As physics p1 MCQS YEARLY ONLY.

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Ahmed Aqdam

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Santiago Carthorla said:
Can anyone explain to me Q8,12,14,18,19,30 and 38. Thanks in advance! http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w12_qp_12.pdf
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8: s=ut + 1/2at^2
u=0 so s=1/2at^2
a=9.81
Put the values in the equation

12: F=m(v-u)/t
mass is 0.5, v and u are 12 and -8 as they are in opposite directions and t is 0.10 so ans is 100N.

14: Upward forces are 4sin37 and 3sin53=4.8N
12-4.8=7.2N
Divide by mass so ans is 6 ms^-2.

18: Weight is 4.2*9.81=41.2
Divide by sin25 to get the total tension=97.5, divide by 2 as force is on both sides.

19: Area=500/10000=0.05m^2 and distance is 0.3m.
Volume=0.015, Multiply by pressure to get energy.

30: n*lamda/d*sin theta
lamda is 690*10^-9 and d is 3.33*10^-6
n is 4.83
4 maxima on one side 4 on the other and 1 zero order so total 9.

38: p.d. will be 1/6*1.5 when illuminated and 1000/1005*1.5 when dark so almost 1.5. pd will vary between these two.
 
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Hanona

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Santiago Carthorla said:
Can anyone explain to me Q8,12,14,18,19,30 and 38. Thanks in advance! http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w12_qp_12.pdf
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q 8 s= 1/2 g t^2 = 0.5 x 9.8 x 0.5^2 = 1.225 = approx. 1.3 m
q 12 force = rate of change of momentum = (12-8)/0.1 = 20 N
q 14 resolving the forces 3c0s53 + 4 cos 37 = 4.99N
net force = 12-4.99 = 7
f = ma .. mass = 12/9.8 =1.224 a= f/m =5.7 = 6 m/s

q 18 resolving vertically where T is the tension in the string = 2Tsin25 = Weight which is 4.2 x 9.8
solving it u get T as48.6 = 49 N
q 19 work done = f x d = f = pressure x area = 4000 x 500 x (10^-2)^2 ( because its area) =200 N
f x d = 200 x 30 x 10^-2 = 60 N
 
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