OCR Physics A G484 Jan 2012 - The Newtonian World

It's not wrong, because 'a' was negative (as it was a deceleration) so F would be negative too.

But you can't really have a negative force acting on it surely?

I put the positive value. I also got a positive value for deceleration... if it had asked for acceleration I would have put the minus value but if you've already specified it's deceleration surely you don't need the minus sign.

Grrrrr I found the paper quite difficult compared to past papers ;/ I think I did okay though, I really hope it's gone alright :frown:

Reply 781

Shack93

6

Original post by _Jasmine

Aww, shucks! You charmer! :colondollar:

lol

Reply 782

godkid308

Original post by Brutal Chav

I got 4.32W as well :biggrin:

I reckon the frequency was 2.4kHz, not 2.5

did you get a monstrous acceleration for the previous question? I got 4.09x10^5 which seems too huge

4.32W? I got half that, because i thought for one oscillation, it goes to the amplitude twice, so energy is used to put it to the amplitude, but while it is at the amplitude and returns to the equilibrium, no energy is used for that. So dW=Fdx = 0.25 (amplitude x 2), then for one second, I did the time period which is 1/f, and the dW/T.
...so not sure if it used energy going back down to the equilibrium point. What do you think?

Reply 783

Shack93

6

Original post by Toshiya

But you can't really have a negative force acting on it surely?

I put the positive value. I also got a positive value for deceleration... if it had asked for acceleration I would have put the minus value but if you've already specified it's deceleration surely you don't need the minus sign.

Grrrrr I found the paper quite difficult compared to past papers ;/ I think I did okay though, I really hope it's gone alright :frown:

Yeah but the next bit you would use acceleration to calculate force which was the negative version of the deceleration

Reply 784

Protosfear

Original post by Toshiya

But you can't really have a negative force acting on it surely?

I put the positive value. I also got a positive value for deceleration... if it had asked for acceleration I would have put the minus value but if you've already specified it's deceleration surely you don't need the minus sign.

Grrrrr I found the paper quite difficult compared to past papers ;/ I think I did okay though, I really hope it's gone alright :frown:

Well the force acting on it was in the opposite direction to motion, so the force was negative. However I think this is one of those questions where if you put + or - you'd still get the marks.

Reply 785

teachercol

9

Just been through the paper with my classes - I'll post the answers later this evening.

They were pretty pleased with this paper - it was very similar to last June's . I'd expect similar grade boundaries.

They didnt change anything for the extra 15mins which I'm pleased about.
Very few describe questions - no experiments. Nothing on impulse or area under hraphs and counting squares :wink:

Reply 786

_Jasmine

5

Original post by Toshiya

But you can't really have a negative force acting on it surely?

I put the positive value. I also got a positive value for deceleration... if it had asked for acceleration I would have put the minus value but if you've already specified it's deceleration surely you don't need the minus sign.

Grrrrr I found the paper quite difficult compared to past papers ;/ I think I did okay though, I really hope it's gone alright :frown:

Force is a vector quantity so you can have a negative force :smile:

Hmm, well F=ma uses acceleration and if you have deceleration, your acceleration is negative...

REGARDLESS, I am positive that a negative or positive value would get the marks. On previous mark schemes they always accept both :biggrin:

Reply 787

Shack93

6

Original post by teachercol

Just been through the paper with my classes - I'll post the answers later this evening.

They were pretty pleased with this paper - it was very similar to last June's . I'd expect similar grade boundaries.

They didnt change anything for the extra 15mins which I'm pleased about.
Very few describe questions - no experiments. Nothing on impulse or area under hraphs and counting squares :wink:

Awesome thanks :smile:

Reply 788

Toshiya

1

Original post by Protosfear

Well the force acting on it was in the opposite direction to motion, so the force was negative. However I think this is one of those questions where if you put + or - you'd still get the marks.

I hope so. Really hoping I don't just lose a load of marks due to stupid mistakes, really don't want to have to retake this paper.

Reply 789

Stevo F

18

Original post by sweetascandy

I thought it went quite well until I looked on here. :ninja:

How did you find it? :smile:

Pretty good, didnt get the power on the damping question though, knew it was W=Fxdistance but didnt get to the right answer, maybe 1 mark for that. Other than that went well i thought :smile:

Also just realised I misread the why is momentum a VECTOR quantity question :'( oh well, cant be helped

(edited 12 years ago)

Reply 790

OllieReynolds

Original post by Shack93

I got all at zero mm

No, i thought 2 were on zero, but the max acceleration is at max displacement... because acceleration is proportional to displacement... so acceleration is zero at the equilibrium point... so the other point was at a max displacement point

Reply 791

_Jasmine

5

Can someone pleaaaaaaaaaaaaaaaase tell me what the answer to the very last question is.
It's killing me :frown:

Reply 792

SharpSchool

i think i got 10.8 watts as the power for one of the question? just by using P= E/T

And on the acceleration in SHM question, could you use the formula in the book and replace x with the Amplitude?

(edited 12 years ago)

Reply 793

Toshiya

1

Original post by _Jasmine

Force is a vector quantity so you can have a negative force :smile:

Hmm, well F=ma uses acceleration and if you have deceleration, your acceleration is negative...

REGARDLESS, I am positive that a negative or positive value would get the marks. On previous mark schemes they always accept both :biggrin:

Yeah I get that accel would be negative. Like, if they'd asking for the value of the acceleration I would have put negative :tongue:

I hope you're right though, I can't really remember the markschemes for other papers!

Reply 794

Megatron94

I liked this exam as teachercol has said nothing on experiments or impulse came up lol, but all in all I liked it, some of my class mates were ok with it to

Reply 795

_Jasmine

5

Original post by OllieReynolds

No, i thought 2 were on zero, but the max acceleration is at max displacement... because acceleration is proportional to displacement... so acceleration is zero at the equilibrium point... so the other point was at a max displacement point

I'm pretty sure it said zero acceleration, not maximum acceleration.

Reply 796

Protosfear

Original post by _Jasmine

I'm pretty sure it said zero acceleration, not maximum acceleration.

That's what I thought, it wouldn't be the first time I've completely mis-read a question though.

Reply 797

Shack93

6

Original post by OllieReynolds

No, i thought 2 were on zero, but the max acceleration is at max displacement... because acceleration is proportional to displacement... so acceleration is zero at the equilibrium point... so the other point was at a max displacement point

Didn't it ask for Min acceleration, Min PE and max velocity?

Reply 798

Megatron94

Original post by _Jasmine

I'm pretty sure it said zero acceleration, not maximum acceleration.

It was zero acceleration and minimum P.E

Reply 799

Stevo F

18

Original post by OllieReynolds

No, i thought 2 were on zero, but the max acceleration is at max displacement... because acceleration is proportional to displacement... so acceleration is zero at the equilibrium point... so the other point was at a max displacement point

no they were all at 0, it asked for MIN acceleration not MAX, sorry

OCR Physics A G484 Jan 2012 - The Newtonian World

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