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    How do we know if F = ma or if it's equal to 0?
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    (Original post by peacefully_)
    yeah I noticed straight away, don't worry about it
    (Original post by amelienine)
    How do we know if F = ma or if it's equal to 0?
    If the particle is said to be in equilibrium or I guess if it's travelling at a constant speed as well because when it's travelling at a constant speed, it's acceleration = 0 ms^-2.
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    Anyone got any advice on how to do those "find the force exerted on the pulley by the string" questions? They're the only ones which stump me...

    THanks in advance
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    Does anyone have the Edexcel January 2016 IAL paper? Would really appreciate it if you could send me a link


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    (Original post by dodgewc25)
    Anyone got any advice on how to do those "find the force exerted on the pulley by the string" questions? They're the only ones which stump me...

    THanks in advance
    Watch this https://www.youtube.com/watch?v=rC2IQ3R5im4
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    Can anyone help me with this Q? For Q I thought that time = t+1, however the mark scheme says t-1?

    I don't understand :/
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    M1 June 2012 --> 66/75 = 80% = A. (I should have scored 71/75 raw marks) :sigh:

    Mistakes & tips for future past papers:
    - Understand how to resolve if a direct force is acting into the plane & split it up into horizontal/vertical components.
    - Sub the value of T back into the vector position equation to make comparisons.
    - Draw a SUVAT diagram carefully & understand when the displacement would be negative.

    I'll be recording every past paper score that I do and my overall notes/opinion on that past paper.

    Past papers completed = 2/13. (edexcel regular + edexcel IAL) [excluding x2 MAM pp closer to exam date]
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    When do you know when to make u and s negative when using suvat equations? i.e in what context?
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    (Original post by nicoledsk)
    When do you know when to make u and s negative when using suvat equations? i.e in what context?
    Well let's say you have a particle being projected upwards with a speed of u ms^-1 5 metres above the ground. If you were then to do a calculation which involves you working out at what speed the particle's travelling when the particle hits the ground, you COULD take u as being -u ms^-1 if you choose the DOWNWARDS direction to be positive. In this case, u would be negative because u ms^-1 is the speed when the particle is being projected up. Does that make sense?

    Similarly for displacement in the scenario above, displacement would be -5 metres since the particle is ending up 5m below its starting position (imagine the particle going up from the point it was projected from, coming back down, going past this point, and then hitting the ground 5 metres below. The total displacement = -5m as a result).

    Hope that explained it !
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    Can anyone help me with a few questions from the M1 edexcel book please:
    Page 152: questions 6 and 7 ( i dont understand those type of questions in general)
    Page 113: Question 16? I dont understand how the horizontal force works
    Page 78 : Question 26c
    Page 51: Example 15c
    Also, the topic where there are 2 forces acting on a particle (plus its weight), I find it difficult to draw the triangle and use cosine rule. I can do it through resolving (e.g. Tsinx , Tcosx) but that makes it much more time consuming. So anyone got any tips for that.
    Thats it for the timebeing. Thanks
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    (Original post by Don Pedro K.)
    Well let's say you have a particle being projected upwards with a speed of u ms^-1 5 metres above the ground. If you were then to do a calculation which involves you working out at what speed the particle's travelling when the particle hits the ground, you COULD take u as being -u ms^-1 if you choose the DOWNWARDS direction to be positive. In this case, u would be negative because u ms^-1 is the speed when the particle is being projected up. Does that make sense?

    Similarly for displacement in the scenario above, displacement would be -5 metres since the particle is ending up 5m below its starting position (imagine the particle going up from the point it was projected from, coming back down, going past this point, and then hitting the ground 5 metres below. The total displacement = -5m as a result).

    Hope that explained it !
    Thank you so much! it helps so much when you can actually understand the principles being applied to equations, thanks once again!
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    (Original post by nicoledsk)
    Thank you so much! it helps so much when you can actually understand the principles being applied to equations, thanks once again!
    It certainly does haha ! No problem ^_^
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    Hey, Can you or anyone explain to me how the diagrams work on pages 58 and 59 from the m1 book. Since the two blocks on top of each other are inside the lift, how would u draw the forces on each box. The book isnt very clear on what force represents what.
    Thanks.
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    Is 3 papers a good amount for today? Or should I do another ?


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    I NEED HELP WITH QUESTION 6 :https://3b0a7b1bc87f5381e60f8f717510...%20Edexcel.pdf
    It says to find the greatest possible value of x - and in order to do that you take moments about B and assume the reaction force at A is 0 - but WHY?!
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    If the time that P has spent past A is t, Q has spent 1 less second than P past A (as it passed A 1 second later than P) so it's time is t-1.

    (Original post by KINGYusuf)
    Can anyone help me with this Q? For Q I thought that time = t+1, however the mark scheme says t-1?

    I don't understand :/
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    3 papers enough to call it a day guys?


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    (Original post by Don Joiner)
    3 papers enough to call it a day guys?


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    It depends on the person to be honest.
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    (Original post by Don Joiner)
    3 papers enough to call it a day guys?


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    lol srs? I haven't even done one paper in the past three days. :rofl:
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    (Original post by Zacken)
    lol srs? I haven't even done one paper in the past three days. :rofl:
 
 
 
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