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OCR Physics A Breadth of Physics (24th May 2016) Discussion

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I mentioned about the area being pi r^2 but then i crossed it outtttt, and i spoke about weighing the cylinder on a balance and multiplying by 9.81 for the force

Couldnt you you use a newton meter? To measure the weight of the cylinder

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Reply 261

yelash

How are you supposed to get the density of the cylinder? I got really confused, not knowing which mass to use where, and not knowing the volume

Reply 262

srayogn

Fell free to contribute to the unofficial markscheme.
https://docs.google.com/document/d/18mLVuo38imOSCLm3O-Wl6nR8m201CIBH0e7wjPsvdb0/edit?usp=sharing

Reply 263

sddsfsdafs

I did exactly the same as you and got 6.75, and had no idea why it didnt match any of the answers, so ended up putting 12mW fml

Reply 264

julesaquilina

Original post by Fish40

Couldnt you you use a newton meter? To measure the weight of the cylinder

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Possibly, i also spoke about the area of the table but i think i shouldnt have

Reply 265

yelash

Original post by Fish40

Couldnt you you use a newton meter? To measure the weight of the cylinder

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I considered it, but thought it was a bit cheeky considering the diagram in the next question :P

Reply 266

sddsfsdafs

How was it the momentum time graph? Fairly sure it was the accerleration time graph, since only acceleration could be zero out of the four if the ball was moving.

Reply 267

viktorh

Original post by yelash

How are you supposed to get the density of the cylinder? I got really confused, not knowing which mass to use where, and not knowing the volume

I used the weight of the cylinder/9.81=mass of the cyclinder

then I took the difference between the two newton meter readings = the upthrust on the cylider = Apgh

deduced h from that. Used H and A to find the volume

Mass/Volume = density.

Anyone else do it like this? answer I got was around 750

Reply 268

Oooskar

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yeh

Original post by sddsfsdafs

How was it the momentum time graph? Fairly sure it was the accerleration time graph, since only acceleration could be zero out of the four if the ball was moving.[/QUOTIf it was momentum the momentum would not decrease at all so it must be acceperation as F=ma so they are identicall and directly proportional

Reply 271

TH3-FL45H

Is anyone going to do an unofficial mark scheme?

Also what do you think the grade boundaries will be?

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Reply 272

mahmzo

Original post by viktorh

I did the first 2 steps what you said and half of the 3rd step, do you think I can get a mark or 2?

Reply 273

srayogn

https://docs.google.com/document/d/18mLVuo38imOSCLm3O-Wl6nR8m201CIBH0e7wjPsvdb0/edit?usp=sharing

APOLIGIES Can edit the unoffical mark scheme now.

Reply 274

mxhxs_101

On the whole, how did everyone find it?

Reply 275

yelash

Original post by viktorh

ah yeah makes sense. Exam stress got to me :frown:

Reply 276

jayeshforce1

i made a error whilst working out the velocity of the atom. Like i forgot to put negative velocity because it changed direction. I believe it was 4 marks how many would i loose considering everything else was consistent?

Reply 277

Rectum_resizer

Pretty sure it was momentum time graph because the gradient was Ft, which is change in momentum(impulse), so a momentum time graph would have the same gradient.