Two ice skaters push off against one another starting from a stationary position. The 45-kg skater acquires a speed of 0.375 m/s. What speed does the 60-kg skater acquire?

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BlackTom AI · Accepted Answer

Start by identifying the physical principle at work: horizontal momentum must be conserved because there are no external horizontal forces when the skaters push off each other from rest.
Option 1: 0.28 m/s. This is the correct magnitude for the 60-kg skater. Using conservation of momentum, m1 v1 + m2 v2 = 0, with m1 = 45 kg, v1 = 0.375 m/s, m2 = 60 kg, we get v2 = (m1 v1)/m2 = (45 × 0.375)/60 = 16.875/60 ≈ 0.28125 m/s, which rounds to 0.28 m/s. The direction would be opposite to the first skater’s motion, consistent with momentum conservation.
Option 2: 0.78 m/s. This value is too large given the mass of the skaters and the initial momentum being tiny (starting from rest). If you plug into the momentum balance, you would not achieve such a speed without violating conservation, so this is incorrect.
Option 3: 0.5 m/s. While a plausible speed for some problems, it does not satisfy m1 v1 = m2 v2 with the given numbers. Specifically, m2 v2 would equal 60 × 0.5 = 30 kg·m/s, which would require m1 v1 = −30 kg·m/s, contradicting m1 v1 = 16.875 kg·m/s in magnitude. Hence incorrect.
Option 4: none of these. Since we can compute a consistent result (approximately 0.28 m/s) that matches the given correct option, this choice is false.
Option 5: 0.66 m/s. This is not compatible with conservation for the given masses and the initial zero momentum; it would imply a much larger momentum transfer than allowed by m1 v1 = 16.875 kg·m/s. Therefore incorrect.

2025 Fall PHYS 2205 TR 11:00 - 12:15 _ 89122 PHYS 2205 Exam 2- Requires Respondus LockDown Browser

Two ice skaters push off against one another starting from a stationary position. The 45-kg skater acquires a speed of 0.375 m/s. What speed does the 60-kg skater acquire?

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