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A 68.0-kg person throws a 0.0405-kg snowball forward with a ground speed of 31.5 m/s. A second person, with a mass of 56.0 kg, catches the snowball. Both people are on skates. The first person is initially moving forward with a speed of 2.05 m/s, and the second person is initially at rest. What are the velocities of the two people after the snowball is exchanged? Disregard the friction between the skates and the ice. (Take the direction the snowball is thrown to be the positive direction. Indicate the direction with the sign of your answer.)

Respuesta :

davidlcastiblanco

Answer:

vf1=2.03 m/s

vf2=0.026 m/s

Explanation:

[tex]m_{p1}=68 kg[/tex]

[tex]m_{s}=0.045 kg[/tex]

[tex]m_{p2}=56 kg[/tex]

[tex]v_{s1}=31.5\frac{m}{s}[/tex]

[tex]v_{p1}=2.05\frac{m}{s}[/tex]

[tex]v_{p2}=0\frac{m}{s}[/tex]

Collision elastic in the moment thrown the snowball

[tex](m_{p1}+m_{s})*v_{p1}=m_{p1}*v_{f1}+m_{s1}*v_{s1}[/tex]

[tex]v_{f1}=\frac{(68kg+0.045kg)*2.05\frac{m}{s}-0.045kg*31.5\frac{m}{s}}{68kg}[/tex]

[tex]v_{f1}=2.03\frac{m}{s}[/tex]

[tex]m_{s1}*v_{s1}+m_{p2}*v_{2}=(m_{s}+m_{2})*v_{f2}[/tex]

[tex]v_{f2}=\frac{0.045kg*31.5\frac{m}{s}}{56kg+0.045kg}[/tex]

[tex]v_{f1}=0.0262\frac{m}{s}[/tex]

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