Physics: Post your doubts here!

[unique840]

http://www.xtremepapers.com/CIE/International%20A%20And%20AS%20Level/9702%20-%20Physics/9702_w03_qp_1.pdf

anybody explain Q11 please

Thank you

is it D?
momentum change = m * change in velocity
initial velocity is = u
final is = -u (-ve sign is due to direction change)
change is [u-(-u)]
change is 2u
momentum change is 2mu

 

[unique840]

hi dears have no fears. kindly help
http://www.xtremepapers.com/CIE/Interna ... 7_qp_2.pdf

Q.5 b i , 4 c plz figure out the relevant concept which is missing and make me clear. thanks

4c) the initial speed is all horizontal. the horizontal component of speed remains constant through out the motion. so the horizontal speed is 18m/s

 

[Silent Hunter]

http://www.xtremepapers.com/CIE/index.p ... 5_qp_1.pdf

help required in Q7

 

[abcde]

http://www.xtremepapers.com/CIE/index.php?dir=International%20A%20And%20AS%20Level/9702%20-%20Physics/&file=9702_s05_qp_1.pdf

help required in Q7

There is only one force acting on the ball; it's weight so the resultant force is constant. Using F=ma, acceleration will also be constant. Choices narrowed down to B and C. The reason B is correct is this: "Take vertically upwards as positive". You can work that acceleration will be negative in this case by taking arbitrary values for speed. Try it.

 

[Silent Hunter]

"Take vertically upwards as positive".

doesnt this mean that rising column should have negative sign with 9.8 ?
i mean rising means its going upwards

 

[abcde]

All columns will have a negative sign with 9.81m/s^2.

 

[Silent Hunter]

but going down means it should be positive.......... shouldn't the falling one should be positive

 

[abcde]

but going down means it should be positive.......... shouldn't the falling one should be positive

Consider a falling object: Let's say it's speed increases from 3 m/s to 13 m/s in 1s. Since downward motion is negative, we would take the speed values as -3 and -13. So, a = -13 - (-3)/1 = - 10 m/s^2.
In upward movement, speed decreases due to the downward gravitational force. Say it decreases from 13 m/s to 3 m/s in 1 s. So, a = 3 - 13/1 = - 10 m/s^2.
Hence, in all cases, acceleration is negative and constant. (-9.81 m/s^2 throughout)

 

[Silent Hunter]

but going down means it should be positive.......... shouldn't the falling one should be positive

Consider a falling object: Let's say it's speed increases from 3 m/s to 13 m/s in 1s. Since downward motion is negative, we would take the speed values as -3 and -13. So, a = -13 - (-3)/1 = - 10 m/s^2.
In upward movement, speed decreases due to the downward gravitational force. Say it decreases from 13 m/s to 3 m/s in 1 s. So, a = 3 - 13/1 = - 10 m/s^2.
Hence, in all cases, acceleration is negative and constant. (-9.81 m/s^2 throughout)

GOT IT............ thanks alot

 

[Silent Hunter]

http://www.xtremepapers.com/CIE/index.p ... 7_qp_1.pdf

Q12 explanation is required

Thank You

 

[rz123]

hi dears have no fears. kindly help
http://www.xtremepapers.com/CIE/Interna ... 7_qp_2.pdf

Q.5 b i , 4 c plz figure out the relevant concept which is missing and make me clear. thanks

4c) the initial speed is all horizontal. the horizontal component of speed remains constant through out the motion. so the horizontal speed is 18m/s

ooh yeah. thanks nd what about 5 b i)...

 

[rz123]

http://www.xtremepapers.com/CIE/index.php?dir=International%20A%20And%20AS%20Level/9702%20-%20Physics/&file=9702_s07_qp_1.pdf

Q12 explanation is required

Thank You

welcome....
.
.
.
see the total momentum remains same after the collision right? so momentum before the collision is 4 X 2=8 + 4 X 1 =4 total p equals to 12. total mass after the collision 6kg , find the velocity now.
12 = 1/2 X 6 X v^2 , the velocity will be 2m/s ok so now calculate the total kinetic energy = 1/2mv^2 u will get 12 J. agar nai samajh aya toh i can explain in some other way.

 

[Silent Hunter]

http://www.xtremepapers.com/CIE/index.php?dir=International%20A%20And%20AS%20Level/9702%20-%20Physics/&file=9702_s07_qp_1.pdf

Q12 explanation is required

Thank You

welcome....
.
.
.
see the total momentum remains same after the collision right? so momentum before the collision is 4 X 2=8 + 4 X 1 =4 total p equals to 12. total mass after the collision 6kg , find the velocity now.
12 = 1/2 X 6 X v^2 , the velocity will be 2m/s ok so now calculate the total kinetic energy = 1/2mv^2 u will get 12 J. agar nai samajh aya toh i can explain in some other way.

12 = momentum not energy then why you have equaled KE to 12 ?

 

[rz123]

yes my mistake. we use the law of conservation of momentum to calculate the velocity of the two bodies after collision. which is p = mv so 12= 6 into v, v= 2, put it in kinetic energy equation. B is the ans

 

[smartangel]

please explain where do we get the length in NOV 2010 PAPER 43 Q5C while using the formula F= BIL.
also, please let me know if anyone knows what do we have to cover up in the points mentioned in p4 syllabus which i posted previously on page 40 of this thread.
thanks a lot!

 

[Silent Hunter]

Asalamoalikum

Wanted some help regarding phase difference.What is it? What is constant phase difference?

EXAM TOMORROW

Thank You

 

[rz123]

Phase Difference between two particles or two waves tells us how much a particle (or wave) is in front or behind another particle (or wave).or the amount by which one oscillation leads or lags behind another. for constructive interference the phase difference is multiples of n lambda, and for destructive interference the phase difference is n+1/2 lambda.

P.S use David sang, Graham Jones... physics Cambridge endorsed book, explains quite well. Good luck

 

[rz123]

http://www.xtremepapers.com/CIE/Interna ... 7_qp_1.pdf

40, how its lithium?..thanks

 

[ad98ify]

hi guys. i m having sm conceptual problems in basic concepts in chapters like gravity and electric field like :-

• why is gravitational field strength maximum at poles rather than at equator when it should be vice versa as seen by the formula. g=GM/r^2 that is g is directly proportional to 1/r^2?
• in electric field, why do we say that inside a point charge, potential is constant that is an equipotential surface exists?

please please answer them

 

[JD REBORN]

Hi can plz somebody explain to me how to answer the following two questions from physics 9702 AS level May/June 2010 Paper 23:

Question 2)Part C
Question 3)Part A

??????