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Find velocity of gun and bullet on frictionless surface

  1. Mar 25, 2012 #1
    1. The problem statement, all variables and given/known data
    A 30.0-kg gun is standing on a frictionless surface. The gun fires a 50.0-g bullet with a muzzle velocity of 310 m/s.

    a) calculate the momenta of the bullet and the gun after the fun was fired.

    b) calculate the kinetic energy of both the bullet and the gun just after firing.


    2. Relevant equations

    Mgun*Vgun + Mbullet*Vbullet= 0 before firing

    KE= 1/2 M*V^2

    3. The attempt at a solution

    a) Before the firing, the system is at rest and therefore the momentum is ZERO. After firing, the total momentum is still ZERO. There is no net momentum gained or lost in this system. ----> Here, I am just looking if anyone could let me know if my reasoning is correct, and I was also wondering if there are anything else I should show in order to get full mark value?

    b) After firing
    0= (30 kg) Vgun+ (0.05kg) (310 m/s)
    Vgun=[(0.05kg)(310m/s)]/(30kg) = 0.517m/s

    Ke=1/2 M*V^2
    for bullet after firing: KE= 1/2 (0.05kg)(310m/s)^2=2402.5J= 2.40x10^3J

    for the gun after firing: KE=1/2 (30kg)(0.517m/s)^2=4.01J

    Also here, I am just looking for someone to check that I did this problem correctly and also to check that the SIGNIFICANT FIGURES are correct. Thank you so much in advance.


    b)
     
  2. jcsd
  3. Mar 25, 2012 #2
    a) You are correct that the NET momentum is zero, but the bullet and the gun still have momentum. They are simply acting in opposite directions, so you can still calculate the momentum.

    b) Your answers look right to me, although without any rounding at all, I calculated the KE for the gun to be 4.00J. Your sig figs are fine.
     
  4. Mar 25, 2012 #3

    gneill

    User Avatar

    Staff: Mentor

    Yes, that describes the situation adequately. You might at this point introduce the coordinate system you are going to use for the next part; is the bullet's direction of motion taken to be positive or negative?
    The momenta are vector quantities, so you should indicate their directions by appropriate sign. Note that when you solved for Vgun using the momentum law you dropped the sign change that should have happened when one term was moved to the other side of the equation.

    Your significant figures look okay.
     
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