### Physics Problem: Motion and Kinetic Energy of Figure Skaters(a) Two figure skaters, one weighing 625 N and the other 725 N, push off against each other on frictionless ice. 1. If the heavier skater travels at 1.50 m/s, how fast will the lighter one travel?(b) How much kinetic energy is "created" during the skaters' maneuver, and where does this energy come from?---#### Explanation:In this problem, we analyze the motion of two skaters pushing off each other and the resulting kinetic energy. **Part A: Determining the Speed of the Lighter Skater**- **Weight of skater 1:** 625 N- **Weight of skater 2:** 725 N- **Speed of heavier skater (skater 2):** 1.50 m/sTo find the mass (m) from weight (W), use the relation: \( W = mg \), where \( g = 9.8 \, \text{m/s}^2 \) (acceleration due to gravity).Weight (N) to mass (kg) conversion:- \( \text{Mass of skater 1} = \frac{625 \, \text{N}}{9.8 \, \text{m/s}^2} \approx 63.78 \, \text{kg} \)- \( \text{Mass of skater 2} = \frac{725 \, \text{N}}{9.8 \, \text{m/s}^2} \approx 73.98 \, \text{kg} \)Using the conservation of momentum:\[ m_1 v_1 = m_2 v_2 \]Where:- \( m_1 \) = mass of lighter skater- \( v_1 \) = speed of lighter skater- \( m_2 \) = mass of heavier skater- \( v_2 \) = speed of heavier skater (1.50 m/s)Solve for \( v_1 \):\[ 63.78 \, \text{kg} \times v_1 = 73.98 \, \text{kg} \times 1.50 \, \text{m/s} \]\[ v_1 = \frac{73.

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) Two figure skaters, one weighing 625 N and the other 725 N, push off against each other on frictionless ice. (a) If the heavier skater travels at 1.50 m/s, how fast will the lighter one travel? (b) How much kinetic energy is "created" during the skaters' maneuver, and where does this energy come from?

) Two figure skaters, one weighing 625 N and the other 725 N, push off against each other on frictionless ice. (a) If the heavier skater travels at 1.50 m/s, how fast will the lighter one travel? (b) How much kinetic energy is "created" during the skaters' maneuver, and where does this energy come from?

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### Physics Problem: Motion and Kinetic Energy of Figure Skaters

(a) Two figure skaters, one weighing 625 N and the other 725 N, push off against each other on frictionless ice.
1. If the heavier skater travels at 1.50 m/s, how fast will the lighter one travel?

(b) How much kinetic energy is "created" during the skaters' maneuver, and where does this energy come from?

---

#### Explanation:
In this problem, we analyze the motion of two skaters pushing off each other and the resulting kinetic energy.

**Part A: Determining the Speed of the Lighter Skater**
- **Weight of skater 1:** 625 N
- **Weight of skater 2:** 725 N
- **Speed of heavier skater (skater 2):** 1.50 m/s

To find the mass (m) from weight (W), use the relation: \( W = mg \), where \( g = 9.8 \, \text{m/s}^2 \) (acceleration due to gravity).

Weight (N) to mass (kg) conversion:
- \( \text{Mass of skater 1} = \frac{625 \, \text{N}}{9.8 \, \text{m/s}^2} \approx 63.78 \, \text{kg} \)
- \( \text{Mass of skater 2} = \frac{725 \, \text{N}}{9.8 \, \text{m/s}^2} \approx 73.98 \, \text{kg} \)

Using the conservation of momentum:
\[ m_1 v_1 = m_2 v_2 \]

Where:
- \( m_1 \) = mass of lighter skater
- \( v_1 \) = speed of lighter skater
- \( m_2 \) = mass of heavier skater
- \( v_2 \) = speed of heavier skater (1.50 m/s)

Solve for \( v_1 \):
\[ 63.78 \, \text{kg} \times v_1 = 73.98 \, \text{kg} \times 1.50 \, \text{m/s} \]

\[ v_1 = \frac{73.

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