Elastic Collision in 2D between 2 Hard Spheres

[captain.joco]

Homework Statement

By resolving the momentum equations parallel and perpendicular to the incident beam direction, show that α is related to the speed v2 of the recoiling particle by:

cos α = m2*v2/(2μ*u1) where μ=m1m2/(m1+m2)

Homework Equations

See attachment for diagram.


m1v1cosθ +m2v2 cosψ = m1u1 ( parallel equation )

m1v1sinθ + m2v2 sinΨ =0 ( perpendicular equation )

The Attempt at a Solution

expressing v1 from the perpendicular equation, then v1 was substituted into the parallel equation, and rearranging for v2 i got
cos α = m2v2/m1u1 , which is not the right thing. What am I doing wrong?

 

[Jhero]

Thank you for your post. It seems like you are on the right track, but there may be a small error in your calculations. Let's take a closer look at your steps:

1. Express v1 from the perpendicular equation:

m1v1sinθ = -m2v2sinψ

v1 = -(m2v2sinψ)/m1sinθ

2. Substitute v1 into the parallel equation:

m1(-(m2v2sinψ)/m1sinθ)cosθ + m2v2cosψ = m1u1

3. Simplify and rearrange:

-m2v2cosψcosθ + m2v2cosψ = m1u1

m2v2(cosψ - cosψcosθ) = m1u1

m2v2(cosψ(1-cosθ)) = m1u1

4. Use the cosine double-angle identity to simplify:

m2v2(2cos^2(ψ/2) - 1) = m1u1

5. Substitute in the definition of μ:

m2v2(2cos^2(ψ/2) - 1) = m1u1

m2v2(2(cos(ψ/2))^2 - 1) = m1u1

m2v2(2(μ/m2)^2 - 1) = m1u1

6. Simplify and solve for v2:

cos α = m2v2/m1u1

cos α = m2v2/(m1u1 - m2v2)

cos α = m2v2/(m1u1(1 - μ/m2))

cos α = m2v2/(m1u1(1 - m1/(m1+m2)))

cos α = m2v2/(m1u1(m2/(m1+m2)))

cos α = m2v2/(m1m2u1/(m1+m2))

cos α = m2v2/(μu1)

v2 = (μu1*cos α)/m2

v2 = (m1m2u1*cos α)/(m2(m1+m2))

v2 = (m1u