Answer:
3.416 m/s
Explanation:
Given that:
mass of cannonball [tex]m_A[/tex] = 72.0 kg
mass of performer [tex]m_B[/tex] = 65.0 kg
The horizontal component of the ball initially [tex]\mu_{xA}[/tex] = 6.50 m/s
the final velocity of the combined system v = ????
By applying the linear momentum of conservation:
[tex]m_A \mu_{xA}+m_B \mu_{xB} = (m_A+m_B) v[/tex]
[tex]72.0 \ kg \times 6.50 \ m/s+65.0 \ kg \times 0 = (72.0 \ kg+65.0 \ kg) v[/tex]
[tex]468 kg m/s + 0 = (137 kg)v[/tex]
[tex]v = \dfrac{468\ kg m/s }{137 \ kg}[/tex]
v = 3.416 m/s