May 2015- Physics Paper 1 and Paper 2

[ibcanada]

Hey everyone, I'm kind of in a tricky situation. I wrote TZ1 SL Papers 1 and 2 on Friday; I answered one of the P2 questions in an answer booklet. I specified the question number in the LCD box thing to the left, but I forgot to indicate the letter for the particular part. 

 

I did indicate in the box that the question was answered in the booklet, and that was the only thing in the booklet. Will the examiner overlook this, or am I screwed? 

[Megamind]

Guys what about the potential divider question in p1 , and the projectile, and the equipotential, equiptential was for same masses and same charges, projectile was vertically upward since initial vertical v is not 0, as some times when it is launched horizontally initial v in the vertical component is 0 as in May tz1 hp2, potential divider it had a 0 but don't know at the beginning or at the end, electron was accelerating at a decreasing rate and speed was increasing, and this was related to the question about the average speed in 9 in paper 2. And it was in m13 hp2 tz1, equations of motion, constant net force and fixed mass so that a is constant, graph always accelerating since the slope the graph is never parallel to the x axis but v is 0 at an instant so it is not always moving, first question is 3 as 10^_-27 and 10^ -30 , and it is not unfamiliar, the girl's gain in me is 600 areas of the three triangles, entropy of the universe increasing for the melting ice, for isothermal work is greater than adiabatic and delta u is 0, for the gravitational potential it is - 2GM / R +r since it is a scalar, if the question was asking what is the gravitational field strength which is a vector it would be 0, emissivity different temperature same, greenhouses gases was last one, resolution was violet, circle was frequency constantly varying, wave velocity to the right was A, when the slope is increasing, u tube I think it is d as initial displacement doesn't mean no 0 initial velocity and the liquid's motion is damped but I chose a as it is oscillating so v is positive and negative, for the momentum the last one was not valid, for the electric field strength it was b , closest to - 20 , for the maximum power in one of the resistors it is VI not 2VI, for the artificial transmutation it was easy, c I think as I do not remember the choices, for the kinetic model assumption it is elasticity of collisions, for the equilibrium temperature between water and paraffin it is 60 degrees, for the centripetal acceleration it was very easy, for the higher frequency couloir a higher photo current and a larger (more negative) stopping potential, for the evidence of nuclear energy levels it is the discrete nature of alpha and gamma, for the power of the wave it is p over 2, for the wavelenth of 4 V it was half the de broglie wavelenth, for the mass spectrometer it is half v and twice the mass, for the resistance the ratio was 8:1, as d/2 will multiply the area by 0.25, for the alpha particles deflection it is maximum for n at theta equal 0 and minimum when theta is near 170, for the frequency and the halved le th it is still 500 Hz as th the closed end will give wavelenth = 4 * L/2 , which is still 2 L, for the coil I chose the first one, positive sine for the current, for the direction of the force I chose A,

Concerning the question that asked what is not possible is momentum: momentum is given by h/ lambda, electron box model states that the wavelengths can be 2L/n where n is an integer, so 2l, l, 2/3, L/2, 2l/5' 2L/6 etc......,

For the sl, the question about gravitational field strength it is 20,

What do you guys think, plz share with me

 

 

Hassan, I don't remember the potential divider question, emmisivity,or the greenhouse gases one. Could you recap it for me once and I'll tell you my answers. 

 

Our answers are almost similiar otherwise. Except for that power of the waves one, I think I got Power = P in that. 

Edited May 10, 2015 by Megamind

[Ossih]

 

Frequency in Q4 is supposed to be calculated by using the path difference for constructive interference, I think, i.e. even multiples of half wavelengths, given the wavelength and probably dividing it by the speed to get the frequency values. 

 

But I couldn't figure it out and didn't have time so I simply just did v/lambda which is wrong

Frequency in Q4 is supposed to be calculated by using the path difference for constructive interference, I think, i.e. even multiples of half wavelengths, given the wavelength and probably dividing it by the speed to get the frequency values. 

 

But I couldn't figure it out and didn't have time so I simply just did v/lambda which is wrong

We didn't have information about whether it is the first ,ms econo, third .... Minimum to know n, also, these equations are in the options and in the core for the new syllabus 2016/

 

 

Hey guys, for this question all I did is the path difference i.e. the difference between consecutive maxima is lambda (0.032m). So the time between two consecutive maxima is lambda/v (where v is the speed of B). So the frequency is v/lamba :$

[Ossih]

 

 

In the data analysis question, I got my slope around 1 and the y intercept aro0und 0.97. Anyone else? Also, when you took out the x intercept by equation y=mx + c to zero, did anyone find the value of x weird? Because I got some -0.98 or something and by observing and extrapolating the graph mentally to the minus x axis, it didn't look more than a -0.10. Did anyone get L as 0.93??

 

In paper 1 (Tz2) there was a question about green light being emitted from two point sources which was just NOT resolved. So the question asked how to resolve it and there were two plausible options: use red light (higher wavelength so bigger theta) and use a circular aperture (again bigger theta since 1.22 is there instead of 1). What'd you guys write? I chose red light. 

 

Ossih, what did you write for the possible binary states? 15 or 16? 

 

I had no idea what the "s" in capacitance meant, so I had to skip that question . I mean I knew it was the Farad but I forgot the definition :/ and in all my years of Physics, I have never seen a freaking s as a fundamental unit . Randomnly ticked something and probably got it wrong. 

 

In Paper 2 (tz2), for question 6, everyone found the gravitational field strength as Del V/ Del T right? And therefore from the calculation the rocket cannot escape without further engine use. 

 

Sorry about the blast of questions in your face, I'm just too excited to know everybody's else's answers . 

Also, any other Paper 1 questions anyone found challenging? I keep forgetting :/

 

I put 16 binary states! 

 

Also for the one with finding the induced emf, i started with NBA cos theta but it struck me that it's wrong.

