A 0.1 kg bullet traveling horizontally at 400 m/s embeds into a 3.9 kg wooden block initially at rest on a frictionless table. What fraction of the initial kinetic energy is dissipated as heat and deformation?

Answer and explanation

Correct answer: C. 97.5%

  1. Step 1: By momentum conservation, common speed V = (m / (m + M)) v_0 = (0.1 / 4.0) 400 = 10 m/s.
  2. Step 2: Initial kinetic energy: K_i = 0.5 0.1 400^2 = 8000 J.
  3. Step 3: Final kinetic energy: K_f = 0.5 4.0 10^2 = 200 J.
  4. Step 4: Fraction remaining: K_f / K_i = 200 / 8000 = 1/40 = 0.025 (2.5%).
  5. Step 5: Fraction dissipated: 1 - 0.025 = 0.975 (97.5%).
Answer
97.5%.
Why not the others: 90.0% is an arbitrary underestimate; 2.5% is the fraction of energy conserved, not dissipated; 50.0% assumes equal partition.

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