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Absorbtion of kinetic energy or momentum

  1. Oct 17, 2009 #1
    IF you throw metal or wood or whatever material ball into soft bed then it almost do not reflecting from bed as you can try. So where goes such ball momentum and kinetic energy? Need to solve such collision problem.
    Metal ball of mass 1 kg moving at velocity 100 m/s hit not moving bed and hit into soft side of bed into center of bed. Bed mass is 40 kg. And in another collision variant the same ball mass of 1 kg moving at 100 m/s hit metal box (box is not empty but all filled with metal or just another ball) of mass 40 kg into center. What would be velocity of bed and of metal box after collision. Or more simple: is velocities will be same of bed and box or different?
    Because if ball very weakly reflecting from bed soft part then maybe it means that ball to bed giving more of his momentum or kinetic energy?
     
  2. jcsd
  3. Oct 17, 2009 #2

    Doc Al

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    In an inelastic collision, such as you describe, the kinetic energy of the ball is transformed into thermal energy of the ball/bed. Momentum is still conserved, but presumably the bed is fixed to the floor so the momentum of the ball is added to the momentum of the ball/bed/earth system.

    If the bed were free to slide on a frictionless floor, then you can calculate the final speed of the bed+ball using conservation of momentum. By comparing the initial KE of the ball with the final KE of the ball+bed, you can determine how much of the original energy was transformed into thermal energy (or other forms).
     
  4. Oct 17, 2009 #3
    Sorry for not clearafication all to end. More precise conditions is those:
    Iron ball of mass m_1=1 kg stroke to into do not moving bed (u_2=0 m/s) soft side of bed into it center and bed have mass m_2=40 kg. Ball speed u_1=10 m/s. Ball and bed is in space without gravity. If instead be there would be iron box or iron ball of 40 kg mass then after collision ball and bed speeds should be those:
    v_1 =[u_1*(m_1-m_2)+2*m_2*u_2]/(m_1+m_2)= [10*(1-40)+2*40*0]/(1+40)= -390/41= -9.512 m/s.
    v_2= (u_2*(m+2-m_1)+2*m_1*u_1)/(m_1+m_2) = (0*(40-1)+2*1*10)/(1+40) = 20/41 = 0.4878 m/s.

    So ball speed v_1 do not much changed from primary speed 10 m/s, but only direction. So no doubt such speeds would be after collision, if instead bed there would be iron box of 40 kg mass or 40 kg metal ball. But if ball stroke into soft bed, then maybe ball give more momentum or kinetic energy and accelerate to bigger speed bed, than same mass metal cube?
     
  5. Oct 17, 2009 #4

    Doc Al

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    Here you use conservation of kinetic energy and momentum, treating the collision as completely elastic.
    I'm not quite sure I understand your question. I think you are asking about what happens in an inelastic collision of a ball with a soft bed. In that case momentum is still conserved, but kinetic energy is not. (Use conservation of momentum to find the final velocity of the bed+ball.)

    Note that when the ball bounces off its target, more momentum is transferred to the target.
     
  6. Oct 17, 2009 #5
    So you suggesting that bed speed still be the same as if it would be 40 kg metal box. Your point of view is that bed velocity will be 0.4878 m/s and not some 0.6 m/s etc, because ball had not reflected with full speed (-9.512 m/s).
    In over words, give me simple answer to this question: what is bed speed after collision?
    1) 20/41 m/s;
    2) > 20/41 m/s;
    3) < 20/41 m/s.

    What variant is your answer 1), 2) or 3)?
     
  7. Oct 17, 2009 #6

    Doc Al

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    No. Why would you think that?

    Assuming a perfectly inelastic collision, the final speed of the bed (and ball) will be 10/41 m/s.
     
  8. Oct 17, 2009 #7
    It's funny, but then almost such inelastic collision can be calculated with forumala as if all momentum is transmited. Ball momentum before collision was m_1*v_1=1*10=10 kg*m/s. In inelastic collision if bed absorb half energy or momentum (how properly to say?), then bed speed will be (20/41)/2=10/41. So it's almost the same as from equation u_2*m_2=m_1*v_1; u_2=m_1*v_1/m_2=10*1/40=0.4 m/s. So 10/41 almost same like 10/40 and such approximations maybe could be used...

