Unit 5 Multiple choice & Unit 4 question help

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Bella_R95
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Hi,

Can someone please explain to me how to do the following multiple choice questions in this unit 5 paper

https://googledrive.com/host/0B1ZiqB...%20Physics.pdf

questions

4) 5) 7) 8) and 9)

and

why the answers for

question 12b) and 13 c i) on this paper are so?

https://googledrive.com/host/0B1ZiqB...%20Physics.pdf

Thanks a lot !
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Bella_R95
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I understand questions 5 & 7 in the Unit 5 part now, can somebody please explain to me

12b) 13 ci) in the unit 4

and 4, 8 & 9 in the unit 5 multiple choice

Thanks !
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krisshP
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(Original post by Bella_R95)
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Unit 4 Q12b:

Read the first few line of the q, it says atom size about 10^-10. The size of nucleus is 10^-15 approximately, so is much smaller. Using lambda=h/p much smaller lambda means much greater momentum p. using KE=(p^2)/2m, much greater momentum p means much greater KE.

Unit 4 Q13ci:

Centripetal force always acts towards the centre of the circular path.

Unit 5 Q4:
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Unit 5 Q8:
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Unit 5 Q9:
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Krishna
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Freddy-Francis
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(Original post by Bella_R95)
Hi,

Can someone please explain to me how to do the following multiple choice questions in this unit 5 paper

https://googledrive.com/host/0B1ZiqB...%20Physics.pdf

questions

4) 5) 7) 8) and 9)

and

why the answers for

question 12b) and 13 c i) on this paper are so?

https://googledrive.com/host/0B1ZiqB...%20Physics.pdf

Thanks a lot !
check the revision notes
These are the revision notes i found which will help.. Good luck with revision.
Hope it helps.
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Bella_R95
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(Original post by krisshP)
Unit 4 Q12b:

Read the first few line of the q, it says atom size about 10^-10. The size of nucleus is 10^-15 approximately, so is much smaller. Using lambda=h/p much smaller lambda means much greater momentum p. using KE=(p^2)/2m, much greater momentum p means much greater KE.

Unit 4 Q13ci:

Centripetal force always acts towards the centre of the circular path.

Unit 5 Q4:
Name:  image.jpg
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Size:  106.9 KB

Unit 5 Q8:
Name:  image.jpg
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Size:  110.3 KB

Unit 5 Q9:
Name:  image.jpg
Views: 92
Size:  63.5 KB

Krishna
Thanks so much krishna,

can you help me with these questions as well


i don't need help on images 3, 4 & 5 they are just there for reference


http://imgur.com/qqqEVi0,8JlpwAK,V2a...ACny,k6jufj9#0
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krisshP
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(Original post by Freddy-Francis)
check the revision notes
These are the revision notes i found which will help.. Good luck with revision.
Hope it helps.
THANK YOU VERY MUCH!

Do you mind if I add them to a post I made for resources on physics? The second post on this thread

http://www.thestudentroom.co.uk/show...2312846&page=2

Thanks
Krishna
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Freddy-Francis
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(Original post by krisshP)
THANK YOU VERY MUCH!

Do you mind if I add them to a post I made for resources on physics? The second post on this thread

http://www.thestudentroom.co.uk/show...2312846&page=2

Thanks
Krishna
No prob buddy
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krisshP
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Good luck to both of you
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krisshP
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(Original post by Bella_R95)
Thanks so much krishna,

can you help me with these questions as well


i don't need help on images 3, 4 & 5 they are just there for reference


http://imgur.com/qqqEVi0,8JlpwAK,V2a...ACny,k6jufj9#0
Q2- The answer is C.

Total initial momentum=total final momentum, as total momentum is conserved

Since you have two particles,

Momentum of P before (1) + momentum of Q before (2)

= momentum of P after (3) + momentum of Q after (4)

Add all the vectors for initial momentum (1)+(2).

Add all the vectors for the final momentum (3)+(4).

In both cases, you start from the same position and also end at the same position. The length of this resultant momentum vector [(1)+(2) or even (2)+(3), both give the same resultant momentum vector] is the same, showing how the magnitude of initial momentum=magnitude of final momentum as expected. The direction of this resultant momentum vector is the same, as expected.


Q10:

A tennis ball can never pass through a tiny narrow slit, so this is silly nonsense.

A tennis ball already is a particle, so this is just contradictory to the obvious, so seems wrong and silly nonsense.

This makes sense. The incredibly tiny wavelength of it means that WHATEVER it passes through, its diffraction will be negligible and the gap it passes through has a size definitely not similar to the wavelength of the ball.

Lambda=h/p
p=h/lambda
mv=h/lambda
v=h/(m)(lambda)

Let m be say 0.1kg (dont worry, just order of magnitude is important here, not whether it is 0.1,0.5 or 0.9kg)

v=6.63ms^-1, definitely not the speed of light, so silly nonsense again. Besides, a textbook says this
"No object can break the 'light barrier', I.e. travel faster than the speed of light,c" and there's a graph of m against velocity which curve where the assymptote is v=c=3 X 10^8 ms^-1 and as v approached c, the curve's gradient rises and the curve shoots up. So I doubt anything can even reach c=3 X 10 ms^-1.


Q18ci see this
http://www.thestudentroom.co.uk/show....php?t=2703238

Krishna
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