!! Physics HL LC '16 - Predictions, guesses, discussion, etc.

bookworm20

How may neutrons are neccesary for a safe self sustaining fissin reaction?

adam240610

Caoimh1997 wrote: »

I think coherent light sources can be got by shining a monochromatic light source through young's slits, as it's a diffraction grating with two slits in it. Could be wrong but that's what I put

That is correct.

alaandocs

DownOneTourist wrote: »

I got lead for an isotope somewhere in the paper

Ya for q.12(a)

iFergal

bookworm20 wrote: »

How did you find max and min tension in q12c?

A string will be max and min tension at the top and bottom of the swing.

The tension will act towards the center of the swing and the weight (mg) vertically downwards. So at the top, F = T + mg and the bottom F = T - mg (as they are opposite directions).

Find centripetal force using F = mv^2 / r and equate.

So for maximum tension (at bottom): T - mg = (mv^2 / r). Therefore, T = (mv^2 / r) + mg and just fill in the values (in their correct units!).

Same idea for minimum tension at top.

Imaginary Friend

Caoimh1997 wrote: »

Does anyone remember what they got for the velocity of the daughter nucleus roughly? For Q12a)

I thought it was the velocity of the alpha particle? I got something by ten to the power of 7 m/s.

adam240610

Caoimh1997 wrote: »

I was running out of time so I said the velocity of the daughter nucleus was the same as the velocity of the alpha particle from the first part, to conserve momentum but I'm probably wrong...

Mass was different, had to do Ke of alpha = 1/2 mv^2 where m was the mass of daughter nucleus

Seaaan

adam240610 wrote: »

I can barely remember that question but O want to say it was in the region of 50000 m/s? No idea

I got 5.875x10^7

DownOneTourist

Seaaan wrote: »

I got 5.875x10^7

yeah that's what i got also.

Seaaan

DownOneTourist wrote: »

yeah that's what i got also.

Oh wow good, I had to use unified atomic mass and stuff was well out of the way for about 6 marks I thought..

For the coherent source one I put that the slits made two coherent sources out of light beam but I then put brackets around that and wrote lasers as they're coherent..

Is physics marked like biology where they cancel out right answers with wrong answers?

adam240610

Seaaan wrote: »

I got 5.875x10^7

Yeah that's what I got, knew it started with a five but couldn't remember the power haha

Also they accept the right answer, even if you cross it out and write the wrong method.
I abuse this too often for maths haha

Imaginary Friend

Just realised you had to get the velocity of the daughter nucleus as well...Well that was an absolute write-off. Can't believe I studied so hard and kept a 100% average for most of 5th Year, not to mention it being my favourite subject, for it to go like this on the day

Seaaan

adam240610 wrote: »

Yeah that's what I got, knew it started with a five but couldn't remember the power haha

Also they accept the right answer, even if you cross it out and write the wrong method.
I abuse this too often for maths haha

Oh good :P


Anyone else do question 10? Wasn't sure if I did it right; got .5A for current in coil, 2.4A for current in 50 Ohm resistor.

Also got 22.5V for EMF and .1125A for current flowing.. May have just embarassed myself with awfully wrong answers just now, but did it as an extra question

Danny_Boi

Seaaan wrote: »

Oh good :P


Anyone else do question 10? Wasn't sure if I did it right; got .5A for current in coil, 2.4A for current in 50 Ohm resistor.

Also got 22.5V for EMF and .1125A for current flowing.. May have just embarassed myself with awfully wrong answers just now, but did it as an extra question

What method did you use for these?I can't remember my answers but I used E=dmagnetic flux/dt and the rms formula.

Also how did people calculate the energy released in Q9?

Caoimh1997

Danny_Boi wrote: »

What method did you use for these?I can't remember my answers but I used E=dmagnetic flux/dt and the rms formula.

Also how did people calculate the energy released in Q9?

Calculate the mass before and after, subtract to get the difference between masses. Then use E=mc^2 to get energy I think

Danny_Boi

Caoimh1997 wrote: »

Calculate the mass before and after, subtract to get the difference between masses. Then use E=mc^2 to get energy I think

Thanks, that's what I did but the answer seemed really small.9 was a lovely question and the semi-conductor question was beautiful.

alaandocs

Seaaan wrote: »

Oh good :P


Anyone else do question 10? Wasn't sure if I did it right; got .5A for current in coil, 2.4A for current in 50 Ohm resistor.

Also got 22.5V for EMF and .1125A for current flowing.. May have just embarassed myself with awfully wrong answers just now, but did it as an extra question

I got .12 amps for the 50 ohm coil as that was in parallel with another 200one. Using the derived formula above 1-200=1/50+1/200 which means it's in the ratio of 1:4 so there it's fifths . I'm remembering this off the top of my head but I think I got .6 amps for the first part so I did .2 x .6 =.12 amps

If i did it wrong please correct me.

Robbiert

Seaaan wrote: »

adam240610 wrote: »

Yeah that's what I got, knew it started with a five but couldn't remember the power haha

Also they accept the right answer, even if you cross it out and write the wrong method.
I abuse this too often for maths haha

Oh good :P


Anyone else do question 10? Wasn't sure if I did it right; got .5A for current in coil, 2.4A for current in 50 Ohm resistor.

Also got 22.5V for EMF and .1125A for current flowing.. May have just embarassed myself with awfully wrong answers just now, but did it as an extra question

I got the 22.5 emf
But thats just the back emf from the coil, so as the average voltage in the circuit is 120v from the power supply the actual voltage of the circuit is 97.5v. The total ohmic resistance of the circuit is 240ohms so I = 97.5/240 = 0.40625a
Could be horribly wrong but My physics teacher on the way out said it was the right method.

