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    A particle A starts at the point with position vector 12i+12j. The initial velocity of A is (-i+j) ms^-1, and it has constant acceleration (2i-4j)ms^-2. Another particle B has initial velocity i ms^-1 and constant acceleration 2j ms^-2. After 3 seconds the two particles collide.

    Find
    A. The position vector of the point where the two particles collide
    B. The position vector of B at the starting point

    =========

    I know where in the case velocity is constant the position vector is (initial position vector +vt) but in this case there is an acceleration and I dont know how to take it into account

    This is supposed to be done without any calculus by the way
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    (Original post by L'Evil Wolf)
    A particle A starts at the point with position vector 12i+12j. The initial velocity of A is (-i+j) ms^-1, and it has constant acceleration (2i-4j)ms^-2. Another particle B has initial velocity i ms^-1 and constant acceleration 2j ms^-2. After 3 seconds the two particles collide.

    Find
    A. The position vector of the point where the two particles collide
    B. The position vector of B at the starting point

    =========

    I know where in the case velocity is constant the position vector is (initial position vector +vt) but in this case there is an acceleration and I dont know how to take it into account

    This is supposed to be done without any calculus by the way
    Step1. Find the distance travelled by A in terms of i and j using suvat
    s=ut +1/2at^2
    =3(-i+j) + .5(2i-4j)*3^2
    =-3i+3j+9i-18j
    =6i+15j
    Add this to the original position of 12i+12j to get 18i+33j
    Part B is simply position vector eqn using this result as you have v and t
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    what paper?
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    (Original post by L'Evil Wolf)
    A particle A starts at the point with position vector 12i+12j. The initial velocity of A is (-i+j) ms^-1, and it has constant acceleration (2i-4j)ms^-2. Another particle B has initial velocity i ms^-1 and constant acceleration 2j ms^-2. After 3 seconds the two particles collide.

    Find
    A. The position vector of the point where the two particles collide
    B. The position vector of B at the starting point

    =========

    I know where in the case velocity is constant the position vector is (initial position vector +vt) but in this case there is an acceleration and I dont know how to take it into account

    This is supposed to be done without any calculus by the way
    What paper was this? Link pls?
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    (Original post by thelegend99)
    What paper was this? Link pls?
    Edexcel 6D Q12 or omething
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    Hi, Just wondering if anyone can help with Jan06 Q1!

    It says to find the value of h, which is a distance above the ground, I've attached a picture showing this, However what I don't understand is, the mark scheme says to use the suvat shown on the right (shown in pic), but doesn't that suvat correspond to the area highlighted in red I've selected in the pic? How does it find h? Thanks
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    (Original post by ryandaniels2015)
    Hi, Just wondering if anyone can help with Jan06 Q1!

    It says to find the value of h, which is a distance above the ground, I've attached a picture showing this, However what I don't understand is, the mark scheme says to use the suvat shown on the right (shown in pic), but doesn't that suvat correspond to the area highlighted in red I've selected in the pic? How does it find h? Thanks
    S is displacement, so the displacement is just h and not the whole of the red thing as the displacement when it goes to its peak and goes to the same kevel as it started cancel.



    Also another way yoy can think of it is that when it goes to its peak and falls to the same level as it was thrown from, it would have the same speed
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    (Original post by metrize)
    S is displacement, so the displacement is just h and not the whole of the red thing as the displacement when it goes to its peak and goes to the same kevel as it started cancel.



    Also another way yoy can think of it is that when it goes to its peak and falls to the same level as it was thrown from, it would have the same speed
    Thanks for the reply, I've attached a pic, so do the 2 displacement cancel (shown in red) so you are left with the blue h shown in pic?
    Attached Images
     
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    (Original post by ryandaniels2015)
    Thanks for the reply, I've attached a pic, so do them 2 displacement cancel (shown in red) so you are left with the blue h shown in pic?
    Yes, displacement is a vector quantity so let us model down as the positive direction, as it goes up, it would have a negative displacement going up and when it goes down to the same level it would have the positive displacement of the same value. When it falls it will have a further displacement of h, so when added all you have left is h.

    Hope this makes sense, if it doesnt i can try to explain it mkre clearly
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    (Original post by metrize)
    Yes, displacement is a vector quantity so let us model down as the positive direction, as it goes up, it would have a negative displacement going up and when it goes down to the same level it would have the positive displacement of the same value. When it falls it will have a further displacement of h, so when added all you have left is h.

    Hope this makes sense, if it doesnt i can try to explain it mkre clearly
    Makes a lot of sense, Thanks a lot!!
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    Can someone please tell me any hard questions in M1 past papers? I don't think I'll have time to go through all of them
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    Hi
    Can anyone help me with regards to solving questions about connected particles on a pulley or on a table and when one falls and the rope becomes slack or breaks. I get very confused with how to start to approach the problem to work out maximum height, total distance etc
    Thank You
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    (Original post by Marccs)
    Hi
    Can anyone help me with regards to solving questions about connected particles on a pulley or on a table and when one falls and the rope becomes slack or breaks. I get very confused with how to start to approach the problem to work out maximum height, total distance etc
    Thank You
    General idea is this.
    Stage 1. The distance the particle travels until it hits the floor is the same as the other particle will travel along the table
    Stage 2. The speed that the particle hits the floor (v) will be the initial speed for the other particle's next stage of travel
    This next stage also needs a new calculation for a, as F=ma no longer includes the force from the other particle
    Use suvat to calculate new distance (v=0, u=old v ,a = new, s=what we are looking for
    Stage 3 Finally add both distances travelled etc

    hope this helps
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    (Original post by candol)
    General idea is this.
    Stage 1. The distance the particle travels until it hits the floor is the same as the other particle will travel along the table
    Stage 2. The speed that the particle hits the floor (v) will be the initial speed for the other particle's next stage of travel
    This next stage also needs a new calculation for a, as F=ma no longer includes the force from the other particle
    Use suvat to calculate new distance (v=0, u=old v ,a = new, s=what we are looking for
    Stage 3 Finally add both distances travelled etc

    hope this helps
    Ohh so as the first particle drops the second particle will move up at velocity of 0 and then final velocity of the other particle?
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    (Original post by Marccs)
    Ohh so as the first particle drops the second particle will move up at velocity of 0 and then final velocity of the other particle?
    The particle will move up with same velocity as the other is going down. The moment the second particle hits the floor, the velocity at that moment becomes the starting velocity (u) for the next stage of motion
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    (Original post by candol)
    The particle will move up with same velocity as the other is going down. The moment the second particle hits the floor, the velocity at that moment becomes the starting velocity (u) for the next stage of motion
    But sometimes the othher particles velocity is immediately the final velocity of the particle that fell and not the second stage ?
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    Name:  f88a22f91af8423ffef551e0f2aea0a3.png
Views: 106
Size:  57.5 KBJune 2015, not sure how to do part b.
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    (Original post by Marccs)
    But sometimes the othher particles velocity is immediately the final velocity of the particle that fell and not the second stage ?
    the key is that final velocity becomes the initial velocity for the next stage of motion
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    (Original post by Thomith)
    Name:  f88a22f91af8423ffef551e0f2aea0a3.png
Views: 106
Size:  57.5 KBJune 2015, not sure how to do part b.
    Watch this
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    Somone please help with this q

    http://www.thestudentroom.co.uk/show....php?t=1255487
 
 
 
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