Given, $m=0.5 \mathrm{kg}$ and $u=20{ \mathrm{m} \mathrm{s}}^{-1}$

Initial kinetic energy $\left(k_i\right)=\frac{1}{2}m{u}^2$

$=\frac{1}{2}\times 0.5\times 20\times 20=100 \mathrm{J}$

After deflection it moves with $5\%$ of $k_i$

$\therefore k_f=\frac{5}{100}\times k_i\Rightarrow \frac{5}{100}\times 100$

$\Rightarrow k_f=5 \mathrm{J}$

Now, let the final speed be $\text{'}v\text{'} { \mathrm{m} \mathrm{s}}^{-1},$ then:

$k_f=5=\frac{1}{2}m{v}^2$

$\Rightarrow v^2=20$

$\Rightarrow v=\sqrt{20}=4.47{ \mathrm{m} \mathrm{s}}^{-1}$