A 2,700-kg truck runs into the rear of a 1,000-kg car that was stationary. The truck and car are locked together after the collision and move with speed 3 m/s. Compute how much kinetic energy was "lost" in this inelastic collision.

Respuesta :

Answer:

6200 J

Explanation:

Momentum is conserved.

m₁ u₁ + mā‚‚ uā‚‚ = m₁ v₁ + mā‚‚ vā‚‚

The car is initially stationary. Ā The truck and car stick together after the collision, so they have the same final velocity. Ā Therefore:

m₁ u₁ = (m₁ + mā‚‚) v

Solving for the truck's initial velocity:

(2700 kg) u = (2700 kg + 1000 kg) (3 m/s)

u = 4.11 m/s

The change in kinetic energy is therefore:

Ī”KE = ½ (m₁ + mā‚‚) v² āˆ’ ½ m₁ u²

Ī”KE = ½ (2700 kg + 1000 kg) (3 m/s)² āˆ’ ½ (2700 kg) (4.11 m/s)²

ΔKE = -6200 J

6200 J of kinetic energy is "lost".