A 1.0 kg rock is dropped from a clifftop and falls 6.3 m to the ground. How much momentum did it have just before it hit the ground? Give your answer to the nearest whole number.

Question

BlackTom AI · Accepted Answer

Momentum p = m v, so first find the rock’s speed just before impact using v^2 = v0^2 + 2 g h (kinematics) or by energy conservation (mgh = 1/2 m v^2).

1. Relevant knowledge points and concepts
- Momentum: p = m v (vector), here we want the magnitude.
- Constant acceleration kinematics: for an object dropped from rest, v^2 = 2 g h.
- Gravitational acceleration: use g = 9.8 m/s^2 (standard near-Earth value).
- Mass m = 1.0 kg, drop height h = 6.3 m.

2. Step-by-step reasoning
- The rock is dropped from rest, so initial velocity v0 = 0.
- Use v^2 = v0^2 + 2 g h to find v just before impact.
- Then compute p = m v.

3. Calculations with intermediate steps
- v^2 = 0 + 2 × 9.8 m/s^2 × 6.3 m = 19.6 × 6.3 = 123.48 (m^2/s^2).
- v = sqrt(123.48) = 11.109... m/s.
- Momentum p = m v = 1.0 kg × 11.109... m/s = 11.109... kg·m/s.
- Rounded to the nearest whole number: p ≈ 11 kg·m/s.

4. Why the correct answer is correct
- The kinematic relation directly gives the impact speed from the drop height; multiplying by mass yields momentum. Numeric substitution and standard rounding produce 11 kg·m/s.

5. Why other answers would be incorrect
- Any different whole-number result would come from using an incorrect formula (for example forgetting the factor 2 in v^2 = 2 g h), using an incorrect g without justification, or arithmetic mistakes. The correct procedure and arithmetic lead uniquely to ~11.11 kg·m/s → 11 kg·m/s.

6. Conclusion
- The calculated momentum just before impact is approximately 11 kg·m/s, which matches the provided correct answer: 11.

A 1.0 kg rock is dropped from a clifftop and falls 6.3 m to the ground. How much momentum did it have just before it hit the ground? Give your answer to the nearest whole number.数值题

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