An air-track cart with mass m1=0.31kg and initial speed v0=0.90m/s collides with and sticks to a second cart that is at rest initially. if the mass of the second cart is m2=0.50kg, how much kinetic energy is lost as a result of the collision?

Since the two charts after the collision stick together, we are dealing with a perfectly inelastic collision.

First, we need to find the speed of the two charts after the collision. In order to do so, we consider the conservation of momentum:
m₁·v₁ + m₂·v₂ = (m₁ + m₂)·v

We can solve for v, considering also that v₂=0
v = m₁·v₁ / (m₁ + m₂)
   = 0.31 · 0.90 / (0.31 + 0.50)
   = 0.34 m/s

The kinetic energy lost (which is transformed into bounding energy between the two charts) will be the difference between the total kinetic energy before the collision and after the collision:
ΔE = E₁ - E₂ = 1/2·m₁·v₁ - 1/2·(m₁ + m₂)·v
      = 1/2(0.31)(0.90) - 1/2(0.81)(0.34)
      = 0.1395 - 0.1377
      = 0.0018J

Hence, the correct answer is ΔE = 0.0018J

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