Answer:
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
