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A 57 kg ice skater moving to the right with a velocity of 1.2 m/s throws a 0.11 kg snowball to the right with a velocity of 31.5 m/s relative to the ground. What is the velocity of the ice skater after throwing the snowball? Disregard the friction between the skates and the ice.

Respuesta :

Answer:

 v_f = 1.14 m / s

Explanation:

This is an exercise of conservation of momentum, let's define a system formed by the skater and the snowball, therefore the forces during the launch are internal and the moment is conserved.

Initial instant. Before throwing the ball

        p₀ = M v₀

Final moment. After throwing the snowball

       p_f = (M-m) v_f + m v

the moment is preserved

        p₀ = p_f

        M v₀ = (M-m) v_f + m v

        v_f = [tex]\frac{M v_o - m v}{M-m}[/tex]

let's calculate

       v_f = [tex]\frac{57 \ 1.2 - 0.11 \ 31.5}{ 57 - 0.11}[/tex]

       v_f = 64.936 / 56.89

       v_f = 1.14 m / s

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