Answer:
t = 20 s
Explanation:
- Assuming no other forces acting on the skaters when they push off against each other, and that we can neglect friction, total momentum must be conserved.
- The initial momentum is just zero, because both skaters are at rest.
- So, when both are gliding to opposite edges of the rink, at any moment, we can write the following expression:
[tex]p_{f} = m_{1} * v_{1} = m_{2} * v_{2} (1)[/tex]
- where m₁ = 50 kg, m₂ = 75 kg.
- We know that the heavier skater reaches the edge in 30 s.
- Since the distance from the center to any point on the edge is just half the diameter, we can find the speed of the heavier skater as follows:
[tex]v_{2} = \frac{15m}{30s} = 0.5 m/s (2)[/tex]
- Replacing m₁, m₂ and v₂ in (1), we can solve for the only unknown (v₁) as follows:
[tex]v_{1} = \frac{m_{2}*v_{2}}{m_{1} } = \frac{75 kg*0.5m/s}{50kg} = 0.75 m/s (3)[/tex]
- Since the distance to the opposite edge from the center is the same than for the heavier skater, we can find the time needed for the lighter one to reach the edge as follows:
- [tex]t_{1} = \frac{15m}{0.75m/s} = 20 s (4)[/tex]
