(conservation of momentum)

Transcription

1 Dynamis of Binay Collisions Assumptions fo elasti ollisions: a) Eletially neutal moleules fo whih the foe between moleules depends only on the distane between thei entes. b) No intehange between tanslational and intenal enegy modes. Had-Sphee Elasti Collisions = ˆ i i i + = + + = + (onsevation of momentum) (onsevation of enegy) 0

2 The elative veloities ae given by = = Let ˆk be an abitay unit veto. Then the solution of the onsevation equations is ( ˆ ) = + k kˆ ( ˆ ) = k kˆ (3.) By substitution we an show these equations satisfy onsevation of momentum and enegy.

3 Fom ealie = Now fo, ( ˆ ) ( ) ( ) = = kˆ kˆ kˆ kˆ = k kˆ Dot multiply by kˆ kˆ = kˆ = =.

4 Fom (3.), the hange in momentum fo eah moleule is ( ˆ ) = k kˆ ( ˆ ) = k kˆ Theefoe, eah moleule has the same momentum hange, but in opposite dietions. The magnitude of the hange is k ˆ, and the dietion is speified by ˆk. The hange in momentum is pependiula to the tangent plane at the point of ontat. Fom the point of view of moleule, the ollision appeas as shown hee. ( kˆ ) k ˆ ental axis o apse line 3

5 Cente of Mass Coodinates Continue the analysis now with a swith to ente-of-mass oodinates. The CM oodinate system moves with onstant veloity, and = ( + ) m = m = ( + ) = ( ) = = ( + ) = ( ) = m = = + ( kˆ ) kˆ = + ( kˆ ) kˆ m m = = ( kˆ ) kˆ = [ + ( kˆ ) kˆ] = m m 4

6 We an show that the elative veloities and ae unhanged as ae Eqs. (3.), whih have the fom = + ( kˆ ) kˆ = ( kˆ ) kˆ (3.a) At the moment of impat, the moleules ae onfigued as shown hee. This sketh is like the pevious one, but now defines the impat paamete b and angle ψ, whih is elative to and ˆk. ψ b ˆk 5

7 The CM system is defined using the Catesian oodinates x j and basis vetos ˆi j. If we assume that and ae known, then and ae known, i.e., = ( )ˆi j j j = = ( ) j j j We still must detemine ˆk and k ˆ in the CM system. Even though the motion is fully 3D, the ollision poess is in a plane b apse χ ψ b χ ˆk 6

8 Beause = and =, the sketh is symmeti about the apse line, thus, b b, χ χ = = (3.) whee χ is the defletion angle. Fom the sketh ψ + χ = π π χ ψ =. (3.) Then, k ˆ = os ψ = sin( χ / ). This is the final esult fo k ˆ. 7

9 The final task is to develop an equation fo ˆk. Without loss of geneality, we define a plane sufae that ontains and î, whee we pesume. Sine the ollision plane ontains ˆ i 0, set this plane at an angle ε elative to the and î plane: ollision plane apse line χ ε θ x ˆk î efeene plane 8

10 Wite ˆk in tems of ˆk kˆ = k = sin( χ / ) k = sin( χ / ). The pependiula omponent satisfies Also, k ˆ = kiii k = 0 (3.4) k k + k k = (3.5) kˆ = kˆ = = + + = = + + =. k ψ ψ k ˆk ˆk 9

11 Equations (3.4) and (3.5) ae insuffiient to detemine k, a thid equation is needed. To deive this equation, intodue the veto n whih is pependiula to and lies in the, plane And n= n + n = n ˆ i n (not a unit veto) 0 = sinθ nn = sin θ. ˆ i = = os θ n î θ î θ x Fom the sketh, ˆ n= i. 30

12 This is veified by noting that = = n nn = + i 0 ˆ 4 = + = = o θ = sin θ. Sine we have = ˆ i ii, n= ( ) ˆ ˆ ˆ 3 3. i i i 3

13 k A thid elation fo is based on the following sketh, whih is in the plane pependiula to. Fom the sketh, ( ) ( ) k n= k k n n os ε. ( ) ( ) k n= k k n n k ˆ sin ( χ / ) os ( χ / ) ( ) os os( / ). os ε. k k = k k = = nn = k n= ε χ ε n 3

14 A seond elation fo k ˆ n is k n= ki ni = ( ) k k k 3 3. Then we have fo a thid equation ( ) k k k = ( ) osεos( χ / ) 3 3 k k = ( ) osεos( χ / ). Fom Eq. (3.4) this gives k = ( ) os os( / ), ε = osεos( χ / ). ( ) χ 33

15 Equations (3.4) and (3.5) beome ( ) k k k = k 3 3 = ( ) osεos( χ / ). (3.6a) k k = k k k 3 = osε os ( χ / ). (3.6b) The solution of Eqs. (3.6) gives osεos( χ / ) ˆ ˆ ˆ k = i i + i + ( ) 3 tanε 3 tan ε 3. (3.7) 34

16 Now we wite k = k + k = + k ˆ sin( χ / ) kˆ = k = sin( χ / ). Finally, fo Eqs. (3.) = + sin( χ / ) sin( χ / ) + k, (3.) = sin( χ / ) sin( χ / ) + k, whee k is given by Eq. (3.7). Thus, we have aomplished the task of witing the post-ollision veloities and as funtions of,, χ, ε, and,, χ, ε, espetively. 35