Physics collision problem?

Question

A 0.500 kg sphere moving with a velocity (2.00î - 3.50ĵ + 1.00k) m/s strikes another sphere of mass 1.50 kg moving with a velocity (-1.00î + 2.00ĵ - 2.40k) m/s.

If the velocity of the 0.500 kg sphere after the collision is (-1.00î + 2.50ĵ + ak) m/s, find the value of a and the velocity of the 1.50 kg sphere after an elastic collision. (Two values of a are possible, a negative value and a positive value. Report each with their corresponding final velocities.)

Been trying for hours and can't seem to get it.

Thanks for your time and assistance.

Answer 1

With an elastic collision you have

1. conservation of momentum: p_{before collision} = p_{after collision} (notated p' in following equations)
2. conservation of kinetich energy: KE_{before collision} = KE_{after collision}

1.Conservation of momentum:

p = p'

p₁ + p₂ = p₁' + p₂'

m₁v₁ + m₂v₂ = m₁v₁' + m₂v₂'

Scalar multiplication with an euclidean vector:

(m₁i₁, m₁j₁, m₁k₁) + (m₂i₂, m₂j₂, m₂k₂) = (m₁i₁', m₁j₁', m₁k₁') + (m₂i₂', m₂j₂', m₂k₂')

Addition of euclidean vectors:

(m₁i₁ + m₂i₂, m₁j₁ + m₂j₂, m₁k₁ + m₂k₂) = (m₁i₁' + m₂i₂', m₁j₁' + m₂j₂', m₁k₁' + m₂k₂')

Equal vectors:

m₁i₁ + m₂i₂ = m₁i₁' + m₂i₂'

m₁j₁ + m₂j₂ = m₁j₁' + m₂j₂'

m₁k₁ + m₂k₂ = m₁k₁' + m₂k₂'

Solving this gives you:

i₂' = 0

j₂' = 0

k₂' = (-6.2 - a)/3

2.Conservation of kinetic energy (the collision is elastic):

m₁v₁² / 2 + m₂v₂² / 2 = m₁v₁'² / 2 + m₂v₂'² / 2

With v² = i² + j² + k² (length of a euclidean vector).

Inserting the known values and the ones obtained in (1.), and after some arithmetic:

a² + 3.1a - 22.1 = 0

a = -6.50 or a = 3.40

k₂' is respectively:

k₂' = 0.10 or k₂' = -3.20

So v₂' = (0i + 0j + 0.10k) or (0i + 0j - 3.20k)