Question

A 0.311 kg tennis racket moving 30.3 m/s east makes an elastic collision with a 0.0570 kg ball moving 19.2 m/s east find the velocity of the tennis ball after the collision

Answer 1

Answer:

38.0 m/s east

Explanation:

Momentum is conserved.

m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂

(0.311) (30.3) + (0.0570) (19.2) = (0.311) v₁ + (0.0570) v₂

10.52 = 0.311 v₁ + 0.0570 v₂

In an elastic collision, kinetic energy is conserved.

½ m₁u₁² + ½ m₂u₂² = ½ m₁v₁² + ½ m₂v₂²

m₁u₁² + m₂u₂² = m₁v₁² + m₂v₂²

(0.311) (30.3)² + (0.0570) (19.2)² = (0.311) v₁² + (0.0570) v₂²

306.5 = 0.311 v₁² + 0.0570 v₂²

Solve the system of equations.

0.311 v₁ = 10.52 − 0.0570 v₂

v₁ = 33.82 − 0.1833 v₂

306.5 = 0.311 (33.82 − 0.1833 v₂)² + 0.0570 v₂²

306.5 = 0.311 (1144 − 12.40 v₂ + 0.03360 v₂²) + 0.0570 v₂²

306.5 = 355.7 − 3.856 v₂ + 0.01045 v₂² + 0.0570 v₂²

0 = 0.06745 v₂² − 3.856 v₂ + 49.16

Use quadratic formula.

v₂ = [ 3.856 ± √(14.87 − 13.26) ] / 0.1349

v₂ = 19.2 or 38.0

We know v₂ isn't 19.2 m/s, so v₂ = 38.0 m/s.