Example

Question:

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

Solution:

The initial kinetic energy of the bullet is \[ \frac{1}{2}mv^2 = \frac{1}{2} \times 0.05 \times 200^2 = 1000\,\mathrm{J} \] The final kinetic energy is \(0.1 \times 1000 = 100\,\mathrm{J}\).
Let \(v_f\) be the emergent speed of the bullet: \[ \frac{1}{2}mv_f^2 = 100\,\mathrm{J} \] \[ v_f = \sqrt{\frac{2 \times 100}{0.05}} = \sqrt{4000} = 63.2\,\mathrm{m\,s^{-1}} \]
The speed is reduced by approximately 68% (not 90%).