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AS Physics P1 MCQs Preparation Thread.

studen12345

studen12345

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Raghad Dia

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Can someone explain for me October 2009 varyant 11 # 12, 14, 15, 22, 26 and 28. I know i'm asking for a lot of questions but i really need help x_x thank you in advance.
 
h4rriet

h4rriet

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salvatore said:
Thank you for your reply..

I haven't understood qn no. 13, could you please show me the working?

For qn no. 31, I found the rate of flow as t = Q/I = (1.6 x 10^-19) / 4.8 = 3.3 x 10^-20. That is not correct right?
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13. For the 5 N, the perpendicular distance from the pivot is 2 metres. For the 15 N, it's 3 metres and for the 10 N, it's 2 m.
In 31, you have to find the rate of flow of ELECTRONS, not charge. So Q=It will turn into Ne=It (N being the number of electrons and e being the charge on one electron).
 
h4rriet

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Jasmine 12 said:
I am having problem in november 09/12 can anyone help ,e in these questions q8,9,12,13,14,15
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8. In an elastic collision, the relative speed of approach = the relative speed of separation.
9. Use F=m(v-u)/t.
12. All you gotta do is locate distances and determine forces.
13. Vertical component of the velocity of a body in projectile motion is 0 at the highest point, so the K.E. can come only from the horizontal component, which is constant throughout.
14. K.E. of the trolleys when they're moving = the E.P.E. stored in the string initially.
15. Power output/Power input=Efficiency. You have Po and Efficiency; Pi will be x.
 
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sagar65265

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itallion stallion said:
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s12_qp_11.pdf plz help with q12 .thanks
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If you write down the equations of motion for the two objects (barrel coming down and man going up), you get:

Barrel:

Tension = 120g - 120 * acceleration
Tension = 1200 - 120a

Man:

Tension = 80g + 80a
Tension = 800 + 80a

Equating the two,

1200 - 120a = 800 + 80a
400 = 200a
a = 2 ms^-2.

Since the two objects both start from rest and have the same accelerations (they are connected by the string, so if one moves a distance x, so does the other, in the same time interval), the mans head will be in line with the bottom of the barrel after travelling 9 meters. Therefore,

u = 0 ms^-1 , a = 2 ms^-2 , s = 9m:

v^2 = 0 + 2 * 2 * 9
v^2 = 36
v = 6 ms^-1 = A

Hope this helped!
Good Luck for all your exams!
 
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sagar65265

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abruzzi said:
Someone please pleaseee help me with qn no. 10 and 20 of the following paper:
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s07_qp_1.pdf

Waiting for a reply..

Thanks!
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Q. 10 :

If a net force acts on an object, it will cause a change in momentum of the object (usually the change in is in velocity, either in terms of direction or magnitude). The average force between any two times is the total change in momentum that occurs between those two times divided by the time taken for the change in momentum (just like the average speed between two points is the total distance divided by the time taken to travel that distance). In general, the force at any one time is the gradient of the momentum-time graph at that point, but since the average force is being asked for, it's just the gradient of the line joining those two points, in other words, it is B, since that shows the change in momentum divided by the time taken.

Q. 20

Hmmmm. How to explain this? Okay, imagine that the beam actually bends under the stress applied by the hanging load. In that case, can you see that the top of the rod stretches and the bottom of the rod decreases in length - hopefully the diagram attached will help, so this basically means that both X and Y are going to be pulled on, towards the load, while Z will be pushed backwards, i.e. compression - sorry, but I'm not sure how else to explain this!

Hope this helped!
Good Luck for all your exams!
 

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