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Bullet in Block- 2D Projectiles w/ Motion

  1. Nov 30, 2011 #1
    1. The problem statement, all variables and given/known data

    A(n) 9 g bullet is fired into a 396 g block
    that is initially at rest at the edge of a frictionless table of height 0.9 m. The bullet remains
    in the block, and after impact the block lands
    2.3 m from the bottom of the table.
    The acceleration of gravity is 9.8 m/s.

    2. Relevant equations

    ΔX = Vi(t) + (1/2)(a)(t^2)

    m(Vf-Vi) = -m(Vf-Vi)

    3. The attempt at a solution
    I separated the equation into vertical and horizontal parts.

    Horizontal: ΔX=2.3, a = 0 m/s/s
    Vertical: ΔY= .9, a= -9.8 m/s/s, Vi = 0 m/s (because it is launched horizontally).

    Using the kinematic equation ΔX = Vi(t) + (1/2)(a)(t^2) to solve for the vertical time:
    t = 0.4285714286 seconds

    Since the vertical time is equal to the horizontal time, I plugged this time into the same kinematic equation to solve for the horizontal Vi:
    Vi = 5.366666667 m/s

    Now, I know you need to use the collision equation to solve for the initial velocity in the bullet, but I do not know how to set it up...
    I have this so far

    9(5.366666667-vi)=-____?_____

    Thank you for the help :)
     
  2. jcsd
  3. Nov 30, 2011 #2
    m(Vf-Vi) = -m(Vf-Vi) That is not the collision equation I am familiar with. With just a single m, it suggests a collision involving a single object, an oxymoron if I've ever heard one. Try m1v1i+m2v2i=m1v1f+m2v2f
    You should then be able to quickly simplify one side by equating one velocity with another, and it will all fall out from there.
     
  4. Nov 30, 2011 #3
    What would m2, v2i, and v2f be?
     
  5. Nov 30, 2011 #4
    m2 would be mass of the block, v2i is the initial velocity of the block, and v2f is the final velocity of the block
     
  6. Nov 30, 2011 #5
    I got it, thanks!
     
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