Lecture 17 Collisions. Chapter

Transcription

1 PHYS 172H: Modern Mechanics Fall 2010 Lecture 17 Collisions Chater

2 Chater 10: Collisions htt://

3 Elastic and inelastic collisions ELASTIC COLLISION: the internal energy of the objects in the system does not change: ΔE int =0 INELASTIC COLLISION: the internal energy of the objects in the system changes: ΔE int = 0 MAXIMALLY INELASTIC COLLISION: Objects stick together maximum allowed energy dissiation Note: only the K of RELATIVE motion can be killed. If the system has overall momentum, it will still be resent after the collision and K f = 2 /2M tot 3

4 Momentum and imulse r 1 r 2 r d dt = F r r d = Fdt r r r Δ = F Δt 1 2on1 r r Δ = F Δt 2 1on2 r r r r Δ P Clicker: sys =Δ Ptot =Δ ( 1+ 2) = 0 The changes in the momenta of orange and green balls are: A)Equal in magnitude and direction Conservation of momentum B)Equal in magnitude, oosite in direction r r Δ 1+Δ 2 = 0 C)Larger in magnitude for the lighter small ball D)Smaller in magnitude for the lighter small ball 4

5 Collisions: conservation of momentum system During collision Δt is small: r r r r Δ P = Δ +Δ = F Δt sys ( ) 1 2 external 0 Collision with contact r Δ In this case, Gravity r Δ 1 2 Collision without contact: Long range force acting at a distance 5

6 A head-on collision Ping-ong ball r r 2 = 0 1 Tennis ball r r 4 3 Momentum conservation: r r r Δ Ptot =Δ ( + ) = r + r = r + r = + 1x 3x 4x Two unknowns x Energy conservation: tot ( ) Δ E =Δ K + U + E = No interaction =0 before/after collision Elastic collision: =0 Δ ( K ) = x 3x 4x = + 2m 2m 2M int ernal 0

7 A head-on collision: elastic Ping-ong ball r r 2 = 0 1 r r 4 3 Momentum conservation: Energy conservation: Tennis ball x x = x + x x 3x 4x = + 2m 2m 2M A head-on noncollision (no interaction) is an oxymoron (contradiction in terms) maybe the ing-ong ball is a ghost? We could use instead the examle of a WIMP assing through a roton. Two equations Two unknown m± M = m+ M 3x 1x + no interaction - interaction 7

8 A head-on collision: elastic Note the subscrits on the masses : we DON T want to use M variously for the target article or the incident article. 3x = (m 1 -m 2 )/(m 1 +m 2 ) 1x 4x = 1x 3x = (1- (m 1 -m 2 )/(m 1 +m 2 )) 1x 4x = 1x. 2m 2 /(m 1 +m 2 ) 8

9 Head-on collision: equal masses 1x = 3x + 4x x 3x 4x = + 2m 2m 2m m± m = m+ m 3x 1x m P 4x = 1x. 2M/(m+M) m r r 2 = 0 1 r r 3 4 3x = 0 = 4x 1x 9

10 1x = 3x + 4x x 3x 4x = + 2M 2M 2m 1 M ± m = M + m 3x 1x Heavy rojectile, light target 2 M 1 m After collision: =0 2 Case M > M m = M + m m 1 3x 1x > 0 The rojectile will continue motion in the same direction 10

11 Very heavy rojectile, very light target Case M >> M m = M + m m 3x 1x 1 3x 1x 2 M m 1 2 =0 After collision: 3 4 Projectile continues motion with aroximately the same seed in the same direction M m M + m M + m = = M + m M + m M 2m2 2m 4x 1x mv4x Mv1x M M = 2m 4x 1x 1x 1x v 2v 4x 1x The same result: chose reference frame that moves with M 11

12 Change reference frame Imagine a very heavy ball (VHB) sitting at rest, and a light ball hitting it squarely with velocity v. ELASTIC collision, none of the KE is lost. The very heavy ball will gain negligible KE. The light ball will be reflected off the heavy ball with almost all the original KE and hence with almost the original seed, but reversed velocity. v v This icture is in the Red (VHB) rest system We re going to use this for the next slide, but first let s aly it to the receding slide s assertion: go to a system where the blue (light) ball is at rest and the red (VHB) is aroaching it with velocity v. After the collision, the VHB lows ahead almost unaffected, and the light ball (which has velocity v relative to the VHB) will have velocity 2v in the lab system of the receding slide 12

