Physics: Post your doubts here!

[raamish]

hmm.. well you can't add them cause if the resistive force was equal to the downward forces, the dude would be hanging in midair... I don't have a very good explaination for this but think of it like this.. since the guy is falling downwards and there is no force pushing him downwards, so no force causing acceleration to make him go towards the ground except for gravity. Gravity is the only force providing accelertion... so shouldn't the accelerating force be 880N itself...? shouldn't the acceleration be 9.81m/s/s?

The logical answer is no, nothing falls with the acceleration of gravity in our world because of air resistance! Why is the accelerating force less than the weight ...because there's air resisting acting against weight! and hence decreasing the accelerating force... which technically is the same force as weight but it's being suppressed. So air resistance has to be the difference between weight and the accelerating force cause that's what come in betweeen them?

If I confused you further, really sorry..

i think if the forces equalled he would be travelling with constant velocity no?

 

[alim]

hw to find the distance moved during the first 0.60s of the motion ???........(4a(ii))
http://www.xtremepapers.com/papers/...and AS Level/Physics (9702)/9702_w05_qp_2.pdf

 

[Mahnoor Khanm]

For uncertainties do this 5/400*answer of 1/h

Yeah but is the logic behind it. really confused!!!

 

[shan5674]

http://www.xtremepapers.com/papers/...nd AS Level/Physics (9702)/9702_w09_qp_22.pdf

4C(ii) all 3 parts please?

 

[Most_UniQue]

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

Hmm yeah

 

[Abdullah syed]

Yeah but is the logic behind it. really confused!!!

Yeah the formula is
U/Answer =DeltaU/1Uncertainty +DeltaU/2Uncertainty
as in this case Their is only one oncertainty we will use U/Answer =DeltaU/1Uncertainty
We can obtain our U from this Eqn

 

[Famous4it]

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.

THanks a lot!!!

 

[KurayamiKimmi]

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

thank you ^^ !

 

[jsg]

http://www.xtremepapers.com/papers/...and AS Level/Physics (9702)/9702_w02_qp_2.pdf

Help needed in Q1 part c and Q5 part b(i) .... !!

 

[KurayamiKimmi]

Initial momentum = mv = 0.035 x 4.5 = 0.1575 kgm/s (downwards)
Final momentum = mu= 0.035 x -3.5 = -0.1225 kgm/s
Change in momentum= mu-mv = -0.1225 - 0.1575 = -0.28 so now it's upwards
Force = change in momentum / time
Force= -0.28/0.14 = 2N Upwards

thanks!

 

[KurayamiKimmi]

a)ii)
p=mv
v=p/m


KE=(mv²)/2 = mp²/m²2 = p²/2m

b)
Change is momentum (Δp) = momentum of ball before collision (m₁v₁) - momentum of ball after collision (m₂v₂)
= 0.035 x 4.5 - 0.035 x -3.5 [don't forget the negative sign here; we're putting it because the ball moves in the opposite direction after collision]
= 0.28

F = Δp/t = 0.28/0.14 = 1.75N

thanks!!^^

 

[Wanzi21]

hi guys.. can u help me with this?

 

[angelicsuccubus]

i think if the forces equalled he would be travelling with constant velocity no?

you mean terminal velocity? well yeah at that point acceleration of gravity will become zero.. that's why its called free falling.. cause there isn't actually a force left that's making the person fall. But even with terminal velocity to work with.. its the same thing... if resistance was the same acceleration would have been zero.. but its not it went from 9.81 to whatever.. so if we subtract them we'll get the resisitive forces.. yeah?

 

[angelicsuccubus]

THanks a lot!!!

anytime ..

 

[applepie1996]

http://www.xtremepapers.com/papers/...and AS Level/Physics (9702)/9702_w05_qp_2.pdf
could some1 solve 4)a)ii

 

[leosco1995]

http://www.xtremepapers.com/papers/...nd AS Level/Physics (9702)/9702_s11_qp_23.pdf
Q6 c (iv)

How do we make the sketch?

 

[angelicsuccubus]

http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w05_qp_2.pdf
could some1 solve 4)a)ii

• distance moved= area under the graph
• area under the graph approximates = 1550 squares
• every 100 squares= 0.2 cm
• so 1550/100= 15.5
• 15.5 x 0.2= 3.1cm =0.031m

 

[angelicsuccubus]

http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s11_qp_23.pdf
Q6 c (iv)

How do we make the sketch?

the new period is 0.25t where t is the old period.. so that means the new wave would occur four times within the old wave.. yeah? cause frequency=1/time period.. so you just draw a wave with the same amplitude but 4 cycles within one wave cycle on fig 6.2... and that would mean that your wave has a quarter of the period of the old wave.

 

[applepie1996]

• distance moved= area under the graph
• area under the graph approximates = 1550 squares
• every 100 squares= 0.2 cm
• so 1550/100= 15.5
• 15.5 x 0.2= 3.1cm =0.031m

where did u get 100 squares=0.2cm from

 

[mariamalik]

ANY idea whats coming in the phys practical ?????.
Does anyone know any good website for revision or test which might be helpful before i go for the exams ??
As level !!!!!