    Now if object m_1=40 kg is bed which flying with speed u_1=10 m/s and hitting metal ball with bed soft side. metal ball of mass m_2=1 kg have no velocity u_2=0 m/s.
    With official calculatation speeds after collision of bed and ball will be those:
    v_1 =[u_1*(m_1-m_2)+2*m_2*u_2]/(m_1+m_2)= [10*(40-1)+2*1*0]/(1+40)= 390/41= 9.512195 m/s.
    v_2= (u_2*(m+2-m_1)+2*m_1*u_1)/(m_1+m_2) = (0*(1-40)+2*40*10)/(1+40) = 800/41 = 19.512195 m/s.
    How we can replace those formuls to get almost same result? Thats how, if assuming, that metal ball slowly moving with bed - with previous bed speed and bet slighly decreasing speed little bit.
    So then ball speed u_2=0 becoming bed speed u_1=10 before collision: v_2=u_1=10 m/s. And u_1 simple calculated with substraction momentum of ball after collision v_2*m_2=10*1=10 from momentum of bed before collision and all dividing by mass of bed: v_2=(u_1*m_1-v_2*m_2)/m_1=(10*40-10*1)/40=(400-10)/40=390/40=9.75 m/s. So calculation with prety good precision, because 9.75 almost same like 9.5. Ball speed do not need to calculate, because it stop at hit time and only bed pushing ball, because ball on bed way.

    EDIT: Only one "BUT". Why in my experiments if I puting small metal ball on the table (of mass about 5 g) and with constant speed moving another say kinda metal box of mass about 10 times bigger than ball mass, then according those formuls:
    v_1 =[u_1*(m_1-m_2)+2*m_2*u_2]/(m_1+m_2)
    v_2= (u_2*(m+2-m_1)+2*m_1*u_1)/(m_1+m_2)
    ball should move with 1.(81) bigger speed and metal box deacelerate to speed 0.(81) m/s, if metal box hit ball at 1 m/s speed. But I do not see, that ball aquaring almost two times bigger speed than metal box which I moving with same speed even after box hit ball. Similar experiments I try in air with heavier metal objects 10-100 times, which hitting ball and seems no ball get 1.9 bigger speed than moving metal object on string. But on table it's visible much better that ball do not moving twice faster but with same speed, with what moving object wich hit ball (initial ball velocity always 0 m/s). On the over hand, if ball hitting 10-100 times heavier object then ball velocity (we talking about same small metal ball of mass about 5-10 grams) almost don't changing or at least deacrease two times, which shouldn't be according those formuls. In short result is second: if small ball hitting 10 times heavier object then ball behaving according calculations "where I get 9.75 m/s". And if metal ball hitting some big 10-50 times heavier object then ball behave according calculations which I show first time, when small ball hit big bed and which are in physics textbooks. On the over hand again when small ball of momentum p_1=100x hit not bed but metal box, metal box then metal box momentum become 200x and ball momentum after collision almost -100x, so momentum from 100x rise to 300x if don't count sign. If (metalic) bed hit ball then if bed momentum was x, bed momentum after collision becoming somthing like 0.9x and ball momentum 0.1x so it is safed even without using sign, but if metal box is heavier than ball and box stroke to ball then this result do not much with my observation of reality. (In both cases kinetic energy is safed with sign or without).
     
    Last edited: Oct 18, 2009
  9. Oct 19, 2009 #8
    I try to let fall metal small ball onto metal object and it always reflecting with 1/2 height of height from what ball fall. So if you fall ball from 1 meter it will jump to the height of 0.5 meter. So this shows law, that any object in elastic colision transmiting not full momentum, but half momentum if it is lighter than over object. And about air friction a little bit. I try to let to fall from about 1 m height both for small little metal ball and for ping pong ball of plastic and within full air so there almost invisible diference, maybe only 0.1-0.05 second pingpong ball fall with delay to ground. So air friction at small speed is almost invisible so can be ignored if whatever object except paper falling from 1-2 m height. So I do not see sense in twice momentum kinetic energy transmition and when big object hiting to small I would easier agree, that was transmited either speed of big object or 1.5 speed of big object (about what I now trying to post), but not 2x speed of big object like states official formuls.
    For example if I through in cosmos 2 kg object at 100 m/s speed into 10 kg object with zero speed, then after collision 10 kg object flying with 10 m/s speed and 2 kg object flying with 50 m/s in oposit direction. 2*100+0*0=2*50+10*10 so moments conserved.
    And if m_1=10 kg, u_1=100 m/s; m_2=2 kg, u_2=0 m/s. Then after collision: v_1=90 m/s, v_2=150 m/s.
    And the similar result will be if two objects have same mass, but remeber that if smaller object hitting bigger it transmiting to bigger half of his momentum and if bigger object hitting smaller then it transmiting 1.5 his speed to smaller object, so if two objects have same mass: m_1=2 kg, u_1=100 m/s; m_2=2 kg, u_2=0 m/s, then after collision v_1=-50 m/s, v_2=50 m/s.

    edit: ping pong ball from table reflecting some times with 60-80 precent of previous height unlike metal ball which from metal reflecting 50 %, maybe 60%. So imagine if in pinpong ball there pressed air then acceleration of unpressing wall to who more will give momentum to metal table or to ball so it's like twice aquaring momentum of ball, first time ball give half momentum then like spring reflecting, but still should be so mayb particularly pinpong ball reflecting from air (remeber air friction), but no, so reflection depending on who can faster move after walls of objects geting them previous phorm and of course lighter objects have better such priority.

