Physics: Post your doubts here!

[a_wiserME!!]

Oh sorry, I thought you didn't get only the energy part.
Anyway, see solution 559 on this page : http://physics-ref.blogspot.in/2015/04/physics-9702-doubts-help-page-109.html
They've given the calculations with the explanation.

thank you!

 

[Konstantino Nikolas]

thank you!

no problem

 

[ashcull14]

Which part exactly do you not get?

b(1) and c

 

[nehaoscar]

For complete AS and A2 Physics Summary check this out:
https://www.xtremepapers.com/community/threads/as-and-a2-physics-summaries.42827/
For more revision notes check out my signature below

 

[Konstantino Nikolas]

b(1) and c

b)(i) is a direct question. The plates must be parallel so that the field is uniform; and horizontal so that the field is vertical. (Since the experiment involves finding the charge on the oil drop by balancing the electric force and gravitational force acting on it.)

c) All these values are small integer multiples of a certain base value, which is equal to 1.6*10^-19 and is the negative of the charge of an electron. (This was proposed by the Millikan oil drop experiment. I am not sure of the details to this.)

 

[The Sarcastic Retard]

http://studyguide.pk/Past Papers/CIE/International A And AS Level/9702 - Physics/9702_s10_qp_42.pdf
Q4(c) Help required.

 

[The Sarcastic Retard]

http://studyguide.pk/Past Papers/CIE/International A And AS Level/9702 - Physics/9702_w10_qp_42.pdf
Q3(c) Anyone?

 

[The Sarcastic Retard]

http://maxpapers.com/syllabus-materials/physics-9702-a-level/attachment/9702_s12_qp_41/
Q4(b) please how can I know the period and amplitude?

 

[ahmedish]

What are the things that were previously removed from the syllabus (not just this session)
I mean when solving 2005 and so. Was there anything that was later removed?

 

[qwertypoiu]

http://studyguide.pk/Past Papers/CIE/International A And AS Level/9702 - Physics/9702_s10_qp_42.pdf
Q4(c) Help required.

Both A and B charge are producing an electric field around themselves.
This means both would affect a unit positive charge.
Let's think about a unit positive charge first.

To move a unit positive charge from point M to P, work has to be done against A's force of repulsion, and B will naturally push the unit positive charge towards P. So:
Effect of A:
Work done = V(3) - V(6)
Where V(r) is the electric potential a particle has for being r micrometers away from A:
V(r) = Q/4πęr

Effect of B:
"Work done" = V(9) - V(6)
Because final position (P) is 9 micrometers away from B. V here should be defined separately for B, but since it has same charge (Q) as A, the formula would be same.

So overall energy to move +1C test charge from M to P is their sum:
W = V(3) + V(9) - V(6) - V(6)
Calculate this by substituting into formula above (remember r is in micro)

And then the answer you get, you have to multiply by charge on an electron, -1.6*10^-19. This is because previous answer was for unit charge. The negative sign is important.

I don't have a calculator now so sorry I can't do the calculation. Hopefully you get right answer.

 

[Xylferion]

http://maxpapers.com/syllabus-materials/physics-9702-a-level/attachment/9702_s12_qp_41/
Q4(b) please how can I know the period and amplitude?

The amplitude's already been given to you, that is, E(k).
What you need to pay attention to is the way kinetic energy behaves in relation to the displacement of the ball.

When the ball is at its amplitude, as shown in the diagram. The kinetic energy is zero because the velocity is zero at that point.
Kinetic energy is maximum when the velocity is maximum. This is at the equilibrium point a.k.a, 0 cm.

One oscillation means the ball starts at 4 cm, goes to -4 cm and then comes back to 4 cm.
If you observe the graph, it's 4 at 0 seconds and 4 at 0.6 seconds. So the period of one oscillation is 0.6 seconds.

If you look more closely, the ball goes from 4 ---> 0 ---> -4 ---> 0 ---> 4

The ball passes the equilibrium point, twice, in one oscillation.
We know that at the equilibrium point, the kinetic energy is maximum because the velocity is maximum at that point.

So the kinetic energy goes through two cycles in one period, hence why the period is 0.3 seconds.

 

[Xylferion]

http://studyguide.pk/Past Papers/CIE/International A And AS Level/9702 - Physics/9702_w10_qp_42.pdf
Q3(c) Anyone?

Attaching the light feathers = reducing weight of metal cube.

