Problem 3/180 Page 196, Engineering Mechanics - Dynamics, Meriam and Kraige, 4th Edition

Given: A bullet of mass m traveling at speed v0 strikes and becomes embedded in a block of mass M.  The block is initially stationary.

Find: Compute the energy lost during the impact (converted to other forms).

Parameter Values:
m = 75 g = 0.075 kg
v0 = 600 m/s
M = 50 kg

1. Mechanical Systems: System 1 is the bullet during the impact.  System 2 is the block during the impact.

2. Free Body Diagrams: System 1 free body shows the bullet and a large horizontal force F.  System 2 free body shows the block and the horizontal force F acting oppositely from its direction on the bullet.

3. Equations:
System 1 Impulse = -Integral of F dt i
System 1 D Linear Momentum = m (vF - v0) i
System 1 D Kinetic Energy = 1/2 m (vF² - v0²)
System 1 Initial Kinetic Energy = 1/2 m v0²

System 2 Impulse = Integral of F dt i
System 2 D Linear Momentum = M (vF - 0 ) i
System 2 D Kinetic Energy = 1/2 M (vF² - 0)
System 2 Initial Kinetic Energy = 0

4. Solve:
-Integral of F dt = m(vF - v0)
+Integral of F dt = M vF

Adding equations:
0 = (m+M)vF - m v0

vF = m v0 / (m+M)

Total change in kinetic energy = 1/2 m [ (m v0 / (m+M))² - v0² ] + 1/2 M ( m v0 / (m+M) )²
    = 1/2 m v0² { -M/(m+M) } = 13480 N m

The percentage of energy lost is given by the ratio of M to m+M.

Percentage loss = 99.9%