G481- mechanics 19th May OCR physics A

Last years was the hardest I've seen, this years will be harder.

Wouldn't that suggest we will have a hard question on another topic, as people may be anticipating a similar one.

Do you guys think that the paper will be easier than last year's? I feel last years paper was more difficult than previous years ones, as there there we more confusing questions that could have easily lost marks. Normally G481 papers are relatively easy but that was just odd.

It will be harder? But wasnt it that not as many people did well last year because it was too difficult?

I thought the trend was that if one year the paper is difficult the next years is slightly easier?

That's it. When students do well on a certain topic or type of question they replace them with something we'd be less familiar with. It has happened in maths too; those doing edexcel Mechanics 1 in AL maths may have noticed that conservation of momentum questions have been discontinued since 2013. This might be standard practice throughout all boards and subjects.

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Thats true, so for that reason it must mean that they are going to do some other silly maths question, and i don't know why, but I feel it will have more writing to do. like even last year they didnt have many definitions, so hopefully they will test us on them.
Who knows, maybe science exams will go light this year.

How do you do inclined plane questions? I am wondering if anyone has any methods to answer them or any resources that help

Cheers

mgSin(theta) for finding the component of the weight parallel to the plane. mgcos(theta) for perpendicular to plane, or the normal.

Hi, could you help me with a question? I don't know why I can't do it, maybe I am thinking too much into it.

(http://www.st-ambrosecollege.org.uk/library/client/documents/Science/AS%20Physics_/G481%201.3.2%20%20Kinetic%20and%20Potential%20Energy.pdf)

It is question 3 on page 5 of the document.

To calculate the speed at B I did v=(2*g*30)^0.5

I am confused as to how to do the second part, I calculated the speed at O as v=(2*g*50)^0.5. I don't know what to do now!

Thanks for any help

Heyho. I have no idea if it is correct, but I thought that you were supposed to use suvat to find the horizontal displacement. Do you know what the answer is?

If you haven't found the solution yet I have added what I thought was supposed to have be done.

Heyho. I have no idea if it is correct, but I thought that you were supposed to use suvat to find the horizontal displacement. Do you know what the answer is?

If you haven't found the solution yet I have added what I thought was supposed to have been done.

Wouldn't that be the vertical distance travelled, which makes sense because the track is frictionless so in theory transfer of energy should be 100% efficient. The only thing I can think to do is to use trigonometric ratios to work out s. tan(30)=50/s.

i got a distance of 87.2m anyone have the mark scheme? speed at o 31.32ms-1

Could you explain how you got that answer? I was unsure of how to proceed with this question too.

sure but i must say i am known for my wacky ideas, so at A skier has max GPE, at B some gpe has been converted into Ke, i used the equation mgh=0.5xmxv^2 to work out the speed.
for part 2 i thought at O all the gpe must be now ke so i found the gpe at A and made that the ke rearranged the equation to get the initial resultant velocity.
the i have a standard projectile motion scenario. with 31ms-1 as the resultant velocity, worked out then the total air time via suvat then used d=sxt to work out the distance

When you worked out the time, did you do the time for a "full parabola" or half a parabola. I would use suvat with u=31ms-1 and v=0 as this would be the maximum distance travelled before the skier would come back down the slope again.

yeah i multiplied the time by 2 i first work out half the parabola, anyone have the correct answer