    Edit2: Maybe for such objects like ping pong ball or over very good jumping balls there is some tricks with sinchronization of air back reflection. I mean once small or light ping pong ball hit heavy metal ball or object then this metal object instanty giving part his momenum to air and air wave reflecting from over air and fast stroking bigger ball back so that it to smaller or lighter ball giving not his momentum little bit more. From thery big objects reflectivity can be about 75%, but seems from small it's similar but even 5-10 times heavier objects in most cases have also about 5-10 times bigger value and area. This area cousing bigger resistivity of air and from there can come those 75% if object which hit is prety light. 50% by definition and aditional up to 30% from countervibration of air. In though if ping pong ball falling from height together with small metal ball or any over metal objects then diference in hitting ground is not more than 0.5-1 cm. And let's calculate what should be speed when ball fall one 1 m without friction with acceleration 9.8 m/s/s. So speed after one second is 9.8 m/s and average speed is 9.8/2=4.9 m/s, so ball without friction should fall 4.9 m in one second. If after two seconds speed is twice bigger and after 3 seconds 3 times bigger then after half second 0.5 times so 0.5*9.8=4.9 m/s and average speed (0+4.9)/2=2.45 m/s. So in about half second they both falling on the ground from 2 m height, but since I do not sow that ping pong ball significantly fall later then say from 2.3 m height any less or more heavy object falling on ground in 0.5 s. So looks like air resistance at such speeds is almost invisible, but it still may act in some waves and reflection time can be not so short. So this waves hitting over air giving momentum to second waves but half still reflecting and giving back to heavy ball which again giving back to small bakc same waves can be true and for small ball which hit, but here is the trick that he is much smaller and waves less act upon him, but on the over hand bigger harder to accelerate, but still somthing is there.
     
    Last edited: Oct 20, 2009
  10. Oct 20, 2009 #9
    Maybe when becoming comprresion of ping pong ball or over very reflective ball this time is by gravity forced too and then reflection is higher but seems his time is too sort to be it true and at all should also be force stoping when it becoming to not deformated shape, but this energy may still be as bonus, I do not get here to the end, but still it seems unrealistic for too sort time of deformation-underformation.
    And here is a trick. If there is in cosmos one 1kg ball and another 10 kg ball then if 1 kg ball hiting with 100 m/s speed in 10 kg ball then acording official theory, 1 kg ball reflecting with 1.9 speed and 10 kg ball aquaring almost 2x=190 momentum of small ball. And when big ball hitting another 1 kg ball then it giving speed to 1 kg ball of agains about 1.8*100 m/s=180 m/s. This ball hits another big ball and reflects with speed 180*1.7=306 m/s and same momentum. So with one samll ball and with another 10 times heavier and if they in universe will be puted one by one I can make universe boil to infinity balls and to infinity energy! Because if oposit balls again will bounce they will still reflect, but no energy still will be of this one mvv/2. But momentum will raise to infinity, but with diferent directions, just think 100+190+306+405 and so on and it's only from small bals. So why do not raise momentum in all directions from single do not moving ball?
     
  11. Oct 20, 2009 #10

    Doc Al

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    Please rephrase your question and example more clearly. As it is, I really don't know what you're trying to say.
    :confused: The small ball has an initial momentum of +100 kg-m/s. If it ends up with -1.9 kg-m/s, then the large ball ends up with +101.9 kg-m/s. But the final kinetic energy is greater than the initial, so your example is not realistic.
     
