Two ice skaters, with masses of 40.0 kg and 65.0 kg , are at the center of a 50.0 m -diameter circular rink. The skaters push off against each other and glide to opposite edges of the rink. Part A If the heavier skater reaches the edge in 10.0 s , how long does the lighter skater take to reach the edge

6.15 s

Explanation:

From the question,

Applying the law of conservation of momentum

Momentum of the Heavier skater = Momentum of the lighter skater.

Mv = mV Equation 1

Where M = mass of the heavier skater, m = mass of the lighter skater, v = Velocity of heavier skater, V = velocity of the lighter skater.

But,

v = r/t Equation 2

V = r/t' Equation 3

Where r = radius of the circular rink, t = time taken for the heavier skater to reach the edge, t' = time taken for the lighter skater to reach the edge.

Substitute equation 2 and equation 3 into equation 1

M(r/t) = m(r/t') Equation 4

Given: M = 65 kg, m = 40 kg, r = 50/2 = 25 m, t = 10 s.

Substitute into equation 4 and solve for t'

65(25/10) = 40(25/t')

162.5 = 1000/t'

t' = 1000/162.5

t' = 6.15 seconds

speed of light in vacuum is 300,000 km/s

speed of light in other mediums is given by the expression v = c/n, where c is the speed of light in vacuum and n is the refractive index of medium.

for the first medium v = 200,000 km/s

  so, refractive index of medium = 300000/200000 = 1.5

    1.5 refractive index is usually for glass. so first medium is glass.

for the second medium v = 250,00 km/s

so, refractive index of medium = 300000/250000 = 1.2

    1.2 refractive index is usually for water. so second medium is water.

for the third medium v = 0 km/s

so, refractive index of medium = 300000/0 = ∞

    ∞ refractive index is usually for opaque mediums. so third medium is opaque medium.