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?

22.2 m/s

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:

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

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:

the answer is d. 4

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