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Impulse and forces

  1. Jun 29, 2004 #1
    According to what I've read, the reason we has safetly belts in cars is because "the collision is made longer, therefore force is decreased according to the impulse equation" and hammers are made of metal because "this is hard and makes collisions short, therefore force is large". Is it really as simple as that? Why does changing the collision time affect the force? Is it just that the equation says so if change in momentum is the same?
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  3. Jun 29, 2004 #2


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    Well, let's say you were rigidly fixed to a skateboard, and wouldn't fall over. What would hurt more: if you were pushed along so that you gradually reached a moderate speed, or if you were violently thrust forward, reaching that speed almost immediately? To change the momentum of an object without changing it's mass, you have to accelerate it, and the less the force, the less the acceleration, the longer it takes for the net momentum change to occur. It's kind of strange to think of the time affecting the force, there's probably a better way the stuff like "this is hard and makes the collision short," could be worded, i.e. they could describe why it makes the collisions short, and this may have something more to do with how the material transfers forces or energy. But it should make sense to say that if a collision went very slowly, you would expect to feel much less force.
  4. Jun 29, 2004 #3
    Newton's second law:

    F = ma

    A safety belt decreases your deceleration, lessening the force of impact. A soft hammer would take longer to decelerate when you hit something with it, so it would exert a smaller force.

    The 'impulse equation' is I = mv - mu (i.e. impulse = change in momentum). Also, I = Ft. From this you derive F = ma:

    I = mv - mu
    I = Ft
    Ft = mv - mu
    F = (mv - mu)/t
    F = m((v-u)/t)
    F = ma
  5. Jun 29, 2004 #4
    Ok, so if you just take it from a mathematical perspective, this would appear to make sense. ie - if we taker the case of the hammer, change in momentum is constant (cause if deaccelerates to 0 ) and if it is harder then the collision is shorter. Therefore using the equation you could conclude that if the collision is shorter then the force exerted on the hammer must be larger. Hence, as a result of Newtons 3rd law the hammer would exert the same large force on a nail. However, if you actually think about it, why should the force alter depedning on how long the collision lasts? If a force is a push or a pull, then why should the strength of this change as the length of collision change?
  6. Jun 29, 2004 #5
    RE: "However, if you actually think about it, why should the force alter depedning on how long the collision lasts?"

    Now you are trying to hold the force down to a constant value and simply apply it for a longer time. However, this is not the situation you described in your original question, where you considered holding the change in momentum to a constant value and changing the time of impact.

    So if you are comparing two situations, consider what is held constant between the two.

    Hammer striking nail versus hammer striking pillow: Here the change in momentum is the same between the two situations, but the impact time varies considerably. So the forces between the two objects is much higher in the first situation. This is not a case where the force is the same between the two situations, but merely applied through a longer time. Not at all.
  7. Jun 29, 2004 #6


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    Think of catching a baseball in two different ways.

    1. Hold your glove with a straight arm rigidly in front of you and let the ball slam into the glove. Back the glove up with your other hand so it won't recoil. Ouch.

    2. Catch in the normal way, with the glove being allowed to recoil a foot or so before stopping.

    1=>short distance, large force.
    2=>longer distance, smaller force.
  8. Jun 29, 2004 #7


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    Another way to think about it might be to compare with the decomposition of garbage, like paper for example. If you have ever noticed it, sometimes a waste paper basket (especially the kind with the lid over a) can get fairly warm. This warmth is energy being released as the paper decomposes. Another way to get the paper to decompose would be to light it on fire. This composition would be this decomposition would be much more rapid, and if you place your hand and the paper, it would get burnt.

    This is because, in order to release the same amount of energy in a shorter period of time, more energy must be released at any given moment.
  9. Jun 29, 2004 #8


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    Or think of the staircase. In climbing up the staircase, and jumping up 2 storeys, you expend the same amount of energy. But in climbing the staircase, you take your time, and so you need to generate less power.

    This isn't really the right question. The way the seat belt works is to exert a relatively small force on you. Because of its design, it can exert this force at any place - it is not just a solid barrier. Because it exerts a small force, it hurts less. But in the end, you still have to lose the same amount of momentum, to bring you to a (hopefully non-dead) stop, and because of its small force, the seat belts has to take a little longer.

