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Need quick help with this applied maths Q !

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  • 29-04-2012 7:46pm
    #1
    MannyFagnet
    Registered Users Posts: 32 ✭


    (a) A particle of mass 4 kg moves under a force: F = 6t^2i - tj - 4k in Newtons.
    Assuming that the particle is initially at the origin with velocity v = i + k in m/s, find its position after 1s.

    (b) A ball of mass m travels with velocity 3 m/s and collides with a second ball with mass 2m at rest. After the collision the 2m ball moves with speed 1 m/s in a direction 60degrees to the original direction of the ball m.
    (i) What is the final velocity of the ball m ?
    (ii) What is the angle between the paths of the two balls after the collision ?

    Exam is tommorow so really need to understand this !
    Thanks
    Tagged:


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    #2
    bnt
    Registered Users Posts: 12,962 ✭✭✭✭bnt


    (a) F=m a , and you need the acceleration if you're going to calculate motion, so a = F / 4 = 1/4 F, where F is the formula you're given.
    Then it's just the standard formula for motion: s = vt + ½ a t². You have all the bits you need, and t = 1 sec. Lots of algebraic manipulation.

    (b) is a conservation of momentum question, where Q (momentum) = m v. The system will have the same amount before and after the collision. So you calculate Q of the moving ball #1 before the collision, subtract Q of ball #2 after, which leaves the Q of ball #2 after the collision.
    Then you can calculate its speed (Q/m), and the sign will tell you its direction e.g. it's negative, it's moving away from ball #2, so the angle would be 180°.

    From out there on the moon, international politics look so petty. You want to grab a politician by the scruff of the neck and drag him a quarter of a million miles out and say, ‘Look at that, you son of a bitch’.

    — Edgar Mitchell, Apollo 14 Astronaut



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    #3
    ride-the-spiral
    Registered Users Posts: 1,263 ✭✭✭ride-the-spiral


    Is this LC applied maths? I don't remember having to deal with non-constant acceleration in anything other than the differential equations section.

    But in the force expression, if t stands for time as it usually does then you get a non-constant acceleration. a=F/m => a=(3/2)t^2 i - t/4 j - k.

    The s=vt + 1/2at^2 formula is only valid of a is constant, so to calculate distance you should have to integrate. The way that I've learnt is simply

    v= (initial velocity) + (integral from 0 to t) a dt
    r= (initial position) + (integral from 0 to t) v dt

    Using this way means you can leave everything in it's vector form and integrate each component.

    or alternatively.

    (integral from v(0) to v(t)) dv = (integral from 0 t) a dt

    Apologies for the lack of latex and horrible layout (It's quite late) and if I'm assuming you know too much, but I'm not sure where you stand on integration or anything.


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    #4
    bnt
    Registered Users Posts: 12,962 ✭✭✭✭bnt


    Hmm. I didn't spot the t in the force formula, which means that F (and therefore a) is increasing from zero as t increases from zero. So I agree that the solution is going to involve calculus on the acceleration formula, as noted: two integration steps, adding in the V formula after the first (a -> v). After the second step (v -> s) your constant of integration is zero, because it starts at the origin (0). What pedantic sadist comes up with these questions? :o

    From out there on the moon, international politics look so petty. You want to grab a politician by the scruff of the neck and drag him a quarter of a million miles out and say, ‘Look at that, you son of a bitch’.

    — Edgar Mitchell, Apollo 14 Astronaut



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    #5
    MathsArthur
    Registered Users Posts: 9 ✭MathsArthur


    a = 6t^2i - tj - 4k

    v=v(0) + INT (a) dt = v(0) + 2t^3i - t^2/2j - 4tk

    =(2t^3+1)i - (t^2/2)j - (4t-1)k

    s=s(0) + INT (v) dt = s(0) + (t^4/2+t)i - (t^3/6)j - (2t^2-t)k

    s(0)=0 so

    s(1)= (1/2 + 1)i - (1/6)j - (2-1)k

    =(3/2)i - (1/6)j - k


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