Answer to Question #172698 in Physics for rhjh

Question

a 5g bullet, traveling at 300 m/s strike 0.495 kg wooden block how much mechanical energy is lost during the collision center they calculate a difference in energy rounded to two decimal places

Answer 1

$$mV=(m+M)v\\(0.005)(300)=(0.005+0.495)v\\v=3\frac{m}{s}$$

$$\Delta K=0.5(mV^2-(m+M)v^2)\\\Delta K=0.5((0.005)(300)^2-(0.5)3^2)=222.75\ J$$