Physics: Post your doubts here!

[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 !!!!!

 

[Arjun Dhanak]

See 9702/22/M/J/10 question 4 c. here's the link:
http://www.xtremepapers.com/papers/...nd AS Level/Physics (9702)/9702_s10_qp_22.pdf
Is "Length= λ/4" always used to find the lowest frequency?
How do we know how many times the A and N are found?
Lastly, how can we know how many times we draw the whole spiral (with antinode at opening and node at piston) in the tube?
Thanks in advance.
BTW, only 4 days to go so GOOD LUCK TO EVERYONE!!

 

[leosco1995]

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.

By cycle you mean wavelength right? Does that mean within one wavelength of the old wave, there has to be 4 wavelengths of the new wave? I'm kind of confused.

 

[angelicsuccubus]

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

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

q1 doesn't have a part c and for 5b(i) .. you just make it lag behind 1/3 of half the wavelength. .. cause 60degrees is pi/3 so its should look like..



sorry, shakey hand on the touchpad

 

[angelicsuccubus]

By cycle you mean wavelength right? Does that mean within one wavelength of the old wave, there has to be 4 wavelengths of the new wave? I'm kind of confused.

yeh.. do you want me to draw it?

 

[leosco1995]

yeh.. do you want me to draw it?

Yeah, please. I want to check if I drew it right.

 

[angelicsuccubus]

where did u get 100 squares=0.2cm from

velocity times time.. gives you cm from the graph..