Question

Two solid marbles A and B with a mass of 3.00 kg and 6.50 kg respectively have an elastic collision in one dimension. Before collision solid marble A (3.00 kg) was at rest and the other solid marble (6.50 kg) had a speed of 3.50 m/s. Calculate the magnitudes of velocities of two solid marbles vA and vB after collision.

Answer 1

Answer:

va = 4.79 m/s

vb = 1.29 m/s

Explanation:

Momentum is conserved:

m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂

(3.00) (0) + (6.50) (3.50) = (3.00) v₁ + (6.50) v₂

22.75 = 3v₁ + 6.5v₂

For an elastic collision, kinetic energy is conserved.

½ m₁u₁² + ½ m₂u₂² = ½ m₁v₁² + ½ m₂v₂²

m₁u₁² + m₂u₂² = m₁v₁² + m₂v₂²

(3.00) (0)² + (6.50) (3.50)² = (3.00) v₁² + (6.50) v₂²

79.625 = 3v₁² + 6.5v₂²

Two equations, two variables.  Solve with substitution:

22.75 = 3v₁ + 6.5v₂

22.75 − 3v₁ = 6.5v₂

v₂ = (22.75 − 3v₁) / 6.5

79.625 = 3v₁² + 6.5v₂²

79.625 = 3v₁² + 6.5 ((22.75 − 3v₁) / 6.5)²

79.625 = 3v₁² + (22.75 − 3v₁)² / 6.5

517.5625 = 19.5v₁² + (22.75 − 3v₁)²

517.5625 = 19.5v₁² + 517.5625 − 136.5v₁ + 9v₁²

0 = 28.5v₁² − 136.5v₁

0 = v₁ (28.5v₁ − 136.5)

v₁ = 0 or 4.79

We know v₁ isn't 0, so v₁ = 4.79 m/s.

Solving for v₂:

v₂ = (22.75 − 3v₁) / 6.5

v₂ = 1.29 m/s