OCR Physics A G484 - The Newtonian World - 11th June 2015 Watch

Tazmain
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#781
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Also, has anyone got notes with markscheme answers from past papers?
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BrokenS0ulz
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#782
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This isn't from OCR so maybe it's not relevant to our spec, but just in case could someone explain this?
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BrokenS0ulz
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(Original post by Tazmain)
Do you guys use any other past papers other than the standard ones from our spec?
I've run out
Try AQA and Edexcel's papers, they're ok, format is weird in comparison to OCR though.
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sagar448
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(Original post by BrokenS0ulz)
This isn't from OCR so maybe it's not relevant to our spec, but just in case could someone explain this?
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For a pendulum T = 2pi x sqrt(L/g) where L is the length of the string
For a mass spring T = 2pi x sqrt(m/k) where m is the mass and k is the spring constant

Since in mass-spring system the mass doesn't give a **** about the gravitational pull it's period isn't affected. While the period of the pendulum it includes "g" in it's equation and since "g" is less and T is proportional to 1/g. Period increases.
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Tazmain
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(Original post by sagar448)
For a pendulum T = 2pi x sqrt(L/g) where L is the length of the string
For a mass spring T = 2pi x sqrt(m/k) where m is the mass and k is the spring constant

Since in mass-spring system the mass doesn't give a **** about the gravitational pull it's period isn't affected. While the period of the pendulum it includes "g" in it's equation and since "g" is less and T is proportional to 1/g. Period increases.
I get the theory, but are those equations in our spec? I've never learnt/seen them before. :confused:
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shorty.loves.angels
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(Original post by sagar448)
For a pendulum T = 2pi x sqrt(L/g) where L is the length of the string
For a mass spring T = 2pi x sqrt(m/k) where m is the mass and k is the spring constant

Since in mass-spring system the mass doesn't give a **** about the gravitational pull it's period isn't affected. While the period of the pendulum it includes "g" in it's equation and since "g" is less and T is proportional to 1/g. Period increases.
Nicely put :cookie:
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L'Evil Fish
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(Original post by Tazmain)
I get the theory, but are those equations in our spec? I've never learnt/seen them before. :confused:
Not in our spec

However

F = ma
kx = ma
kx = m w² x
k = m w²
(2pi/T)² = k/m

Etc

And for simple pendulum, you just swap k with g, and m for L, I thini
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Tiwa
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Could someone please give me a list of the "describe an experiment" topics we need to learn? I would really appreciate it.
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sagar448
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(Original post by shorty.loves.angels)
Nicely put :cookie:
Thanks
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L'Evil Fish
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(Original post by sagar448)
Thanks
F = ma
Tn-mgcostheta = mw²x

Tn - mg cos theta = m (2pi/T)² L sin theta

T² = (2pi)² (L sin theta)/(Tn-mgcostheta)

Not working how do you get simple pendulum formula?

Dw, I Googled, it's an approximations

And I was resolving to circle
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Raizel
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Impulse = FT = m(v-u)
so why is the definition on this source "Product of force acting on a body and the time its acting;equal to the rate of change of momentum"
the problem im having is the "equal to the rate of change of momentum".
https://b1a3c7824d6ff508e2bd2668b57b...%20A-Level.pdf
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BrokenS0ulz
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I get the contact force is least at A because weight is acting in same direction as centripetal force, and I get that it's the most at C because weight is acting in the opposite direction.
But at B, isn't the contact force also 0, my diagram could just be wrong?
EDIT what is the reaction force, I think I might be misunderstanding.
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Raizel
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Anyone wondering on the impulse issue.
Attached files
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ImLiterally12
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Hey guys, what experiments do we need to know for this one? I know the specific heat capacity of a liquid/solid came up last year.
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sagar448
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(Original post by BrokenS0ulz)
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I get the contact force is least at A because weight is acting in same direction as centripetal force, and I get that it's the most at C because weight is acting in the opposite direction.
But at B, isn't the contact force also 0, my diagram could just be wrong?
EDIT what is the reaction force, I think I might be misunderstanding.
Centripetal force is not just created because something moves in a circle. Centripetal force is equal to either the weight, reaction force, thrusting force etc...
At point B, the contact force isn't zero since the weight of the sock is acting angle to the surface of the drum. So the horizontal component of the weight is pushing the surface and the surface reacts with a normal force which is also providing for the centripetal force.

Reaction force is basically the force created in reaction to an incident force. SO me pushing on a desk is incident force and desk pushing me is reaction force, in this case they are equal so I won't move but in other cases such as the horse and the cart situation, according to newtons law the horsey moves.

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seizetoday
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Could someone list all the experiments we need to know. Thanks.
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sagar448
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(Original post by seizetoday)
Could someone list all the experiments we need to know. Thanks.
(Original post by ImLiterally12)
Hey guys, what experiments do we need to know for this one? I know the specific heat capacity of a liquid/solid came up last year.
We went through this earlier there are a few that come to mind....

(Original post by ReeceG46)
What examples of practicals do they think they could ask us to describe ? (4-6 marks maybe)The ones I could think of are:
-pendulum bob
-bung on a string
-specific heat capacity
-resonance and damping
-brownian motion smoke cell
-boyle's law with the oil and pressure gauge
Can anyone think of any others?Thanks!
(Original post by randlemcmurphy)
Obtaining a cooling curve possibly.
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Raizel
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3bi) why does F=mg where m is mass of the slotted masses.
https://6d58568f77fa37b9c685ab8ce928...%20A-level.pdf
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sagar448
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(Original post by Raizel)
3bi) why does F=mg where m is mass of the slotted masses.
https://6d58568f77fa37b9c685ab8ce928...%20A-level.pdf
The F=mg gives you the tension. Tension is the same along the string.
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wonderkid14
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hey guys hope revision is going well!

does anyone know where i can find the june 2014 paper?
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