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How to find the acceleration of 2 objects under same force

  1. Oct 16, 2007 #1


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    1. The problem statement, all variables and given/known data
    A force F to an object m1 produces an acceleration of 3m/s^2. The same force is applied to a second object of mass m2 produces an acceleration of 1.00m/s^2

    a.) value ratio of m1/m2

    b.) If m1 and m2 are combined into 1 object, what is the acceleration under the action of force [tex]\vec{} F [/tex]?

    2. Relevant equations

    3. The attempt at a solution

    a.) m1/m2= a2/a1

    m1/m2= 1.00m/s^2/ 3.00m/s^2

    b.) for b I'm not sure what they mean by combine the masses but m1 + m2 = ?

    I'm not sure if I can just combine the accelerations since they are vector quantities but mass is scalar so it can be combined...

    I need help with part b.

    Thanks :smile:
  2. jcsd
  3. Oct 16, 2007 #2
    For part a, don't forget that the units cancel out.
    For part b, solve for m1 and m2 in terms of F. Then you put (m1+m2) in terms of F, which you plug back into F=ma, then solve for a.
  4. Oct 16, 2007 #3


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    well for part
    a) m1/m2= .333

    b) F= m1(3.00m/s^2)
    m1= F/(3.00m/s^2)
    F= m2(1.00m/s^2)
    m2= F/(1.00m/s^2)

    m1 + m2= F/(3.00m/s^2) + F/ (1.00m/s^2) = F/(3.00m/s^2) + 3F/(3.00m/s^2) = 4F/(3.00m/s^2)

    then if I did that right..I plug that into F=ma...

    F= 4F/3.00m/s^2 * a

    3.00m/s^2(F)= 4F*a
    a= (3.00m/s^2)(F)/4F

    Is this fine??
  5. Oct 16, 2007 #4
    The F's cancel, and its neater if you write 3/4 m/s^2, imo. But yes, that's right
  6. Oct 16, 2007 #5


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    Oh..okay..Thank You =D
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