    What this does not mean is that if you touch the hammer to the nail after you hit it, thus increasing the time, you would reduce the force.
  10. Jun 29, 2004 #9


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    To say the same thing everyone else is saying (hopefully in a slightly different way), force= mass times acceleration. The longer you can extend the time taken to change from v1 to v0, the lower the acceleration and, so, the lower the force.
  11. Jun 29, 2004 #10
    Consider a thousand BB's, dropped one after another with a slight delay from a height of, say, 1-foot onto a sheet of writing paper held taught between your hands.
    What happens to that sheet of paper? Does it tear from impact?
    Now, consider the same thousand BB's molded into a single mass and dropped onto that same paper.
  12. Jun 29, 2004 #11
    Yes, but that is not the purpose of seat belts : you can brace for impact with your arms. The purpose of seat belt is to keep you in front of the controls, so that you can use them and possibly reduce the problem. Ever pulled yourself out of a 360/720 spin ? Without the belt you will be in the passenger seat.

  13. Jun 30, 2004 #12
    True, but the slight elasticity of the belt(and body) affords an impact force time differential.
    You brought-up a good point though, and maybe selt belts should be re-designed to more effectively utilize this potential.
  14. Jul 1, 2004 #13
    Dude, I'd like to see you "brace for the impact with your arms"... just do the math:
    You're an average dude of something like 80 kg body mass (you're traveling at 50 km/h (30 mph). Suddenly you hit a concrete wall. You now have like 0.7 meters to absorb the impact with your arms. Lets use the energy equation:

    You'll find that the average force on your arms would be like 11.2 kN (2500 pounds)... deceleration will be 15 g.
    It's like falling from the 4th floor dude, you ain't gonna make it. In fact a collision of just 5 mph is like falling face forward form a stool onto a concrete floor. Most people can't "brace with their arms" for that impact either.
    More numbers: 25 kg at 30 mph will be thrown forward with the force of about one metric tonne in a collision. Anchor your payload!

    That's why we have seatbelts.
  15. Jul 1, 2004 #14
    Here's what I think is a much easier way to understand the time dependence of force. What Newton really said in his Second Law was that F = dp/dt. In other words, force is equal to the time rate of change of momentum. The dp and the dt in this equation are finite quantities of momentum and time respectively, and we can legitimately consider this as a fraction with a numerator (dp) and a denominator (dt). So, if we make the time interval (dt) large for a given desired momentum change (dp); dp/dt = F is small, if we make, for the same given desired momentum change, the time interval (dt) small; dp/dt = F is large.

    The common way of writing Newton's Second Law as F = ma is valid only if mass is constant, and it is less illuminating than Newton's actual statement in this case.
  16. Jul 2, 2004 #15
    RE: "
    The common way of writing Newton's Second Law as F = ma is valid only if mass is constant, "

    Even though everyone is probably sick of me saying it, I know of no physical system that simply changes mass without a collision occuring. So the acceleration version, in my opinion, is perfectly valid and identical to Newton's version.

    And one is no more intuitive than the other. Acceleration is certainly easy to understand, since changing a velocity vector with time is well-defined and measurable. Mass? Well, if you have a hard time understanding mass then you will probably also have a hard time understanding momentum.

    So which version is better? That depends on each instructor's approach and chronological ordering of topics. I use the acceleration version because it fits my approach.
  17. Jul 3, 2004 #16
    The driver has a certain mass that mass will be decelerated by the collision, so the driver will change velocity. When the velocity change happens in a short time the deceleration will be larger than when it happens in a longer time. Thus the force, which equals the driver's mass times the deceleration, will be lower when the velocity change takes longer.
    Last edited: Jul 3, 2004
  18. Jul 3, 2004 #17


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    Rocket motor.
    Movement at relativitistic velocities.

    dp/dt is better, because it works as a general case, and you can reduce it to F = ma if you should ever wish.
  19. Jul 3, 2004 #18
    RE: "Rocket motor."

    A collision process. This system is no different than two blocks compressed by a spring and released. The mass of the system does not change, because whichever system you choose (rocket, propellent, rocket+propellant), the mass simply moves to a new location. It does not change in value.

    RE: "Movement at relativitistic velocities."

    Mass is a Lorentzian invariant. Regardless, this system cannot be described by Newton's second law anyway.
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