Question

A 22-g bullet traveling 265 m/s penetrates a 1.9 kg block of wood and emerges going 125 m/s .

If the block is stationary on a frictionless surface when hit, how fast does it move after the bullet emerges?
Express your answer to two significant figures and include the appropriate units.

Answer 1

The final velocity of the block of wood after the bullet emerges is 1.6 m/s. This was determined by using the law of conservation of momentum.

Step-by-step explanation:

To solve this problem, we need to apply the law of conservation of momentum. The total momentum before and after the event remains the same because no external force is acting on the bullet-block system. Before the collision, the block is stationary while the bullet is moving, so the initial momentum of the system is just the momentum of the bullet. After the bullet emerges from the block, both the bullet and the block have momentum.

The initial momentum of the bullet can be calculated as follows:

Initial momentum of the bullet = mass of bullet × velocity of bullet = 0.022 kg × 265 m/s

Using the law of conservation of momentum:

(Mass of bullet × initial velocity of bullet) + (Mass of block × initial velocity of block) = (Mass of bullet × final velocity of bullet) + (Mass of block × final velocity of block)

0.022 kg × 265 m/s + 0 kg × 0 m/s = 0.022 kg × 125 m/s + 1.9 kg × final velocity of block

After calculating, the final velocity of the block emerges as:

Final velocity of block = (0.022 kg × 265 m/s - 0.022 kg × 125 m/s) / 1.9 kg

After solving, we get:

Final velocity of block = 1.6 m/s