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Original post by 123bensufc
didnt they ask for the change in velocity on this question? i got 11/180 as the change in velocity 11Ns for area under graph

That's what I did cos I swear I= mass times change in velocity lol
I got I as 10.8, Deffo will be a +-2 range I guess
Reply 1641
For the resonance graph what did people draw? Isn't the resonance graph Amp vs Freq but it said that Amp was constant????
Original post by Elcor
For the last part of the first question I ended up with

d=2u2gMsinθcosθd=\frac{2u^2}{g_M} \sin \theta \cos \theta


I got

it to be -sin2theta/a *

Where a would have been a negative value as I took up to be positive

My double angle formula ye
Actually wanna dissapear. Bye bye medicine


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Original post by Elcor
For the last part of the first question I ended up with

d=2u2gMsinθcosθd=\frac{2u^2}{g_M} \sin \theta \cos \theta


I had this too. I was a little confused when I had the product of sin and cos, I didn't see how that would be constant. I just wrote a little note and said "product of sin and cos is constant here"

Did they say to assume the angle of trajectory was constant?
Original post by MMB07
For the resonance graph what did people draw? Isn't the resonance graph Amp vs Freq but it said that Amp was constant????


It meant the amplitude of the oscillator was constant
A* will be lower that 43 guaranteed.
Original post by L'Evil Fish
I got

it to be -sin2theta/a *

Where a would have been a negative value as I took up to be positive

My double angle formula ye


Same as me then, but I got it as positive because I took into account directions in my suvat.

Mate don't confuse the examiner with double angles, pretty sure the most complicated equation they know is speed=distance/time
Original post by MMB07
For the resonance graph what did people draw? Isn't the resonance graph Amp vs Freq but it said that Amp was constant????


It said the amplitude of the oscillator oscillating it was constant, not the spring/plate thing itself.
Original post by Elcor
For the last part of the first question I ended up with

d=2u2gMsinθcosθd=\frac{2u^2}{g_M} \sin \theta \cos \theta


Yeah I got this, so s was proportional to u^2 with the other stuff as constant
Original post by ibanezmatt13
I had this too. I was a little confused when I had the product of sin and cos, I didn't see how that would be constant. I just wrote a little note and said "product of sin and cos is constant here"

Did they say to assume the angle of trajectory was constant?


I didn't say that, I just assumed it was obvious and concluded with 'Hence du2d \propto u^2'

You can't have the golf ball being hit, then its initial angle of trajectory changing mid-flight lol. Good of you to state it though, hopefully I haven't lost a mark there.
Reply 1651
Original post by ellisvlad
I put that Newton's law of gravitation states that any two point masses attract each other with a force directly proportional to the product of their masses and inversely proportional to the square of their distance. I'm sure that'll be fine for one mark... right?


no idea
i'm sure ocr is gonna find a way to mess us up even more with their crap
Original post by Flux_Dubstep
Yeah I got this, so s was proportional to u^2 with the other stuff as constant


At first I wrote my suvat variables as g=9.81ms^-2, then I was like hold on a minute...
Original post by maattwileman
A* will be lower that 43 guaranteed.


That seems slightly too low.
Unofficial mark scheme?
Original post by Elcor
I didn't say that, I just assumed it was obvious and concluded with 'Hence du2d \propto u^2'

You can't have the golf ball being hit, then its initial angle of trajectory changing mid-flight lol. Good of you to state it though, hopefully I haven't lost a mark there.


Yeah hopefully :smile:

Theta varies throughout motion though unless they say to assume it doesn't. Need an unofficial ms
Original post by Elcor
At first I wrote my suvat variables as g=9.81ms^-2, then I was like hold on a minute...


Yeah I reckon a few people will have done that
Has anyone got the paper itself?
Original post by ibanezmatt13
Yeah hopefully :smile:

Theta varies throughout motion though unless they say to assume it doesn't. Need an unofficial ms



Theta was the angle at launch
Original post by SH0405
That seems slightly too low.


that is what it was last year

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