An air-track cart with mass m1=0.28kg and initial speed v0=0.75m/s collides with and sticks to a second cart that is at rest initially. if the mass of the second cart is m2=0.43kg, how much kinetic energy is lost as a result of the collision?

Kinetic energy is calculated through the equation,

KE = 0.5mv²

At initial conditions,

m₁: KE = 0.5 (0.28 kg) (0.75 m/s) ² = 0.07875 J

m₂ : KE = 0.5 (0.45 kg) (0 m/s) ² = 0 J

Due to the momentum balance,

m₁v₁ + m₂v₂ = (m₁ + m₂) (V)

Substituting the known values,

(0.29 kg) (0.75 m/s) + (0.43 kg) (0 m/s) = (0.28 kg + 0.43 kg) (V)

V = 0.2977 m/s

The kinetic energy is,

KE = (0.5) (0.28 kg + 0.43 kg) (0.2977 m/s) ²

KE = 0.03146 J

The difference between the kinetic energies is 0.0473 J.