A boy is rolling a 0.5 kg ball on the frictionless floor with the speed of 20 m s$$^{-1}$$. The ball gets deflected by an obstacle on the way. After deflection it moves with 5% of its initial kinetic energy. What is the speed of the ball now?

The initial kinetic energy of the ball is $$K_i = \frac{1}{2}mv^2 = \frac{1}{2}(0.5)(20)^2 = 100$$ J. After deflection the ball retains 5% of this energy, so $$K_f = 0.05 \times 100 = 5$$ J.

Using $$K_f = \frac{1}{2}mv_f^2$$, we get $$5 = \frac{1}{2}(0.5)v_f^2$$, which gives $$v_f^2 = 20$$, so $$v_f = \sqrt{20} = 2\sqrt{5} \approx 4.47$$ m/s.

This is closest to 4.4 m s$$^{-1}$$, so the correct answer is option 2.