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Mechanics Understanding Help Watch

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    (Original post by Mr T Pities You)
    Acceleration due to gravity always acts down, and if you always pick up as being positive, g is always negative.
    Take what as positive?

    Q1b) shouldn't it be 90+tan-1(4/3)

    http://michaelfloyd.ashbournecollege...lomon-B-MS.pdf
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    (Original post by ghostwalker)
    Normal reaction where?

    You can take moments about wherever you want. Taking it about a support is usually more useful, as eliminates the forces acting at the support from your equations.
    https://d26f54f108f25b7a7b38959c5e1b...%20Edexcel.pdf

    Could you help me for 5c)

    My equations:
    For the motorbike, S=S t=t u=30 v=30, therefore S=(60)t/2

    For car, S=S t=t u=0 v=30 a=4 therefore S=0.5(4)t^2

    So t=15
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    (Original post by ps1265A)
    https://d26f54f108f25b7a7b38959c5e1b...%20Edexcel.pdf

    Could you help me for 5c)

    My equations:
    For the motorbike, S=S t=t u=30 v=30, therefore S=(60)t/2

    For car, S=S t=t u=0 v=30 a=4 therefore S=0.5(4)t^2

    So t=15
    Your equations assume constant acceleration throughout, but you're told in the question that's not the case. They stop accelerating at a certain point. Also you have the equations for the displacement for when they're travelling at constant speed in the markscheme.
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    (Original post by ghostwalker)
    Your equations assume constant acceleration throughout, but you're told in the question that's not the case. They stop accelerating at a certain point. Also you have the equations for the displacement for when they're travelling at constant speed in the markscheme.
    My equations:
    For the motorbike, S=S t=t u=30 v=30, therefore S=(60)t/2
    I have used this equation at the portion where there's no acceleration, as stated in the question

    For car, S=S t=t u=0 v=30 a=4 therefore S=0.5(4)t^2
    I have used the points where they said "acceleration was uniform"
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    (Original post by ps1265A)
    My equations:
    For the motorbike, S=S t=t u=30 v=30, therefore S=(60)t/2
    I have used this equation at the portion where there's no acceleration, as stated in the question

    For car, S=S t=t u=0 v=30 a=4 therefore S=0.5(4)t^2
    I have used the points where they said "acceleration was uniform"
    In both cases, you want the displacement once it is moving with constant speed.

    This will equal the distance it moved whilst it was accelerating plus the distance it travels whilst at constant speed.

    Say for example your object had accelerated for 20 seconds,

    Then at time "t", it will have been travelling at a constant speed for t-20 seconds.

    So, displacement at time t = "distance travelled whilst accelerating" + (t-20) x "constant speed" and this will be valid as long as t>=20.
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    (Original post by ghostwalker)
    In both cases, you want the displacement once it is moving with constant speed.

    This will equal the distance it moved whilst it was accelerating plus the distance it travels whilst at constant speed.

    Say for example your object had accelerated for 20 seconds,

    Then at time "t", it will have been travelling at a constant speed for t-20 seconds.

    So, displacement at time t = "distance travelled whilst accelerating" + (t-20) x "constant speed" and this will be valid as long as t>=20.
    Thanks! So I just got my times messed up right?

    I need help with question 7d)
    https://3b0a7b1bc87f5381e60f8f717510...%20Edexcel.pdf

    My working out:
    Because there's no longer tension, -mg = 3ma and so, acceleration is -g/3
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    (Original post by ps1265A)
    Thanks! So I just got my times messed up right?

    I need help with question 7d)
    https://3b0a7b1bc87f5381e60f8f717510...%20Edexcel.pdf

    My working out:
    Because there's no longer tension, -mg = 3ma and so, acceleration is -g/3
    Why do you say there is no tension? C has dropped off, but A and B are still there.
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    (Original post by ghostwalker)
    Why do you say there is no tension? C has dropped off, but A and B are still there.
    Thanks! Got it!