 

The reason is, they ask for the induced emf JUST when the coil is about to completely enter. And so you'd so induced emf = ∆flux/∆t (emf is not NBA cos theta, that's flux ) and you'd find ∆t using suvat equations for free fall. The problem is, that's not the instantaneous change, that's average, that's not specific to when the coil just enters the field.

 

Instead, I did this. I used e = BLv, i said L was the length of one side (cause only one side is moving perpendicular to the field) and v = √2gh because its falling freely, where h is the length of the side. This made sense to me cause thats the instantaneous velocity at the time it just almost completely enters the field! What do you think?

Ossih, what was the answer to the question about frequency in section A ? And please tell me how you got the reading on the voltmeter in q 9

 

 

I answered the frequency one right now. For the reading on the voltmeter, it was really straightforward. The resistance in the external circuit was the same as the resistance in the internal circuit. The voltmeter measures the voltage in the external circuit, which would be half of total voltage i.e. 6 V.

[Ossih]

Guys what about the potential divider question in p1 , and the projectile, and the equipotential, equiptential was for same masses and same charges, projectile was vertically upward since initial vertical v is not 0, as some times when it is launched horizontally initial v in the vertical component is 0 as in May tz1 hp2, potential divider it had a 0 but don't know at the beginning or at the end, electron was accelerating at a decreasing rate and speed was increasing, and this was related to the question about the average speed in 9 in paper 2. And it was in m13 hp2 tz1, equations of motion, constant net force and fixed mass so that a is constant, graph always accelerating since the slope the graph is never parallel to the x axis but v is 0 at an instant so it is not always moving, first question is 3 as 10^_-27 and 10^ -30 , and it is not unfamiliar, the girl's gain in me is 600 areas of the three triangles, entropy of the universe increasing for the melting ice, for isothermal work is greater than adiabatic and delta u is 0, for the gravitational potential it is - 2GM / R +r since it is a scalar, if the question was asking what is the gravitational field strength which is a vector it would be 0, emissivity different temperature same, greenhouses gases was last one, resolution was violet, circle was frequency constantly varying, wave velocity to the right was A, when the slope is increasing, u tube I think it is d as initial displacement doesn't mean no 0 initial velocity and the liquid's motion is damped but I chose a as it is oscillating so v is positive and negative, for the momentum the last one was not valid, for the electric field strength it was b , closest to - 20 , for the maximum power in one of the resistors it is VI not 2VI, for the artificial transmutation it was easy, c I think as I do not remember the choices, for the kinetic model assumption it is elasticity of collisions, for the equilibrium temperature between water and paraffin it is 60 degrees, for the centripetal acceleration it was very easy, for the higher frequency couloir a higher photo current and a larger (more negative) stopping potential, for the evidence of nuclear energy levels it is the discrete nature of alpha and gamma, for the power of the wave it is p over 2, for the wavelenth of 4 V it was half the de broglie wavelenth, for the mass spectrometer it is half v and twice the mass, for the resistance the ratio was 8:1, as d/2 will multiply the area by 0.25, for the alpha particles deflection it is maximum for n at theta equal 0 and minimum when theta is near 170, for the frequency and the halved le th it is still 500 Hz as th the closed end will give wavelenth = 4 * L/2 , which is still 2 L, for the coil I chose the first one, positive sine for the current, for the direction of the force I chose A,

Concerning the question that asked what is not possible is momentum: momentum is given by h/ lambda, electron box model states that the wavelengths can be 2L/n where n is an integer, so 2l, l, 2/3, L/2, 2l/5' 2L/6 etc......,

For the sl, the question about gravitational field strength it is 20,

What do you guys think, plz share with me

 

 

1.The potential divider was actually a potentiometer this year which I've never seen them ask before. It's the graph going from 0 to 3 as a straight line.

 

2.For the projectile, it was fired above the horizontal.

 

3.For the equipotentials, yeah, same mass and same charges. 

 

4. The electron accelerating thing, damn you're right haha I did that one wrong.

 

5. Yeah, constant net force and fixed mass.

 

6. Yeah the first question, the ratio of the orders of magnitude was 3.

 

7. The girl's gain in energy is 600J yup

 

8. Entropy of the universe increases, yea

 

9. For isothermal, ∆U = 0 and work done is greater than adiabatic yup

 

10. Yup, its -2GM/(R+r) cause you add potentials as scalars.

 

11. I don't remember the emissivity question hm.

 

12. Greenhouse gases, they transmit radiation from the sun but absorb radiation from the Earth.

 

13. Resolution was to replace with a violet light yea

 

14. The moving sound source in a circle would be heard with constantly varying frequency, yeah

 

15. Yeah the wave velocity to the right was A.

 

16. U tube was D, the damped one. 

 

17. Hm, for the electric field strength one, I chose A, the one furthest from both. 

 

18. Yeah its VI

 

19. Yes, artificial transmutations was C.

 

20. For kinetic model, I didn't put elasticity of the collisions because they didn't say the it was perfectly elastic. Instead, I chose that the temperature is proportional to the average kinetic energy of the molecules. According to Wikipedia (not that it's the best source haha), one of the assumptions of the kinetic model is that the absolute temperature is a measure of the average kinetic energy of the particles.

 

21. For the water and paraffin, it was 60º yup

 

22. For the photocurrent one, I disagree. The question said that they fired a higher frequency BUT the same number of photons, which means that the number of electrons are ejected and hence the photocurrent is the same. All it is that the stopping potential is now more negative.

 

23. For evidence of nuclear energy levels, its alpha and gamma yup

 

I can't remember the power of the wave and de Broglie wavelength questions.

