Qualification > Sciences
Edexcel PHYSICS Doubts
astarmathsandphysics:
sqrt(22^2+11^2)=24.6m/s
xainer:
Thankss a lot!! :) realised later that it was so easy...
I have a few more questions..
http://www.xtremepapers.com/papers/Edexcel/Advanced%20Level/Physics/2008%20Jan/6731_01_que_20080116.pdf
- Question no. 3) b)i) and 5) c)ii)
hope you can give the answers for these two questions....:)
thanks once again!
astarmathsandphysics:
--- Quote from: xainer on August 22, 2012, 09:08:28 am ---Thankss a lot!! :) realised later that it was so easy...
I have a few more questions..
http://www.xtremepapers.com/papers/Edexcel/Advanced%20Level/Physics/2008%20Jan/6731_01_que_20080116.pdf
- Question no. 3) b)i) and 5) c)ii)
hope you can give the answers for these two questions....:)
thanks once again!
--- End quote ---
Max speed will be at bottom. In travelling from P to Q the work done against friction is Force times distance =60F where F is the average force. This is the difference between the initial and final energy
60F=mgh (at P) -1/2mv^2 at Q
60F=750*9.8*50-1/2*750*27^2 =94125 so F=94125/60=1568.75N
astarmathsandphysics:
And for 5cii constant deceleration so use suvat
s
u=0.95
v=0
a
t=9.3*10^-2
s=(u+v)/2*t=0.95/2*9.3*10^-2=0.044m
xainer:
--- Quote from: astarmathsandphysics on August 23, 2012, 07:00:31 pm ---Max speed will be at bottom. In travelling from P to Q the work done against friction is Force times distance =60F where F is the average force. This is the difference between the initial and final energy
60F=mgh (at P) -1/2mv^2 at Q
60F=750*9.8*50-1/2*750*27^2 =94125 so F=94125/60=1568.75N
--- End quote ---
why is the workdone against friction taken as 60F?
and, the answer isn't matching with the marking scheme. it says, answers in the range of 1100N-1300N
here's the link
http://www.xtremepapers.com/papers/Edexcel/Advanced%20Level/Physics/2008%20Jan/6731_01_rms_20080306.pdf
Q3) b)i)
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