A boy is rolling a 0 . 5kg ball on the frictionless floor with the...

Given, $m = 0 . 5kg and u = 20{ms}^{- 1}$

Initial kinetic energy \left({k}_{i}\right) = \dfrac{1}{2}m{u}^{2}$</p><p>$= \dfrac{1}{2} \times 0 . 5 \times 20 \times 20 = 100J$</p><p>After deflection it moves with$5 % of {k}_{i}$</p><p>$∴ {k}_{f} = \dfrac{5}{100} \times {k}_{i} ⇒ \dfrac{5}{100} \times 100$</p><p>$⇒ {k}_{f} = 5J$</p><p>Now, let the final speed be$' v ' {ms}^{- 1},$then:</p><p>${k}_{f} = 5 = \dfrac{1}{2}m{v}^{2}$</p><p>$⇒ {v}^{2} = 20$</p><p>$⇒ v = \sqrt{20} = 4 . 47{ms}^{- 1}$