Physics: Post your doubts here!

[emkay]

http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w09_qp_12.pdf
please help with question 12

the torque between the pulley Q and P in the upper wire is the torque at Q.
as it is causing the driving force for Q.
i.e T=F*d
F= T/d
F=3/100*10^-3
F=30*2 (since, the formula suggests force into perpendicular distance between the two forces)
F=60 N.
Torque on P,
T=F*d
T=30*150*10^-2 (Force remains the same, i.e 30N)
T=4.5 N m
(ps. Credit to geek101 for doing this)

 

[akshay.RONALDO]

i

http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s11_qp_12.pdf
question 17...please

i'm also stuck with this number. Please, do inbox me, if somebody replies.

 

[Kumkum]

the torque between the pulley Q and P in the upper wire is the torque at Q.
as it is causing the driving force for Q.
i.e T=F*d
F= T/d
F=3/100*10^-3
F=30*2 (since, the formula suggests force into perpendicular distance between the two forces)
F=60 N.
Torque on P,
T=F*d
T=30*150*10^-2 (Force remains the same, i.e 30N)
T=4.5 N m

thank you!

 

[Kumkum]

i
i'm also stuck with this number. Please, do inbox me, if somebody replies.

ya sure

 

[mustafa shakeel]

Aoa
any idea what the threshold for for As p2 and p3 might be?

 

[emkay]

http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s11_qp_12.pdf
question 17...please

i
i'm also stuck with this number. Please, do inbox me, if somebody replies.

sin theta = 40/50
theta = 53
F= 8Cos53 (Since, there are two components of 4N i multiplied it with 2, 4*2=8)

 

[Kumkum]

sin theta = 40/50
theta = 53
F= 8Cos53 (Since, there are two components of 4N i multiplied it with 2, 4*2=8)

thank you !

 

[Khalifa]

Oct09
Q12. since the btm belt is slacked, so tension is only at the upper belt, torque at Q = 3.0 N m,
so 3.0 = T x (100 x 10^-3)/2
T = 60N
Torque at P = T x (150 x 10^-3)/2 = 4.5
Q13. assuming it is projected with velocity v, so the initial KE, E= (mv^2)/2
at the highest point, the vertical component of the veolcity = 0, so its left with v cos 45,
so final KE = 1/2 x m x (v cos 45)^2 = 0.5E
Q14. By conservation of momentum, 2 x 2 = 1 x v, v =4
so the 1kg trolley moves 4ms-1 to the right,
calculating the total KE of the both trolleys, which is 1/2 x 2 x 2^2 + 1/2 x 1 x 4^2 = 12J
assuming no energy is lost, the energy stored in the spring is converted to KE gained by the trolleys, so its 12J

Oct11/12
Q23. the area under graph is the elastic PE, so we can c that the answer is obviously (i hope its obvious for u too) B. not much that i could explain here

In question 12, why did you divide the diameter by 2 each time since torque=fd

 

[queen of the legend]

oct/nov 2009 : q 22 , 23, 26, 28

oct/nov 2010: q 13

 

[geek101]

bestaa luck everybody! hope todays paper is a piece of chocolate fudge cake for ya!

 

[InnocentAngel]

 

[queen of the legend]

View attachment 18675

everything is given ...taking moments ..clockwise moments wa+fh=2Wa..... and in the question the corner of the wall is considered as the pivot

 

[geek101]

View attachment 18675

ok for this simply see, first the total clockwise moments and the the total anticlockwise moments, take the moments about the corner of the wall, that being ur pivot

clockwise moments = F h + W a
anticlockwise = W 2 a
anticlockwise moments = clockwise moments
F h + W a = 2 Wa

 

[Goku]

Thank you SOOOOOOO much! That made perfect sense! Thanks a lot! :')

question 7 is direct application of formulas
for vertical displacement s= ut + o.5 gt^2 and horizontal is S=ut

question 10 : since f and m are directly proportinal ...heavier mass need more force to move .. so , masses given m and 3 m ...if F=1 N then 3/4 the force goes to 3m and 1/4th to m ( well this is how i could explain it )

q 14 : you need to focus on what question asks for
vertical component of force = 10000-9000=1000N

question 17 : take average current ...since its reduced over a period of time ...at one instant the current is not constant
so 100+20/2 = 60 x 8 = 480mC

question 36: earthed plate is considered positive so electric field is directed upwards and E=V/d (dnt forget to convert the units)

question 40 : here is a reply from one of the members (Taci12) :

Question 40:​

​

According to syllabus, we must know that a proton has a charge +e (the basic/elementary charge)​

​

Now according to this question, protons themselves are made up of particles called quarks. There are different types of quarks, each one with its own charge.​

​

Whatever the combination of quarks, the total charge on a proton must be +e. Hence the sum of the charges of all the constituent quarks in a proton must be + e.​

​

So simply check the sum of charges of each option to see if it is +e.​

​

option A: 3 x (-e/3) = -e​

option B: (2e/3) + (-e/3) = +(e/3)​

option C: (2e/3) + 2(-e/3) = o​

option D: 2(2e/3) + (-e/3) = +(3e/3) = +e​

​

answer is D​

 

[emkay]

Graphs Graphs Graphs Graphs, STUPID GRAPHS!

 

[geek101]

Graphs Graphs Graphs Graphs, STUPID GRAPHS!

what about them??

 

[emkay]

what about them??

itney saarey hain, confuse ker k rakh diya hae ;(

 

[geek101]

itney saarey hain, confuse ker k rakh diya hae ;(

dont memorize them all...see the formula, find the relationship between the 2 quantities and the graph will be a piece of cake!

 

[Yousif Mukkhtar]

http://www.xtremepapers.com/papers/...and AS Level/Physics (9702)/9702_s04_qp_2.pdf
Can someone explain
q4 a ii) and for b)i) What is the direction of the force?

 

[emkay]

http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s04_qp_2.pdf
Can someone explain
q4 a ii) and for b)i) What is the direction of the force?

for Q4 a ii)
Ek = p^2/2m
1/2mv^2 = (mv)^2/2m (since, p=mv)
1/2mv^2= mv^2/2 (m and m cancels out)
1/2mv^2 = 1/2mv^2. (Hence, Proven)

for Q4 b i)
find change in momentum first,
delta p = m*delta v
p= 0.035*(4.5+3.5)
p=0.28.
F=change in momentum/change in time
F=0.28/0.14
F=2.0N.