Question

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?

Answer 1

Answer:

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