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Newton's Third Law Problem

  1. Oct 8, 2013 #1
    1. The problem statement, all variables and given/known data
    A paintball with a mass of 0.15 kg is fired from a paintball gun that has a mass of 5.5kg. The paintball leaves the gun with a velocity of 45 m/s [N] having accelerated for 0.10s. Calculate the acceleration and the final velocity of the paintball gun.
     
  2. jcsd
  3. Oct 9, 2013 #2
    Here, you would use Newton's third law, and that the forces should equal. "For every action, there is an equal, but opposite reaction." The paintball gun forces the paintball in the +x direction, the ball pushes the gun in the -x direction. (this is why you feel recoil when shooting).

    Find the force that's acting on the paintball.

    Fball = mballaball

    aball = (vf - vi)/t

    This is all given to you in your question. simply solve for the Force. Since the total force must equal 0

    Fball + Fgun = 0

    Fball = - Fgun

    Here, you can equate the force that's on the gun. Since you know the mass of the gun, you can find its acceleration. Then using the same equation as before,

    agun = (vf - vi)/t

    And just solve for the final velocity. (the initial should be zero as it starts from rest, before the gun is fired) (and the t value is the same as before. It's only accelerating as long as the ball enacts on the gun, so the time is 0.10s)
     
  4. Oct 9, 2013 #3
    When calculating the acceleration of the ball would I use 0 as the final velocity?
     
  5. Oct 9, 2013 #4
    The final velocity would be the velocity it has when it leaves the barrel. So in your initial question, you say the ball leaves the gun with a velocity of 45 m/s . The initial velocity of the ball would be zero as both the gun and the ball are at rest, before the gun is shot.
     
  6. Oct 9, 2013 #5
    aball= Vf-Vi/t
    = 45-0/0.10
    = 450m/s^2

    Fball=ma
    =(0.15)(450)
    =67.5 N

    Fgun= -67.5N

    Fnet=ma
    -67.5=(5.5)(-a)
    agun=12.27 m/s^2

    agun=Vf-Vi/t
    12.27 =Vf-0 / 0.10
    Vf = 1.2 m/s

    Is this correct?
     
  7. Oct 9, 2013 #6
    Looks correct! Just remember that the velocity of the gun is in the opposite direction of the ball. Just like when you shoot a bullet, the bullet goes one way, and the gun goes in the other (the recoil). (If it's going to be graded, it might be beneficial to write down the direction of the velocity, in case they look for that).
     
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