This results in heavy damping. Damping occurs because of resistive forces in the air, if you have something that is very light, the effect of these forces will be higher and the amplitude of oscillation decreases at all frequencies.

The curve you already see in the graph is an example of light damping with resonance. If I had to ask you to draw heavy damping with resonance, what would you expect the curve to look like? Heavy damping = amplitude reduced at all frequencies. This results in a much flatter peak.

 

[The Sarcastic Retard]

Attaching the light feathers = reducing weight of metal cube.

This results in heavy damping. Damping occurs because of resistive forces in the air, if you have something that is very light, the effect of these forces will be higher and the amplitude of oscillation decreases at all frequencies.

The curve you already see in the graph is an example of light damping with resonance. If I had to ask you to draw heavy damping with resonance, what would you expect the curve to look like? Heavy damping = amplitude reduced at all frequencies. This results in a much flatter peak.

View attachment 59081

so it reduces approx how much?
Thanks.

 

[The Sarcastic Retard]

The amplitude's already been given to you, that is, E(k).
What you need to pay attention to is the way kinetic energy behaves in relation to the displacement of the ball.

When the ball is at its amplitude, as shown in the diagram. The kinetic energy is zero because the velocity is zero at that point.
Kinetic energy is maximum when the velocity is maximum. This is at the equilibrium point a.k.a, 0 cm.

One oscillation means the ball starts at 4 cm, goes to -4 cm and then comes back to 4 cm.
If you observe the graph, it's 4 at 0 seconds and 4 at 0.6 seconds. So the period of one oscillation is 0.6 seconds.

If you look more closely, the ball goes from 4 ---> 0 ---> -4 ---> 0 ---> 4

The ball passes the equilibrium point, twice, in one oscillation.
We know that at the equilibrium point, the kinetic energy is maximum because the velocity is maximum at that point.

So the kinetic energy goes through two cycles in one period, hence why the period is 0.3 seconds.

View attachment 59080

How amplitude = E(k) ?
How u got 4?

 

[The Sarcastic Retard]

Both A and B charge are producing an electric field around themselves.
This means both would affect a unit positive charge.
Let's think about a unit positive charge first.

To move a unit positive charge from point M to P, work has to be done against A's force of repulsion, and B will naturally push the unit positive charge towards P. So:
Effect of A:
Work done = V(3) - V(6)
Where V(r) is the electric potential a particle has for being r micrometers away from A:
V(r) = Q/4πęr

Effect of B:
"Work done" = V(9) - V(6)
Because final position (P) is 9 micrometers away from B. V here should be defined separately for B, but since it has same charge (Q) as A, the formula would be same.

So overall energy to move +1C test charge from M to P is their sum:
W = V(3) + V(9) - V(6) - V(6)
Calculate this by substituting into formula above (remember r is in micro)

And then the answer you get, you have to multiply by charge on an electron, -1.6*10^-19. This is because previous answer was for unit charge. The negative sign is important.

I don't have a calculator now so sorry I can't do the calculation. Hopefully you get right answer.

Thanks a lot.

 

[Xylferion]

How amplitude = E(k) ?
How u got 4?

Look at the graph. You don't have to write anything on the y-axis, they already have put E(k) as the max value there.

Also what do you mean how did I get 4? Look at the previous graph.. 4 is the amplitude of displacement, the sine wave starts at 4 cm.

 

[Xylferion]

so it reduces approx how much?
Thanks.

There's no fixed value for how much it reduces, but I'd say approx. half the previous amplitudes. Just make sure the heavy damping curve is always flatter and that all the points on the curve are below the points on the light damping curve.

 

[The Sarcastic Retard]

Look at the graph. You don't have to write anything on the y-axis, they already have put E(k) as the max value there.

Also what do you mean how did I get 4? Look at the previous graph.. 4 is the amplitude of displacement, the sine wave starts at 4 cm.

There's no fixed value for how much it reduces, but I'd say approx. half the previous amplitudes. Just make sure the heavy damping curve is always flatter and that all the points on the curve are below the points on the light damping curve.

Thanks.

 

[i_try9621]

How do I solve this? It's from http://papers.xtremepapers.com/CIE/...and AS Level/Physics (9702)/9702_s05_qp_1.pdf

 

[The Sarcastic Retard]

How do I solve this? It's from http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s05_qp_1.pdf
View attachment 59108

Acceleration always acts downwards, so its always -ve as ball is thrown upwards against accn of free fall g.