In a ballistics demonstration a police officer fires a bullet of mass `50.0 g` with speed `200 ms^(-1)` (see Table 6.2) on soft plywood of thickness 2.00 cm. The kinetic energy. What is the emergent speed of the bullet?

Explanation

The initial kinetic energy of the bullet is `mv^(2)"/"2= 1000 J`. It has final kinetic energy of `0.1xx1000= 100 J`. If `v_(f)` is the emergent speed of the bullet,
`(1)/(2) mv_(f)^(2)= 100 J`
`v_(f)= sqrt((2xx100 J)/(0.05 kg))`
`= 63.2 ms^(-1)`.
The speed is reduced by approximately 68% (not 90%).