Oblique Elastic Collision between 2 bodies

GIVEN  
mass m1
mass m2
initial velocity of m1: u1
initial velocity of m2: u2
Angle between u1,u2 in degree (A)
FIND          
m (m2/m1)
cosA
cos2A
Z =u12 +mu22 ;
a=1-m
C=(1-m)u22 -2u1u2cosA
R=Cm - Za
RC
R/C
-Z/C
(-Z/C)2 ;
dis :(-Z/C)2 cos2φ - (R/C) cos2φ-
Angle between v1 and v2 : φ in degree
cosφ
cos2φ
dis3:
dis3r
x =  v1/v2 or
Total kinetic Energy before Collision
Total square of magnitude of total linear momentum before collision
v22 =   or 
v2   or
    or 
v1 or
    or 
Total kinetic Energy after Collision
Total square of magnitude of total linear momentum after collision
   

 

From conservation of energy, m1u12 +m2u22 =m1v12 +m2v22 ; Multiplying by m1 and rearranging,

m12(u12-v12)=m1m2(v22-u22)

From conservation of momentum,  m12u12 +m22u22 +2m1m2u1u2cosA =m12v12 +m22v22 ;+2m1m2v1v2cosφ

or m12(u12 -v12)+2m1m2u1u2cosA = m22(v22 -u22)    +2m1m2v1v2cosφ or

m1m2(v22 -u22)+2m1m2u1u2cosA = m22(v22 -u22)    +2m1m2v1v2cosφ or

av22 -2v1v2cosφ =au22 -2u1u2cosA where a=(m1-m2)/m1 =1-m where m=m2/m1

Let v1=xv2

av22 -2xv22cosφ =au22 -2u1u2cosA=C    or

v22(a-2xcosφ) =C  ------(1)

now  m1u12 +m2u22 =m1v12 +m2v22 ;

 v12=u12 +m(u22-v22) =(u12+mu22) -mv22  =Z -mv22 =  x2 v22 ; here Z=(u12+mu22)

or v22=Z /(m+ x2) Putting this value in eqn 1,

(a-2xcosφ)*[Z /(m+ x2)] =C

or Cx2 +2zcosφ x+(Cm-Za) =0 or Cx2 +2zcosφ x+R =0 where R=(Cm-Za)

x=-(-Z/C)cosφ +√ [(Z2/C2)cos2φ -R/C] or

x=-(-Z/C)cosφ -√ [(Z2/C2)cos2φ -R/C]

Example : take m1=4,m2=3,u1=5,u2=8,A =30, φ =20 and try with other combinations