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=xv2av22 -2xv22cosφ =au22 -2u1u2cosA=C    orv22(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)] =Cor 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