A ball of mass 0.440 kg moving east (+x direction) with a speed of 3.80 m/s collides head-on with a 0.220-kg ball at rest. If the collision is perfectly elastic, what will be the speed and direction of each ball after the collision?

The speed and direction of each ball after the collision is 1.27 m/s to East direction and 5.07 m/s to East direction.

Explanation:

given information

m₁ = 0.440 kg

v₁ = 3.80 m/s

m₂ = 0.220 kg

v₂ = 0

collision is perfectly elastic

v₁ - v₂ = - (v₁' - v₂')

v₁ = - (v₁' - v₂')

v₂' = v₁ + v₁'

according to momentum conservation energy

m₁v₁ + m₂v₂ = m₁v₁' + m₂v₂'

m₁v₁ = m₁v₁' + m₂ (v₁ + v₁')

m₁v₁ = m₁v₁' + m₂v₁ + m₂v₁'

m₁v₁ - m₂v₁ = m₁v₁' + m₂v₁'

v₁ (m₁ - m₂) = (m₁ + m₂) v₁'

v₁' = (m₁ - m₂) v₁ / (m₁ + m₂)

= (0.440 - 0.220) (3.8) / (0.440 + 0.220)

= 1.27 m/s to East direction

v₂' = v₁ + v₁'

= 3.8 + 1.27

= 5.07 m/s to East direction