Physics: Post your doubts here!

[Abdullah syed]

http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s04_qp_2.pdf
please explain question 4 a (ii) and b =)
thank you for any help ^_^

For a(ii) it is Derivation
first we kill take as Ke=0.5mv^2
and then we replace the v² in ke with v=p/m

and it is square we will get p² and m² as m is in Ke ² get canceled
And remaining we get Ke = 1*p²/2*m

 

[angelicsuccubus]

http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w07_qp_2.pdf

how do we do 1b. really confused helpp

i tried to make sense out of it but umm... wth? Ill post again when I figure it out though

 

[Unicorn]

Can someone please explain question 4c(ii)
http://www.xtremepapers.com/papers/...nd AS Level/Physics (9702)/9702_s11_qp_23.pdf
http://www.xtremepapers.com/papers/...nd AS Level/Physics (9702)/9702_s11_ms_23.pdf

I don't get how they are finding the extension

 

[Most_UniQue]

iS IT RIGHT? Or both of them collide at same time?

 

[angelicsuccubus]

iS IT RIGHT? Or both of them collide at same time?
quote]

I think the line continues for the same amount of time but anyway the marks are only for the ball going in the opposite direction so it should be okay

 

[angelicsuccubus]

http://www.xtremepapers.com/papers/...nd AS Level/Physics (9702)/9702_s10_qp_21.pdf
Q2 b(ii) how to do???

I didnt check the ms, but is it 1.375m/s?

 

[Mahnoor Khanm]

How do we find the uncertainties in 2(b)? http://www.xtremepapers.com/papers/...nd AS Level/Physics (9702)/9702_s11_qp_51.pdf

 

[raamish]

http://www.xtremepapers.com/papers/...nd AS Level/Physics (9702)/9702_w09_qp_21.pdf

2c(ii) WHy is acceleration and weight getting subtracted. is the acceleration actually the dec. and given in negative???

 

[iKhaled]

Nope, doesn't fit in.

If we calculate the time taken for the ball to reach the ground using the vertical velocity:
s=ut+(gt²)/2
Since u=0, this gives us t= 1.81 seconds (s=16, g=9.8)

Now if you put in the values in v=u+at:
v = 9.8 * 1.81 = 17.7 ms⁻¹ ≠ 25 ms⁻¹

But √(17.7² + 18²) ≈ 25 ms⁻¹!

thank you bro

 

[Khan_971]

I didnt check the ms, but is it 1.375m/s?

2.8 +- .1

 

[Most_UniQue]

You mean they collide at the same time? Like at 2s?

 

[angelicsuccubus]

2.8 +- .1

got it, its s=1/2(u+v)t... u is 0 cause it falls from rest .. the rest is just reading d.. I took it as 0.55, so 0.55=1/2(u+v)(0.4) gives you 2.75m/s

 

[angelicsuccubus]

You mean they collide at the same time? Like at 2s?

well of course they collide at the same time! they collide with each other, how can they collide at different times -_-"

 

[raamish]

well of course they collide at the same time! they collide with each other, how can they collide at different times -_-"

http://www.xtremepapers.com/papers/...and AS Level/Physics (9702)/9702_w07_qp_2.pdf

how do we do 1b. really confused helpp

 

[iKhaled]

need help in may june 2009
http://www.xtremepapers.com/papers/...and AS Level/Physics (9702)/9702_s09_qp_2.pdf

Q2 d)...

the mark scheme says

(d) either
speed of approach = 4.0 m s-1 and speed of separation = 2.4 m s-1 .............................................. M1 not equal and so inelastic ....................................................... A1
or kinetic energy before = 9.6 J and kinetic energy after collision = 4.99 J ...................................... M1 kinetic energy after is less / not conserved so inelastic .......... A1

how can i get the kinetic energy after collision and what does speed of approach and speed of separation..first time to hear that in phys :O:O

 

[angelicsuccubus]

http://www.xtremepapers.com/papers/...and AS Level/Physics (9702)/9702_w07_qp_2.pdf

how do we do 1b. really confused helpp

uh Im still tryna figure that one out..

 

[angelicsuccubus]

need help in may june 2009
http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s09_qp_2.pdf

Q2 d)...

the mark scheme says

(d) either
speed of approach = 4.0 m s-1 and speed of separation = 2.4 m s-1 .............................................. M1 not equal and so inelastic ....................................................... A1
or kinetic energy before = 9.6 J and kinetic energy after collision = 4.99 J ...................................... M1 kinetic energy after is less / not conserved so inelastic .......... A1

how can i get the kinetic energy after collision and what does speed of approach and speed of separation..first time to hear that in phys :O:O

Have you studied physics before or is today like the first introduction to momentum?
Speed of approach is the speed which with the particles come towards each other and speed of seperation is the speeds they have once they have collided and are moving away from each other.

Kinetic energy is the energy that an object has during any sort of movement, if the kinetic energy of the objects were zero after collison that would have meant they came to a stand still. If the K.E was non zero, that means they kept moving.

If a collision is elastic, the kinetic energy before collision is equal to the kinetic energy after the collison, so it doesn't change. The relative speed of approach is equal to the relative speed of seperation.
If a collision is inelastic, the kinetic energies before and after the collision are different.

Momentum is always conserved.

I need you to watch this:
http://www.khanacademy.org/science/physics/v/introduction-to-momentum

 

[Famous4it]

Does anyone have any notes on projectile motion? Totally lost =/

 

[iKhaled]

Have you studied physics before or is today like the first introduction to momentum?
Speed of approach is the speed which with the particles come towards each other and speed of seperation is the speeds they have once they have collided and are moving away from each other.

Kinetic energy is the energy that an object has during any sort of movement, if the kinetic energy of the objects were zero after collison that would have meant they came to a stand still. If the K.E was non zero, that means they kept moving.

If a collision is elastic, the kinetic energy before collision is equal to the kinetic energy after the collison, so it doesn't change. The relative speed of approach is equal to the relative speed of seperation.
If a collision is inelastic, the kinetic energies before and after the collision are different.

Momentum is always conserved.

I need you to watch this:
http://www.khanacademy.org/science/physics/v/introduction-to-momentum

dude ik about all that kinetic stuff but my question is how did the mark scheme got the kinetic energy after collision..isn't it supposed to be 1/2(1.2)(0.8)^2 ?

 

[angelicsuccubus]

Does anyone have any notes on projectile motion? Totally lost =/

It's really simple... here, watch this:
http://www.khanacademy.org/science/physics/v/projectile-at-an-angle
And here's the notes I once made, the cover everything on projectile motion in the syllabus but if you can't read my hand writing, check the revision notes on xtremepapers under phsyics... and if you have any questions, I'll be here.
https://www.dropbox.com/s/vcfzfuh8dy6tlww/IMAG0317.jpg (pg1)
https://www.dropbox.com/s/8pa9x7ivzc0z2he/IMAG0318.jpg (pg2)
https://www.dropbox.com/s/qjllzifs0cox482/IMAG0319.jpg (pg3)

double click to zoom, or downlaod.