Verified Solution Science How Forces Affect Motion

A 50 g ball is dropped from a height of 10 m. What is its momentum just before it hits the ground? (Assume g = 10 m/s², neglect air resistance)

6 views 1 helpful Updated Jul 5, 2026
Solution ✔ Verified
  • A0.5 kg m/s
  • B5 kg m/s
  • C50 kg m/s
  • D0.05 kg m/s
Explanation

First, find the final velocity using v² = u² + 2gs. v² = 0² + 2 * 10 m/s² * 10 m = 200 m²/s². So, v = sqrt(200) = 10 * sqrt(2) ≈ 14.14 m/s. Now calculate momentum: p = mv = 0.05 kg * 14.14 m/s ≈ 0.707 kg m/s. Let's recheck with options. The options are quite round. Perhaps the 'g' value choice or simplification. Let's try working backward from the options or re-evaluating the desired level of precision. If velocity was simpler, e.g., v=10m/s, then momentum would be 0.05*10=0.5. If g=5, then v²=2*5*10=100, v=10. This is too small for g. If g=10, v = sqrt(200). So, p = 0.05 * sqrt(200) = 0.05 * 14.14 = 0.707. None of the options are exactly 0.707. Let's consider if the question meant a different velocity. What if g=50? Then v² = 2*50*10 = 1000. v=sqrt(1000)~=31.6. p = 0.05*31.6 = 1.58. The options are very specific. Let me reconsider the problem if it's a simplification, or if there's a common approximation.

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