In a ballistics demonstration,a police officer fires a bullet of mass $50.0 \; g$ with a speed of $200 \; m s^{-1}$ into soft plywood of thickness $2.00 \; cm$. The bullet emerges with only $10 \%$ of its initial kinetic energy. What is the emergent speed of the bullet?

(C) The initial kinetic energy $(K_i)$ of the bullet is given by $K_i = \frac{1}{2} m v_i^2$.
Given $m = 50.0 \; g = 0.050 \; kg$ and $v_i = 200 \; m s^{-1}$.
$K_i = \frac{1}{2} \times 0.050 \times (200)^2 = 0.025 \times 40000 = 1000 \; J$.
The bullet emerges with $10 \%$ of its initial kinetic energy,so the final kinetic energy $(K_f)$ is:
$K_f = 0.10 \times 1000 \; J = 100 \; J$.
Let $v_f$ be the emergent speed of the bullet.
Using the formula $K_f = \frac{1}{2} m v_f^2$:
$100 = \frac{1}{2} \times 0.050 \times v_f^2$.
$100 = 0.025 \times v_f^2$.
$v_f^2 = \frac{100}{0.025} = 4000$.
$v_f = \sqrt{4000} \approx 63.25 \; m s^{-1}$.
Thus,the emergent speed of the bullet is approximately $63.2 \; m s^{-1}$.