*MEGATHREAD* - BMAT for 2016 Entry Discussion

Original post by iamlegend96

you're right 11 is D! and 19 is not B, but D.

Care to explain either?

So with Q11, in diagram 1 nothing is moving - the person is just sitting on the chair and not moving downwards in the direction indicated by F. In Diagram 2, the wheelbarrow is moving upwards in the direction of F and in 3, the weight is going upwards in the direction of F. Therefore, work is being done by force F only in diagrams 1 and 2 - just see it as “is stuff moving”? Now for the second part although the equation for work done is force * distance, in diagram 2, the distance that they’ve labelled is NOT the distance that you’d need for the wheel barrow to be lifted upwards; the distance that you’d use would be a vertical distance (as it’s moving upwards, not sideways). In diagram 3, the distance that they’ve labelled on the diagram is correct as the weight is travelling vertically, so the answer is D.

Give me 5 mins for Q19 :P

Reply 1583

michellem3

6

2014 paper section 2 question 19. Can anyone tell me why the wavelength is 1.5? There must be an easier way to work it out than I am doing :/

Reply 1584

suzywithaZY

18

Original post by ToLiveInADream

Did ANYBODY do section 2 2006? Omg my worse yet by FAR! (Do it in timed conditions)

Yeah, I did XD it actually didn't go so badly for me, I got 18/27. Section 2, 2013 though... That was awful :lol:

Reply 1585

jessica ahmed

4

Original post by tehtarik

In one half-life, the no.of Nuclide X atoms halves.

8 years later - there would be 1/4 left - there would be 16+8 = 24 less atoms of Nuclide X, but 24 more atoms of Nuclide Y.

(the unwritten *10^20 is there)

24+4 = 28

Thank you!

Reply 1586

Original post by thechemistress

Yeah, I did XD it actually didn't go so badly for me, I got 18/27. Section 2, 2013 though... That was awful :lol:

2013 section 2 is the hardest paper!
I'm saving that for tuesday night!

Reply 1587

Original post by iamlegend96

you're right 11 is D! and 19 is not B, but D.

Care to explain either?

For Q19 we’re looking at current and I think of it as people travelling around a route. When lamp X blows, that part of the circuit is no longer in use so all the people travelling down that route would now have to travel through the bit of the circuit with Ammeter 2. When you have, say, 5 people travelling down a corridor, travelling is fine, but when you have double the number all trying to push down the same route which hasn't gotten any larger to accommodate the increase, travelling becomes a lot more difficult - this is basically resistance, so the resistance increases. The reading at Ammeter 2 increases for the same reason - you have way more people travelling down the same route (so more current). The reading at Ammeter 1 decreases as current is inversely proportional to resistance - as resistance has increased, current must decrease. I hope that makes sense, physics isn’t my strong point :s-smilie:

Reply 1588

Screen Shot 2015-10-31 at 17.16.41.png48.8KB

does anyone get how to do this question:

Reply 1589

vittoria198

Original post by michellem3

2014 paper section 2 question 19. Can anyone tell me why the wavelength is 1.5? There must be an easier way to work it out than I am doing :/

It might not be what you are looking for but I'll try. I remember an exercice with a wavelenght of 1.5 and I recently did the 2014 papers so it might be the same problem I had! In the corrected answer, I saw they were 1.5 as the value of the whavelenght. I struggled for ages trying to figure out why, when I saw that they had written it in the exercice as an assumed fact! Is this what you were looking for?
By the way, I felt really dumb as I had used 1.5 in my own calculations

Reply 1590

Original post by ilovecake123

does anyone get how to do this question:

Is the answer B?

Reply 1591

Original post by cyberchibii

Is the answer B?

yes please could you explain it

Reply 1592

anyone know how to answer this?

In an experiment concerning radioactive decay, the count rate of radiation 5 cm from asource X was measured as 140 counts per minute. 12 minutes later, with the detector in thesame position, the count rate was measured as 35 counts per minute.Background radiation was recorded as 20 counts per minute.Calculate the half-life of source X. (Give your answer in minutes.)

Reply 1593

Original post by ilovecake123

yes please could you explain it

Sure - I don't think it's actually on the current spec and I only remembered it from when I did M1 but just in case:

Force * distance in the clockwise direction around the pivot = Force * distance in the anticlockwise direction around the pivot. When I say anti/clockwise - think of it as a seesaw; the weights on the right hand side of the pivot would cause the beam to tip to the right (in a clockwise direction) if there were nothing balancing it out and the weight on the left side of the pivot would cause the beam to tip to the left (in an anticlockwise direction) if there were nothing balancing it out. So with this in mind, multiply 500N by 20cm, which gives 10,000. This is your total anticlockwise value. Then add up your clockwise values; multiply 200N by 40cm, which gives 8000 and then multiply 200N by x, so you get 200x. Anticlockwise must equal clockwise for the beam to be balanced! So 10,000 = 8000 + 200x and solve for x, which is 10 :smile:

Reply 1594

michellem3

6

Original post by vittoria198

It might not be what you are looking for but I'll try. I remember an exercice with a wavelenght of 1.5 and I recently did the 2014 papers so it might be the same problem I had! In the corrected answer, I saw they were 1.5 as the value of the whavelenght. I struggled for ages trying to figure out why, when I saw that they had written it in the exercice as an assumed fact! Is this what you were looking for?
By the way, I felt really dumb as I had used 1.5 in my own calculations

omg thank you!!! feel a bit stupid now but has been annoying me for so long. the good thing is I have learnt a lot about wavelengths by trying to figure out the answer lol thanks again :smile:

Reply 1595

GrandMasti