Question

A 2.8 kg block of wood sits on a frictionless table. A 3.0 g bullet, fired horizontally at a speed of 400 m/s , goes completely through the block, emerging at a speed of 220 m/s .

What is the speed of the block immediately after the bullet exits?

Answer 1

Mass of the bullet = m₁ = 3 g = 3 x 10^-3 kg

Mass of the block of wood = m₂ = 2.8 kg

Initial speed of the bullet = V₁ = 400 m/s

Initial speed of the block of wood = V₂ = 0 m/s

Speed of the bullet after it emerges from the block = V₃ = 220 m/s

Speed of the block of wood after the bullet exits = V₄

By conservation of linear momentum,

m₁V₁ + m₂V₂ = m₁V₃ + m₂V₄

(3x10^-3)(400) + (2.8)(0) = (3x10^-3)(220) + (2.8)V₄

V₄ = 0.193 m/s

Speed of the block immediately after the bullet exits = 0.193 m/s