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

  • View Poll Results: What do you think the grade boundary for an A will be?
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    (Original post by Amir52)
    anyone mind sharing the Jan IAL 2015 QP-MS, cause I haven't got them yet!

    Thanks!
    http://56leomessiphotoshop.blogspot....blog-post.html
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    3 questions...

    Anyone know the grade boundaries for 2015 IAL?

    What's the best way to tackle those questions where you get given an equation and are told certain variables double, others half, etc., how does the result change?

    Will the equation sheet be the same as all the previous ones?

    And good luck to everyone tomorrow
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    (Original post by jay_em)
    3 questions...

    Anyone know the grade boundaries for 2015 IAL?

    What's the best way to tackle those questions where you get given an equation and are told certain variables double, others half, etc., how does the result change?

    Will the equation sheet be the same as all the previous ones?

    And good luck to everyone tomorrow
    61 raw marks for an A, 67 for A*.

    I generally just plug in some values to see what changes, if I'm not sure about what will happen.

    You'll get the data sheet, it won't change.
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    Some notes:

    -impulse = change in momentum = force x time
    -magnetic force F = Bqv
    -Centripetal force is the resultant force
    -Magnetic flux density B has the unit T
    -Magnetic flux (phi) = BA, and has the unit Wb
    -Magnetic flux linkage = N x (phi) or NBA, and has the unit Wb also
    -Rate of change of magnetic flux linkage = emf induced (Faraday's law)
    -Ideal transformer equation: Vs/Vp = Ns/Np. Alternatively, IsVs = IpVp (power output = power input), however there are always energy losses so IsVs < IpVp
    -Baryons and Mesons are both examples of Hadrons. Mesons consists of a quark-antiquark pair, whilst baryons consist of 3 quarks or 3 antiquarks
    -If a heavier mass strikes a much lighter mass, the angle between the two after collision will be <90 degrees if the collision is elastic
    -If a lighter mass strikes a much heavier mass, the angle between the two after collision will be >90 degrees if the collision is elastic
    -Joules to eV : divide by 1.6x10^-19, eV to Joules : multiply by 1.6x10^-19
    -MeV/c^2 is a unit of mass, to convert it to kg, multiply by 1.6x10^-13, then divide by 9x10^16
    -In particle collisions, charge, momentum and baryon number are always conserved
    -kg to u: divide by 1.66x10^-27

    Hope this is useful. If any of these are wrong please point it out
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    Could someone please explain the balancing the wire between two electromagnets experiment please? I don't understand how it works!
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    https://ad57adc4f6eb5ea42b541057f162...%20Physics.pdf

    Someone explain 15 please
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    (Original post by ArielHaeems)
    Some notes:

    -impulse = change in momentum = force x time
    -F = Bqv
    -Centripetal force is the resultant force
    -Magnetic flux density B has the unit T
    -Magnetic flux (phi) = BA, and has the unit Wb
    -Magnetic flux linkage = N x (phi) / NBA, and has the unit Wb also
    -Rate of change of magnetic flux linkage = emf induced (Faraday's law)
    -Ideal transformer equation: Vs/Vp = Ns/Np. Alternatively, IsVs = IpVp (power output = power input), however there are always energy losses so IsVs < IpVp
    -Baryons and Mesons are both examples of Hadrons. Mesons consists of a quark-antiquark pair, whilst baryons consist of 3 quarks or 3 antiquarks
    -If a heavier mass strikes a much lighter mass, the angle between the two after collision will be <90 degrees if the collision is elastic
    -If a lighter mass strikes a much heavier mass, the angle between the two after collision will be >90 degrees if the collision is elastic
    -Joules to eV : divide by 1.6x10^-19, eV to Joules : multiply by 1.6x10^-19
    -MeV/c^2 is a unit of mass, to convert it to kg, multiply by 1.6x10^-13, then divide by 9x10^16
    -In particle collisions, charge, momentum and baryon number are always conserved

    Hope this is useful. If any of these are wrong please point it out
    Do you get how to do this question?

    The drum of a washing machine rotates with an angular velocity of 8.5rad per second. The time to complete 10 revolutions is
    The answer is supposed to be 7.4s. Please show me how you work this out?
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    (Original post by sj97)
    Do you get how to do this question?

