A 22-g bullet traveling 240 m/s penetrates a 2.0-kg block of wood and emerges going 150 m/s. If the block is stationary on a frictionless surface when hit, how fast does it move after the bullet emerges?

After the bullet emerges the block moves at 0.99 m/s

Explanation:

Given;

mass of bullet, m₁ = 22 g = 0.022 kg

initial speed of the bullet, u₁ = 240 m/s

final speed of the bullet, v₁ = 150 m/s

mass of block, m₂ = 2.0 kg

initial speed of the block, u₂ = 0

Let the final speed of the block = v₂

Apply principles of conservation of linear momentum;

m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂

0.022 x 240 + 2 x 0 = 0.022 x 150 + 2v₂

5.28 = 3.3 + 2v₂

5.28 - 3.3 = 2v₂

1.98 = 2v₂

v₂ = 1.98 / 2

v₂ = 0.99 m/s

Therefore, after the bullet emerges the block moves at 0.99 m/s