Question

Bumper car A 281 kg moving 2. 82 m/s makes an elastic collision with bumper car B 209 kg moving 1. 72 m/s what is the velocity of car B after the collision.

Answer 1

Let v₁ and v₂ denote the velocity of car A and car B, respectively, after the collision. It sounds like both cars are initially moving in the same direction (since both have positive initial velocity).

Since momentum is conserved,

(281 kg) (2.82 m/s) + (209 kg) (1.72 m/s) = (281 kg) v₁ + (209 kg) v₂

1151.9 kg•m/s = (281 kg) v₁ + (209 kg) v₂

Kinetic energy is also conserved, so

1/2 (281 kg) (2.82 m/s)² + 1/2 (209 kg) (1.72 m/s)² = 1/2 (281 kg) v₁² + 1/2 (209 kg) v₂²

2852.93 m²/s² = (281 kg) v₁² + (209 kg) v₂²

Solve the first equation for v₁ :

v₁ = (1151.9 kg•m/s - (209 kg) v₂) / (281 kg)

Substitute v₁ into the second equation and solve for v₂ :

2852.93 m²/s² = (281 kg) ((1151.9 kg•m/s - (209 kg) v₂) / (281 kg))² + (209 kg) v₂²

⇒ v₂² ≈ 2.96 m²/s or v₂² ≈ 8.89 m²/s²

⇒ v₂ = 1.72 m/s or v₂ ≈ 2.98 m/s

where we ignore the first solution since it corresponds to the initial condition.