 

24. For the mass spectrometer, I chose twice the mass, twice the charge and same v. My reasoning is this: For the mass spectrometer, mv^2/r = Bqv right, which gives me r = (mv)/(Bq). B is a constant. Now, both twice the mass and twice the charge, and twice the mass but half the velocity give me the same value of r. So I was confused. But I reasoned that in a mass spectrometer, it wouldn't even enter the deflection area if it had a velocity of v/2! That's the job of the velocity selector, everything comes out with velocity v. So I chose twice the mass and twice the charge.

 

25. For resistance the ratio was 8:1 yup

 

26. For the alpha particles deflection, you're right, its a peak at 0 and falls down.

 

27. Yeah, for the frequency where the length is halved but closed at one end, it remains the same.

 

28. For the coil, I chose positive cos but I'm not even bothered explaining why haha, but I think you're right.

 

29. For the direction of force, I got C! The current was into the page, B was from left to right, and my left hand rule gives me the force as downwards

 

30. For the electron in a box, momentum cant be h/4L because no n in nh/2L will give you h/4L

 

Btw holy crap haha you have a great memory.

 

Others that I remember

 

31. Albedo was water < desert < snow

 

32. The last question (about the potential difference on a pixel) was nNe/C

 

33. The 2/3rd question (the one with the graph), she is constantly accelerating

 

34. The direction of v and a for the electron orbiting a proton, v was tangential and a was to the centre

 

35. The renewable energy source was 10% :/ I put 30%

 

36. The units of capacitance was the one with s^4 (A I think).

 

I can't remember the remaining 4 haha, what were they?

Edited May 10, 2015 by Ossih

[Vioh]

@Ossih, although I didn't do the exam, I think I might have spotted 2 mistakes of yours.

 

 

20. For kinetic model, I didn't put elasticity of the collisions because they didn't say the it was perfectly elastic. Instead, I chose that the temperature is proportional to the average kinetic energy of the molecules. According to Wikipedia (not that it's the best source haha), one of the assumptions of the kinetic model is that the absolute temperature is a measure of the average kinetic energy of the particles.

 

I think the assumption should be the elasticity of the collisions (that is in agreement with wikipedia).

Although it's true that temperature is proportional to the average kinetic energy of the molecules, but I don't think that it was an assumption; it was actually a mathematical consequence of the kinetic model. Wikipedia only says that:

• The average kinetic energy of the gas particles depends only on the absolute temperature of the system

This means that kinetic energy has a relationship with the temperature, but this assumption isn't very clear on what type of relationship it is. It's only when you work out the mathematics that you find the temperature to be proportional to the kinetic energy.

 

22. For the photocurrent one, I disagree. The question said that they fired a higher frequency BUT the same number of photons, which means that the number of electrons are ejected and hence the photocurrent is the same. All it is that the stopping potential is now more negative

 

I would agree with Hassan76 for this one, though I may be wrong. It's true that the same number of electrons are ejected due to the same intensity of the light that was used. But increasing the frequency of the photons means that you've increased the kinetic energy of the emitted electrons. This will cause the electrons to move faster (since KE is proportional to the speed). Here, you need to realize that current is the measure of charge per unit time. The increase in speed will increase the number of charge per unit time, thus the current must increase.

So the chain of reasoning is as follow:

 

Higher frequency --> larger KE --> electrons gaining speed --> more photocurrent --> more negative stopping potential

 

And damn, it's crazy to see how you guys remembered all that stuff!!! Good luck with the rest of the exams

[Sofia.]

 

 

In the data analysis question, I got my slope around 1 and the y intercept aro0und 0.97. Anyone else? Also, when you took out the x intercept by equation y=mx + c to zero, did anyone find the value of x weird? Because I got some -0.98 or something and by observing and extrapolating the graph mentally to the minus x axis, it didn't look more than a -0.10. Did anyone get L as 0.93??

 

In paper 1 (Tz2) there was a question about green light being emitted from two point sources which was just NOT resolved. So the question asked how to resolve it and there were two plausible options: use red light (higher wavelength so bigger theta) and use a circular aperture (again bigger theta since 1.22 is there instead of 1). What'd you guys write? I chose red light. 

 

Ossih, what did you write for the possible binary states? 15 or 16? 

 

I had no idea what the "s" in capacitance meant, so I had to skip that question . I mean I knew it was the Farad but I forgot the definition :/ and in all my years of Physics, I have never seen a freaking s as a fundamental unit . Randomnly ticked something and probably got it wrong. 

 

In Paper 2 (tz2), for question 6, everyone found the gravitational field strength as Del V/ Del T right? And therefore from the calculation the rocket cannot escape without further engine use. 

 

Sorry about the blast of questions in your face, I'm just too excited to know everybody's else's answers . 

Also, any other Paper 1 questions anyone found challenging? I keep forgetting :/ 

 

We got the same gradient, and I think my intercept was 0.92.

For question 6, I used g = F/m, but since the force is also ma, g = ma/m = a. Not sure that was correct, though. Then I simply did the calculation - found the needed escape speed, and compared it to the speed of the rocket. Also got that it couldn't escape, at least. 

 

 

But Sofia, I believe this particular formula and your final answer g=a would only apply if the spaceship was on the surface of Earth itself. Over here, we have a spaceship that is far away from an unknown planet, therefore I think a = change in velocity over change in time = gravitational field strength should apply here. This came to around 2.7 m/s squared. However, escape velocity still was not achieved in either case. 

 

 

Wait, you are saying that the gravitational field strength is equal to acceleration, and that is what I'm saying?

[Sofia.]

 

 

 

again to those who did thermodynamics, there was this last question saying that in adiabatic change work done was i think 210J, what was the change in internal energy.. so since that was adiabatic there was no heat exchange so the change in internal energy was -210J, did i get this one right?

 

Yep, just used the first law of thermodynamics for that one.

Q = 0, Q=dU+W, W=210, -210=U

 

Damn, I really can't read the questions carefully. I somehow skipped the adiabatic part hehe... And here I've been thinking P2 went well. 

