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B Momentum Transfer: Elastic Collision vs Inelastic Collision

  1. Nov 10, 2016 #1

    morrobay

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    In elastic collision momentum and kinetic energy are conserved. Where as in inelastic collision K.E. and P are not conserved since K.E. lost during deformation in target object is consistent with momentum conservation.
    The question here is in context of working out with a 70kg heavy bag (m2 )
    Since heavy bag is stationary before punch then m1u1 = m1v1 + m2v2
    And then v2 in inelastic collision is less the v2 during elastic collision.
    So is it correct to say that momentum transfer in working out with a dense heavy bag that approximates an elastic collision, is greater and a "better " workout . Ie doing more work displacing the bag.
    As compared to working out with less dense bag ?

    ΔP = ∫t2-t1 F dt
    ΔKE = ∫x2-x1 Fdx
     
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  3. Nov 10, 2016 #2

    Simon Bridge

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    An elastic collision would involve working out with a bag that does not deform under impact and a fist that also does not deform when you hit the bag.
    So attach a steel ball to the end of your arm and swing it at a wrecking ball ... is that a better workout?

    Well it kinda depends on what you want to acheive - ie: how do you define "better" in this case?
    You have suggested that you define better as the amount of work done by the person in displacing the bag.
    This is going to be determined by the amount of resistance you work out against.
    When you hit the bag, energy goes into lifting the bag against gravity (it swings a bit), into deforming the bag, and into friction if the hook it hangs from is a bit stiff.
    If the bag does not deform much, and we ignore the friction, then the same work will just lift the bag higher.
    So for the same displacement, ignoring friction, the bag that deforms requires the most work.

    But do you get a harder workout when the bag is free to swing or when someone holds it?
     
  4. Nov 11, 2016 #3

    morrobay

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    When bag is free there is the ballistic pendulum effect.. In inelastic collision the softer bag is
    deformed more v2 is less and bag is not lifted as high. So in that case it seems there is less work.
     
  5. Nov 11, 2016 #4

    Simon Bridge

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    For the same strength of punch - the deformed bag is not lifted as high as the non-deformed bag.
    So the mechanical work against gravity is less.
    But there is also energy going into deforming the bag - that is also work.
    If the same energy goes into the bag - the same work is done overall - it just gets apportioned differently.
    If you use the same force each time, then ... equal total work gets a smaller overall displacement for the softer bag.

    The question about what happens when the bag is being held still is important for thinking about this.
    Here the bag has no displacement, all the force of the punch goes into deforming the bag.
    Clearly you are working out right?
     
  6. Nov 11, 2016 #5

    morrobay

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    IMG_20160730_084912.jpg
    Indeed about 2x /wk. After this conversation altering somewhat.
     
  7. Nov 11, 2016 #6
    I don't think this is germane to the original post, but I thought I'd mention that elastic doesn't mean lack of deformation. A spring or a rubber ball deform quite a lot under impact. Just watch a super slow-mo of a major leaguer crushing a baseball. Collision always involves deformation. The magnitude of the deformation does not determine elasticity. The question of elastic or inelastic depends on whether or not the deformation is reversible. Will it spring back or not.
     
  8. Nov 11, 2016 #7
    I think the last part of the statement may indicate that you understand momentum is always conserved, but I mention it because the first part of the statement contradicts this stating that P is not conserved.
     
  9. Nov 11, 2016 #8
    The density and weight of the bag do not determine whether the collision is elastic or inelastic. Hitting a 70 kg rubber bag while wearing rubber gloves would be pretty elastic.
     
  10. Nov 12, 2016 #9

    Simon Bridge

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    Rubber ball collisions tend not to be very elastic - you exceed the elastic limit very quickly, so some energy gets absorbed heating the ball. Steel ball bearings off a steel surface can be very elastic. There is this common conflation between the word "elastic" used in "elastic collisions" vs "elastic material". Admitedly some rubber balls can be pretty bouncy.

    The spring example for high deformation elastic is pretty good though ... punching bags tend not to bounce back, so, off the context, we are talking about the kins of deformations you get in a bag when you punch it and not elastic vs inelastic collisions in general.
     
  11. Nov 12, 2016 #10

    sophiecentaur

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    We get frequent threads concerning the application of 'Pure Physics' to the way the human body performs. It appears to be very difficult to convince people that it's a rather fruitless exercise but our physiology is so complicated and much of the energy we expend is dissipated internally that it has always seem obvious to me that it's a lost cause to make assumptions about 'ideal' equipment.
    Our bodies are just not designed to deal with 100% efficient mechanical systems and situations where the efficiency is high (trampoline / running blades / ice skates etc.) are very difficult to cope with, without a lot of practice. A good workout needs equipment that will dissipate your energy quickly and not throw it back at you.
     
  12. Nov 12, 2016 #11

    Simon Bridge

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    Hence the low elasticity but solid punching bags ... though iirc the heavy bag workouts are more about developing timing and endurance. Strength gets built in resistance training or something - it is literally 3 decades since... but different kinds of workouts do different things.

    ... there are good applications of physics to sports workouts, but they do tend to be more complicated than the kind of arguments above would suggest. A quick google throws up several discussions on punching physics and a youtube video. There's probably peer reviewed stuff in fitness lit on the subject too.
     
  13. Nov 17, 2016 #12

    morrobay

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    Your right my error/typo. With inelastic collision KE is not conserved and P is conserved. When heavier object (m2) is stationary then u2 = 0
    So in inelastic collision m1 u1 = (m1 +m2 ) v
    m1 = 10kg
    u1 = 10m/sec
    m2 = 15 kg
    v = 4m/sec
    P : 100 kg m/sec = 100 kg m/sec
    KE : 500 J ≠ 200 J
     
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