A boy is rolling a $0.5 kg$ ball on the frictionless floor with the speed of $20 {ms}^{- 1}$ . 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?

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$19 .0 {ms}^{- 1}$

$4 .47 {ms}^{- 1}$

$14 .41 {ms}^{- 1}$

$1 .00 {ms}^{- 1}$

answer is D.

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Detailed Solution

The new kinetic energy is 5% of initial KE
$KE' = \frac{5}{100} (KE)$

$\Rightarrow \frac{1}{2} mv'^{2} (1 + \frac{k^{2}}{R^{2}}) = \frac{5}{100} \times \frac{1}{2} {mv}^{2} (1 + \frac{k^{2}}{R^{2}})$
$\Rightarrow v'^{2} = \frac{5}{100} {(20)}^{2}$

$\Rightarrow v' = \sqrt{\frac{5 \times 20 \times 20}{100}} = \frac{20}{10} \sqrt{5}$ $= 2 \sqrt{5} = 4 .47 {ms}^{- 1}$
[Note: We have assumed ball is still rolling after deflection]