13 Droed heavy & light balls mgh = ½ mv 2 h PE to = KE at floor v v 3v At seed 3v, blue ball has 9 times the KE, will rise to 9 times higher than it fell. v v v 4 snashots of the balls droing, 3d view the big ball has reflected from the floor. Blue ball is aroaching red ball at 2v 4 th view, the blue ball has reflected from the red ball, is receding at 2v RELATIVE to red ball, therefore at 3v relative to the floor. 13

14 m Elastic scattering: Collisions in 2D and 3D M θ 3 1 y x 2 =0 r r r 1 = = + 2m 2m 2M searate for x and y comonents: 1 = 3cosθ + 4cosφ 0 = 3sinθ 4sinφ = + 2m 2m 2M φ (ositive angle) 4 Three equations Four unknowns: 3, 4, θ, φ 14

15 Elastic scattering: equal masses one initially at rest r r r 1 = = + 2m 2m 2m r r = + = cosa ( ) A = ± π / cosA = + + 2m 2m 2m 2m cos A = 0 A =± π /2 or: 3 4 =0 =0 What case, exactly, has 3 4 =0? 15

16 Elastic scattering: equal masses 1 A = ± π /2 2 CLICKER: Can you ocket both balls in one shot? (Ignore sin) A) No B) Yes, if you use ball 1 as a cue ball C) Yes, if you use ball 2 as a cue ball D) Yes, you can do it using any ball as a cue ball 16

17 Imact arameter -the distance between the centers erendicular to the incoming velocity Too far for interaction 17

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19 Discovery of the nucleus: early models 1903, Kelvin: electrons are embedded in a shere of uniform ositive charge 1906, Thomson: lum udding model Ernst Rutherford, former student of Thomson, tested lum udding model in

20 What s inside? Suose you have a Christmas resent and you would like to find out what is inside it without oening it The contents of a Christmas resent could be robed by firing a rifle into it and noting how bullets are scattered by the contents. 20

21 What s inside? Of course, if the contents are fragile, a rifle bullet is a stuid robe. Use X- rays instead! And hoe the contents aren t a kitten or hotograhic film. When we bombard rotons with high energy electrons (sort of like bullets) as a robe of the quarks inside, if the electron energy is great enough the roton can be altered or disruted 21

22 Rutherford: discovery of the nucleus (undergrad Ernest Marsden) 1910 He nucleus (2 rotons + 2 neutrons) Alha articles are too heavy to be scattered by electrons The mass must be concentrated in small regions - nuclei Plum udding model is incorrect Nucleus is ~10-15 m in diameter Electrons roam emty sace ~10-10 m across (1 Å) Alhas collide rather rarely with the small nuclei htt://galileo.hys.virginia.edu/classes/252/rutherford_scattering/rutherford_scattering.html 22

23 Rutherford: discovery of the nucleus Rutherford (in one of the later lectures): (Geiger counter) "I had observed the scattering of alha-articles, and Dr. Geiger in my laboratory had examined it in detail. He found, in thin ieces of heavy metal, that the scattering was usually small, of the order of one degree. One day Geiger came to me and said, "Don't you think that young Marsden, whom I am training in radioactive methods, ought to begin a small research?" Now I had thought that, too, so I said, " Why not let him see if any alha-articles can be scattered through a large angle?" I may tell you in confidence that I did not believe that they would be, since we knew the alha-article was a very fast, massive article with a great deal of energy, and you could show that if the scattering was due to the accumulated effect of a number of small scatterings, the chance of an alhaarticle's being scattered backward was very small. Then I remember two or three days later Geiger coming to me in great excitement and saying "We have been able to get some of the alha-articles coming backward " It was quite the most incredible event that ever haened to me in my life. It was almost as incredible as if you fired a 15-inch shell at a iece of tissue aer and it came back and hit you." htt://galileo.hys.virginia.edu/classes/252/rutherford_scattering/rutherford_scattering.html 23