  12. Oct 20, 2009 #11
    Initial conditions:
    m_1=1 kg, u_1=100 m/s, r_1=1*100=10 kg*m/s;
    m_2=10 kg, u_2=0 m/s, r_2=10*0=0 kg*m/s.
    u defines velocity before collision and r defines momentum before collision.
    After collision:
    v_1 =[u_1*(m_1-m_2)+2*m_2*u_2]/(m_1+m_2)= [100*(1-10)+2*10*0]/(1+10)= -900/11= 81.81(81) m/s.
    v_2= (u_2*(m_2-m_1)+2*m_1*u_1)/(m_1+m_2) = (0*(10-1)+2*1*100)/(1+10) = 200/11 = 18.(18) m/s.
    v_1 is speed of 1 kg ball after collision and v_2 is speed of 10 kg ball after collision.
    r_1=100 and p_1=81.81818181*1=81.(81) kg*m/s; and p_2=18.(18)*10=181.(81) kg*m/s. So p_1+p_2=900/11+2000/11=2900/11=263.(63) kg*m/s.
    Now at second hit big ball which now have momentum 181.(81) hitting again some over small ball on his way. This time:
    m_1=10 (kg), u_1=200/11 (m/s), r_1=m_1*u_1=10*200/11=2000/11 (kg*m/s).
    m_2=1 kg, u_2=0 m/s, r_2=0 kg*m/s.
    After collision:
    v_1 =[u_1*(m_1-m_2)+2*m_2*u_2]/(m_1+m_2)= [(200/11)*(10-1)+2*1*0]/(1+10)= (1800/11)/11=1800/121= 14.876 m/s.
    v_2= (u_2*(m_2-m_1)+2*m_1*u_1)/(m_1+m_2) = (0*(1-10)+2*10*200/11)/(1+10) = 4000/121 = 33.05785 m/s.
    This time momentum r_1 before collision was r_1=2000/11=181.(81) kg*m/s and after collision this time moments are safed p_1+p_2=10*1800/121+1*4000/121=181.(81) kg*m/s. But thats ok then big stroking to small then moments safed, but only then reflected (first one) stroke again to big one then moments again rise or like this one from big one in second collision with speed 33.05785 m/s if stroke to big one then total moments rise again almost 3x. Just watch:
    m_1=1 (kg), u_1=4000/121 (m/s), r_1=m_1*u_1=1*4000/121 (kg*m/s).
    m_2=10 kg, u_2=0 m/s, r_2=0 kg*m/s.
    After collision:
    v_1 =[u_1*(m_1-m_2)+2*m_2*u_2]/(m_1+m_2)= [(4000/121)*(1-10)+2*10*0]/(1+10)= -36000/1331= 27.04733 m/s.
    v_2= (u_2*(m_2-m_1)+2*m_1*u_1)/(m_1+m_2) = (0*(10-1)+2*1*4000/121)/(1+10) = 8000/1331 = 6.0105184 m/s.
    So now from this third collision moment was before collision r_1=4000/121=33.05785 kg*m/s. And after collision p_1+p_2=36000/1331+10*8000/1331=87.1525 N.
    So Idea was that moment after this calculations should raise 6x, but because of divergence it is now 87.1525+181.(81)+263.(63)=532.6070623 kg*m/s. So moment from 100 kg*m/s grew to 532.6 N. So more than 5x. Maybe I am here incalculate some moments twice, but some 4x grow at least should be.
    So 263.(63) N from just 100 N is guaranted. Now 900/11+1800/121+87.1525=183.8462 N. Even 2x do not get, how it can be? Do they farther deacreasing or what?
    But we do not use yet first time reflected ball with moment -900/11, so here new conditions:
    m_1=1 kg, u_1=900/11 m/s, r_1=900/11 N.
    m_2=10 kg, u_2=0, r_2=0.
    v_1 =[u_1*(m_1-m_2)+2*m_2*u_2]/(m_1+m_2)= [(900/11)*(1-10)+2*10*0]/(1+10)= -8100/121= -66.942 m/s.
    v_2= (u_2*(m_2-m_1)+2*m_1*u_1)/(m_1+m_2) = (0*(10-1)+2*1*900/11)/(1+10) = 1800/121 = 14.876 m/s.
    p=8100/121+10*1800/121=215.7 N.
    So tota moment now is P=215.7+1800/121+87.1525=317.73 N. So I am not certainly sure 100%, but seems farther and farther ir will grow slower and slower, but still to infinity, but maybe and not like me confusing why not 215.7+263.63=479.336 N, so maybe farther even it deacreasing, but at least about 4.8x moment raise I get and sow to you from initial 100 N.
    edit: In one place where was 14.876 N should be 148.76 N and thats why I get deacreasing of moment so teoreticly it will rise to infinity if to ignore sign with which we will get old 100 N.
     
    Last edited: Oct 21, 2009
  13. Oct 20, 2009 #12

    Doc Al

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    So? What's your point? (Here you assume an elastic collision.)
     
  14. Oct 20, 2009 #13

    Doc Al

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    How about this. Just state your point in words, leaving out the equations and momentum/energy calculations.

