Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Impulse-Momentum Theorem Q.

  1. Jun 30, 2005 #1
    I have this problem that says a 3 kg object has a velocity of 7.00m/s j (vertical, right?). Then a total of 12.0Ni (horizontal) acts on it for 5 seconds. What is the objects final velocity.

    I know that the equation is Ft=mvf-mvi

    I could just plug everything in, but what's bothering me is the fat that the mass has an vertical velocity and the force is acting on it horizontally. Am I reading this wrong or is this the case?

    If so, how do I go avout finding what I need to plug in.

    My work:

    Ft=mvf-mvi
    (12N)(5s)=(5)(vf)-(5)(7)... then solve for vf, but that seems too easy.
     
  2. jcsd
  3. Jun 30, 2005 #2
    The object is moving in the vertical direction with that velocity, yes and then the force acts on it for 5 seconds. So yeah, that is the case.

    You still use [itex]\vec{F}t = m(\vec{v_{f}} - \vec{v_{i}})[/tex] and solve for [itex]\vec{v_{f}}[/itex], but writing the force and inital velocity in terms of vectors and not just their magnitudes.
     
    Last edited: Jun 30, 2005
  4. Jun 30, 2005 #3

    OlderDan

    User Avatar
    Science Advisor
    Homework Helper

    The motion in the x direction is independent of the motion in the y direction. Since there is no y-directed force, the y-velocity will be unchanged. In you last equation, you should only have x-directed quantities. You substituted the wrong number for mass, and you have the wrong initial velocity.

    After you get the final x-velocity, you need to add it to the y-velocity by vector addition.

    OOPs- scooped again :smile:
     
  5. Jun 30, 2005 #4
    ohhh i didn't mean to put in 5 as the mass :)
    so i have to find the initial velocity first? it's not just 7 huh.....
    then find the final V and doo Vector addition,

    ok thank you all very much!!!
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook