Physics: Post your doubts here!

[angelicsuccubus]

need help in may june 2008

question 5 c) why they used the tensian as 4.00 ?

cause the weight is 4N .. and since the string remains taught.. it doesnt move horizontally.. the tension is the same as the weight.. doesn't change.

 

[geek101]

Can anyone puhhleeeez post a link of a Cambridge endorsed study guide for physics...(like the one for bio by Mary Jones) EXTREMELY APPRECIATED Good luck to all ya peeps!

 

[KurayamiKimmi]

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 ^_^

 

[raamish]

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

how do we do 1b. really confused helpp

 

[angelicsuccubus]

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 ^_^

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

 

[Jaf]

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 ^_^

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

 

[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