A 2 kg block moving at 3 m/s collides inelastically with a stationary 4 kg block and sticks together. What is their common velocity after collision?

A 2 kg block moving at 3 m/s collides inelastically with a stationary 4 kg block and sticks together. What is their common velocity after collision?

• A. 0.5 m/s
• B. 0.9 m/s
• C. 1.0 m/s
• D. 1.5 m/s

Answer: (D)

Momentum is conserved in a collision where the blocks stick together, while kinetic energy is not. The first block has momentum 2 kg × 3 m/s = 6 kg·m/s, and the second block is at rest, so the total initial momentum is 6 kg·m/s. After they stick, the combined mass is 2 kg + 4 kg = 6 kg, moving with velocity v. Setting momentum before equals momentum after gives v = 6 kg·m/s ÷ 6 kg = 1 m/s. The kinetic energy drops (inefficiency of inelastic collisions): initial KE is 0.5×2×3^2 = 9 J, final KE is 0.5×6×1^2 = 3 J. So the common velocity is 1 m/s.

Momentum is conserved in a collision where the blocks stick together, while kinetic energy is not. The first block has momentum 2 kg × 3 m/s = 6 kg·m/s, and the second block is at rest, so the total initial momentum is 6 kg·m/s. After they stick, the combined mass is 2 kg + 4 kg = 6 kg, moving with velocity v. Setting momentum before equals momentum after gives v = 6 kg·m/s ÷ 6 kg = 1 m/s. The kinetic energy drops (inefficiency of inelastic collisions): initial KE is 0.5×2×3^2 = 9 J, final KE is 0.5×6×1^2 = 3 J. So the common velocity is 1 m/s.