*Please help, Brainliest to first correct answer*

A Roller Derby Exhibition recently came to town. They packed the gym for two
consecutive weekend nights at South's field house. On Saturday evening, the
68-kg Anna Mosity was moving at 17 m/s when she collided with 76-kg Sandra
Day O'Klobber who was moving forward at 12 m/s and directly in Anna's path.
Anna jumped onto Sandra's back and the two continued moving together at
the same speed. Determine their speed immediately after the collision.

Please Show your work, so i can follow through

Respuesta :

The final velocity is 14.4 m/s

Explanation:

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 two girls 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 = 68 kg[/tex] is the mass of Anna

[tex]u_1 = 17 m/s[/tex] is the initial velocity of Anna

[tex]m_2 = 76 kg[/tex] is the mass Sandra

[tex]u_2 = 12 m/s[/tex] is the initial velocity of Sandra

[tex]v[/tex] is the final combined velocity of the two girls

By re-arranging the equation and solving for v, we find the  final velocity of the two girls:

[tex]v=\frac{m_1 u_1 + m_2 u_2}{m_1+m_2}=\frac{(68)(17)+(76)(12)}{68+76}=14.4 m/s[/tex]

Learn more about momentum here:

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