Physics: Post your doubts here!

[applepie1996]

can any one explain me what is impulse?

f=mass*acceleration , f=mass*(v-u/t) so force= mv-mu/t rearranging the equation u get force*time=mv-mu and mv-mu is change in momentum and force*time is impulse so impulse=change in momentum

 

[Jaf]

can someone please explain to me the question 6c?? The mark scheme has additional values I don't know where the got them

http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w11_qp_23.pdf
http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w11_ms_23.pdf

for all I know
charge of alpha particle = 2
mass of alpha particle =4

charge of beta particle 1.6x10^-19
mass of beta particle 9.11x10^-31

The charge of the alpha particle is 2x the elementary charge (charge of a single electron/beta particle). [obviously the sign changes but that doesn't matter since it's only a ratio]
The mass of the alpha particle is 4x the rest mass of a proton (1.67 × 10^–27 kg).

Both these^ values are given in the beginning of the question paper.

 

[Jaf]

http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w05_qp_2.pdfQ5 c 1 annd 2 how to show that intensity of B at pont p ( and whaere is poimt p) is 4/9 I ...and the next part also plzz help

The wave is passing through a point, and the displacement-time graph is drawn for that. P is not a point on the wave, rather P is the point the wave is passing through (think about it, there's a difference). Any how, even if you don't get it, it makes no difference to your answers. The intensity is common to all points ON a wave and so is the phase difference between the two waves.

c)i)
I/a^2 = constant (where I is intensity and a is amplitude)
Let's take the intensity of wave xB as x.

So I/3^2 = x/2^2
x = 4I/9

ii) The resultant amplitude when xA and xB meet is 1. Let the new amplitude be y.
So I/3^2 = y/1^2
y = I/9.

 

[Soldier313]

could somebody please help me with question 5 aii.) and iii.)
qn paper: http://www.xtremepapers.com/papers/...nd AS Level/Physics (9702)/9702_w11_qp_22.pdf
ms: http://www.xtremepapers.com/papers/...nd AS Level/Physics (9702)/9702_w11_ms_22.pdf

i just don't understand kirchoff's second law;(

 

[Jaf]

An athlete bends down and then jumps straight up into the air. At the instant he starts his jump
three forces act on him – his weight, an upthrust from the air and the contact force with the
ground. What is the correct order in increasing magnitude of these three forces
A. contact force → weight → upthrust

B. upthrust → contact force → weight

C. upthrust → weight → contact force
D. weight → upthrust → contact force

Provide an explanation if you can answer. It's not from past papers.

From my understanding, the answer should be C.
So first, know that even though the force is called 'up'thrust, it acts against the motion of the body and hence downwards.
The only force from the three forces that IS acting upwards is the contact force. The weight also, obviously, acts downwards.

So since the athelete accelerates (upwards; for a few fractions of a second atleast) the net force acting on the athlete must be upwards. Since there's only one upward force acting on the athlete, this must be the contact force. So we can conclude the contact force is the greatest force acting. A and B can definitely be eliminated.

Now I'm unable to explain why the upthrust should or should not be greater than the weight because of lack of information. I mean there are way to increase/decrease the upthrust. But I'm guessing since this force depends on the velocity, and the initial velocity can't be so high so as to produce a force greater than the weight, the upthrust should have a smaller magnitude than the weight.

 

[Jaf]

could somebody please help me with question 5 aii.) and iii.)
qn paper: http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w11_qp_22.pdf
ms: http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w11_ms_22.pdf

i just don't understand kirchoff's second law;(

I've redrawn the circuit to make it easier to apply Kirchoff's law. Try now (kirchoff's 2nd law is kind of hard to explain on paper )

 

[Soldier313]

I've redrawn the circuit to make it easier to apply Kirchoff's law. Try now (kirchoff's 2nd law is kind of hard to explain on paper )
View attachment 10790

omg tx soo much .....i actually understood now....God bless

 

[applepie1996]

could someone please explain Q1 PART b and Q4 PART c
WHY WON'T ANYONE ANSWER THIS!!!!