    For Q6a), can I put the resultant force anywhere?

    https://3b0a7b1bc87f5381e60f8f717510...%20Edexcel.pdf
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    (Original post by ps1265A)
    Thanks! Got it!

    For Q6a), can I put the resultant force anywhere?

    https://3b0a7b1bc87f5381e60f8f717510...%20Edexcel.pdf
    Well it's going to be coming out of O, and somewhere between P and Q, but other than that any average position will do.
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    (Original post by ghostwalker)
    Well it's going to be coming out of O, and somewhere between P and Q, but other than that any average position will do.
    Thanks
    http://www.thestudentroom.co.uk/show....php?t=2705424
    For Q5b) second attachment, I thought the new acceleration was going to be without tension i.e -3mg=3ma ? Because the string would be slack as B comes down?

    And if not, then why isn't the same acceleration used as that of part a)?
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    (Original post by ps1265A)
    Thanks
    http://www.thestudentroom.co.uk/show....php?t=2705424
    For Q5b) second attachment, I thought the new acceleration was going to be without tension i.e -3mg=3ma ? Because the string would be slack as B comes down?

    And if not, then why isn't the same acceleration used as that of part a)?
    Huh! What am I supposed to be looking at in that 19 page thread?
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    (Original post by ghostwalker)
    Huh! What am I supposed to be looking at in that 19 page thread?
    Look at the original post and its attachments


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    (Original post by ghostwalker)
    Huh! What am I supposed to be looking at in that 19 page thread?
    I've got it, no problems!

    If say two trains reach a destination at the same time. And train B starts 5 seconds after train A. Which do we use T-5 for the time B reaches the destination if they both reach there at the SAME time???
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    (Original post by ps1265A)
    I've got it, no problems!

    If say two trains reach a destination at the same time. And train B starts 5 seconds after train A. Which do we use T-5 for the time B reaches the destination if they both reach there at the SAME time???
    If t is the time, from its start, of train A

    And t' is the time, from its start, of train B.

    Then when B starts (t'=0), A has been running for 5 scononds (t=5)

    So, t' = t-5

    I find it best to remember how the results is derived, rather than the final expression - you can then adapt it more easily to other situations, as necessary.
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    (Original post by ghostwalker)
    The ring is smooth (key point), so there is no additional force acting along the string, and tension is constant throughout.
    Is this true for a plank attached by a string?
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    (Original post by ps1265A)
    Is this true for a plank attached by a string?
    If it's attached you'll then it's not a smooth contact.

    If it's not attached and the contact is smooth then there is no force along the plank at that contact.
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    (Original post by ghostwalker)
    If it's attached you'll then it's not a smooth contact.

    If it's not attached and the contact is smooth then there is no force along the plank at that contact.
    Thanks!

    If a ball is projected upwards with speed of ums-1, then it will fall at ums-1 in my model as well?


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    (Original post by ps1265A)
    Thanks!

    If a ball is projected upwards with speed of ums-1, then it will fall at ums-1 in my model as well?


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    When it's at the same vertical displacment, i.e. s=0, it's velocity will have the same magnitude. You could have worked that out from v^2=u^2+2as
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    (Original post by ghostwalker)
    When it's at the same vertical displacment, i.e. s=0, it's velocity will have the same magnitude. You could have worked that out from v^2=u^2+2as
    Thanks!

    Could you clarify if there's tension/thrust present when a car which is attached to another object by a rope, is breaking? In some questions, there's thrust to consider but in some questions, tension is not considered? It makes sense tension is never considered when decelerating, but what about thrust?


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    (Original post by ps1265A)
    Thanks!

    Could you clarify if there's tension/thrust present when a car which is attached to another object by a rope, is breaking?
    If the car is breaking, there may be tension in the rope, but there can't be thrust, a rope doesn't support compression.

    In some questions, there's thrust to consider but in some questions, tension is not considered? It makes sense tension is never considered when decelerating, but what about thrust?
    Can't make sense of that as it stands.
 
 
 
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