I somehow got Q to 1300 something? Can't even remember how I got that!

You used the values for the net work done given above

 

 

Must have read system instead of that specific process then. Ah well, hopefully I got the other questions right. 

[Megamind]

 

 

 

In the data analysis question, I got my slope around 1 and the y intercept aro0und 0.97. Anyone else? Also, when you took out the x intercept by equation y=mx + c to zero, did anyone find the value of x weird? Because I got some -0.98 or something and by observing and extrapolating the graph mentally to the minus x axis, it didn't look more than a -0.10. Did anyone get L as 0.93??

 

In paper 1 (Tz2) there was a question about green light being emitted from two point sources which was just NOT resolved. So the question asked how to resolve it and there were two plausible options: use red light (higher wavelength so bigger theta) and use a circular aperture (again bigger theta since 1.22 is there instead of 1). What'd you guys write? I chose red light. 

 

Ossih, what did you write for the possible binary states? 15 or 16? 

 

I had no idea what the "s" in capacitance meant, so I had to skip that question . I mean I knew it was the Farad but I forgot the definition :/ and in all my years of Physics, I have never seen a freaking s as a fundamental unit . Randomnly ticked something and probably got it wrong. 

 

In Paper 2 (tz2), for question 6, everyone found the gravitational field strength as Del V/ Del T right? And therefore from the calculation the rocket cannot escape without further engine use. 

 

Sorry about the blast of questions in your face, I'm just too excited to know everybody's else's answers . 

Also, any other Paper 1 questions anyone found challenging? I keep forgetting :/ 

 

We got the same gradient, and I think my intercept was 0.92.

For question 6, I used g = F/m, but since the force is also ma, g = ma/m = a. Not sure that was correct, though. Then I simply did the calculation - found the needed escape speed, and compared it to the speed of the rocket. Also got that it couldn't escape, at least. 

 

 

But Sofia, I believe this particular formula and your final answer g=a would only apply if the spaceship was on the surface of Earth itself. Over here, we have a spaceship that is far away from an unknown planet, therefore I think a = change in velocity over change in time = gravitational field strength should apply here. This came to around 2.7 m/s squared. However, escape velocity still was not achieved in either case. 

 

 

Wait, you are saying that the gravitational field strength is equal to acceleration, and that is what I'm saying?

 

 

Yes, the gravitational field strength is theoretically the acceleration due to gravity g but your answer would only apply on the surface of the Earth itself. The spaceship clearly isn't ( ) so I used a = del V/ del T to calculate the acceleration (or rather deceleration) of the spaceship. 

[Sofia.]

 

 

 

 

In the data analysis question, I got my slope around 1 and the y intercept aro0und 0.97. Anyone else? Also, when you took out the x intercept by equation y=mx + c to zero, did anyone find the value of x weird? Because I got some -0.98 or something and by observing and extrapolating the graph mentally to the minus x axis, it didn't look more than a -0.10. Did anyone get L as 0.93??

 

In paper 1 (Tz2) there was a question about green light being emitted from two point sources which was just NOT resolved. So the question asked how to resolve it and there were two plausible options: use red light (higher wavelength so bigger theta) and use a circular aperture (again bigger theta since 1.22 is there instead of 1). What'd you guys write? I chose red light. 

 

Ossih, what did you write for the possible binary states? 15 or 16? 

 

I had no idea what the "s" in capacitance meant, so I had to skip that question . I mean I knew it was the Farad but I forgot the definition :/ and in all my years of Physics, I have never seen a freaking s as a fundamental unit . Randomnly ticked something and probably got it wrong. 

 

In Paper 2 (tz2), for question 6, everyone found the gravitational field strength as Del V/ Del T right? And therefore from the calculation the rocket cannot escape without further engine use. 

 

Sorry about the blast of questions in your face, I'm just too excited to know everybody's else's answers . 

Also, any other Paper 1 questions anyone found challenging? I keep forgetting :/ 

 

We got the same gradient, and I think my intercept was 0.92.

For question 6, I used g = F/m, but since the force is also ma, g = ma/m = a. Not sure that was correct, though. Then I simply did the calculation - found the needed escape speed, and compared it to the speed of the rocket. Also got that it couldn't escape, at least. 

 

 

But Sofia, I believe this particular formula and your final answer g=a would only apply if the spaceship was on the surface of Earth itself. Over here, we have a spaceship that is far away from an unknown planet, therefore I think a = change in velocity over change in time = gravitational field strength should apply here. This came to around 2.7 m/s squared. However, escape velocity still was not achieved in either case. 

 

 

Wait, you are saying that the gravitational field strength is equal to acceleration, and that is what I'm saying?

 

 

Yes, the gravitational field strength is theoretically the acceleration due to gravity g but your answer would only apply on the surface of the Earth itself. The spaceship clearly isn't ( ) so I used a = del V/ del T to calculate the acceleration (or rather deceleration) of the spaceship. 

 

 

But g is not the gravitational field strength on Earth, it is the gravitational field strength anywhere, including where the spaceship is. As in, my answer was not 9.81 - it was 2.17, I think. I obviously calculated a, not simply said g was that of Earth. 

Edited May 10, 2015 by Sofia.

[Sofia.]

Hey everyone, I'm kind of in a tricky situation. I wrote TZ1 SL Papers 1 and 2 on Friday; I answered one of the P2 questions in an answer booklet. I specified the question number in the LCD box thing to the left, but I forgot to indicate the letter for the particular part. 

 

I did indicate in the box that the question was answered in the booklet, and that was the only thing in the booklet. Will the examiner overlook this, or am I screwed? 

 

Don't worry, I don't think it'll matter if that was the only thing in there. 

