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    Did anyone get work done in Q4 as 3450J ????

    I did work done = frictional force * distance :mad:

    I use that to find velocity which I got as 10.2 and also I found out velocity another way by first finding acceleration and both answers were the same:confused::confused:



    Anyone remember distance which particle travel up and angle slope make with horizontal??????
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    Any got an unofficial mark scheme?
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    (Original post by x10)
    Any got an unofficial mark scheme?
    Arsey should post one up in 24 hours time.
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    (Original post by cazzy-joe)
    Did anyone get work done in Q4 as 3450J ????

    I did work done = frictional force * distance :mad:

    I use that to find velocity which I got as 10.2 and also I found out velocity another way by first finding acceleration and both answers were the same:confused::confused:



    Anyone remember distance which particle travel up and angle slope make with horizontal??????
    I agree weigh that I got 10.2 out using work energy and acelleration.
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    (Original post by Ari Ben Canaan)
    Arsey should post one up in 24 hours time.
    If anyone has the paper I think I got almost alll the questions out except uniform rod.
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    Does anyone remember whether it asked to find the time when the particle was travelling 45 degrees BELOW or ABOVE horizontal?
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    it was 45 degrees below. i tottally ****ed up that question and the paper in general. might have to resit
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    I thought the work done would be the work done against friction + The gain in p.e ?? So I got something like 13 000.
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    Sorry to be a pain and ask the same question again, but does anyone remember the marks for each question? I remember 1a) 3 marks 1b) 3 marks and 7) 10 marks but can't remember the rest! If none of the others the marks for question 4a and b would be helpful because I think I may have got part of this wrong and can then work out my rough mark. Thanks
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    The work done to raise the particle up the slope was the work done against friction and the work done against its weight.

    Annoyingly I got the wrong answer for the ten marker, but retyped my exact workings into my calculator and ended up with the same number as all of you guys. Must have made some silly error when I was punching in the values or using my storage buttons. How many method marks do you think I'll net?
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    How did people get the answer for 45 degrees BELOW horizontal to be around 0.2? Isn't that the value for ABOVE?
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    (Original post by Panda Vinnie)
    Did you consider work done against gravity?

    I think the work done by the pulling force was equal to the total work done (as the pulling force is working against both friction and gravity).

    I did that question twice, first using work done against friction and gravity...and the other using work done by the pulling force. They both gave the same answer ~8040J (or 8480...something like that, I can't remember!)

    What did everybody get for that question?
    Yeh I think that is how you are meant to do it, I got that answer and from what I gathered so did most others sitting the exam with me.
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    (Original post by raggamuffin93)
    Sorry to be a pain and ask the same question again, but does anyone remember the marks for each question? I remember 1a) 3 marks 1b) 3 marks and 7) 10 marks but can't remember the rest! If none of the others the marks for question 4a and b would be helpful because I think I may have got part of this wrong and can then work out my rough mark. Thanks
    what was Q4 about? If it was the Work Energy one i remember 1 part was 5 marks
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    (Original post by Harry S Truman)
    The work done to raise the particle up the slope was the work done against friction and the work done against its weight.

    Annoyingly I got the wrong answer for the ten marker, but retyped my exact workings into my calculator and ended up with the same number as all of you guys. Must have made some silly error when I was punching in the values or using my storage buttons. How many method marks do you think I'll net?
    I thought work done against friction = total lost in energy = lost KE + lost PE
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    (Original post by ChopinLiszt)
    How did people get the answer for 45 degrees BELOW horizontal to be around 0.2? Isn't that the value for ABOVE?
    Yeah, that's why i used -3 instead of 3 and i got the time to be 0.816.
    No idea if that's right.
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    I'm so annoyed at myself! i didnt spot the two triangles in the centre of mas question and ended up using 5 separate shapes and I got (9,9). will I lose all 7 marks or get ft marks?
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    (Original post by AndersonsBentFoot)
    what was Q4 about? If it was the Work Energy one i remember 1 part was 5 marks
    It was to do with pulling something up a rough plane. Part a was to work out the work done I think and part b was to find the speed it reach when released. Do you remember which was the 5 mark part? I think I made a silly mistake on b. thanks
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    I think I did very well on that test. For the centre of mass question, I got centre of mass being (11,11) and angle between AB and vertical when hung from A to be 24 degrees to the nearest degree.

    Time for something to be 45 degrees below the horizontal means it's 135 degrees below the normal. tan 135 is -1, so the j vector divided by the i vector (t o a) would equal -1. As a result, I got time to be 40/49 seconds or 0.816 seconds and the velocity of it would be 3i - 3j so the speed would be 3 root 2. Did anyone else get that?

    Work done against resistances = Work done to over come friction (that was the only resistance) x distance travelled.

    I got 3450 or something like that and the speed going back down the hill to be 10.2.

    For the question with the plank resting on the peg and we had to calculate mu, I got mu to be 0.517 or something close to that.

    Oh yeh, and for the final question, did anyone get the velocity of P after colliding with another ball to be 2/9 ms^-1 towards the wall and so there'll be another collision between P and the wall?
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    (Original post by chaz1992)
    I think I did very well on that test. For the centre of mass question, I got centre of mass being (11,11) and angle between AB and vertical when hung from A to be 24 degrees to the nearest degree.

    Time for something to be 45 degrees below the horizontal means it's 135 degrees below the normal. tan 135 is -1, so the j vector divided by the i vector (t o a) would equal -1. As a result, I got time to be 40/49 seconds or 0.816 seconds and the velocity of it would be 3i - 3j so the speed would be 3 root 2. Did anyone else get that?
    Yes that's what i got
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    (Original post by India13)
    Yes that's what i got
    YAY!!!

    I messed up C4 so badly and I think i've got 100 in M2 and S2 lol.

    Is the A* in Further Maths dependent on an A* in maths or are they completely independent?
 
 
 
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