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Change in Momentum Problem

  1. Oct 20, 2014 #1
    Question
    A 0.0420-kg hollow racquetball with an initial speed of 12.0 m/s collides with a backboard. It rebounds with a speed of 6.0 m/s.
    a. Calculate the total impulse on the ball.
    b. If the contact time lasts for 0.040 s, calculate the average force on the ball.

    What I did on the test:
    (A)Impulse=F x change in T=change in momentum(mass x velocity)

    F= unknown, so use the formula, F=M x A

    _________________________

    Momentum=

    P(momentum)= 0.0420kg x 6m/s

    P= .252N

    Therefore, the impulse on the ball is, .252kg. m/s.

    ****************************************************************

    (B) F= Mass x Acceleration

    change in Velocity= Acceleration x Time

    Acceleration= change in Velocity/ Time

    A= Velocity final - Velocity initial/ Time

    A= 6m/s -12m/s/0.040s

    A= -6m/s/0.040s

    A= - 150 m/s^2

    _____________

    F= M x A

    F= 0.042kg x (-150m/s^2)

    F= -6.3 N
     
    Last edited by a moderator: Oct 20, 2014
  2. jcsd
  3. Oct 20, 2014 #2
    (-6) - 12 = -18
     
  4. Oct 22, 2014 #3
    Why is the "6" negative?
     
  5. Oct 22, 2014 #4
    Because after it bounces off the wall it moves in the opposite direction.
     
  6. Oct 23, 2014 #5
    Ohhh....Okay.
    so, 18 x 0.0420kg
    P = .756N?
     
  7. Oct 24, 2014 #6
    Are you sure about the units of P (if you indeed mean that P is the impulse of the force)?

    Chet
     
  8. Oct 24, 2014 #7
    Nm
     
  9. Oct 24, 2014 #8
    No. You're just guessing.

    What are the units of mass? What are the units of velocity? What are the fundamental units of a Newton?

    Chet
     
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