Seaaan

Danny_Boi wrote: »

What method did you use for these?I can't remember my answers but I used E=dmagnetic flux/dt and the rms formula.

Also how did people calculate the energy released in Q9?

For the first ones I got resistance in parallel then resistance in series, then v=ir to get current using appropriate resistance values.

Second one I did E= number of coils multiplied by flux/dt and last part I think I just used V=IR again..

Electricity never was my strong point..

Robbiert wrote: »

I got the 22.5 emf
But thats just the back emf from the coil, so as the average voltage in the circuit is 120v from the power supply the actual voltage of the circuit is 97.5v. The total ohmic resistance of the circuit is 240ohms so I = 97.5/240 = 0.40625a
Could be horribly wrong but My physics teacher on the way out said it was the right method.

You're very probably right; not depending on question for marks anyway really only did it because I had the time to.. Magnets and stuff, nah.

adam240610

alaandocs wrote: »

I got .12 amps for the 50 ohm coil as that was in parallel with another 200one. Using the derived formula above 1-200=1/50+1/200 which means it's in the ratio of 1:4 so there it's fifths . I'm remembering this off the top of my head but I think I got .6 amps for the first part so I did .2 x .6 =.12 amps

If i did it wrong please correct me.

I had 40 ohms across the series resistors, sa that was I =120/40 = 3. That was split into a ratio of 50/250 which ended up as .6 amps again, same as the 200 ohm resistor, I think.

Danny_Boi

Seaaan wrote: »

For the first ones I got resistance in parallel then resistance in series, then v=ir to get current using appropriate resistance values.

Second one I did E= number of coils multiplied by flux/dt and last part I think I just used V=IR again..

Electricity never was my strong point..

Ah feck I forgot the number of turns on the coil and I'd say I'm completely wrong on the last part.

Robbiert

Seaaan wrote: »

Robbiert wrote: »

I got the 22.5 emf
But thats just the back emf from the coil, so as the average voltage in the circuit is 120v from the power supply the actual voltage of the circuit is 97.5v. The total ohmic resistance of the circuit is 240ohms so I = 97.5/240 = 0.40625a
Could be horribly wrong but My physics teacher on the way out said it was the right method.

You're very probably right; not depending on question for marks anyway really only did it because I had the time to.. Magnets and stuff, nah.

Just realised when doing that out that I made a mathematical slip during that q..., Yeah I did 2 extra questions so hopefully that will counteract a few slips here and there.

adam240610

Robbiert wrote: »

Just realised when doing that out that I made a mathematical slip during that q..., Yeah I did 2 extra questions so hopefully that will counteract a few slips here and there.

I got a bit worried there then I realised I have 3 extra questions done hahaha, think you'll be fine.

Robbiert

Anyone do question 6? what did ye get for the restoring force? thought it was fairly hard if you hadn't done applied maths

adam240610

Robbiert wrote: »

Anyone do question 6? what did ye get for the restoring force? thought it was fairly hard if you hadn't done applied maths

4.125N or thereabouts, wasn't too keen on the question myself

Robbiert

adam240610 wrote: »

4.125N or thereabouts, wasn't too keen on the question myself

Since the distance of the oscillation was 18cm, the amplitude was 0.045m. T=2pi/w =2 so w=pi. Used the applied maths formula Amax=w^2A to get the max acceleration and F=ma to get the maximum force. I got 0.1586N. My method could be flawed though.

adam240610

Robbiert wrote: »

Since the distance of the oscillation was 18cm, the amplitude was 0.045m. T=2pi/w =2 so w=pi. Used the applied maths formula Amax=w^2A to get the max acceleration and F=ma to get the maximum force. I got 0.1586N. My method could be flawed though.

I used the perpendicular distance from equilibrium to get the angle between furthest displacement and equilibrium strings.

I drew the weight of the Bob going vertically down, and took the weight acting along the same line as the string to be the restoring force.

Sane here

How Hugh up was the pendulum for the last part?

Robbiert

adam240610 wrote: »

Robbiert wrote: »

Since the distance of the oscillation was 18cm, the amplitude was 0.045m. T=2pi/w =2 so w=pi. Used the applied maths formula Amax=w^2A to get the max acceleration and F=ma to get the maximum force. I got 0.1586N. My method could be flawed though.

I used the perpendicular distance from equilibrium to get the angle between furthest displacement and equilibrium strings.

I drew the weight of the Bob going vertically down, and took the weight acting along the same line as the string to be the restoring force.

Sane here

How Hugh up was the pendulum for the last part?

127.42km

TSMGUY

Just put used a sodium vapour lamp for Young's slits lol. Would that get 4 marks at least? Imagine they didn't have hi-tech lasers back in his day:D

For the velocity of the lead nucleus, it was 1/52.5ish multiplied by the velocity of the daughter alpha particle.

You basically find the ratio of masses between lead and helium (51.5:1), and then because kinetic energy is 1/2mv2 and momentum has to be conserved, the alpha particle. which is 51.5 times lighter, has 51.5 times more velocity.

adam240610

Robbiert wrote: »

127.42km

Thank god, same answer here

legocrazy505

Wasn't a bad paper. Had accounting afterwards which was worse. Would be happy enough with some sort of B in physics after that and in Accounting I'm not sure what I got....