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During a circus act, a performer thrills the crowd by catching a cannon ball shot at him. The cannon ball has a mass of m1 = 8.5 kg and the horizontal component of its velocity is v = 8.1 m/s when the performer, of mass m2 = 66 kg, catches it. If the performer is on nearly frictionless roller skates, what is his recoil velocity?

Respuesta :

Answer:

[tex]v = 0.92 m/s[/tex]

Explanation:

As we know that there is no external horizontal force on the cannon ball + performer

So here total momentum of the system must be conserved

so we will have

[tex]m_1v_1 = (m_1 + m_2) v_f[/tex]

[tex]m_1 = 8.5 kg[/tex]

[tex]v_1 = 8.1 m/s[/tex]

[tex]m_2 = 66 kg[/tex]

now from above equation we have

[tex]8.5 \times 8.1 = (8.5 + 66) v[/tex]

[tex]v = 0.92 m/s[/tex]

pstnonsonjoku

The recoil velocity of the performer on stage is 0.92 m/s.

Following from the principle of conservation of linear momentum, the total momentum of a system is constant. This implies that momentum before collision is equal to momentum after collision.

Given that;

( 8.5 × 8.1) + (66 × 0) = (8.5 + 66) v

v = ( 8.5 × 8.1) + (66 × 0)/(8.5 + 66)

v = 68.85/74.5

v = 0.92 m/s

Learn more about recoil velocity:https://brainly.com/question/953809

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