1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Equal and Opposite Force Help

  1. Jul 25, 2011 #1
    So, this isn't actually homework, but I figured since it is so low-leveled, it qualifies.

    I am a programmer writing a little physics engine, just for fun. So this is all theoretical physics I guess. I have very little physics training above high school physics, so bare with me.

    Basically, pretend there is a box flying towards a wall at a certain velocity. When it hits the wall, let's call the velocity it was at when it hit it, v. And since the wall is not moving, the velocity of impact would be v, correct? So, an equal and opposite force acts on the box. So I assumed I would just subtract its current velocity v by that force, also v. But then, that would zero out and it wouldn't bounce back. So, then I thought, well maybe its current velocity gets reset to 0, and THEN we subtract v, ending up with -v. But that seems too strong.. I mean, in real life, if you throw something at a wall, it doesn't bounce back at the velocity that it hit the wall (I tried just a few minutes ago to check).

    So what happens to the velocity of an object as it hits another object?
     
  2. jcsd
  3. Jul 25, 2011 #2

    Redbelly98

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper

    Welcome to Physics Forums :smile:

    I have moved your question to the General Physics subforum since it is not homework, or a problem from a textbook.

    You seem to have a misconception about what forces are. They are conceptually different than velocity, and cannot be added to or subtracted from velocity. More to the point, the force does not "equal" the velocity v.

    It will bounce back (as you're aware already). If the other object is a lot more more massive and at rest, it will bounce back (as you're probably aware already). The speed it has after bouncing can be anything from zero up to the speed it initially had. Just what speed it has depends on how much of the original energy (in the form of kinetic energy) gets converted to heat and/or sound upon colliding. And that depends on the materials the object and wall are made of: a clay ball might stick to the wall (have zero speed after bouncing), whereas a highly elastic ball will bounce back with nearly the same speed it had before.

    For a simple programming algorithm, you could just have the speed after the collision be some fraction of the initial speed, depending on how elastic you want the collision to be. Eg., perhaps make it bounce back at 80% of the speed it had initially. You would set that percentage as a parameter in the code.

    Collisions in 2D or 3D get tricky, since you have to worry about components of the velocity both normal and parallel to the wall. Not my forte, so I'll just leave it at that.
     
  4. Jul 25, 2011 #3

    BruceW

    User Avatar
    Homework Helper

    Redbelly98 has written some useful stuff.
    I'd like to add that the energy lost wouldn't necessarily be linearly dependent on the initial kinetic energy, but I suppose a linear dependence would be a first approximation.
    Another issue, if this physics engine is being written for a game, the most realistic equation isn't necessarily the one you should go for.

    EDIT: I now realise Redbelly98 wasn't proposing using a linear dependence of energy lost on initial energy. He was thinking of a linear dependence of final speed on initial speed. I would personally guess that energy lost goes with initial energy. But that's just a guess, really.
     
    Last edited: Jul 25, 2011
  5. Jul 25, 2011 #4
    Thanks for the replies, guys!

    Firstly, no, this isn't being written for a game. Just playing around with physics, sort of creating a scientific simulation zone, with moderate accuracy.

    So you're saying that it is more difficult to accurately measure that force, because it completely depends on the material? So, I can just throw stuff at walls in real life and try to use that to measure how much my virtual objects should bounce back!

    Thanks!
     
  6. Jul 25, 2011 #5

    BruceW

    User Avatar
    Homework Helper

    Yes, if you're going for realism, that's probably the best way to do it
     
  7. Jul 25, 2011 #6
    Exactly!It depends upon the material just like friction.Google Co-efficient of restitution.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Equal and Opposite Force Help
  1. Equal but opposite force (Replies: 14)

Loading...