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Question about impulse

  1. Jan 2, 2015 #1
    Hey all! I was thinking about a problem the other day. (For some background, I'm a geology student - I took A-level Physics in high school, a few years ago, and have done a few physics modules since then at University, so I'm probably better educated in physics than the average person, but it's definitely not my speciality! I feel like this is a really basic problem and the answer's something someone with my education should already know, but I can't figure it out. So that's why I'm here on PF - to try and clear up my dumb geologist's questions about basic physics problems and get my understanding of the Universe straightened out!)

    The problem I was thinking about was this: why is kicking a hard rock a lot more painful than kicking a soft pillow when swinging your foot at the same speed?

    Going back to my school days I remembered textbooks that talk about impulse - a quantity defined as force * time and equal to the change in momentum a body experiences. They say that by prolonging the time taken for a momentum transfer you can lessen the forces involved. So in this case, swinging your foot with the same velocity at the rock and the pillow will result in the same momentum change (since (in Newtonian physics at least!) your foot's mass is constant), but it will take a longer time for the soft pillow to deform and slow your foot to zero than for the hard rock to do it. Therefore the soft pillow exerts less force on your foot than the hard rock, which is why kicking the pillow is a much less painful option.

    That's all well and good, but... how does the surface of the rock or pillow "know" upon first contact how long the collision's going to last? How does it "know" to reduce the forces immediately upon contact between your body and the mat because the collision's going to take a while? Other examples given in introductory textbooks to illustrate the momentum-impulse relationship, such as falling on a soft mat rather than a hard floor, using an airbag, rolling your knees to prolong the time of contact with the ground at the end of a parachute landing, etc. all seem to suffer from the same problem to me. How do the two colliding surfaces "know" how long the collision will last upon first contact and therefore how much force they need to exert on each other?
     
  2. jcsd
  3. Jan 2, 2015 #2

    Dale

    Staff: Mentor

    The surfaces don't know anything. They exert a force based on their stiffness and deformation (think of them as springs of differing stiffness). That force is applied over time to give an impulse. It requires a lot of force to deform a very stiff material even a small amount. It requires very little force to deform a low-stiffness material the same amount. The surfaces do not exert forces based on some supernatural anticipation of future durations, they exert forces based on their current deformation.
     
  4. Jan 2, 2015 #3
    Of course there's no "supernatural anticipation," I was using the word "know" in the loose/metaphorical sense physicists use it when they say things like "it's not just that the Heisenberg Uncertainty Principle is caused by our inability to measure the position and momentum of a particle, but that the Universe itself does not simultaneously 'know' these quantities". Sorry if that word caused any confusion, I wasn't sure how better to phrase it!

    That makes sense for the stone/pillow and airbag examples, but what about the "bending your knees and rolling to prolong the contact time" example? Is it simply that any action you take to prolong the contact time (whether that be bending your knees and rolling for the parachute example or hitting a soft surface for the pillow example) would by necessity be an action that lowers the current amount of force being used?
     
  5. Jan 2, 2015 #4

    Dale

    Staff: Mentor

    Same thing. The only difference is that the force vs deformation relationship is much more complicated.

    Your other comment is also correct. For a fixed impulse increasing the time is the same as decreasing the force.
     
  6. Jan 3, 2015 #5
    Right. Question resolved, thank you!
     
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