    The drum of a washing machine rotates with an angular velocity of 8.5rad per second. The time to complete 10 revolutions is
    The answer is supposed to be 7.4s. Please show me how you work this out?
    You know that

    Omega = 8.5 rads^-1

    and Omega = 2pi/T. Rearranging, this gives us T=2pi/Omega

    2pi is one cycle, we're looking for the time in 10 cycles, so 10x2pi = 20pi

    so T=20pi/8.5 = 7.39s
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    (Original post by ArielHaeems)
    You know that

    Omega = 8.5 rads^-1

    and Omega = 2pi/T. Rearranging, this gives us T=2pi/Omega

    2pi is one cycle, we're looking for the time in 10 cycles, so 10x2pi = 20pi

    so T=20pi/8.5 = 7.39s (make sure you're in radian mode)
    donesnt matter if u are in rad mode or not. Does it
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    (Original post by Freddy-Francis)
    donesnt matter if u are in rad mode or not. Does it
    Na it doesnt lol


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    (Original post by ArielHaeems)
    Some notes:

    -impulse = change in momentum = force x time
    -F = Bqv
    -Centripetal force is the resultant force
    -Magnetic flux density B has the unit T
    -Magnetic flux (phi) = BA, and has the unit Wb
    -Magnetic flux linkage = N x (phi) / NBA, and has the unit Wb also
    -Rate of change of magnetic flux linkage = emf induced (Faraday's law)
    -Ideal transformer equation: Vs/Vp = Ns/Np. Alternatively, IsVs = IpVp (power output = power input), however there are always energy losses so IsVs < IpVp
    -Baryons and Mesons are both examples of Hadrons. Mesons consists of a quark-antiquark pair, whilst baryons consist of 3 quarks or 3 antiquarks
    -If a heavier mass strikes a much lighter mass, the angle between the two after collision will be <90 degrees if the collision is elastic
    -If a lighter mass strikes a much heavier mass, the angle between the two after collision will be >90 degrees if the collision is elastic
    -Joules to eV : divide by 1.6x10^-19, eV to Joules : multiply by 1.6x10^-19
    -MeV/c^2 is a unit of mass, to convert it to kg, multiply by 1.6x10^-13, then divide by 9x10^16
    -In particle collisions, charge, momentum and baryon number are always conserved

    Hope this is useful. If any of these are wrong please point it out
    Add conversion of kg to u whoch is divide by 1.66x10^-27


    Posted from TSR Mobile
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    (Original post by Freddy-Francis)
    donesnt matter if u are in rad mode or not. Does it
    You're right, I'm getting confused, apologies
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    (Original post by ArielHaeems)
    You know that

    Omega = 8.5 rads^-1

    and Omega = 2pi/T. Rearranging, this gives us T=2pi/Omega

    2pi is one cycle, we're looking for the time in 10 cycles, so 10x2pi = 20pi

    so T=20pi/8.5 = 7.39s
    Thanks! DO you also get this one:

    http://qualifications.pearson.com/co...-June-2014.pdf

    number 9?
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    (Original post by sj97)
    Thanks! DO you also get this one:

    http://qualifications.pearson.com/co...-June-2014.pdf

    number 9?
    Impulse = force x time

    since impulse = change in momentum;

    mv = force x time

    v = (force x time) / m

    m = 1.5kg
    force x time = area under graph = 5x8/2 = 20

    so v = 20/1.5 = 13.3 ~13ms^-1 - so B
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    Is the formula for flux linkage Nphi=NBA, or phi=NBA ?

    Posted from TSR Mobile
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    (Original post by BP_Tranquility)
    Is the formula for flux linkage Nphi=NBA, or phi=NBA ?

    Posted from TSR Mobile
    First one


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    Can someone please explain Question 18 c)ii + iii on the June 2012 paper to me?
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    (Original post by ArielHaeems)
    Impulse = force x time

    since impulse = change in momentum;

    mv = force x time

    v = (force x time) / m

    m = 1.5kg
    force x time = area under graph = 5x8/2 = 20

    so v = 20/1.5 = 13.3 ~13ms^-1 - so B
    Oh, i accidentally divided by time. My last question is on question 15 of that same paper/ WHy are you multiplying the electric field strength by two? Also why wouldn't you use the inverse square relationship? Could you explain both a and b please?
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    (Original post by sj97)
    Oh, i accidentally divided by time. My last question is on question 15 of that same paper/ WHy are you multiplying the electric field strength by two? Also why wouldn't you use the inverse square relationship? Could you explain both a and b please?
    (a) Electric field strength is multiplied by two because there are two charges both contributing equally to E at point B.

    (b) A is a neutral point, both charges exert an equal and opposite force, which cancel out, so the net field strength is zero.Going along the horizontal axis, E decreases since it follows an inverse square law, because electric fields are radial, and their intensity decreases with increasing distance (so you would use the inverse square relationship here)
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    magnetic field questions will be the death of me!!!!!!!!!!
 
 
 
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