Question

Derive the expression for the elastic collisions in two dimension

Answer 1

From the conservation laws, we have the three equations

m1v1i−m1v1fcosθm1v1fsinθm1v21i−m1v21f=m2v2fcosϕ,=m2v2fsinϕ,=m2v22f.(1)(2)(3)(1)m1v1i−m1v1fcos⁡θ=m2v2fcos⁡ϕ,(2)m1v1fsin⁡θ=m2v2fsin⁡ϕ,(3)m1v1i2−m1v1f2=m2v2f2.

Summing the squares of (1) and (2) eliminates ϕϕ. The RHS of the resultant equation contains v22fv2f2 which can be eliminated using (3). Then, one would obtain a quadratic equation in terms of v1fv1iv1fv1i, which can be solved to obtain the desired equation

v1fv1i=m1m1+m2[cosθ±cos2θ−m21−m22m21−−−−−−−−−−−−−−√].v1fv1i=m1m1+m2[cos⁡θ±cos2⁡θ−m12−m22m12].

For the next equation, we rotate the axes to obtain the angle θ+ϕθ+ϕ more easily. Here, the conservation laws are

m1v1icosϕ−m1v1fcos(θ+ϕ)m1v1isinϕm1v21i−m1v21f=m2v2f,=m1v1fsin(θ+ϕ),=m2v22f.(4)(5)(6)(4)m1v1icos⁡ϕ−m1v1fcos⁡(θ+ϕ)=m2v2f,(5)m1v1isin⁡ϕ=m1v1fsin⁡(θ+ϕ),(6)m1v1i2−m1v1f2=m2v2f2.
First, we square (4) and use (6) to eliminate v2fv2f. Then, we use (5) to eliminate v1i,vifv1i,vif from the resultant equation, obtaining an equation in terms of ϕϕ and θ+ϕθ+ϕ. Using trigonometric identities, we get an equation in terms of tanϕtan⁡ϕ and tan(θ+ϕ)tan⁡(θ+ϕ) only. Then, this equation can be rewritten as a quadratic equation in tanϕtan(θ+ϕ)tan⁡ϕtan⁡(θ+ϕ):
[1−tanϕtan(θ+ϕ)]2=m2m1[1−tan2ϕtan2(θ+ϕ)],[1−tan⁡ϕtan⁡(θ+ϕ)]2=m2m1[1−tan2⁡ϕtan2⁡(θ+ϕ)],
which can be solved to obtain the desired equation tan(θ+ϕ)tan(ϕ)=m1+m2m1−m2.