Physics: Post your doubts here!

[Unicorn]

since the ball's initial velocity after collision is perfectly horizontal, take the horizontal distance moved and divide it by the time above(1.333)
horizontal velocty stays constant throughout travel.

3.b.i v^2=u^2 - 2gs
--> 0= 13^2 - (2*9.81 s)
--> s= 8.61 ~ 8.61m
3.b.ii v= u -gt Alternatively, you can also use horizontal velocity to calculate the time.
--> 0= 13 - 9.81*t time = distance/ horizontal velocity = 9.95/7.5 = 1.33 s
--> t= 1.325 s ~ 1.33 s

3. b.iii using vertical velocity, The horizontal velocity IS THE SPEED after rebounding from P. There is no vertical speed at P since in the beginning of the question
--> s= ut + .5 gt^2 it was said "The ball hits the wall at P with a velocity that is at right angles to the wall" This means there is no vertical velocity at P.
--> 8.61 = 4.905 t^2
--> t = 1.324 s

so horizontal velocity = 6.15/1.324 = 4.64 ~ 4.6 m/s

Hope that helps man.

ohh ok

thanks

 

[leosco1995]

www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w07_qp_2.pdf
Q2 d (ii)

The force calculated in (i) is 75 N and the resistance force is 23 N.. so shouldn't the power be the net force * velocity? In the marking scheme, the forces have been added instead. Why?

 

[hussamh10]

http://www.xtremepapers.com/papers/...and AS Level/Physics (9702)/9702_w05_qp_2.pdfQ5 c 1 annd 2 how to show that intensity of B at pont p ( and whaere is poimt p) is 4/9 I ...and the next part also plzz help

 

[Saad (سعد)]

www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w07_qp_2.pdf
Q2 d (ii)

The force calculated in (i) is 75 N and the resistance force is 23 N.. so shouldn't the power be the net force * velocity? In the marking scheme, the forces have been added instead. Why?

They are added, because the question asks for the total output power supplied by the boy's legs. The boy's legs first provide a force equal to the resistive force (23 N) in order to overcome the resistance. Then a further 75 N is supplied by the boy's legs, which causes his acceleration of 1.12 m/s. So the total force supplied by the boy's legs is = 23 + 75 N = 98 N.

Power = Force x velocity
= 98 x 4.5
= 441 W.

Hope that helps insha' Allah!

 

[leosco1995]

They are added, because the question asks for the total output power supplied by the boy's legs. The boy's legs first provide a force equal to the resistive force (23 N) in order to overcome the resistance. Then a further 75 N is supplied by the boy's legs, which causes his acceleration of 1.12 m/s. So the total force supplied by the boy's legs is = 23 + 75 N = 98 N.

Power = Force x velocity
= 98 x 4.5
= 441 W.

Hope that helps insha' Allah!

It did help. Thanks a lot.

 

[hussamh10]

www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w07_qp_2.pdf
Q2 d (ii)

The force calculated in (i) is 75 N and the resistance force is 23 N.. so shouldn't the power be the net force * velocity? In the marking scheme, the forces have been added instead. Why?

Because the boy had to do work against the resistance as well as to maintain the speed till 4.5 so the both forces will add (the one which cause acceleration and against the friction).Hope i helped

 

[mariamalik]

Any guess about the phys practical ??

 

[code11]

hey i wana know wat distance to be used in order to determin initial horizontal velocity in projectile motion ? if the time is given to us

 

[hussamh10]

hey i wana know wat distance to be used in order to determin initial horizontal velocity in projectile motion ? if the time is given to us

the horizontal distance travelled .......

 

[hussamh10]

http://www.xtremepapers.com/papers/...and AS Level/Physics (9702)/9702_w07_qp_2.pdf
Q1 (ii) part why the absolute obtained error is halved?

 

[iKhaled]

help me in a question may/june 2003 Q3 b) i. :S:S:S:S

http://www.xtremepapers.com/papers/...and AS Level/Physics (9702)/9702_s03_qp_2.pdf

 

[hussamh10]

help me in a question may/june 2003 Q3 b) i. :S:S:S:S

http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s03_qp_2.pdf

In first part pointer has a deflection of 6.5 degrees so by finding out the arc lenth we will get the extension by
(6.5/360)x2pi(1.5)=0.17
ii)in second part we have to find out the strain which is
Extension/Original length
iii)stess
Force/area=(6.0x9.81)/(7.9x10^-7)=7.44x10^7 Pa
Then we have to find the Young modulus which =Stress/strain
in the last part their is NO extension if the pointer comes back to its position....

 

[iKhaled]

In first part pointer has a deflection of 6.5 degrees so by finding out the arc lenth we will get the extension by
(6.5/360)x2pi(1.5)=0.17
ii)in second part we have to find out the strain which is
Extension/Original length
iii)stess
Force/area=(6.0x9.81)/(7.9x10^-7)=7.44x10^7 Pa
Then we have to find the Young modulus which =Stress/strain
in the last part their is NO extension if the pointer comes back to its position....

dude, for the first part y we get the arc length? isn't it curved and extension can't be curved? it should be a straight line? :/

 

[hussamh10]

dude, for the first part y we get the arc length? isn't it curved and extension can't be curved? it should be a straight line? :/

extension is curved here the movement of pointer shows the extension so the only way to do is to get the arc lenth....
If you did this paper how did you got the absolute error in the first question of this paper??/

 

[iKhaled]

extension is curved here the movement of pointer shows the extension so the only way to do is to get the arc lenth....
If you did this paper how did you got the absolute error in the first question of this paper??/

i didn't do it skipped the q .. but dude this question we r talking about is confusing my ass..in the mark scheme it says u can use s = r x theta...y did they use sin ?

 

[hussamh10]

i didn't do it skipped the q .. but dude this question we r talking about is confusing my ass..in the mark scheme it says u can use s = r x theta...y did they use sin ?

sin is an alternative way to do this by by pyrhgras theoram it is not related with Rtheta

 

[iKhaled]

sin is an alternative way to do this by by pyrhgras theoram it is not related with Rtheta

thanks broo

 

[Unicorn]

can someone please explain to me the question 6c?? The mark scheme has additional values I don't know where the got them

http://www.xtremepapers.com/papers/...nd AS Level/Physics (9702)/9702_w11_qp_23.pdf
http://www.xtremepapers.com/papers/...nd AS Level/Physics (9702)/9702_w11_ms_23.pdf

for all I know
charge of alpha particle = 2
mass of alpha particle =4

charge of beta particle 1.6x10^-19
mass of beta particle 9.11x10^-31

 

[iKhaled]

need help with may june 2006

question 7 a) nature of fault : the lamp is shorted....what does that mean?

http://www.xtremepapers.com/papers/...and AS Level/Physics (9702)/9702_s06_qp_2.pdf

 

[raamish]

http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w07_qp_2.pdf
Q1 (ii) part why the absolute obtained error is halved?

THe error is halved because look at the equation: V=pier2l. So r is 2 so that means we will have to divide by 2 to get the answer of R only not 2r. But i dont get why they are adding the uncertainties and then dividing them. Shouldnt they subtract the uncertainties like this: V-L=3.3-1.5/2 to get answer????