A 65.0 kg ice skater moving forward with a velocity of 1.40 m/s throws a 0.280 kg snowball forward with a velocity of 38.0 m/s.

Question

(a) What is the velocity of the ice skater after throwing the snowball? Disregard the friction between the skates and the ice.

(b) A second skater initially at rest with a mass of 60.0 kg catches the snowball. What is the velocity of the second skater after catching the snowball in a perfectly inelastic collision?

Answer 1

A) Let's use a nice linear formula from conservation of momentum:

V(m₁+m₂) = V₁m₁+V₂m₂

Read this as "Momentum of ball and skater together = Summed momentum of each after the throw"

1.4 * 65.28 = 38 * 0.28 + V₂ * 65

V₂ = 1.24

The skater lost speed by throwing the ball.

B) Same formula:

V(m₁+m₂) = V₁m₁+V₂m₂

Read this as "Momentum of ball and skater together = Summed momentum of each before the throw" And in this case only the ball had momentum before.

V = V₁m₁ / (m₁+m₂)

V = 38 * 0.28 / 60.28

V = 0.177 m/s

The two together make sense. You'll see that the lighter skater's speed changed more than the heavier skater's did, in proportion to their weight.

IMPORTANT NOTE: You might think conservation of kinetic energy equations should be able to solve this one, but they do not and here's why. For the first skater to throw the ball, he must add energy or do work to get it to accelerate. You would have to add a term to the energy balance equation to accommodate this. The same thing is true when the second skater absorbs energy from the moving ball.