Someone fires a 0.04 kg bullet at a block of wood that has a mass of 0.5 kg. (The block of wood is sitting on a frictionless surface, so it moves freely when the bullet hits it). The wood block is initially at rest. The bullet is traveling 300 m/s when it hits the wood block and sticks inside it. Now the bullet and the wood block move together as one object. How fast are they traveling?

We can answer this question by using the law of conservation of momentum. In fact, the total momentum before and after the collision must be conserved. Therefore, we can write:

[tex]p_i = p_f[/tex]

where [tex]p_i[/tex] is the total momentum before the collision and [tex]p_f[/tex] the total momentum after the collision. Re-writing the equation,

[tex]mu+MU = (m+M)v[/tex]

where:

m = 0.04 kg is the mass of the bullet

u = 300 m/s is the initial velocity of the bullet

M = 0.5 kg is the mass of the block of wood

U = 0 is the initial velocity of the block (it is at rest)

v is the final velocity of the bullet+block combined, after the collision

Solving for u, we find the final velocity of the bullet+block after the collision:

[tex]v=\frac{mu}{(M+m)}=\frac{(0.04)(300)}{0.04+0.5}=22.2 m/s[/tex]