A 72.0 kg ice skater is moving at 3.1 m/s on frictionless ice throws a 0.21 kg snowball horizontally at a speed of 28.0 m/s. What is the final velocity of the skater?

the final velocity of the skater after throwing the snowball is 3.17 m/s.

Explanation:

Given;

mass of the ice skater, m₁ = 72 kg

initial velocity of the ice skater, u₁ = 3.1 m/s

mass of the snowball, m₂ = 0.21 kg

initial speed of the snowball, u₂ = 28.0 m/s

Let the final velocity of the skater after throwing the snowball = v

Apply the principle of conservation of linear momentum to determine v;

m₁u₁ + m₂u₂ = v(m₁ + m₂)

72 x 3.1   +  0.21 x 28  = v(72 + 0.21)

229.08 = v(72.21)

v = 229.08 / 72.21

v = 3.17 m/s

Therefore, the final velocity of the skater after throwing the snowball is 3.17 m/s.

a. is the answer because they are not shared

pe = weightxheight = 2000x1000=2,000,000joules