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Another Mechanics Problem Using Vectors

  1. May 22, 2013 #1
    1. The problem statement, all variables and given/known data
    the unit vectors i and j are parallel to the cordinate axey Ox and Oy. Two particles A and B each have a mass of 0.4kg, they both leave the origin O at the same time, and move in the x-y plane. The velocity vectors at t time for A and B are [itex](4t\hat{i} + 6t\hat{j})m/s[/itex] and [itex](3t^2\hat{i}-2\hat{j})m/s[/itex] respectively.

    A: calculate in Joules the kinetic energy of B when t=2
    B: Find the acceleration vector of A and prove the force acting on A is constant
    C: Calcualte to the nearest 0.1N the magntiude of F and find to the nearest degree the angly between Oy and the line of action of F.
    D:Find the distance between A and B when t=2.

    I am unsure on all of them to be honest, I am not sure if I am using the vectors correctly and especially my integration as I cant see where any integration constants are.


    2. Relevant equations

    KE=0.5mv^2
    F=ma

    3. The attempt at a solution
    Part A:
    First I find the magnitude of the velocity of B
    [tex]
    \dot{r_b}= 3(2)^2\hat{i} - 2\hat{j} \\
    \dot{r_b} = 12\hat{i} - 2\hat{j} \\
    |\dot{r_b}|= \sqrt{12^2 + 2^2} = 12.17m/s
    [/tex]
    and then use that in the KE formula
    [tex]
    KE=\frac{1}{2}mv^2 \\
    KE=\frac{1}{2}0.4(12.17)^2 \\
    KE=0.2 \times 148.11 = 29.62J \\
    [/tex]

    Part B:
    I differentiate to get the acceleration vector
    [tex]
    \ddot{r_a}=4\hat{i}+6\hat{j}
    [/tex]
    and as the I and J components are not functions of time that means the force must be constant as the acceleration is constant. Is that correct?

    Part C:
    I use newtons second law to find F
    [tex]
    F=0.4(4\hat{i}+6\hat{j}) \\
    F=1.6\hat{i}+2.4\hat{j} \\
    |F|=\sqrt{1.6^2+2.4^2}=2.9N
    [/tex]
    And then to get the angle i did..
    [tex]
    arctan(\frac{2.4}{1.6})=56.31 \\
    90-56.31=34°
    [/tex]
    to the nearest degree, I took the 56.31 from 90 as the question wants the angle between F and the y-axis, i hope I done it correct.

    Part D:
    To find the distance I first found both of the position vectors by integrating.
    [tex]
    r_a=∫(4t\hat{i}+6t\hat{j}) \ dt = 2t^2\hat{i}+3t^2\hat{j} \\
    r_b=∫(3t^2\hat{i}+2\hat{j}) \ dt = t^3\hat{i}+2t\hat{j} \\
    [/tex]
    And it asks when t is 2 so then i found the position vectors for both when t=2
    [tex]
    r_a=2(2)^2\hat{i}+3(2)^2\hat{j} = 8\hat{i}+12\hat{j} \\
    r_b=(2)^3\hat{i}+2(2)\hat{j} = 8\hat{i}+4\hat{j}
    [/tex]
    And then taking b from a
    [tex]
    (8\hat{i}+12\hat{j})-(8\hat{i}+4\hat{j})=8\hat{j}
    [/tex]
    and then the magnitude
    [tex]
    \sqrt{8^2}=8m
    [/tex]

    Any help is appreciated. :)
     
    Last edited: May 22, 2013
  2. jcsd
  3. May 22, 2013 #2

    BruceW

    User Avatar
    Homework Helper

    your answers to parts A,B,C all look correct. But in part D, you use
    [tex]r_b=∫(3t^2\hat{i}+2\hat{j}) \ dt = t^3\hat{i}+2t\hat{j} \\[/tex]
    But I don't think this is right, because the velocity vector that was given is:
    [tex](3t^2\hat{i}-2\hat{j})m/s[/tex]
    Maybe you accidentally forgot the minus sign?
    p.s. excellent name :)
     
  4. May 23, 2013 #3
    Ah thanks for point that out!

    So should it be...

    [tex]r_b=∫(3t^2\hat{i}-2\hat{j}) \ dt = t^3\hat{i}-2t\hat{j} \\[/tex]
    and then taking the difference and magnitude then becomes..
    [tex]
    (8\hat{i}+12\hat{j})-(8\hat{i}-4\hat{j})=16\hat{j} \\
    \sqrt{16^2}=16m
    [/tex]
     
  5. May 23, 2013 #4

    BruceW

    User Avatar
    Homework Helper

    yep. looks good, nice work! pretty neat too. it makes it easy to read, so thanks!
     
  6. May 23, 2013 #5
    Thanks for your help :)
     
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