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    (Original post by Hep123)
    I got 13.2
    that was for the front hitting the back would be that plus 3 although i got 13.6 for the front
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    Surely for the 10 marker. If the velocties are parallel you equate i of vector A to j of vector B. Solve for t then sub into displacements.
    Subtraction B-A displacements then find magnitude for distance.

    Iol i ended up equating i together and j together to get t=5 and t=25
    Then using both to get 118m but that is wrong imo. I think id get 4 method marks?!

    I think grade boundaries for this will be similar to june 14.
    You could trip up on a few questions e.g.
    7 marker projectile, 10 marker parallel velocities and that resultant velocity one was a bit funny. You could argue to toy car one was a bit funny too.

    Overall i think for full ums you could drop ~8 marks.
    Generous but considering june 14 you could drop 8 marks for full in that paper.
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    (Original post by smitherito)
    Surely parallel means in the same plane. If the I and j equated each other they would be travelling in the same line not parallel?
    Parallel just means they have the same direction and not necessarily the same speed.
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    any one else get 48.75 for the 10 marker
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    (Original post by Hep123)
    did people get 8 for the bearing?
    Yeah I got 008 too.
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    (Original post by girl :D)
    any one else get 48.75 for the 10 marker
    I got 60 point something hence why im keeping quit aboit this question
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    (Original post by jtebbbs)
    The question definitely said "model the ball as a particle with no air resistance" so whilst that's correct, it wouldn't be a valid answer. I don't know what the right answer is, but I put that you can ignore the ball's tendency to rotate, cause that's come up as an advantage before. Also, in real life, if you put a spin on the ball it changes its motion
    I definitely agree with you that it said model the ball as a particle in the text but I swear it did not talk about ignoring the effect of air resistance. Either way spin is also correct and would have been my second choice, although I would not be surprised if you are actually right, a classic AQA move tbh.
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    (Original post by RorMur2992)
    I got 106m for the 10 marker, anyone else?
    Yes friend, I did, here's the catch though, we're wrong.
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    (Original post by Examguy123)
    I got 60 point something hence why im keeping quit aboit this question
    most ppl got 60.8 i think
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    (Original post by reuel)
    Attachment 554069 Question 7 well most of it
    Well I got T right!
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    Does any1 know how u find angle of ramp. thx
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    So what was the answer to the last question??
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    For question 7c) I used speed=distance/time for the horizontal component to find the time as (13.6+3)/Vcos50. Then use this as the value for t in S=UT+1/2T^2. Rearrange to find V

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    I got 106m for Q8
    The components must give the same angle of the resultant velocity in order to be parallel, no?
    Therefore (ia + ja) = x(ib +jb)?
    You can then equate the i and j components to form a pair of simultaneous equations.
    Solve for t (from v = u +at)
    x = 2/3 or 3/2 (can't remember)
    t=12 and find displacement and distance from there?
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    (Original post by QuantumSylar)
    Does any1 know how u find angle of ramp. thx
    i didnt know but someone explained to me that you know the resultant acceleration =1.5 from one of the earlier answers and the vertical acceleration is 9.8 as its due to gravity or sumthing. then you use Pythagoras to find the horizontal component and then tan with the horizontal and vertical to find the angle hope this is clear
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    Parallel velocities means A's velocity is a multiple of B's. That's the only correct way to do it. If you think about Core 4, two vectors are parallel when one is a multiple of the other, the same applies here.

    Once you've got an expression for the displacement and velocity of both A and B, one method is to set velocity of A = k * velocity of B. This gives you two equations in terms of k and t, one for the i component and one for the j component. You then solve for k and t, which eventually gives you t=12.

    Another method is to set the angles between the two resultant velocities and the vertical equal to each other, then solve for t which also gives t=12.

    You put t=12 into your expression for the displacements of A and B, then you find how to get from A to B in terms of i and j. This gives you a vector, which you then find the magnitude of, leading to the final answer of 106m.
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    (Original post by LiesandAlibis)
    I got 106m for Q8
    The components must give the same angle of the resultant velocity in order to be parallel, no?
    Therefore (ia + ja) = x(ib +jb)?
    You can then equate the i and j components to form a pair of simultaneous equations.
    Solve for t (from v = u +at)
    x = 2/3 or 3/2 (can't remember)
    t=12 and find displacement and distance from there?
    Though, thinking about it this seems wrong...somehow...
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    (Original post by LiesandAlibis)
    Surely you cannot keep time as the same?
    It would take a different amount of time to reach the back of the net, no?
    Well I thought this but it is going faster so will travel more distance in same so I'm not sure
    Is there a detailed explanation of the way?
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    (Original post by Eliottooooo)
    Yes friend, I did, here's the catch though, we're wrong.
    i did! i felt quite confident with it. t=12s. because they are parallel, the vectors are multiples of eachother - not equal cause that would mean they are at the same point/intersecting. right?
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    (Original post by Eliottooooo)
    Yes friend, I did, here's the catch though, we're wrong.
    No we're not
 
 
 
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