    You seem to be saying that somehow in a series of elastic collisions, momentum ends up being created. Is that what you're saying?
     
    Last edited: Oct 20, 2009
  15. Oct 21, 2009 #14
    Actualy momentum raise not to infinity but to mvv/2=1*100*100/2=5000 kg*m/s or 5000 J, because total momentum if ignore signs cann't raise higher than mvv/2=5000, because such kinetic energy and farther and farther momentum divergence infinitly to 5000 kg*m/s. So speed of balls will deacrease to infinity and momentum will try to be as close as possible to 5000 kg*m/s from initial 100 kg*m/s.

    And seems if there would be only same mass all bals, then momentum do not increase to 5000 kg*m/s, but always would be 100 kg*m/s if first ball mass m=1 kg, u=100 m/s.

    By the way, If bullets at shooting time flying from 200 to 800 m/s according official indications then if in 1 second object falling (0+10)/2=5 m/s average speed so 5 meters. So if you holding pistol at height 1 m then then flying distance of bullet will be from 200/5=40 to 800/5=160 m. But on the over hand acceleration becoming slowly so need to calculate after what part of second bullet will fall 1 m. So in half second gravity will accelerate bullet down to 5 m/s, so at that time bullet average speed will be 2.5 m/s. So bullet sould fall 2.5 m in 0.5 s. But to us need 1 m. So let's try 0.25 s. In 0.25 s bullet speed to the ground will be 2.5 m/s and average speed wil be 1.25 m/s so almost good for as. So if you holding pistol at 1.52 m then in 1/4 second bullet will slide on the ground. So bullet do not gliding on ground will fly distance from 200/4=50 m to 800/4=200 m. You can add little bit air friction but seems it is not too significant, so substract maybe 5 metters from 50 m and 20 meters from 200 m. Will be 45 m and 180 m. And deaceleration for falling bullet will be almost invisble because of air friction, just because distance which fall is about 100 times smaller, so it just about (1*(1/((50+200)/2))/10=(1/125)/10=0.008/10=0.0008 s will fall longer on ground because of air friction so 0.25+0.008=0.2508 s.

    Edit: Let's see what results will be in those conditions:
    m_1=1 kg, u_1=100 m/s;
    m_2=10 kg, u_2=-100 m/s.
    v_1 =[u_1*(m_1-m_2)+2*m_2*u_2]/(m_1+m_2)= [100*(1-10)+2*10*(-100)]/(1+10)=(-900+(-2000))/11=-2900/11=-263.(63) m/s.
    v_2= (u_2*(m_2-m_1)+2*m_1*u_1)/(m_1+m_2) = (-100*(10-1)+2*1*100)/(1+10) = (-900+200)/11 = -700/11=-63.(63) m/s.

    Another conditions:
    m_1=1 kg, u_1=100 m/s;
    m_2=100 kg, u_2=-100 m/s.
    v_1 =[u_1*(m_1-m_2)+2*m_2*u_2]/(m_1+m_2)= [100*(1-100)+2*100*(-100)]/(1+100)=(-9900+(-20000))/101=-29900/101=-296.0396 m/s.
    v_2= (u_2*(m_2-m_1)+2*m_1*u_1)/(m_1+m_2) = (-100*(100-1)+2*1*100)/(1+100) = (-9900+200)/101 = -9700/101=-96.0396 m/s.

    How could be explained of giving of 2x or so moment by small to big and reflecting with x? Assume that particle or ball which of 1 kg stroke into 10 on another ball. Then small ball deformatings and very slowly pushing big ball, but first of all all energy going to deformation of small ball (or big ball, it's do not matter, or both), so in this time big ball almost do not moving, because small ball very deformative so big ball almost do not moving and so small ball becoming fast to become of previous shape (not deformed) and at this time undeforming energy faster acting upon small ball. And from small ball perspective it's like big ball almost static and do not move from place so after becoming normal shape small ball get acceleration and only small underfoming force was mist, because big ball little bit move. So I don't know how much momentum small ball transmit to big ball, maybe x, maybe 1.5x maybe 1.9x of his momentum before collision, but from small ball position it's like big ball is some big wall which is frozen in time and almost can't move so like all energy which deformed returns again to small ball reflecting energy from big ball. So it's possible that say small ball to big ball give 0.01x and itself reflex with 0.99x. It's not quite explanalable, becouse deformation force when small ball hit big ball should push big ball from Niuton law. So maybe realy if to ignore signs momentum can increase up to somthing or kinetic energy?
     
    Last edited: Oct 21, 2009
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