Question

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?

Answer 1

22.2 m/s

Step-by-step explanation:

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:

`p_i = p_f`

where
`p_i` is the total momentum before the collision and
`p_f` the total momentum after the collision. Re-writing the equation,

`mu+MU = (m+M)v`

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:

`v=(mu)/((M+m))=((0.04)(300))/(0.04+0.5)=22.2 m/s`