[Megamind]

 

 

 

 

 

In the data analysis question, I got my slope around 1 and the y intercept aro0und 0.97. Anyone else? Also, when you took out the x intercept by equation y=mx + c to zero, did anyone find the value of x weird? Because I got some -0.98 or something and by observing and extrapolating the graph mentally to the minus x axis, it didn't look more than a -0.10. Did anyone get L as 0.93??

 

In paper 1 (Tz2) there was a question about green light being emitted from two point sources which was just NOT resolved. So the question asked how to resolve it and there were two plausible options: use red light (higher wavelength so bigger theta) and use a circular aperture (again bigger theta since 1.22 is there instead of 1). What'd you guys write? I chose red light. 

 

Ossih, what did you write for the possible binary states? 15 or 16? 

 

I had no idea what the "s" in capacitance meant, so I had to skip that question . I mean I knew it was the Farad but I forgot the definition :/ and in all my years of Physics, I have never seen a freaking s as a fundamental unit . Randomnly ticked something and probably got it wrong. 

 

In Paper 2 (tz2), for question 6, everyone found the gravitational field strength as Del V/ Del T right? And therefore from the calculation the rocket cannot escape without further engine use. 

 

Sorry about the blast of questions in your face, I'm just too excited to know everybody's else's answers . 

Also, any other Paper 1 questions anyone found challenging? I keep forgetting :/ 

 

We got the same gradient, and I think my intercept was 0.92.

For question 6, I used g = F/m, but since the force is also ma, g = ma/m = a. Not sure that was correct, though. Then I simply did the calculation - found the needed escape speed, and compared it to the speed of the rocket. Also got that it couldn't escape, at least. 

 

 

But Sofia, I believe this particular formula and your final answer g=a would only apply if the spaceship was on the surface of Earth itself. Over here, we have a spaceship that is far away from an unknown planet, therefore I think a = change in velocity over change in time = gravitational field strength should apply here. This came to around 2.7 m/s squared. However, escape velocity still was not achieved in either case. 

 

 

Wait, you are saying that the gravitational field strength is equal to acceleration, and that is what I'm saying?

 

 

Yes, the gravitational field strength is theoretically the acceleration due to gravity g but your answer would only apply on the surface of the Earth itself. The spaceship clearly isn't ( ) so I used a = del V/ del T to calculate the acceleration (or rather deceleration) of the spaceship. 

 

 

But g is not the gravitational field strength on Earth, it is the gravitational field strength anywhere, including where the spaceship is. As in, my answer was not 9.81 - it was 2.17, I think. I obviously calculated a, not simply said g was that of Earth. 

 

 

Oh, alright, I probably misunderstood what you meant then. We both got the same answers then, I remember mine being very close to 2.2. 

[Sofia.]

 

 

 

 

 

 

In the data analysis question, I got my slope around 1 and the y intercept aro0und 0.97. Anyone else? Also, when you took out the x intercept by equation y=mx + c to zero, did anyone find the value of x weird? Because I got some -0.98 or something and by observing and extrapolating the graph mentally to the minus x axis, it didn't look more than a -0.10. Did anyone get L as 0.93??

 

In paper 1 (Tz2) there was a question about green light being emitted from two point sources which was just NOT resolved. So the question asked how to resolve it and there were two plausible options: use red light (higher wavelength so bigger theta) and use a circular aperture (again bigger theta since 1.22 is there instead of 1). What'd you guys write? I chose red light. 

 

Ossih, what did you write for the possible binary states? 15 or 16? 

 

I had no idea what the "s" in capacitance meant, so I had to skip that question . I mean I knew it was the Farad but I forgot the definition :/ and in all my years of Physics, I have never seen a freaking s as a fundamental unit . Randomnly ticked something and probably got it wrong. 

 

In Paper 2 (tz2), for question 6, everyone found the gravitational field strength as Del V/ Del T right? And therefore from the calculation the rocket cannot escape without further engine use. 

 

Sorry about the blast of questions in your face, I'm just too excited to know everybody's else's answers . 

Also, any other Paper 1 questions anyone found challenging? I keep forgetting :/ 

 

We got the same gradient, and I think my intercept was 0.92.

For question 6, I used g = F/m, but since the force is also ma, g = ma/m = a. Not sure that was correct, though. Then I simply did the calculation - found the needed escape speed, and compared it to the speed of the rocket. Also got that it couldn't escape, at least. 

 

 

But Sofia, I believe this particular formula and your final answer g=a would only apply if the spaceship was on the surface of Earth itself. Over here, we have a spaceship that is far away from an unknown planet, therefore I think a = change in velocity over change in time = gravitational field strength should apply here. This came to around 2.7 m/s squared. However, escape velocity still was not achieved in either case. 

 

 

Wait, you are saying that the gravitational field strength is equal to acceleration, and that is what I'm saying?

 

 

Yes, the gravitational field strength is theoretically the acceleration due to gravity g but your answer would only apply on the surface of the Earth itself. The spaceship clearly isn't ( ) so I used a = del V/ del T to calculate the acceleration (or rather deceleration) of the spaceship. 

 

 

But g is not the gravitational field strength on Earth, it is the gravitational field strength anywhere, including where the spaceship is. As in, my answer was not 9.81 - it was 2.17, I think. I obviously calculated a, not simply said g was that of Earth. 

 

 

Oh, alright, I probably misunderstood what you meant then. We both got the same answers then, I remember mine being very close to 2.2. 

 

 

Haha, at least we both got the right answer - best end to any argument

[Ossih]

@Ossih, although I didn't do the exam, I think I might have spotted 2 mistakes of yours.

 

 

20. For kinetic model, I didn't put elasticity of the collisions because they didn't say the it was perfectly elastic. Instead, I chose that the temperature is proportional to the average kinetic energy of the molecules. According to Wikipedia (not that it's the best source haha), one of the assumptions of the kinetic model is that the absolute temperature is a measure of the average kinetic energy of the particles.

 

I think the assumption should be the elasticity of the collisions (that is in agreement with wikipedia).

