Jose runs behind his friend, Jeff, at 8 m/s and jumps on his back shouting
"Surprise piggy back ride!" If Jose if 45 kg and Jeff is 55kg and was standing at
rest when Jose jumped on his back, what is their final velocity?

We can solve this problem by using the law of conservation of momentum. In fact, in absence of external forces, the total momentum of the system (Jose+Jeff) must be conserved before and after the collision. Therefore we can write:

[tex]p_i = p_f\\m_1 u_1 + m_2 u_2 = (m_1+m_2)v[/tex]

where:

[tex]m_1 = 45 kg[/tex] is the mass of Jose

[tex]u_1 = 8 m/s[/tex] is the initial velocity of Jose

[tex]m_2 = 55 kg[/tex] is the mass of Jeff

[tex]u_2 = 0 m/s[/tex] is the initial velocity of Jeff, which is standing

[tex]v[/tex] is the final combined velocity

Re-arranging the equation and solving for v, we find their final velocity:

[tex]v=\frac{m_1 u_1 + m_2 u_2}{m_1+m_2}=\frac{(45)(8)+0}{45+55}=3.6 m/s[/tex]

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Jose runs behind his friend, Jeff, at 8 m/s and jumps on his back shouting "Surprise piggy back ride!" If Jose if 45 kg and Jeff is 55kg and was standing at rest when Jose jumped on his back, what is their final velocity?

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Jose runs behind his friend, Jeff, at 8 m/s and jumps on his back shouting
"Surprise piggy back ride!" If Jose if 45 kg and Jeff is 55kg and was standing at
rest when Jose jumped on his back, what is their final velocity?