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    I said that if you model the ball as a particle then the size of the ball won't affect whether it hits the box or not.
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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?
    i did exactly that - seems right to me
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    (Original post by Eliottooooo)
    Yes friend, I did, here's the catch though, we're wrong.
    nah we're right
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    (Original post by con.brown)
    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.
    I'm certain it already mentioned air resistance, it's always my go to answer for these questions so I made sure to check it was an option, which it turned out not to be
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    (Original post by Chaotix28)
    I said that if you model the ball as a particle then the size of the ball won't affect whether it hits the box or not.
    Sounds like a good answer mate
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    (Original post by georgiashayes)
    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?
    thats what one of the other people doing the excam said he does A2 and is very good at maths so its probably right
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    Well that was ****
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    (Original post by reuel)
    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?
    I personally did:
    (Horizontal):
    13.6 + 3 = Vcos(50)*t +0
    V = 16.6/(cos(50)*t)

    (Vertical):
    1 = Vsin(50)*t + 0.5(-9.8)t^2
    Sub in V?

    This is probably wrong, tbh.
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    (Original post by georgiashayes)
    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?
    If the velocities were the same they wouldn't be at the same point because they have different origins and initial speeds. They'd only be at the same point if the displacements from the origin were the same?

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    (Original post by QuantumSylar)
    Does any1 know how u find angle of ramp. thx
    I modelled the sides of the triangle in seconds
    1.2 for the hypotenuse and get the the distance travelled by a vertical drop using 9.8ms-2 and how lon it takes to get to the same speed
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    I got t=25 but got the distance between as 161(double the actual answer) so how many marks do you think I would have got??
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    wait wtf
    1) the first question i got like 6.8 for the velocity and 335kg for the mass lol what did i do wrong
    i did momentum a + momentum b = momentum of particle -> 3(6) + 2(8) = 5(v) v = 6.8m/s
    then the second part i done momentum particle = 5+m(0.1) -> 5(6.8) = 0.5 + 0.1m -> m = 335kg

    5) how did u wiork out max height of connected particle? is that when v=0?

    7) i got 12.3m/s for the speed. i used horizontal displacement formula -> v=d/t and re-arragned to get in terms of t(since t is same for horizontal and vertical_, i ended up with like 16.6m = ucos50 * t -> u(speed) = 16.6/tcos50 and subbed that into s=ut+1/2at^2 to work out what the time was
    then used that time in 16.6m = ucos50 * t to get u = 12.3

    8) i got t=5 and t=25 but i thought time isnt a vector so i minused them and used that in the equation i ended up with like 138m or 132m, something like that


    how many marks would i have lost on the 10 marker, 7 marker if my answer is wrong?
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    (Original post by Hep123)
    I got 12s sorry Im stressing now haha because vectors will be parallel when vector a is a multiple of vector b
    same!
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    (Original post by jtebbbs)
    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.
    Most people that did this exam haven't done C4 yet though.
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    (Original post by LiesandAlibis)
    I personally did:
    (Horizontal):
    13.6 + 3 = Vcos(50)*t +0
    V = 16.6/(cos(50)*t)

    (Vertical):
    1 = Vsin(50)*t + 0.5(-9.8)t^2
    Sub in V?

    This is probably wrong, tbh.
    i done the same and worked out time to be like 1.96 or 1.7 something eded up getting 12.3 for the speed, which makes sense cuz it moved by 3m so ofc the speeds wont be so different nor will the times
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    (Original post by matt5822)
    Most people that did this exam haven't done C4 yet though.
    yeah but you learn at gcse that parallel vectors are multiples of each other
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    (Original post by QuantumSylar)
    Does any1 know how u find angle of ramp. thx
    Not sure if its remotely correct but the method I used:

    •No resisitive forces so acceleration comes soley from component of toys car weight.
    •The acceleration down the slope from the previous SUVAT was 1.5ms^-2
    •If the slope was vertical (at 90° to horizontal) the acceleration in the SUVAT would be 9.8
    • 9.8sinTHETA=1.5
    •Angle = sin^-1(1.5/9.8)


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    (Original post by johnapplebottom)
    wait wtf
    1) the first question i got like 6.8 for the velocity and 335kg for the mass lol what did i do wrong
    i did momentum a + momentum b = momentum of particle -> 3(6) + 2(8) = 5(v) v = 6.8m/s
    then the second part i done momentum particle = 5+m(0.1) -> 5(6.8) = 0.5 + 0.1m -> m = 335kg

    5) how did u wiork out max height of connected particle? is that when v=0?

    7) i got 12.3m/s for the speed. i used horizontal displacement formula -> v=d/t and re-arragned to get in terms of t(since t is same for horizontal and vertical_, i ended up with like 16.6m = ucos50 * t -> u(speed) = 16.6/tcos50 and subbed that into s=ut+1/2at^2 to work out what the time was
    then used that time in 16.6m = ucos50 * t to get u = 12.3

    8) i got t=5 and t=25 but i thought time isnt a vector so i minused them and used that in the equation i ended up with like 138m or 132m, something like that


    how many marks would i have lost on the 10 marker, 7 marker if my answer is wrong?
    question 1, they were heading to eachother so the velocity of b is negative (-8)

    question 5 - yes v = 0 a = -9.8 u = 2.5 (from previous answer) then youre finding s
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    (Original post by GabbytheGreek_48)
    yeah but you learn at gcse that parallel vectors are multiples of each other
    I didn't know that until I did M3
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    (Original post by johnapplebottom)
    wait wtf
    1) the first question i got like 6.8 for the velocity and 335kg for the mass lol what did i do wrong
    i did momentum a + momentum b = momentum of particle -> 3(6) + 2(8) = 5(v) v = 6.8m/s
    then the second part i done momentum particle = 5+m(0.1) -> 5(6.8) = 0.5 + 0.1m -> m = 335kg

    5) how did u wiork out max height of connected particle? is that when v=0?

    7) i got 12.3m/s for the speed. i used horizontal displacement formula -> v=d/t and re-arragned to get in terms of t(since t is same for horizontal and vertical_, i ended up with like 16.6m = ucos50 * t -> u(speed) = 16.6/tcos50 and subbed that into s=ut+1/2at^2 to work out what the time was
    then used that time in 16.6m = ucos50 * t to get u = 12.3

    8) i got t=5 and t=25 but i thought time isnt a vector so i minused them and used that in the equation i ended up with like 138m or 132m, something like that


    how many marks would i have lost on the 10 marker, 7 marker if my answer is wrong?
    I had 335kg!!!!! But what is wrong with that?
 
 
 
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