Physics: Post your doubts here!

[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.

 

[xxxtoughxxx]

q(7) b......in the ms and er it says for damage of supply to take place lamp A must be shorted ....why especially lamp A wasnt lamp c da shorted one in the quetion b4 ??? any 1 plzz thx in adv

 

[angelicsuccubus]

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 ?

no, total k.e after collision is the kinetic energy of ball B after collision plus the kinetic energy of ball S after collision, that's why they said to use your answer in(c).

its 1/2(1.2)(0.8)^2 + 1/2(3.6)(1.6)^2

sorry, for the earlier long explaination.. its cause you said you'd never heard of speed of approach/seperation so I freaked..

 

[Jaf]

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

how do we do 1b. really confused helpp

What's so difficult about this one? :S

R² = V/πL
So by calculation, R = 0.489 cm

For the uncertainties since the values are multiplied/divided you're to add the fractional uncertainties. So you get
(ΔR/R)x2 = ΔV/V + ΔL/L = 0.5/15 + 0.1/20 = 23/600
ΔR/0.489 = 23/1200
ΔR = 0.00937 cm

So R = 0.489 ± 0.009 cm
---------------------------------------------------

Another way to do this would be when R = √(V/πL) = (V/πL)^(1/2) = [V^(1/2)]/[(πL)^(1/2)]
Now the fractional uncertainties would be written as:
ΔR/R = (ΔV/V) x 1/2 + (ΔL/L) x 1/2
...and so on.

 

[angelicsuccubus]

What's so difficult about this one? :S

R² = V/πL
So by calculation, R = 0.489 cm

For the uncertainties since the values are multiplied/divided you're to add the fractional uncertainties. So you get
(ΔR/R)x2 = ΔV/V + ΔL/L = 0.5/15 + 0.1/20 = 23/600
ΔR/0.489 = 23/1200
ΔR = 0.00937 cm

So R = 0.489 ± 0.009 cm
---------------------------------------------------

Another way to do this would be when R = √(V/πL) = (V/πL)^(1/2) = [V^(1/2)]/[(πL)^(1/2)]
Now the fractional uncertainties would be written as:
ΔR/R = (ΔV/V) x 1/2 + (ΔL/L) x 1/2
...and so on.

that's what I was tryna do! .. except here:
ΔR/0.489 = 23/1200 I wasn't multiplying the 0.489... donno why I didnt realise that's what happens lol

 

[angelicsuccubus]

Jaf there's loads of other unanswered questions here though.. and Im getting really sleepy.. i think you should take over

 

[iKhaled]

no, total k.e after collision is the kinetic energy of ball B after collision plus the kinetic energy of ball S after collision, that's why they said to use your answer in(c).

its 1/2(1.2)(0.8)^2 + 1/2(3.6)(1.6)^2

sorry, for the earlier long explaination.. its cause you said you'd never heard of speed of approach/seperation so I freaked..

thanks bro

 

[Jaf]

Jaf there's loads of other unanswered questions here though.. and Im getting really sleepy.. i think you should take over

Me going too - need to study bio now. You're on your own everyone.

 

[angelicsuccubus]

q(7) b......in the ms and er it says for damage of supply to take place lamp A must be shorted ....why especially lamp A wasnt lamp c da shorted one in the quetion b4 ??? any 1 plzz thx in adv

cause lamp A isn't in parallel.. if it's shorted, there would be no resistance in its spot and loads of current will flow to lamp c and lamp b blowing them too. Part b is not based on a, its just a general suggestion based question on the circuit given.... that if you wanted to test the circuit by placing an ohm-meter/power supply between X and Y .. which would you choose? At this point in time, you don't know that lamp C is shorted, you just want to test the circuit and you know that if you played something between X and Y, that's where the current would flow from, and the first device it would flow through is lamp A... and if A is dead... C and B would get blown up by the power supply's excessive current.

 

[angelicsuccubus]

thanks bro

anytime.. so long as I don't fall asleep on the laptop .....

 

[angelicsuccubus]

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

I don't get how they are finding the extension

it's just 1/2kx^2=1/2mv^2 and the speed used is twice than the speed found in c(i) so.. 16m/s and k is from the beginning 1250N/m and m is 25g and x you're going to get as 72mm

 

[angelicsuccubus]

http://www.xtremepapers.com/papers/CIE/Cambridge International A and 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???

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..

 

[Abdullah syed]

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

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

 

[mady666]

someone plz give me a link to the marking scheme for paper 2 oct/nov 2001 and may/june 2002

 

[raamish]

What's so difficult about this one? :S

R² = V/πL
So by calculation, R = 0.489 cm

For the uncertainties since the values are multiplied/divided you're to add the fractional uncertainties. So you get
(ΔR/R)x2 = ΔV/V + ΔL/L = 0.5/15 + 0.1/20 = 23/600
ΔR/0.489 = 23/1200
ΔR = 0.00937 cm

So R = 0.489 ± 0.009 cm
---------------------------------------------------

Another way to do this would be when R = √(V/πL) = (V/πL)^(1/2) = [V^(1/2)]/[(πL)^(1/2)]
Now the fractional uncertainties would be written as:
ΔR/R = (ΔV/V) x 1/2 + (ΔL/L) x 1/2
...and so on.

Man i had solved everything. Just the part: So R = 0.489 ± 0.009 cm. i dont get how u get 0.009 cm in your answer. That is the only step i dont understand so plzz explain how u wrote it thnx