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    (Original post by RKM21)
    You got this final answers for this, I did A and B - got the same answers as you.
    My answers are in the spoiler for C&D.
    Spoiler:
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    C)18.9N
    D)1.93kg
    how did you solve 7(b)?
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    (Original post by KINGYusuf)
    I mean why for OQ can I not do like OB + BQ?

    This gives me the wrong answer though
    You can use \overrightarrow{OQ}= \overrightarrow{OB} +\overrightarrow{BQ}

    Using
    \overrightarrow{OB}=2\underline{  r}
    \overrightarrow{BQ}=-\underline{r}+\underline{p}
    Hence:
    \overrightarrow{OQ}=2\underline{  r}-\underline{r}+\underline{p}= \underline{r} + \underline{p}
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    (Original post by Cryptokyo)
    You can use \overrightarrow{OQ}= \overrightarrow{OB} +\overrightarrow{BQ}

    Using
    \overrightarrow{OB}=2\underline{  r}
    \overrightarrow{BQ}=-\underline{r}+\underline{p}
    Hence:
    \overrightarrow{OQ}=2\underline{  r}-\underline{r}+\underline{p}= \underline{r} + \underline{p}
    I'm sorry if I'm being annoying, I just don't get this part of vectors

    Isn't BQ the same as OP, but in the opposite direction?
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    (Original post by Elaria)
    how did you solve 7(b)?

    tension acts away from the pulley so resolve the forces of tension then add them together

    the resultant force of the tension acting upon the pulley is equal to the weight of the particle on the rod i think
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    (Original post by Elaria)
    how did you solve 7(b)?
    You know that the tension in the string is 12.6N from part a and this is used to find the force exerted on the pulley.
    Name:  Diagram 2.png
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    The force exerted on the pulley is the vector sum of the tensions acting upon it hence the force exerted on the pulley is 2T\cos(90-\alpha)=2T\sin(\alpha). So the force exerted on the pulley is 2\times12.6\times\frac{3}{4}=18.  6.
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    Marxist RKM21 KloppOClock

    A pulley question.
    Name:  Pulley Q.png
Views: 86
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    Answer in spoiler

    [edit: answer now correct]
    Spoiler:
    Show
    t=\frac{u}{2g}(3+\sqrt{13})
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    Can anyone explain to me how the forces on the blocks in scale pans work in general. E.g.
    |Block A|
    | BlockB |
    What will be the forces acting on Block A an B assuming Block A is above block B and block B is on the scale Pan
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    (Original post by kkboyk)
    Recent AQA papers, and also Delphis ones.
    Can I theoretically do them with Edexcel m1 and m2 knowledge? And do they have 3d vectors like with OCR MEI?
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    (Original post by 83457)
    Can I theoretically do them with Edexcel m1 and m2 knowledge? And do they have 3d vectors like with OCR MEI?
    Delphis has certain papers for Edexcel. AQA is less in the style of an Edexcel paper but still useful. You should be able to do nearly all the questions in the AQA papers and all questions in the Delphis for Edexcel papers.
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    Someone explain to me how the forces work here pls. I dont understand how the horizontal force works.
    http://postimg.org/image/phad5dk3f/bb77638f/
    • Very Important Poster
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    (Original post by fpmaniac)
    Someone explain to me how the forces work here pls. I dont understand how the horizontal force works.
    http://postimg.org/image/phad5dk3f/bb77638f/
    Think of the 30N force as being a z angle, so the angle between the line representing 30N and the force is alpha as well.
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    (Original post by KINGYusuf)
    I'm sorry if I'm being annoying, I just don't get this part of vectors

    Isn't BQ the same as OP, but in the opposite direction?
    No it is not. They have different directions and are therefore different vectors.


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    (Original post by SeanFM)
    Think of the 30N force as being a z angle, so the angle between the line representing 30N and the force is alpha as well.
    Thanx. Can you or anyone else. also help me with a few more queries I have. What direction does thrust act in. I thought it was in direction of motion but they did something different on question 1c on M1 specimen paper.
    Also what do you do for questions where theres a pulley and then it says the string breaks. Do you not consider the tension anymore? There was a question which said the string broke then immediately the particle stopped You had to find out the time between when the string broke and the particle stopped. However they took the forward force as 0 but even if the particle moved slightly wouldnt there be a forward force? (Jan 2011 question 7c)
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    i'd really appreciate some help with part (c) i'm a bit confused on what happens to direction of thrust when the forward force is removed. is it equal and opposite because i tried that and got a different answer to the markscheme?
    Name:  2016-06-05.png
Views: 76
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    Spoiler:
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    the answer is 900N
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    Whilst looking over the edge of a vertical cliff, 122.5 metres in height, Jim dislodges a stone.The stone falls freely from rest towards the sea below.Ignoring the effect of air resistance,

    (a) calculate the time it would take for the stone to reach the sea, (3 marks)

    (b) find the speed with which the stone would hit the water. (2 marks)Two seconds after the stone begins to fall, Jim throws a tennis ball downwards at the stone.The tennis ball’s initial speed is u ms-1 and it hits the stone before they both reach the water.

    (c) Find the minimum value of u. (5 marks)

    (d) If you had taken air resistance into account in your calculations, what effect would thishave had on your answer to part (c)? Explain your answer. (2 marks)

    Can anyone help with part C, no matter what I did I had 2 variables and one equation, then the answer got rid of a variable by using the original 'S' as distance in the S=ut+(at^2)/2.

    I get that they did T-2 as it was thrown 2 seconds after, but id the balls collided before hitting the water how can they still use the original distance when 'S' has to be less as they collide before reaching the water? Madness!?
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    (Original post by nicoledsk)
    i'd really appreciate some help with part (c) i'm a bit confused on what happens to direction of thrust when the forward force is removed. is it equal and opposite because i tried that and got a different answer to the markscheme?
    Name:  2016-06-05.png
Views: 76
Size:  105.7 KB
    Spoiler:
    Show
    the answer is 900N

    [Trailer] <--------> [Car]

    the arrow is the direction of thrust
    also it says the magnitude of thrust is 100N

    if you dont get that I can draw a diagram for you
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    (Original post by Cryptokyo)
    Marxist RKM21 KloppOClock

    A pulley question.
    Name:  Pulley Q.png
Views: 86
Size:  23.0 KB
    Answer in spoiler

    [edit: answer now correct]
    Spoiler:
    Show
    t=\frac{u}{2g}(3+\sqrt{13})
    would there be tension when A and B are back to there original position, or if A and B reach a height of 2L? Also if A gets tension before B does after reaching its original position first, does it rebound?
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    When doing velocity vector questions, do we have to underline the vector components such as i and j constantly and would we lose marks if we didn't?
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    (Original post by 99joey)
    When doing velocity vector questions, do we have to underline the vector components such as i and j constantly and would we lose marks if we didn't?
    no, just use the bracket method, they dont penalise you for using this method
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    (Original post by philo-jitsu)
    no, just use the bracket method, they dont penalise you for using this method
    oh ok, however if they asked us present it in the format i and j, would I then be penalised if I didn't underline the vectors?
 
 
 
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