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A 1.00-kg duck is flying overhead at 1.50 m/s when a hunter fires straight up. The 0.010 0-kg bullet is moving 100 m/s when it hits the duck and stays lodged in the duck's body. What is the speed of the duck and bullet immediately after the hit?

  • physics -

    m1= 1kg
    v1= 1.50 m/s
    m2= 0.010kg
    v2= 100m/s

    m1*v1= 1kg*1.50kg= 1.50kg-m/s
    m2*v2= 0.010kg*100m/s= 1.0kg-m/s

    mTotal= 1.010kg

    Since the bullet is shot in a vertical movement, and the duck was flying horizontally, then we can use Pythagorean to solve for the final velocity(vF).

    Velocity in the x-component(vx)
    -> (mT)(vx)= momentum of duck.
    --> (1.010kg((vx)= 1.50kg-m/s
    ---> (vx)= (1.50kg-m/s)/(1.010kg-m/s)
    ----> (vx)= 1.485m/s
    Velocity in the y-component (vy)
    -> (mT)(vy)= 1.0kg-m/s
    --> (1.010kg)(vy)= (1.0kg-m/s)
    ---> (vy)= (1.0kg-m/s)/(1.010kg-m/s)
    ----> (vy)= (0.99009m/s)

    *Now we can use Pythagorean to solve for the final velocity!
    -> a^2 + b^2 = c^2
    --> (1.485m/s)^2 + (0.99009m/s)^2 = c^2
    ---> 2.2055m/s + 0.9818m/s = c^2
    ----> 3.1871m/s = c^2
    -----> c = √(3.1871m/s)
    ------> c = 1.785 m/s!
    * Your final velocity is 1.785m/s!

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