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    Name:  image.jpg
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Size:  488.7 KB Question 7 well most of it
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    Exactly, I got two value for t, t=5 AND t=25.k assumed one of them was correct so I solved for the displacement using t=5 and 25 and got two different answers, when t=25 I got around the 330m areas and when t=5 I got around 80m(i cant remember exact value but I think the correct answer was around 80m between them)

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    How do you guys think it was compared to previous years?
    Any thoughts on grade boundaries/full UMS?
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    (Original post by reuel)
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    i got the same!!
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    This paper was tooo easy defo 70 for an A, the 10 mark was joke clocked it in 3 minutes had 35 minutes to check answers. WOOOOAHHH, some of you man are **** cambizzy here i come gunna celebrate 100 ums with some cheeky nandos see ya ******s
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    These are the answer that me and my friends got, let me know if you can't read it/if any of them are wrong 😊
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    (Original post by RorMur2992)
    I got 106m for the 10 marker, anyone else?
    Yes I got this too
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    U got Identical answers to you

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    (Original post by matt5822)
    But the i components were different
    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?
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    No air resistance for the assumption is correct and valid, as implying it is a particle means its surface area is negligible and therefore is not affected by the effect of drag.
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    (Original post by Minnie1998)
    i got the same!!
    Surely you cannot keep time as the same?
    It would take a different amount of time to reach the back of the net, no?
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    (Original post by reuel)
    Name:  image.jpg
Views: 418
Size:  488.7 KB Question 7 well most of it
    You can't use 1.76 for the time in the second part, the horizontal distance travelled has changed so the time of flight will also have changed.
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    How many marks would I lose in Q8 would you guys think? I got T = 12 but i calculated the difference of speed but not distance
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    For Q8 I only did it for the i components and got 80.5m with t=5, but I think people who did it for j components got t=12 and 106m, which is why there are 2 answers many are getting. Not sure which is right.
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    (Original post by con.brown)
    No air resistance for the assumption is correct and valid, as implying it is a particle means its surface area is negligible and therefore is not affected by the effect of drag.
    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
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    ah finding the values of P & q wasn't it just comparing I + J components separately?
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    (Original post by LiesandAlibis)
    How do you guys think it was compared to previous years?
    Any thoughts on grade boundaries/full UMS?
    I think it was comparable to june 13 paper in which case the marks needed for an A were 53
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    This is what I thought which Is why I couldn't figure it out

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    (Original post by smartsy)
    My answers (might have misremembered some, and others are probably wrong):
    1) 15 kg
    2) p = 14, q = 17
    3) 9 degrees
    4) V = 12, bearing = 008
    5) Max heigh 1.2 m (can't really remember)
    6) 43.8 N
    7) Velocity around 13, modelling as particle means air resistance can be ignored)
    8) I got up to finding t = 5 and t = 25, but then did a load of meaningless BS and got the wrong answer around 28 lol.
    I got all of these
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    (Original post by Minnie1998)
    i got the same!!
    Shame I didn't put that in the paper
    I forgot this at the time 🙁
 
 
 
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