 

[Peter Check]

I have a serious problem with uncertainties, almost all the sums i solve are wrong! For instance in question 1b(ii) 21/M/J/11 http://www.xtremepapers.com/papers/...nd AS Level/Physics (9702)/9702_s11_qp_21.pdf
I do not get how the uncertainty in A = 2 × (0.01 / 0.38) × 100 = 5.3 % I guess we multiply by 2 because of pie r square.
But because it is the radius why don't we halve the diameter's percentage uncertainty?
Also how do i decide between which significant figures, if nothing is mentioned, for instance in the question where the percentage uncertainties are .17% and .27, why is it not .17% and .267% or .20% and .27%

 

[Jaf]

I'll post the answers. You'll be surprised just as I was.

a. horizontal velocity= s/t = 2.5/0.6= 4.17m/s

b. Vertically final velocity is zero as ball passes over cross-bar.
v = u + at
0 = u - (9.81*0.6)
u = 5.89m/s (ans)

c. u = root over(4.17^2 + 5.89^2)
= 7.21m/s

d. angle = tan^-1 (5.89/4.17)
= 55degrees

The mark scheme totally trolled me!!! I don't get it. How can horizontal velocity be 2.5/0.6 ??

The mark scheme is incorrect.
Put in the values in the range formula, and you don't get your original question.
range = [v^2 * sin2(theta)]/g
You get 4.98 m when you put in the values from the answers when it's really supposed to be 40 m.

 

[Saad (سعد)]

Can someone please tell me what definitions are commonly asked in questions, so that I can prepare them? Jazak Allah.

 

[hussamh10]

I have a serious problem with uncertainties, almost all the sums i solve are wrong! For instance in question 1b(ii) 21/M/J/11 http://www.xtremepapers.com/papers/...nd AS Level/Physics (9702)/9702_s11_qp_21.pdf
I do not get how the uncertainty in A = 2 × (0.01 / 0.38) × 100 = 5.3 % I guess we multiply by 2 because of pie r square.
But because it is the radius why don't we halve the diameter's percentage uncertainty?
Also how do i decide between which significant figures, if nothing is mentioned, for instance in the question where the percentage uncertainties are .17% and .27, why is it not .17% and .267% or .20% and .27%

Always use the formula Pi (D)^2/4 donot use Pi R^2 it adds complicatons i have been through this problem many times and my teacher reccomended to use the other formula .......

 

[Unicorn]

The charge of the alpha particle is 2x the elementary charge (charge of a single electron/beta particle). [obviously the sign changes but that doesn't matter since it's only a ratio]
The mass of the alpha particle is 4x the rest mass of a proton (1.67 × 10^–27 kg).

Both these^ values are given in the beginning of the question paper.

ohh thanks a lot you saved me ♥

 

[Unicorn]

Can someone please tell me what definitions are commonly asked in questions, so that I can prepare them? Jazak Allah.

just read these notes and they cover everything

 

[hussamh10]

I have a serious problem with uncertainties, almost all the sums i solve are wrong! For instance in question 1b(ii) 21/M/J/11 http://www.xtremepapers.com/papers/...nd AS Level/Physics (9702)/9702_s11_qp_21.pdf
I do not get how the uncertainty in A = 2 × (0.01 / 0.38) × 100 = 5.3 % I guess we multiply by 2 because of pie r square.
But because it is the radius why don't we halve the diameter's percentage uncertainty?
Also how do i decide between which significant figures, if nothing is mentioned, for instance in the question where the percentage uncertainties are .17% and .27, why is it not .17% and .267% or .20% and .27%

final answer is always in 1 S.f both of absolute error and the value

 

[applepie1996]

CAN SOMEONE PLEASE ANSWER QUESTIONS 6b and 6c

 

[raamish]

How do we take out speed of seperation and speed of approach. Its required to tell if collisions are inelastic or not

 

[Yousuf Ykr]

Can anyone help me with these past papers' questions
M/J 02- 7b
O/N 02- 9b, 2c
O/N 03- 3 c
M/J 04- 1b
O/N 04- 1b (ii)
M/J 05- 5c, 6b(iii)
O/N 05- 3c

 

[Peter Check]

final answer is always in 1 S.f both of absolute error and the value

the final answer right? But what if I use my percentage uncertainty as lets say 2.6666667 instead of 2.7(this is not the final answer) so when I use 2.666667 to add to the other two percentages, I will get a different answer!

 

[Peter Check]

Always use the formula Pi (D)^2/4 donot use Pi R^2 it adds complicatons i have been through this problem many times and my teacher reccomended to use the other formula .......

What is that formula for? If it is for area, I am getting a different answer for the area, using that formula.