Question

A 55 kg skater is gliding along the ice at a velocity of 8 m/s to the right while holding a 3 kg ball. The skater throws the ball at a velocity of 4 m/s to the right. What will be the skaters velocity after throwing the ball?

Answer 1

Speed of Skater = 8.16 m/s

Step-by-step explanation:

Using kinetic energy:

`M_(t) = M_(skater) + m_(ball)\\(1)/(2)M_(t)V_(i)^2 = (1)/(2)*M*V_(s) ^2+(1)/(2)*m*V_(b)^2\\ M_(t)V_(i)^2 = M_(s)*V_(s) ^2+m_(b)*V_(b)^2\\M_(t)V_(i)^2-m_(b)*V_(b)^2 = M_(s)*V_(s) ^2\\(M_(t)V_(i)^2-m_(b)*V_(b)^2)/M_(s) = V_(s) ^2\\V_(s) = \sqrt{((M_(t)V_(i)^2-m_(b)*V_(b)^2))/(M_(s)) } \\`

This gives the skater a velocity of 8.16 m/s after throwing the ball