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Please help me with this applied maths question?

  1. Apr 29, 2012 #1
    (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 ?

    If you could answer and explain how to do either or even both of these it will possibly save me from failing an exam tommorow .Hope you can help, Thanks !
     
  2. jcsd
  3. Apr 29, 2012 #2

    tiny-tim

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    Welcome to PF!

    Hi Appliedmaths! Welcome to PF! :wink:

    For (a), integrate; for (b), use conservation of momentum

    show us what yout get. :smile:
     
  4. Apr 29, 2012 #3
    Re: Welcome to PF!

    I dont know how to integrate so is there any way around that for part 1 ?

    I know how to do conservation of momentum but how do i take into account the angle 60 degrees ?

    I never should have chosen applied maths as I have discovered maths is definately not my thing but inh this test tommorow I have to get at least 30 percent to pass and I can drop the subject. If I fail its 200euro to repeat. I can do all the short questions and a one dimensional kinematic question on my test but I dont know how to do this one.
    It has come up every year for the last 4 years so if i can understand this it will probably enable me to pass. Please understand where I am coming from. Thats why I dont have a good attempt at the question for you. Its less than a day to the exam.
     
  5. Apr 29, 2012 #4
    (a) F=ma, so find the acceleration of the particle by dividing your equation for F by m.
    You know that d = d0 + v0*t + (1/2)*a*t^2, depending upon which direction you're considering. In this case, you must consider every direction separately, at least that's what I think the equation for F hints at. So take every direction and use the formula for d and insert t=1. This will give you the end coordinates in x, y and z.

    (b)
    (i) Remember that momentum is conserved, so the momentum of the ball of mass m must equal the momentum of both of the balls afterwards, so mv = mv' + 2mv_2. Find v'!
    (ii) Use the fact that some vertical motion has been initiated from the collision. Before impact the vertical motion was zero, so the momentum for each ball and energy for each ball must be equal in the vertical motion. The vertical velocity is given by v*sin(angle). Equate the two expressions and solve for the angle.

    Hope this could help a little.
     
  6. Apr 29, 2012 #5

    tiny-tim

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    Hi Appliedmaths! :smile:
    Nooo. :redface:
    You need to do it twice, once for the x direction and once for the y direction.

    That gives you two equations for two unknowns, so you can solve it! :wink:

    Please show us what you get …

    you obviously need the practice. :smile:

    no, that's only for constant acceleration

    this isn't constant :redface:

     
  7. Apr 29, 2012 #6
    Ah, that's right.. Integration from t=0 to t=1 it is. Sorry for the misleading previous comment :/
     
  8. Apr 29, 2012 #7
    So is this wrong then ??

    F = ma
    F = ti - 6t^2j + 3k
    v = j + 2k
    mass = 4 kg

    ti - 6t^2j + 3k = (4)a

    ti - 6t^2j + 3k / 4 = a

    V = u + at

    j + 2k = 0 + (ti - 6t^2j + 3k / 4) 1

    4(j + 2k) = ti - 6t^2j + 3k
    4j + 8k = ti - 6t^2j + 3k
    = ti - 6t^2j - 4j + 3k - 8k
    = ti - 6t^2j - 4j - 5k
     
  9. Apr 30, 2012 #8

    tiny-tim

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