Although it's true that temperature is proportional to the average kinetic energy of the molecules, but I don't think that it was an assumption; it was actually a mathematical consequence of the kinetic model. Wikipedia only says that:

• The average kinetic energy of the gas particles depends only on the absolute temperature of the system

This means that kinetic energy has a relationship with the temperature, but this assumption isn't very clear on what type of relationship it is. It's only when you work out the mathematics that you find the temperature to be proportional to the kinetic energy.

 

22. For the photocurrent one, I disagree. The question said that they fired a higher frequency BUT the same number of photons, which means that the number of electrons are ejected and hence the photocurrent is the same. All it is that the stopping potential is now more negative

 

I would agree with Hassan76 for this one, though I may be wrong. It's true that the same number of electrons are ejected due to the same intensity of the light that was used. But increasing the frequency of the photons means that you've increased the kinetic energy of the emitted electrons. This will cause the electrons to move faster (since KE is proportional to the speed). Here, you need to realize that current is the measure of charge per unit time. The increase in speed will increase the number of charge per unit time, thus the current must increase.

So the chain of reasoning is as follow:

 

Higher frequency --> larger KE --> electrons gaining speed --> more photocurrent --> more negative stopping potential

 

And damn, it's crazy to see how you guys remembered all that stuff!!! Good luck with the rest of the exams

 

Thanks Vioh! I think you may be right for the temperature one. But this site http://chemwiki.ucdavis.edu/Physical_Chemistry/Physical_Properties_of_Matter/Phases_of_Matter/Gases/Kinetic_Theory_of_Gases/Kinetic_Theory_of_Gases states both as assumptions of the Kinetic theory so I really don't know. My book only states elastic though, so maybe that's what it is. Oh well, we'll find out soon enough ahha

 

However, for the photocurrent one, I'm sure, because it said that the number of photons per second is the same, (it doesn't say the same intensity of light, that would give a different result) implying that the number of electrons per second is the same (assuming that we're above threshold frequency). Because of that photocurrent must stay the same. 

 

If it were same intensity of light, then I agree, the photocurrent would change

 

I just hope I get a 7. I think it was a pretty think-out-of-the-box paper so a boundary for a 7 will be 32-34ish, so I hope I'll be fine

Edited May 10, 2015 by Ossih

[Vioh]

Thanks Vioh! I think you may be right for the temperature one. But this site http://chemwiki.ucdavis.edu/Physical_Chemistry/Physical_Properties_of_Matter/Phases_of_Matter/Gases/Kinetic_Theory_of_Gases/Kinetic_Theory_of_Gases states both as assumptions of the Kinetic theory so I really don't know. My book only states elastic though, so maybe that's what it is. Oh well, we'll find out soon enough ahha

 

Yeah, I can't deny the fact that many sources consider the proportionality relationship as an assumption of the kinetic theory. So perhaps you are also correct. You know what, maybe this is one of the questions where there are 2 correct answers Hopefully that is the case!!!

 

However, for the photocurrent one, I'm sure, because it said that the number of photons per second is the same, (it doesn't say the same intensity of light, that would give a different result) implying that the number of electrons per second is the same (assuming that we're above threshold frequency). Because of that photocurrent must stay the same. 

 

If it were same intensity of light, then I agree, the photocurrent would change

 

Not really! We know that the intensity of light is the same. This is because intensity depends only on the number of photons; and since earlier you said that the same number of photons was used, that's why the intensity must be the same!

 

Now, there're 2 ways that the current can increase in a photoelectric experiment, either by:

• Increasing the intensity of light (i.e. the number of photons) --> This works because with more photons, more electrons would get emitted (as each electron can only absorb one single photon). However, since the question says that the number of photons (i.e. intensity) is kept the same, we need not worry about this case.
• OR by increasing the frequency of light --> This works because increasing the frequency means an increase in the energy of the photon (as E=hf), leading to an increase in the kinetic energy (and hence, the velocity) of the electrons. Here, you must realize that current is the amount of charge passing through a point in space per second. So as the electrons have more kinetic energy and velocity, more electrons would pass through that point in space per second, resulting in a larger current. In other words, even though the total number of electrons emitted is the same (as indicated by the same number of photons), but this has nothing to do with the current, because the current is defined as charge passing through a point per unit time.

[Hassan76]

Guys what about the potential divider question in p1 , and the projectile, and the equipotential, equiptential was for same masses and same charges, projectile was vertically upward since initial vertical v is not 0, as some times when it is launched horizontally initial v in the vertical component is 0 as in May tz1 hp2, potential divider it had a 0 but don't know at the beginning or at the end, electron was accelerating at a decreasing rate and speed was increasing, and this was related to the question about the average speed in 9 in paper 2. And it was in m13 hp2 tz1, equations of motion, constant net force and fixed mass so that a is constant, graph always accelerating since the slope the graph is never parallel to the x axis but v is 0 at an instant so it is not always moving, first question is 3 as 10^_-27 and 10^ -30 , and it is not unfamiliar, the girl's gain in me is 600 areas of the three triangles, entropy of the universe increasing for the melting ice, for isothermal work is greater than adiabatic and delta u is 0, for the gravitational potential it is - 2GM / R +r since it is a scalar, if the question was asking what is the gravitational field strength which is a vector it would be 0, emissivity different temperature same, greenhouses gases was last one, resolution was violet, circle was frequency constantly varying, wave velocity to the right was A, when the slope is increasing, u tube I think it is d as initial displacement doesn't mean no 0 initial velocity and the liquid's motion is damped but I chose a as it is oscillating so v is positive and negative, for the momentum the last one was not valid, for the electric field strength it was b , closest to - 20 , for the maximum power in one of the resistors it is VI not 2VI, for the artificial transmutation it was easy, c I think as I do not remember the choices, for the kinetic model assumption it is elasticity of collisions, for the equilibrium temperature between water and paraffin it is 60 degrees, for the centripetal acceleration it was very easy, for the higher frequency couloir a higher photo current and a larger (more negative) stopping potential, for the evidence of nuclear energy levels it is the discrete nature of alpha and gamma, for the power of the wave it is p over 2, for the wavelenth of 4 V it was half the de broglie wavelenth, for the mass spectrometer it is half v and twice the mass, for the resistance the ratio was 8:1, as d/2 will multiply the area by 0.25, for the alpha particles deflection it is maximum for n at theta equal 0 and minimum when theta is near 170, for the frequency and the halved le th it is still 500 Hz as th the closed end will give wavelenth = 4 * L/2 , which is still 2 L, for the coil I chose the first one, positive sine for the current, for the direction of the force I chose A,

