Two ice skaters, with masses of 50 kg and 75 kg, are at the center of a 50 m - diameter circular rink. The skaters push off against each other and glide to opposite edges of the rink. If the heavier skater reaches the edge in 10 s, how long does the lighter skater take to reach the edge?

It takes the lighter skater 6.7 s to reach the edge.

Explanation:

Hi there!

Due to the conservation of momentum, the momentum of the system formed by the 2 skaters after the skaters push off against each other is the same as the momentum of the system when the skaters are at the center.

The momentum of the system is calculated as the sum of the momenta of each skater. The equation of momentum is the following:

p = m · v

Where:

p = momentum.

m = mass.

v = velocity.

Initially, both skaters are at rest (v = 0), so, the momentum of the system is zero.

The final momentum is also zero because the momentum of the system does not change (it is conserved). Then, we can express the momentum of the system as follows:

m1 · v1 + m2 · v2 = 0

Where m1 and v1 are the mass and velocity of one skater and m2 and v2 are the mass and velocity of the other skater.

We know that the heavier skater reaches the edge (located at 25 m from the center) in 10 s. So, we can calculate his velocity:

v = d/t

Where:

v = velocity.

d = traveled distance.

t = time.

v = 25 m / 10 s

v = 2.5 m/s

Now, we can obtain the velocity of the lighter skater usig the equation of momentum of the systen:

m1 · v1 + m2 · v2 = 0

75 kg · 2.5 m/s + 50 kg · v2 = 0

v2 = - 75 kg · 2.5 m/s / 50 kg

v2 = - 3.75 m/s

The velocity of the lighter skater is 3.75 m/s.

Now, let's find the time it takes the lighter skater to reach the edge traveling at that velocity:

v = d/t

t = d/v

t = 25 m / 3.75 m/s

t = 6.7 s

It takes the lighter skater 6.7 s to reach the edge.