A small bomb, of mass 10 kg, is moving toward the North with a velocity of 4.0 m/s. It explodes into three fragments: a 5.0 kg fragment moving west with a speed of 8.0 m/s; a 4.0 kg fragment moving east with a speed of 10 m/s; and a third fragment with a mass of 1.0 kg. What is the velocity of the third fragment? (Neglect air friction.)

Question

BlackTom AI · Accepted Answer

Conservation of momentum applies: the total momentum of the system before the explosion equals the vector sum of the momenta of the fragments (no external horizontal forces during the explosion).

1. Relevant knowledge points and concepts
- Momentum p = m v is a vector; treat x (east) and y (north) components separately.
- Total momentum is conserved in the explosion: p_initial = sum p_fragments.
- Choose coordinate axes: +y = north, +x = east.

2. Step-by-step reasoning
- Initial momentum: the bomb (mass 10 kg) moves north at 4.0 m/s, so
  p_initial = (p_xi, p_yi) = (0, 10·4.0) = (0, 40) kg·m/s.
- Fragment 1: m1 = 5.0 kg, velocity 8.0 m/s west → vx1 = −8.0, vy1 = 0.
  p1 = m1 v1 = (5·(−8), 0) = (−40, 0) kg·m/s.
- Fragment 2: m2 = 4.0 kg, velocity 10 m/s east → vx2 = +10, vy2 = 0.
  p2 = (4·10, 0) = (40, 0) kg·m/s.
- Fragment 3: m3 = 1.0 kg, unknown velocity v3 = (vx3, vy3). Its momentum is p3 = (1·vx3, 1·vy3) = (vx3, vy3).

3. Calculations (component equations)
- x-components: p_xi = p1_x + p2_x + p3_x
  0 = (−40) + 40 + vx3  → vx3 = 0 m/s.
- y-components: p_yi = p1_y + p2_y + p3_y
  40 = 0 + 0 + vy3  → vy3 = 40 m/s.

4. Why the correct answer is correct
- The third fragment must have velocity (0, 40) m/s, i.e., 40 m/s north, so that the x-momenta of the first two fragments cancel and the net y-momentum equals the initial 40 kg·m/s. This satisfies momentum conservation exactly.

5. Why other options are incorrect
- "zero": would give no momentum from the 1-kg fragment and the y-momentum would be short of the initial 40 kg·m/s.
- "40 m/s south": wrong direction (would give −40 kg·m/s in y, not +40).
- "none of these": incorrect because one of the listed choices (40 m/s north) is the correct result.

6. Conclusion
The velocity of the third fragment is 40 m/s north, which matches the provided correct answer.

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