Concerning the question that asked what is not possible is momentum: momentum is given by h/ lambda, electron box model states that the wavelengths can be 2L/n where n is an integer, so 2l, l, 2/3, L/2, 2l/5' 2L/6 etc......,

For the sl, the question about gravitational field strength it is 20,

What do you guys think, plz share with me

 Potential divider the variation of V with distance p1 and p2 ,

 

Hassan, I don't remember the potential divider question, emmisivity,or the greenhouse gases one. Could you recap it for me once and I'll tell you my answers. 

 

Our answers are almost similiar otherwise. Except for that power of the waves one, I think I got Power = P in that.

In the data analysis question, I got my slope around 1 and the y intercept aro0und 0.97. Anyone else? Also, when you took out the x intercept by equation y=mx + c to zero, did anyone find the value of x weird? Because I got some -0.98 or something and by observing and extrapolating the graph mentally to the minus x axis, it didn't look more than a -0.10. Did anyone get L as 0.93??

 

In paper 1 (Tz2) there was a question about green light being emitted from two point sources which was just NOT resolved. So the question asked how to resolve it and there were two plausible options: use red light (higher wavelength so bigger theta) and use a circular aperture (again bigger theta since 1.22 is there instead of 1). What'd you guys write? I chose red light. 

 

Ossih, what did you write for the possible binary states? 15 or 16? 

 

I had no idea what the "s" in capacitance meant, so I had to skip that question . I mean I knew it was the Farad but I forgot the definition :/ and in all my years of Physics, I have never seen a freaking s as a fundamental unit . Randomnly ticked something and probably got it wrong. 

 

In Paper 2 (tz2), for question 6, everyone found the gravitational field strength as Del V/ Del T right? And therefore from the calculation the rocket cannot escape without further engine use. 

 

Sorry about the blast of questions in your face, I'm just too excited to know everybody's else's answers . 

Also, any other Paper 1 questions anyone found challenging? I keep forgetting :/

 

I put 16 binary states! 

 

Also for the one with finding the induced emf, i started with NBA cos theta but it struck me that it's wrong.

 

The reason is, they ask for the induced emf JUST when the coil is about to completely enter. And so you'd so induced emf = ∆flux/∆t (emf is not NBA cos theta, that's flux ) and you'd find ∆t using suvat equations for free fall. The problem is, that's not the instantaneous change, that's average, that's not specific to when the coil just enters the field.

 

Instead, I did this. I used e = BLv, i said L was the length of one side (cause only one side is moving perpendicular to the field) and v = √2gh because its falling freely, where h is the length of the side. This made sense to me cause thats the instantaneous velocity at the time it just almost completely enters the field! What do you think?

Ossih, what was the answer to the question about frequency in section A ? And please tell me how you got the reading on the voltmeter in q 9

 

I answered the frequency one right now. For the reading on the voltmeter, it was really straightforward. The resistance in the external circuit was the same as the resistance in the internal circuit. The voltmeter measures the voltage in the external circuit, which would be half of total voltage i.e. 6 V.

Why 6, they were in parallel

Frequency in Q4 is supposed to be calculated by using the path difference for constructive interference, I think, i.e. even multiples of half wavelengths, given the wavelength and probably dividing it by the speed to get the frequency values. 

 

But I couldn't figure it out and didn't have time so I simply just did v/lambda which is wrong

Frequency in Q4 is supposed to be calculated by using the path difference for constructive interference, I think, i.e. even multiples of half wavelengths, given the wavelength and probably dividing it by the speed to get the frequency values. 

 

But I couldn't figure it out and didn't have time so I simply just did v/lambda which is wrong

We didn't have information about whether it is the first ,ms econo, third .... Minimum to know n, also, these equations are in the options and in the core for the new syllabus 2016/

 

Hey guys, for this question all I did is the path difference i.e. the difference between consecutive maxima is lambda (0.032m). So the time between two consecutive maxima is lambda/v (where v is the speed of B). So the frequency is v/lamba :$

Was it asking for maxima not minima ?

[Sofia.]

It was asking for minima, I think, but the frequency would be the same. 

[Hassan76]

