A 0.018 kg bullet traveling 222 m/s penetrates a 3.9 kg block of wood, and exits the block with a velocity of 72 m/s. If the block was stationary on a horizontal frictionless surface before impact, how fast does it move after the bullet exits?

momentum before = momentum after

.018*222 = .018*72 + 3.9 * v

thank you

You are welcome :)

To find the velocity of the block after the bullet exits, we can use the principle of conservation of momentum. According to this principle, the total momentum before the collision is equal to the total momentum after the collision.

The momentum (p) of an object is calculated by multiplying its mass (m) by its velocity (v): p = m * v.

Before the bullet enters the block, the block is stationary. Hence, its momentum is zero: p_block_before = 0.

The initial momentum of the bullet is: p_bullet_before = m_bullet * v_bullet_before.
Substituting the given values: p_bullet_before = 0.018 kg * 222 m/s.

After the bullet exits the block, the block and the bullet together have a common final velocity (v_common), and their respective masses are 3.9 kg (block) and 0.018 kg (bullet).

The final momentum of the block is: p_block_after = m_block * v_block_after.
Substituting the given values: p_block_after = 3.9 kg * v_block_after.

The final momentum of the bullet is: p_bullet_after = m_bullet * v_bullet_after.
Substituting the given values: p_bullet_after = 0.018 kg * 72 m/s.

Since momentum is conserved, we can equate the total momentum before the collision with the total momentum after the collision:

p_block_before + p_bullet_before = p_block_after + p_bullet_after.

0 + 0.018 kg * 222 m/s = 3.9 kg * v_block_after + 0.018 kg * 72 m/s.

Simplifying the equation, we have:

0.018 kg * 222 m/s = 3.9 kg * v_block_after + 0.018 kg * 72 m/s.

3.916 kg * v_block_after = 0.018 kg * 150 m/s.

Dividing both sides by 3.916 kg, we get:

v_block_after = (0.018 kg * 150 m/s) / 3.916 kg.

Simplifying the right side:

v_block_after ≈ 0.688 m/s.

Therefore, the block moves with a velocity of approximately 0.688 m/s after the bullet exits.