IBsurvivor2015May Posted May 9, 2015 Report Share Posted May 9, 2015 How did it go for the Tz2 paper Physics guys? I was unsure about quite a few in paper 1 but I think paper 2 was easier. Some paper 1 questions were so different from past papers or is it only me?? Reply Link to post Share on other sites More sharing options...
abrio98 Posted May 9, 2015 Report Share Posted May 9, 2015 In p1 there was a lot of guessing :/ There was one question with two magnets in which I thought the force direction would be out of the page, but none of the four options made sense to me. P2 I did question 5 in section B. SHM was a gift and then as much as I hate electric currents, think I did quite alright in the Part 2. However, question 1 really put me off haha Reply Link to post Share on other sites More sharing options...
IBsurvivor2015May Posted May 9, 2015 Author Report Share Posted May 9, 2015 In p1 there was a lot of guessing :/ There was one question with two magnets in which I thought the force direction would be out of the page, but none of the four options made sense to me.P2 I did question 5 in section B. SHM was a gift and then as much as I hate electric currents, think I did quite alright in the Part 2. However, question 1 really put me off haha Yes I remember that question. I just guessed A but I was so confused with the magnetic field, I couldnt figure it out either. oh nice. I did question 5 as well. Yes, I loved the SHM question as well. I thought it was so simple. I was just a bit scared about the last two questions on electric currents. I got the current to be something like 0.4545 and the volt to be 6 V. What did u get for ii for current and the voltage? And how did u do the last power question? How did you do the displacement question in Section A. What did u get for it? I think i got something like 5.6 m. Reply Link to post Share on other sites More sharing options...
Eid Jazairi Posted May 10, 2015 Report Share Posted May 10, 2015 The magnet one was difficult to understand, the north pole is on the inner side of the magnet (there was a small arrow at the bottom) so force was going down and answer was C. For the Electric currents, I think I got the exact same results, V was 6 and Current at 0.4545 and the power question I just multiplied them. Which is odd cause it was worth 3 marks and the answer was too simple!For the displacement I did one of the Suvat formulas. V^2 = U^2 + 2as. To calculate the acceleration I used the friction minus the result in the Mg sin X question and got a force of 37 or something like that, F=ma, found the a. The final Velocity was 0, and they tell us the initial one, so just calculate for s. The only thing that worries me is that I got the force calculations wrong. Hopefully not. Reply Link to post Share on other sites More sharing options...
Eid Jazairi Posted May 10, 2015 Report Share Posted May 10, 2015 Elevator and Force meter.. someone please confirm. One of the people I asked said that since whatever force the downward acceleration causes on the person, it also causes it on the force meter (given that they are both in the elevator together). So the answer should be Mg only. Some people said mg+ma! Can anyone confirm? Also the Albedo question was tricky What about the graph for the resistance? That had V and I as axis? Being a linear graph seemed too easy an answer so I didnt choose it! lol Reply Link to post Share on other sites More sharing options...
IBsurvivor2015May Posted May 10, 2015 Author Report Share Posted May 10, 2015 (edited) The magnet one was difficult to understand, the north pole is on the inner side of the magnet (there was a small arrow at the bottom) so force was going down and answer was C. For the Electric currents, I think I got the exact same results, V was 6 and Current at 0.4545 and the power question I just multiplied them. Which is odd cause it was worth 3 marks and the answer was too simple!For the displacement I did one of the Suvat formulas. V^2 = U^2 + 2as. To calculate the acceleration I used the friction minus the result in the Mg sin X question and got a force of 37 or something like that, F=ma, found the a. The final Velocity was 0, and they tell us the initial one, so just calculate for s. The only thing that worries me is that I got the force calculations wrong. Hopefully not. I wrote Mg + ma as well because I messed up the sign in the exam but it was (Mg - ma). Edited May 10, 2015 by IBsurvivor2015May Reply Link to post Share on other sites More sharing options...
Eid Jazairi Posted May 10, 2015 Report Share Posted May 10, 2015 Not that one, I mean the MC question with graphing of V against I .. Reply Link to post Share on other sites More sharing options...
IBsurvivor2015May Posted May 10, 2015 Author Report Share Posted May 10, 2015 Not that one, I mean the MC question with graphing of V against I .. Oh I picked A. The linear graph but starting above 0. Reply Link to post Share on other sites More sharing options...
abrio98 Posted May 10, 2015 Report Share Posted May 10, 2015 Not that one, I mean the MC question with graphing of V against I .. Not that one, I mean the MC question with graphing of V against I .. Oh I picked A. The linear graph but starting above 0. The question with the graph of V against I showed up in I believe the may 2012 paper and the answer started at 0,0 and went up sort of curved The displacement for section A, I got 5.something. 1 Reply Link to post Share on other sites More sharing options...