Guys what about the potential divider question in p1 , and the projectile, and the equipotential, equiptential was for same masses and same charges, projectile was vertically upward since initial vertical v is not 0, as some times when it is launched horizontally initial v in the vertical component is 0 as in May tz1 hp2, potential divider it had a 0 but don't know at the beginning or at the end, electron was accelerating at a decreasing rate and speed was increasing, and this was related to the question about the average speed in 9 in paper 2. And it was in m13 hp2 tz1, equations of motion, constant net force and fixed mass so that a is constant, graph always accelerating since the slope the graph is never parallel to the x axis but v is 0 at an instant so it is not always moving, first question is 3 as 10^_-27 and 10^ -30 , and it is not unfamiliar, the girl's gain in me is 600 areas of the three triangles, entropy of the universe increasing for the melting ice, for isothermal work is greater than adiabatic and delta u is 0, for the gravitational potential it is - 2GM / R +r since it is a scalar, if the question was asking what is the gravitational field strength which is a vector it would be 0, emissivity different temperature same, greenhouses gases was last one, resolution was violet, circle was frequency constantly varying, wave velocity to the right was A, when the slope is increasing, u tube I think it is d as initial displacement doesn't mean no 0 initial velocity and the liquid's motion is damped but I chose a as it is oscillating so v is positive and negative, for the momentum the last one was not valid, for the electric field strength it was b , closest to - 20 , for the maximum power in one of the resistors it is VI not 2VI, for the artificial transmutation it was easy, c I think as I do not remember the choices, for the kinetic model assumption it is elasticity of collisions, for the equilibrium temperature between water and paraffin it is 60 degrees, for the centripetal acceleration it was very easy, for the higher frequency couloir a higher photo current and a larger (more negative) stopping potential, for the evidence of nuclear energy levels it is the discrete nature of alpha and gamma, for the power of the wave it is p over 2, for the wavelenth of 4 V it was half the de broglie wavelenth, for the mass spectrometer it is half v and twice the mass, for the resistance the ratio was 8:1, as d/2 will multiply the area by 0.25, for the alpha particles deflection it is maximum for n at theta equal 0 and minimum when theta is near 170, for the frequency and the halved le th it is still 500 Hz as th the closed end will give wavelenth = 4 * L/2 , which is still 2 L, for the coil I chose the first one, positive sine for the current, for the direction of the force I chose A,

Concerning the question that asked what is not possible is momentum: momentum is given by h/ lambda, electron box model states that the wavelengths can be 2L/n where n is an integer, so 2l, l, 2/3, L/2, 2l/5' 2L/6 etc......,

For the sl, the question about gravitational field strength it is 20,

What do you guys think, plz share with me

 

 

Hassan, I don't remember the potential divider question, emmisivity,or the greenhouse gases one. Could you recap it for me once and I'll tell you my answers. 

 

Our answers are almost similiar otherwise. Except for that power of the waves one, I think I got Power = P in that.

Greenhouse asking why 2 gh gases are greenhouse,me missives the graph of intensity vs wavelenth, potential divider was asking how does V varies with distance, and the emf was 6

It was asking for minima, I think, but the frequency would be the same.

Yes but to get the path difference we must get the value of n, whether it is first, 2nd or third .... Minima

[Hassan76]

again to those who did thermodynamics, there was this last question saying that in adiabatic change work done was i think 210J, what was the change in internal energy.. so since that was adiabatic there was no heat exchange so the change in internal energy was -210J, did i get this one right?

 

Yep, just used the first law of thermodynamics for that one.

Q = 0, Q=dU+W, W=210, -210=U

 

Damn, I really can't read the questions carefully. I somehow skipped the adiabatic part hehe... And here I've been thinking P2 went well. 

I somehow got Q to 1300 something? Can't even remember how I got that!

You used the values for the net work done given above

 

Must have read system instead of that specific process then. Ah well, hopefully I got the other questions right.

You'll do don't worry, what options do you study?

[Hassan76]

@Ossih, although I didn't do the exam, I think I might have spotted 2 mistakes of yours.

 

 

20. For kinetic model, I didn't put elasticity of the collisions because they didn't say the it was perfectly elastic. Instead, I chose that the temperature is proportional to the average kinetic energy of the molecules. According to Wikipedia (not that it's the best source haha), one of the assumptions of the kinetic model is that the absolute temperature is a measure of the average kinetic energy of the particles.

 

I think the assumption should be the elasticity of the collisions (that is in agreement with wikipedia).

Although it's true that temperature is proportional to the average kinetic energy of the molecules, but I don't think that it was an assumption; it was actually a mathematical consequence of the kinetic model. Wikipedia only says that:

• The average kinetic energy of the gas particles depends only on the absolute temperature of the system

This means that kinetic energy has a relationship with the temperature, but this assumption isn't very clear on what type of relationship it is. It's only when you work out the mathematics that you find the temperature to be proportional to the kinetic energy.

 

22. For the photocurrent one, I disagree. The question said that they fired a higher frequency BUT the same number of photons, which means that the number of electrons are ejected and hence the photocurrent is the same. All it is that the stopping potential is now more negative

 

I would agree with Hassan76 for this one, though I may be wrong. It's true that the same number of electrons are ejected due to the same intensity of the light that was used. But increasing the frequency of the photons means that you've increased the kinetic energy of the emitted electrons. This will cause the electrons to move faster (since KE is proportional to the speed). Here, you need to realize that current is the measure of charge per unit time. The increase in speed will increase the number of charge per unit time, thus the current must increase.

So the chain of reasoning is as follow:

 

Higher frequency --> larger KE --> electrons gaining speed --> more photocurrent --> more negative stopping potential

 

And damn, it's crazy to see how you guys remembered all that stuff!!! Good luck with the rest of the exams

 

Thanks Vioh! I think you may be right for the temperature one. But this site http://chemwiki.ucdavis.edu/Physical_Chemistry/Physical_Properties_of_Matter/Phases_of_Matter/Gases/Kinetic_Theory_of_Gases/Kinetic_Theory_of_Gases states both as assumptions of the Kinetic theory so I really don't know. My book only states elastic though, so maybe that's what it is. Oh well, we'll find out soon enough ahha

 

However, for the photocurrent one, I'm sure, because it said that the number of photons per second is the same, (it doesn't say the same intensity of light, that would give a different result) implying that the number of electrons per second is the same (assuming that we're above threshold frequency). Because of that photocurrent must stay the same. 

 

If it were same intensity of light, then I agree, the photocurrent would change

 

I just hope I get a 7. I think it was a pretty think-out-of-the-box paper so a boundary for a 7 will be 32-34ish, so I hope I'll be fine

Guys concerning this question, 1- it is low pressure and high temperature, 2- these are not assumptions they are facts otherwise it will be a real gas