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M1 OCR (Not MEI) Exam - 9/06/2015 watch

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    (Original post by Super199)
    Btw how is C3 revision going?
    It's going okay. Just taking longer than I should when doing a past paper. So will be doing few more past papers and increase my speed. Wbu?
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    (Original post by Super199)
    Do you mind explain 7iia and b. For January 2007
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    Hope you understand it.

    So when it moves up and down the plane the friction is changing direction so the acceleration will also change.

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    (Original post by MsFahima)
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    Hope you understand it.

    So when it moves up and down the plane the friction is changing direction so the acceleration will also change.

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    Why is the first one not 0.5gsin40-fr=0.5a?
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    (Original post by Super199)
    Why is the first one not 0.5gsin40-fr=0.5a?
    Because the weight component and fr is in the same direction so it has to be the same sign. Both either positive or negative depending on which way you resolve. And if you resolve down the plane while the particle is moving up, the accelerations sign will also have to be negative.
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    (Original post by MsFahima)
    Because the weight component and fr is in the same direction so it has to be the same sign. Both either positive or negative depending on which way you resolve. And if you resolve down the plane while the particle is moving up, the accelerations sign will also have to be negative.
    ooh I see thanks
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    (Original post by kawehi)
    By the way, I got it!!

    Contact force is the resultant of ALL of the forces acting on the particle from the surface, not the normal force. Should I go through my working?
    Yes please, sorry I wasn't able to reply quickly
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    can anyone help me with the last part of this q from m1 june 06 please?

    A man travels 360 m along a straight road. He walks for the first 120 m at 1.5 m s−1, runs the next180 m at 4.5 m s−1, and then walks the final 60 m at 1.5 m s−1. The man’s displacement from hisstarting point after t seconds is x metres.
    (i) Sketch the (t, x) graph for the journey, showing the values of t for which x = 120, 300 and 360.[5]
    A woman jogs the same 360 m route at constant speed, starting at the same instant as the man andfinishing at the same instant as the man.
    (ii) Draw a dotted line on your (t, x) graph to represent the woman’s journey. [1]
    (iii) Calculate the value of t at which the man overtakes the woman. [5]

    (answer is 106.67)
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    Hi I don't get part ii b of this question. Any help would be appreciated
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    (Original post by steven123456789)
    can anyone help me with the last part of this q from m1 june 06 please?

    A man travels 360 m along a straight road. He walks for the first 120 m at 1.5 m s−1, runs the next180 m at 4.5 m s−1, and then walks the final 60 m at 1.5 m s−1. The man’s displacement from hisstarting point after t seconds is x metres.
    (i) Sketch the (t, x) graph for the journey, showing the values of t for which x = 120, 300 and 360.[5]
    A woman jogs the same 360 m route at constant speed, starting at the same instant as the man andfinishing at the same instant as the man.
    (ii) Draw a dotted line on your (t, x) graph to represent the woman’s journey. [1]
    (iii) Calculate the value of t at which the man overtakes the woman. [5]

    (answer is 106.67)
    Pic might be sideways. Hope it makes sense!
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    can someone explain how to use velocity time and displacement time graphs and what we might be asked about them?
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    (Original post by sykik)
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    Hi I don't get part ii b of this question. Any help would be appreciated
    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?
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    (Original post by buckeybarnes)
    can someone explain how to use velocity time and displacement time graphs and what we might be asked about them?
    Velocity Time graphs:
    Gradient = acceleration
    Area under graph = displacement
    Area under graph represents negative displacement as the particle or whatever it is is traveling in the opposite direction
    Look at past papers to find out what kind of questions they ask, there are quite a few.
    Displacement time graphs:
    Gradient = velocity
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    (Original post by alexann95)
    Velocity Time graphs:
    Gradient = acceleration
    Area under graph = displacement
    Area under graph represents negative displacement as the particle or whatever it is is traveling in the opposite direction
    Look at past papers to find out what kind of questions they ask, there are quite a few.
    Displacement time graphs:
    Gradient = velocity
    i always end up using SUVAT equations but that can sometimes be problematic, would you say most question can be answered by using the gradient and areas under the graph, from what you know?
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    (Original post by buckeybarnes)
    i always end up using SUVAT equations but that can sometimes be problematic, would you say most question can be answered by using the gradient and areas under the graph, from what you know?
    It depends on the question, sometimes using SUVAT or the graph would get you the answer. Having said that, the graph is there for a reason and sometimes its easier and quicker to use it! If you have been staring at the graph for ages and just can't seem to see how you would use it to get to the answer, then give SUVAT a go. But generally, I seem to use the graph to work out the acceleration and displacement etc. Also a good tip for graph questions is the imagine the situation in your head and actually visualise the motion of the car or whatever. I hope this helps a bit, its hard to answer when i don't have any examples of questions! Are you feeling confident in other areas M1? Im sure you will be fine.
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    (Original post by alexann95)
    It depends on the question, sometimes using SUVAT or the graph would get you the answer. Having said that, the graph is there for a reason and sometimes its easier and quicker to use it! If you have been staring at the graph for ages and just can't seem to see how you would use it to get to the answer, then give SUVAT a go. But generally, I seem to use the graph to work out the acceleration and displacement etc. Also a good tip for graph questions is the imagine the situation in your head and actually visualise the motion of the car or whatever. I hope this helps a bit, its hard to answer when i don't have any examples of questions! Are you feeling confident in other areas M1? Im sure you will be fine.
    mostly confident, yeah. Some of the more complex questions that have SUVAT and newtons laws and stuff all mixed in with each other get me confused, but i've learned to split them up into individual components and draw force diagrams and stuff, so theyre getting a bit easier, bout to do a practise paper, so would it be ok for me to ask you any questions if i cant answer something?
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    (Original post by buckeybarnes)
    mostly confident, yeah. Some of the more complex questions that have SUVAT and newtons laws and stuff all mixed in with each other get me confused, but i've learned to split them up into individual components and draw force diagrams and stuff, so theyre getting a bit easier, bout to do a practise paper, so would it be ok for me to ask you any questions if i cant answer something?
    Yeah sure, I'll try my best to answer them
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    (Original post by alexann95)
    Yeah sure, I'll try my best to answer them
    omg, question 1 as well lol.... do you know how to do 1iii. on this ?
    http://www.ocr.org.uk/Images/63190-q...echanics-1.pdf

    never mind, i got it ! i realised if it is decelerating to an eventual velocity of 0, then friction must be the same as the force that is causing it to move, so that it can eventually reach equilibrium (when it stops) so i did f=ma, 3x1.2=3.6
    and R= 3x9.8= 29.4 then did coefficient of friction= 3.6/29.4= 0.122 which was the answer
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    how do u do 2ii. on this paper? http://www.ocr.org.uk/Images/63190-q...echanics-1.pdf
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    (Original post by alexann95)
    Yeah sure, I'll try my best to answer them
    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
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    (Original post by buckeybarnes)
    how do u do 2ii. on this paper? http://www.ocr.org.uk/Images/63190-q...echanics-1.pdf
    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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