Badalyan Posted May 10, 2015 Report Share Posted May 10, 2015 (edited) You know the question when the guy was going down the hill on a bike? When he was breaking, did you get something like 5 meters for him to stop completely? Edited May 10, 2015 by Badalyan 1 Reply Link to post Share on other sites More sharing options...
Hassan76 Posted May 10, 2015 Report Share Posted May 10, 2015 You know the question when the guy was going down the hill on a bike? When he was breaking, did you get something like 5 meters for him to stop completely?You should use F S= mv^2 /2 Reply Link to post Share on other sites More sharing options...
Badalyan Posted May 10, 2015 Report Share Posted May 10, 2015 I was not asking about what you should use tbh:/ I was asking what you got for the answer PS: You were given net breaking force, so you could find acceleration from dividing the force by the mass. Then you could use that acceleration in order to calculate the distance since you know u and v (using v^2 = u^2 + 2as)So not sure about the method you suggested there You know the question when the guy was going down the hill on a bike? When he was breaking, did you get something like 5 meters for him to stop completely?You should use F S= mv^2 /2 1 Reply Link to post Share on other sites More sharing options...
Hassan76 Posted May 10, 2015 Report Share Posted May 10, 2015 Sorry for misunderstanding you Reply Link to post Share on other sites More sharing options...
Eid Jazairi Posted May 10, 2015 Report Share Posted May 10, 2015 I was not asking about what you should use tbh:/ I was asking what you got for the answer PS: You were given net breaking force, so you could find acceleration from dividing the force by the mass. Then you could use that acceleration in order to calculate the distance since you know u and v (using v^2 = u^2 + 2as)So not sure about the method you suggested there You know the question when the guy was going down the hill on a bike? When he was breaking, did you get something like 5 meters for him to stop completely?You should use F S= mv^2 /2 The guy above you said he got 5. I also used the same method (Stated above as well), I just don't remember the exact values, so I can't tell you. Either way don't worry, seems like we got it correct. 1 Reply Link to post Share on other sites More sharing options...
IBsurvivor2015May Posted May 10, 2015 Author Report Share Posted May 10, 2015 You know the question when the guy was going down the hill on a bike? When he was breaking, did you get something like 5 meters for him to stop completely? YES I got that too. I got 5.something as well Reply Link to post Share on other sites More sharing options...
Sanidhya.M Posted May 10, 2015 Report Share Posted May 10, 2015 Why was it mg-ma and also the graph, was it straight or curved like a filament lamp. - for paper 1. Also who else did question 4 and what did you guys get for the halflive problem i got 16.8 but im sure thats wrong. Reply Link to post Share on other sites More sharing options...
Hassan76 Posted May 10, 2015 Report Share Posted May 10, 2015 Why was it mg-ma and also the graph, was it straight or curved like a filament lamp. - for paper 1. Also who else did question 4 and what did you guys get for the halflive problem i got 16.8 but im sure thats wrong.Guys what was the graph for and what was on the y and x axis? Reply Link to post Share on other sites More sharing options...
IBsurvivor2015May Posted May 10, 2015 Author Report Share Posted May 10, 2015 Why was it mg-ma and also the graph, was it straight or curved like a filament lamp. - for paper 1. Also who else did question 4 and what did you guys get for the halflive problem i got 16.8 but im sure thats wrong. Do you remember if the elevator was accelerating downwards or upwards?? Reply Link to post Share on other sites More sharing options...
Sanidhya.M Posted May 10, 2015 Report Share Posted May 10, 2015 @ibsurvivior downwards. My thinking was, that since the elevator is going down, it would add more force due to the accelration which could be shown as MxA and so mg and add the acceleration of elevator going down = mg+ma? tell me what you guys think? Reply Link to post Share on other sites More sharing options...
Sanidhya.M Posted May 10, 2015 Report Share Posted May 10, 2015 Why was it mg-ma and also the graph, was it straight or curved like a filament lamp. - for paper 1. Also who else did question 4 and what did you guys get for the halflive problem i got 16.8 but im sure thats wrong. Do you remember if the elevator was accelerating downwards or upwards?? downwards Reply Link to post Share on other sites More sharing options...
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