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    (Original post by king of swords)
    It would help if you knew whether the collision was inelastic or elastic.
    In the meantime....work out friction on A from F = μR, where R = weight...then work out acceleration of that particle as a result to work out velocity of A just before impact.,
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    (Original post by king of swords)
    In the meantime....work out friction on A from F = μR, where R = weight...then work out acceleration of that particle as a result to work out velocity of A just before impact.,
    F=(0.05)(0.15g)

    F= 0.0735
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    (Original post by 2776)
    F=(0.05)(0.15g)

    F= 0.0735
    So acceleration...is -0.0735/0.15
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    (Original post by king of swords)
    So acceleration...is -0.0735/0.15
    how did u get that from?
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    (Original post by 2776)
    how did u get that from?
    well it's a = F/m.......where F is the sum of forces (just friction) and m is the mass of particle...notice there is a - sign because friction acts in the opposite direction to motion.
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    (Original post by king of swords)
    well it's a = F/m.......where F is the sum of forces (just friction) and m is the mass of particle...notice there is a - sign because friction acts in the opposite direction to motion.
    so acceleration = (-0.49)

    and put that in formulae:

    v sqrd = 4 - (2x (-0.49) x 4)

    v = 0.28

    therefore

    (0.28x0.15) + 0= (0.2 x v) +0

    v of B= 0.212
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    (Original post by 2776)

    and put that in formulae:

    v sqrd = 4 - (2x (-0.49) x 4)
    v sqrd = u sqrd + 2 x a x s
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    (Original post by 2776)
    v sqrd = u sqrd + 2 x a x s
    v = 2.8 not 0.28.i tell a lie sorry lol. i'm sleepy.
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    (Original post by king of swords)
    well it's a = F/m.......where F is the sum of forces (just friction) and m is the mass of particle...notice there is a - sign because friction acts in the opposite direction to motion.
    One thing about this. I have never heard of a=f/m can u explain this bit a bit more. I'll give u rep.
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    (Original post by king of swords)
    v = 2.8 not 0.28.i tell a lie sorry lol. i'm sleepy.
    So now you have "momentum before = momentum after".
    0.15kg * 0.28m/s + 0.2kg * 0m/s = 0.15kg * 0m/s + 0.2kg * V.
    Isolate V.
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    (Original post by king of swords)
    So now you have "momentum before = momentum after".
    0.15kg * 0.28m/s + 0.2kg * 0m/s = 0.15kg * 0m/s + 0.2kg * V.
    Isolate V.
    Ive done that already before, I got 0.212.
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    (Original post by 2776)
    One thing about this. I have never heard of a=f/m can u explain this bit a bit more. I'll give u rep.
    It's from Newton's Second Law: F = ma. So F/m = a....the sum of forces on that particle are reaction and weight (they cancel out) and the friction (which acts against the motion of the particle)...so the sum of the forces (ie. friction) over the mass of the particle gives the acceleration of that particle (it is negative in this case because it is a deceleration).
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    (Original post by king of swords)
    It's from Newton's Second Law: F = ma. So F/m = a....the sum of forces on that particle are reaction and weight (they cancel out) and the friction (which acts against the motion of the particle)...so the sum of the forces (ie. friction) over the mass of the particle gives the acceleration of that particle (it is negative in this case because it is a deceleration).
    A particle p travels in a straight line from O to A. And then back to O. At time t seconds after starting from O, the displacement of P from O is x metres, where x=2 (t cubed) - t (to the power of 4)

    I differentiated:

    so velocity = 6 (t squared) - 4 (t cubed)

    and acceleration = 12t-12 (t squared)

    So the value of t the instant P returns to O:

    t cubed ( 2-t) =0 therefore t=0 or t=2.

    The value of t at the instant when p reaches A: When t=2, v= -8

    (is this correct?)
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    (Original post by 2776)
    A particle p travels in a straight line from O to A. And then back to O. At time t seconds after starting from O, the displacement of P from O is x metres, where x=2 (t cubed) - t (to the power of 4)

    I differentiated:

    so velocity = 6 (t squared) - 4 (t cubed)

    and acceleration = 12t-12 (t squared)

    So the value of t the instant P returns to O:

    t cubed ( 2-t) =0 therefore t=0 or t=2.

    The value of t at the instant when p reaches A: When t=2, v= -8

    (is this correct?)
    If A is the furthest away from O the particle can reach before returning back to O then v at A = 0 not -8. so t = 1½s.
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    (Original post by king of swords)
    If A is the furthest away from O the particle can reach before returning back to O then v at A = 0 not -8. so t = 1½s.
    i.e. since u can't have a negative velocity, t must = 0?
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    (Original post by 2776)
    i.e. since u can't have a negative velocity, t must = 0?
    You can have a -ve velocity...since it is a positive speed just going in the -ve direction...what I'm saying is....when the particle has reached its furthest distance from O before coming back to O (ie at point A)...it must come to instantaneous rest and therefore has a velocity of 0....as you worked out v = 6t^2 - 4t^3 where ^ means "to the power". t is therefore either 0 or 1½ but we know that when t=0 it is at point O not at point A.
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    (Original post by king of swords)
    You can have a -ve velocity...since it is a positive speed just going in the -ve direction...what I'm saying is....when the particle has reached its furthest distance from O before coming back to O (ie at point A)...it must come to instantaneous rest and therefore has a velocity of 0....as you worked out v = 6t^2 - 4t^3 where ^ means "to the power". t is therefore either 0 or 1½ but we know that when t=0 it is at point O not at point A.
    Yes, but I'm tryingt o find the velocity of the particle when it comes back to O, the starting point.
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    (Original post by 2776)
    Yes, but I'm tryingt o find the velocity of the particle when it comes back to O, the starting point.
    Work out the distance O-->A on the way out.
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    (Original post by king of swords)
    Work out the distance O-->A on the way out.
    Ive found the value of t when p returns to O. It is 0 or 2.

    So it must be 2.

    Put that into equation, I get, v= -8.

    Does that make sense? as -8 means it is reversing, so it must be true.
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    (Original post by king of swords)
    Work out the distance O-->A on the way out.
    In fact don't bother doing that: just say x = 0 and work out t....then plug in t into the equatio to find v.
 
 
 
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