An ice skater with a mass of 61 kg pushes off against a second skater with mass 37 kg. If both skaters are initially at rest, and the first (heavier) ice skater leaves with a speed of 2.8 m/s, what is the speed of the second skater in m/s
Question
Answer:
The speed of the second skater is 4.6m/s.
Explanation:
Since there are no external forces acting in the direction of the velocities, we can say that the total linear momentum of the system is conserved. So:
[tex]m_1v_{1o}+m_2v_{2o}=m_1v_{1f}+m_2v_{2f}[/tex]
As the initial velocity is zero, we simplify the equation to:
[tex]m_1v_{1f}+m_2v_{2f}=0\\\\\implies v_{2f}=-\frac{m_1}{m_2} v_{1f}[/tex]
Since we need the speed and not the velocity, the minus sign can be ignored. If we plug in the given values, we obtain:
[tex]v_{2f}=\frac{61kg}{37kg} 2.8m/s=4.6m/s[/tex]
In words, the speed of the second skater is 4.6m/s.