A 2 kg block moving at 6 m/s collides inelastically with a stationary 4 kg block and the two stick together on a rough floor with coefficient of kinetic friction 0.3. Taking g = 10 m/s^2, how far do they slide?

Answer and explanation

Correct answer: C. 0.67 m

  1. Step 1: Momentum is conserved: 2 x 6 = (2 + 4) v, so v = 2 m/s.
  2. Step 2: Friction deceleration a = mu g = 0.3 x 10 = 3 m/s^2.
  3. Step 3: Distance = v^2 / (2a) = 4 / 6 = 0.67 m.
Answer
0.67 m.
Why not the others: 6.00 m uses the initial 6 m/s instead of the common speed; 1.33 m divides by a instead of 2a; 0.33 m divides v^2 by 4a.

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