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    (Original post by MsFahima)
    Basically you take the velocity of p to be negative and then find the velocity of q and then use suvat to find the distance. You have u, v is o, t is 3. You need to find s.

    Try it and let me know if you can't.

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    ya i did the initial momentum and final momentum and got an answer to the velocity of Q, the did s=vt and added them together, that was the right method apparently but it says i worked out the velocity of Q incorrectly :/
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    (Original post by buckeybarnes)
    ya i did the initial momentum and final momentum and got an answer to the velocity of Q, the did s=vt and added them together, that was the right method apparently but it says i worked out the velocity of Q incorrectly :/
    What did you get for q?

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    (Original post by MsFahima)
    What did you get for q?

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    (Original post by buckeybarnes)
    do you know how to do 2ii? http://www.ocr.org.uk/Images/63190-q...echanics-1.pdf
    i've almost got the method but i cant seem to get an appropriate answer
    Ill post a picture of my answer soon, I'm just revising for my exam tomorrow so sorry for the slow replies.
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    I feel too relaxed for this exam but after how FP1 and S2 went I think I only need 45-50 UMS to get an A overall 😄
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    can someone pls help me with q5 http://www.ocr.org.uk/Images/63190-q...echanics-1.pdf
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    (Original post by buckeybarnes)
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    (Original post by buckeybarnes)
    can someone pls help me with q5 http://www.ocr.org.uk/Images/63190-q...echanics-1.pdf
    All of it?
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    (Original post by MsFahima)
    All of it?
    yeah pls i dont get it , and the end part of q4 i got wrong as well, using t=5 instead of t=4, could u explain why in the last part of 4 it's t=4 as well if it's not too much please
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    (Original post by alexann95)
    When P is at rest, the only force acting on P is the weight. Hence the reaction force must be equal and opposite to the weight.
    Does this make sense?
    And angle?
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    (Original post by buckeybarnes)
    yeah pls i dont get it , and the end part of q4 i got wrong as well, using t=5 instead of t=4, could u explain why in the last part of 4 it's t=4 as well if it's not too much please
    D.w I'll get back to you in a few mins!
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    (Original post by MsFahima)
    D.w I'll get back to you in a few mins!
    thanks ! i just tend to generally get very confused and annoyed when i see velocity time graph questions, never really got the grasp of them
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    (Original post by buckeybarnes)
    thanks ! i just tend to generally get very confused and annoyed when i see velocity time graph questions, never really got the grasp of them
    I'm struggling with Q5 ii) unfortunatley. For Q5)i) The area under the graph is the displacement so we can work that out easily. (3*20)/2 = 30.

    I'll tag kawehi to help you out. I'm sorry!
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    huh, this question has got me pretty stumped, it's question 3i on http://www.ocr.org.uk/Images/59970-q...echanics-1.pdf

    apparently you use v^2=u^2+2as, and take u=5, v=0, a=9.8, and s=2.5, but i don't get that. Surely if the particle is projected into the air from a point of 2.5m the displacement is over 2.5m because it is projected into the air? on the ascent the acceleration would be -9.8ms^2, but on the descent it would be 9.8ms^2, so on the descent wouldn't the initial velocity be 0, instead of 5? it's confusing me, it's like it has used the acceleration on the descent but the displacement cant be right....?

    EDIT wait nvm i went through the process and i was right the mark scheme was just being very weird
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    (Original post by MsFahima)
    I'm struggling with Q5 ii) unfortunatley. For Q5)i) The area under the graph is the displacement so we can work that out easily. (3*20)/2 = 30.

    I'll tag kawehi to help you out. I'm sorry!
    thanks
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    (Original post by buckeybarnes)
    thanks
    You're using t = 4 because when it comes to (instantaneous) rest after moving past O, t = 4.
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    (Original post by MsFahima)
    You're using t = 4 because when it comes to (instantaneous) rest after moving past O, t = 4.
    i got as far as thinking that the total time is 60+ the base of the trapezium next to 60, but i was unsure how to work out the area of the trapezium with only the information we were provided with
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    (Original post by buckeybarnes)
    i got as far as thinking that the total time is 60+ the base of the trapezium next to 60, but i was unsure how to work out the area of the trapezium with only the information we were provided with
    I'm so confused!
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    (Original post by MsFahima)
    I'm so confused!
    i think bc it leaves the weighbridge at 0 seconds, and goes back onto it at the furthest along time on the graph?
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    how do you know whether constant of integration =0 when you're integrating the equation if you're doing an SVA question?
 
 
 

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