Answer:
The velocities of the skaters are [tex]v_{1} = 3.280\,\frac{m}{s}[/tex] and [tex]v_{2} = 0.024\,\frac{m}{s}[/tex], respectively.
Explanation:
Each skater is not under the influence of external forces during process, so that Principle of Momentum Conservation can be used on each skater:
First skater
[tex]m_{1} \cdot v_{1, o} = m_{1} \cdot v_{1} + m_{b}\cdot v_{b}[/tex] (1)
Second skater
[tex]m_{b}\cdot v_{b} = (m_{2}+m_{b})\cdot v_{2}[/tex] (2)
Where:
[tex]m_{1}[/tex] - Mass of the first skater, in kilograms.
[tex]m_{2}[/tex] - Mass of the second skater, in kilograms.
[tex]v_{1,o}[/tex] - Initial velocity of the first skater, in meters per second.
[tex]v_{1}[/tex] - Final velocity of the first skater, in meters per second.
[tex]v_{b}[/tex] - Launch velocity of the meter, in meters per second.
[tex]v_{2}[/tex] - Final velocity of the second skater, in meters per second.
If we know that [tex]m_{1} = 70\,kg[/tex], [tex]m_{b} = 0.043\,kg[/tex], [tex]v_{b} = 32\,\frac{m}{s}[/tex], [tex]m_{2} = 58.5\,kg[/tex] and [tex]v_{1,o} = 3.30\,\frac{m}{s}[/tex], then the velocities of the two people after the snowball is exchanged is:
By (1):
[tex]m_{1} \cdot v_{1, o} = m_{1} \cdot v_{1} + m_{b}\cdot v_{b}[/tex]
[tex]m_{1}\cdot v_{1,o} - m_{b}\cdot v_{b} = m_{1}\cdot v_{1}[/tex]
[tex]v_{1} = v_{1,o} - \left(\frac{m_{b}}{m_{1}} \right)\cdot v_{b}[/tex]
[tex]v_{1} = 3.30\,\frac{m}{s} - \left(\frac{0.043\,kg}{70\,kg}\right)\cdot \left(32\,\frac{m}{s} \right)[/tex]
[tex]v_{1} = 3.280\,\frac{m}{s}[/tex]
By (2):
[tex]m_{b}\cdot v_{b} = (m_{2}+m_{b})\cdot v_{2}[/tex]
[tex]v_{2} = \frac{m_{b}\cdot v_{b}}{m_{2}+m_{b}}[/tex]
[tex]v_{2} = \frac{(0.043\,kg)\cdot \left(32\,\frac{m}{s} \right)}{58.5\,kg + 0.043\,kg}[/tex]
[tex]v_{2} = 0.024\,\frac{m}{s}[/tex]
