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# Edexcel Unit 4: Physics on the Move 6PH04 (11th June 2015) Watch

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1. (Original post by AhmedBilal)
Guys, during the anhiliation of electron and positron, do we always multiply by 2 in the calculations?
Because twice the same mass
2. (Original post by AhmedBilal)
Guys, during the anhiliation of electron and positron, do we always multiply by 2 in the calculations?
because twice the same mass
3. (Original post by Ambrina)
because twice the same mass
DO you get how to do this question?

Two protons, eparated by a distance x, experience a repulsive force F. If the serparation is reduced to x/3, the force between the protons will be:

THe answer is 9F, but I keep getting 3F?
4. (Original post by sj97)
DO you get how to do this question?

Two protons, eparated by a distance x, experience a repulsive force F. If the serparation is reduced to x/3, the force between the protons will be:

THe answer is 9F, but I keep getting 3F?
R^2. That is why F=kQQ/(r^2) It is 9F
5. (Original post by sj97)
DO you get how to do this question?

Two protons, eparated by a distance x, experience a repulsive force F. If the serparation is reduced to x/3, the force between the protons will be:

THe answer is 9F, but I keep getting 3F?
F=kQ1Q2/r^2
The distance b/w the two proton is r. And it is an inverse square relationship because F is propotional to 1/r^2 so F also gets squared.
Sorry idk how to explain it. Do you understand tho?
6. (Original post by bbyturtlexox)
could someone please briefly explain magnetic flux?and flux linkage? and what flux cutting is?

also when do you use the thumb rule or the left hand rule
Magnetic Flux Density (Magnetic Field Strength, B) is a measure of the number of magnetic flux 'lines' per unit area.

So if you multiply B by the area you're dealing with:

Magnetic Flux per unit area x Area = Magnetic Flux (ф)
This is then the amount of magnetic flux cutting the area you're dealing with, whether that be a coil or something.

This multiplied by the number of turns in the coil is the Flux Linkage (Փ):

Փ = Nф
= NAB

Then using Faraday's law you can calculate the magnitude of emf induced.

Hope it helped

(Someone correct me if I'm wrong please)
7. (Original post by bbyturtlexox)
could someone please briefly explain magnetic flux?and flux linkage? and what flux cutting is?

also when do you use the thumb rule or the left hand rule
These 3 pages explain everything

https://270dd1fec3d16900cdb57f52003e...c%20Fields.pdf
8. (Original post by LukeBarnett)
Magnetic Flux Density (Magnetic Field Strength, B) is a measure of the number of magnetic flux 'lines' per unit area.

So if you multiply B by the area you're dealing with:

Magnetic Flux per unit area x Area = Magnetic Flux (ф)
This is then the amount of magnetic flux cutting the area you're dealing with, whether that be a coil or something.

This multiplied by the number of turns in the coil is the Flux Linkage (Փ):

Փ = Nф
= NAB

Then using Faraday's law you can calculate the magnitude of emf induced.

Hope it helped

(Someone correct me if I'm wrong please)
9. (Original post by Ambrina)
These 3 pages explain everything

https://270dd1fec3d16900cdb57f52003e...c%20Fields.pdf
tyvm
10. Why dont high speed roads have sharp bends?
Also, what does this mean?
"The direction of Electric field strength at any point is the direction of force experiened by a positive charge at that point".

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11. (Original post by Wannabe007)
Why dont high speed roads have sharp bends?

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As F=mv^2/r, the velocity will be high and the sharp bend will have a small radius. So, the required centripetal force to maintain circular motion, will be greater than the maximum frictional force between the tyres and road of the car. So the car will slide off the road.
12. (Original post by Ambrina)
F=kQ1Q2/r^2
The distance b/w the two proton is r. And it is an inverse square relationship because F is propotional to 1/r^2 so F also gets squared.
Sorry idk how to explain it. Do you understand tho?
(Original post by Gumileva)
R^2. That is why F=kQQ/(r^2) It is 9F
ALo do you guys get why the time ha to be 0.125s for the calculation in 18b?
http://qualifications.pearson.com/co...e_20130613.pdf
13. (Original post by TheBeardedGuy)
As F=mv^2/r, the velocity will be high and the sharp bend will have a small radius. So, the required centripetal force to maintain circular motion, will be greater than the maximum frictional force between the tyres and road of the car. So the car will slide off the road.
Thank you so much

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14. why is it that the total momentum has to be zero when we are colliding particles?
15. (Original post by sj97)
ALo do you guys get why the time ha to be 0.125s for the calculation in 18b?
http://qualifications.pearson.com/co...e_20130613.pdf
2 refs per second so 4 pi per sec then pi/2 is 1/8 s is 0.125

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16. (Original post by Hectors)
why is it that the total momentum has to be zero when we are colliding particles?
Apparently this is the casein jan 14 IAL but then another paper it is not necessary? Edexcel can **** off with there bull****.

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17. When to use KE = p²/2m vs KE = ½mv²?

And any chance of needing protractors or compasses?
18. (Original post by physicsmaths)
Apparently this is the casein jan 14 IAL but then another paper it is not necessary? Edexcel can **** off with there bull****.

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yh i just did that paper and i am confused af. After today's chemistry flop i have to make up for it in physics and this paper makes like 0 sense to me
19. (Original post by Hectors)
yh i just did that paper and i am confused af. After today's chemistry flop i have to make up for it in physics and this paper makes like 0 sense to me
In examiners report no one got it right so I wouldn't worry the rest of the paper was easy though.

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20. (Original post by jay_em)
When to use KE = p²/2m vs KE = ½mv²?

And any chance of needing protractors or compasses?
Maybe, I would bring them incase. Edexcel love testing how good out equipment is,

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