# Speedy Collision Fun!

## How fast will the block be moving if the bullet becomes embedded in the block?

Given that a 4.00 g bullet is fired horizontally at a speed of 148 m/s into a 1.00 kg wooden block resting on a frictionless horizontal surface, determine the speed of the block after the collision.

## Answer:

The block will be moving to the right at a speed of 0.5907 m/s after the collision.

We can use conservation of momentum to solve this problem, assuming that the system (bullet + block) is isolated and there are no external forces acting on it. The momentum of the bullet before it hits the block is:

**p1 = m1 * v1 = (0.004 kg) * (148 m/s) = 0.592 kg m/s**

After the collision, the bullet becomes embedded in the block, so the mass of the system changes to:

**m = m1 + m2 = 0.004 kg + 1.00 kg = 1.004 kg**

Let v be the final velocity of the block + bullet system after the collision. Then, the conservation of momentum gives us:

**p1 = p2**

where p2 is the momentum of the block + bullet system after the collision. We can express p2 in terms of the final velocity v:

**p2 = m * v**

Substituting the values we found earlier, we get:

**0.592 kg m/s = (1.004 kg) * v**

Solving for v, we get:

**v = 0.5907 m/s**