us to draw the position of the wave after 0.5t...so wouldnt it be just opposite to the wave already given by them?
Samwe question :) I did the same :/
 
Messages
43
Reaction score
9
Points
18
I thought we are suppose to discuss instead of criticizing here. What you mean is that they won't accept that answer because it does not answer the question based on equilibrium of forces. So if I answer "The rod exert a horizontal force on A, there would be an equal but opposite force (normal force) by the wall acting on the rod. Therefore, no net force." Is that satisfactory? And the momentum, are you sure it's 16.4 not 16.8?[/quote]
 
Messages
43
Reaction score
9
Points
18
Despite getting another variant, you answered this one well. Did you receive the question paper or something? If you did, can I have it?
 
Messages
43
Reaction score
9
Points
18
Momentum time graph is correct , i think so , because i did the same

the graph was to start from 0 for Accelaration time graph ??
because they said , OBJECT IS INITIALLY AT REST ...
i made the graph start from 0rigin , and look it directly upward , closer to Y Axis , at the point 3.5 ..and then made the graph similar to the graph of Force
That's very detailed of you! Couldn't have come up with that. But acceleration 0 implies force 0, that would contradict the force/time graph despite the statement of initially at rest. CIE is also testing your sensitivity, besides cleverness. The sample of answers given above have so many loose ends in that terms. So don't worry. Be confident if you know you're right. And I don't think they would penalize you for extra workings.
 
Messages
66
Reaction score
20
Points
18
Can somebody please discuss this question paper......i really need this !!!!!
 

Attachments

  • 9702_w10_qp_22.pdf
    147.5 KB · Views: 7
Messages
6
Reaction score
3
Points
3
in Q2.i think a have drawn the graph having +ve gradient upto 2 s
is it correct
can anybody tell me
 
Messages
1,160
Reaction score
3,858
Points
273
yes may be its gt will be also around 35.....the phy gt is always low as comparing chem and bio ...so dw ....if gt of ppr 22 is 35 then gt of 21 will be 35+_ 2 ....* but it also depends on what type of ppr 21 was
 
Messages
253
Reaction score
1,195
Points
153
yes may be its gt will be also around 35.....the phy gt is always low as comparing chem and bio ...so dw ....if gt of ppr 22 is 35 then gt of 21 will be 35+_ 2 ....* but it also depends on what type of ppr 21 was
it was much easier thn 22 but there were 22 part questions which ppl found to b difficult
 
Messages
1,160
Reaction score
3,858
Points
273
it was much easier thn 22 but there were 22 part questions which ppl found to b difficult
if it was easier even then its gt cant exceed more then 40...thats sure ...cz i have seen the gts of last 12 pprs from 10-12 and some were so easy and even then there gts are lying around 30's
 
Messages
253
Reaction score
1,195
Points
153
if it was easier then even then its gt cant exceed more then 40...thats sure ...cz i have seen the gts of last 12 pprs from 10-12 and some were so easy and even then there gts are lying around 30's
just hoping tht thresholds are low enuff for good grades for evry1 InshAllah.........
 
Top