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physics paper 42 was very difficult
thank you so much! it was really good!It is given that Lambda=630nm=630 x 10^-9 m
It is given that the grating is 450 Lines per 1mm
First, we need a, which is distance between 2 lines only.
To get a, we need to use a=d/n, where d is the whole length of grating and n is number of lines.
We need to change 1 mm into m, which is 0.001 m.
Now, a= 0.001/450=2.2 x 10^-6m
We now need to use, asintheta=nLambda where a is separation between 2 lines.
We need to find n, we put theta=90, so (2.2 x 10^-6)sin90=n(630 x 10^-9), notice that all values should be the same unit. I prefer changing them all to meters.
Notice that our answer of n=3.52, but n has to be a whole number, so we round it down to 3.
I hope I helped you, i tried to be as clear as possible.
lol not after physicsKeep smiling ?
lol not after physics
Inshallah we do good in paper 5..hmm its true but there's nothing we all can do now! Except doing good in paper 5! R u maths givin tomorrow?
Inshallah we do good in paper 5..
and no i don't take maths.
I hated it I made lots and lots of mistakes and got the CT scanner question all wrong. If you remember any answers please tell me.How was paper 41 guys? I found it okayish
i can't see the images. they are broken -_-
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s09_qp_31.pdf
Q1d please tell me how to find the units for such questions and what makes it different from other questions !
ok dats fyni can't see the images. they are broken -_-
7a(ii):
1.
(I'm pretty sure the E.g thing is a typing error, since there's no "g" variable in the question - the answer should be E.q or qE).
Since the moving particle is positively charged and the upper plate is negatively charged (not to forget, the lower plate is positively charged), the moving particle will be attracted upwards (by the negative plate) and repelled in the same direction (by the lower plate).
This force does not depend on the position of the particle or it's velocity, and it is constant all throughout.
The magnitude of the force exerted on a particle with charge q when the strength of the electric field at it's location is E, is given by:
F = q * E
So, since the electric field strength is constant and directed upwards (between the plates) all along the length of the plates, this is the force exerted by the plates on the particle. Therefore, F = qE.
2.
Since the force exerted on the particle is always in the vertical plane, the horizontal component of the particle's velocity will remain constant throughout it's journey, because no force component (neither the force of gravity or the electric force have a horizontal component) will influence it's velocity.
So, since the horizontal displacement of the particle as it travels from one side of the plates to the other is equal to L and this is related to the velocity by
v = s / t = L / t
we can write
t = L / v.
7b(i)
The law of conservation of momentum says that "the momentum - or quantity of motion - of a particle or the center of mass of a system of particles will not change unless a net external force acts on that particle or system of particles".
In other words, "the momentum of a closed system will not change unless a net external force acts on that system of particles".
It can be extended to the situation where two or more objects, taken as components of a complete system, interact by collision or action-at-a-distance forces; if no force exerted by a body from outside the system exists, then the momentum of the system (of all the components of the system) summed up can never change.
So, if no external forces act on a system, momentum before interaction = momentum after interaction.
7b(ii)
This question requires the use of the Impulse theorem, which can be written in symbols as
Δ(mv) = F(net) * t
And so, we can write
Δ(mv) = qE * L/v = qLE/v
7b(iii)
The charged particle is not an isolated system; since a net force exerted by a body other than itself acts on the particle, it cannot be considered a close system and as such it's momentum will change, so the conservation of momentum doesn't apply here.
However, if you take all the objects involved to be part of the system (i.e. if you say that the particle as well as the plates form the system) all the forces involved in the diagram are between the objects of the system, and no external force acts on it. Therefore, it's momentum is conserved.
(I'll try the first part after some time, the path thing).
Hope this helped!
Good Luck for all your exams!
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