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    The question is in the Edexcel M2 book, exercise 3E.

    Picture: yfrog.com/h0exisagj

    I worked out a) = 3.92m/s/s and that's correct in the answer book, but I'm completely stumped on b)

    I keep getting statements like 11.78m = 11.78m :facepalm:

    I tried working it out without using energy - and found when the string breaks the velocity of A is 2.8m/s up the slope, and the acceleration to be 7.84m/s/s down the slope.

    Using SUVAT this gives me an answer of 2m, but the book says 1.5m, and I'm not using the correct method anyway. Any ideas?
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    (Original post by Wilko94)
    The question is in the Edexcel M2 book, exercise 3E.

    Picture: yfrog.com/h0exisagj

    I worked out a) = 3.92m/s/s and that's correct in the answer book, but I'm completely stumped on b)

    I keep getting statements like 11.78m = 11.78m :facepalm:

    I tried working it out without using energy - and found when the string breaks the velocity of A is 2.8m/s up the slope, and the acceleration to be 7.84m/s/s down the slope.

    Using SUVAT this gives me an answer of 2m, but the book says 1.5m, and I'm not using the correct method anyway. Any ideas?
    After the string breaks, if there was no friction, the acceleration down the slope would be 0.6g=5.88. And since there is friction, it's going to be less than that, so 7.84 is not going to be correct.

    Your velocity of 2.8m/s up the slope is correct.
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    (Original post by ghostwalker)
    After the string breaks, if there was no friction, the acceleration down the slope would be 0.6g=5.88. And since there is friction, it's going to be less than that, so 7.84 is not going to be correct.

    Your velocity of 2.8m/s up the slope is correct.
    I thought that as it still had velocity going up the slope, it was still moving up, and therefore friction was acting down the slope, leading to an increased acceleration down?

    So Fnet = ma

    0.25*0.8mg (Friction) + 0.6mg (Gravity acting down the slope) = ma

    1/5mg + 0.6mg = ma

    0.8mg = ma
    0.8g = a = 7.84m/s/s
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    (Original post by Wilko94)
    I thought that as it still had velocity going up the slope, it was still moving up, and therefore friction was acting down the slope, leading to an increased acceleration down?

    So Fnet = ma

    0.25*0.8mg (Friction) + 0.6mg (Gravity acting down the slope) = ma

    1/5mg + 0.6mg = ma

    0.8mg = ma
    0.8g = a = 7.84m/s/s
    Yes you're right, my apologies. Brain cell clearly not functioning. :sigh:
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    In which case, applying

    v^2=u^2+2as gives s=0.5, which when you add the initial one, gives the desired result.
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    (Original post by ghostwalker)
    Yes you're right, my apologies. Brain cell clearly not functioning. :sigh:
    No worries, you're the only person to help and you've helped me before.

    Do you have any idea otherwise?
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    (Original post by Wilko94)
    Do you have any idea otherwise?
    See my last post.
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    (Original post by ghostwalker)
    In which case, applying

    v^2=u^2+2as gives s=0.5, which when you add the initial one, gives the desired result.
    :facepalm: I've spent so long on this question - I did 15.68/7.84, not the other way around. Thank you, +rep.

    Do you have any idea how to do it energy wise? I think you have to work out m, but I can't. :confused:
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    Ever since I saw this post I've been trying to do B energy wise and its turning into a mess.
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    (Original post by Wilko94)
    :facepalm: I've spent so long on this question - I did 15.68/7.84, not the other way around. Thank you, +rep.

    Do you have any idea how to do it energy wise? I think you have to work out m, but I can't. :confused:
    Energywise you don't need to work out "m", as it cancels out.

    I'll give you the equations as I balls'd up the previous.

    There are two parts as before; working out the velocity after the masses have gone a metre, and then working out the distance travelled.

    Spoiler:
    Show

    For the velocity:

    Loss in PE = Gain in KE + Work done against friction.

    2mg -mg\sin\theta = \frac{1}{2}3mv^2 +\frac{mg}{4}\cos\theta

    Which yields v^2=0.8g

    And then for the distance; if we call the distance x.

    Loss in KE = gain in PE + work done against friction.

    \frac{1}{2}mv^2 = xmg\sin\theta + x\frac{mg}{4}\cos\theta

    Which boils down to 0.4mg = 0.8xmg

    and x = 0.5

    and add 1.
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    (Original post by ghostwalker)
    For the velocity:

    Loss in PE = Gain in KE + Work done against friction.

    2mg -mg\sin\theta = \frac{1}{2}3mv^2 +\frac{mg}{4}\cos\theta
    Please could you explain this bit? Sorry.
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    (Original post by Wilko94)
    Please could you explain this bit? Sorry.
    What part? The original equation, the values assigned to each part, or both?
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    (Original post by ghostwalker)
    What part? The original equation, the values assigned to each part, or both?
    Both please - I can understand the equation but I don't see where it comes from, and I can see how you've got friction but not anything else.
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    (Original post by Wilko94)
    Both please - I can understand the equation but I don't see where it comes from, and I can see how you've got friction but not anything else.
    Loss is PE, is due to the 2m mass going down by 1 metre, and the m mass going up by 1\times\sin\theta metres

    Gain in KE.
    Masses start from rest, and after moving 1 metre, both are travelling at the same speed v (as yet unknown) and total mass is 3m, hence...
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    Alternative derivation of equation.

    Energy before = energy after + work done.

    So, with the obvious suffixes,

    PEb + KEb = PEa + KEa + Fd

    Re-arranging.

    (PEb-PEa) = (KEa-KEb) +Fd

    KEb = 0, hence...

    PEb-PEa = KEa +Fd

    LHS is just the loss in PE.
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    (Original post by ghostwalker)
    Alternative derivation of equation.

    Energy before = energy after + work done.

    So, with the obvious suffixes,

    PEb + KEb = PEa + KEa + Fd

    Re-arranging.

    (PEb-PEa) = (KEa-KEb) +Fd

    KEb = 0, hence...

    PEb-PEa = KEa +Fd

    LHS is just the loss in PE.
    For which particle are we talking about here? I can see where 2mg comes from, for particle B, but where does mgsintheta (PEa) come from?
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    (Original post by Wilko94)
    For which particle are we talking about here? I can see where 2mg comes from, for particle B, but where does mgsintheta (PEa) come from?
    This first equation covers both particles.

    "mgsintheta (PEa)" is for particle A, the mass on the slope. As it moves one metre up the slope, the gain in height is sintheta, and gain in potential energy is mgsintheta.
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    (Original post by Killjoy-)
    ghostwalker had the distance moved up the plane as 1 and the distance moved down as 0.5...
    Did I?

    Had a mix-up at the start that was corrected.

    Particle A is going UP the plane.
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    (Original post by ghostwalker)
    Did I?

    Had a mix-up at the start that was corrected.

    Particle A is going UP the plane.
    I think I made a mistake, did you get particle A going up the plane 0.5 metres before descending 1 metre- or the other way around?
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    (Original post by Killjoy-)
    I think I made a mistake, did you get particle A going up the plane 0.5 metres before descending 1 metre- or the other way around?
    Neither.

    Particle A goes up the plane by 1 metre before the string breaks (you're told this in the question), and then by another half metre after the string has broken.
 
 
 
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