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.


  1. Answer:

     v_f = 1.14 m / s


    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 = \frac{M v_o - m v}{M-m}

    let’s calculate

           v_f = \frac{57 \ 1.2 - 0.11 \ 31.5}{ 57 - 0.11}

           v_f = 64.936 / 56.89

           v_f = 1.14 m / s

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