Question

A 20-gram bullet traveling at 250 m/s strikes a block of wood that weighs 2 kg. With what velocity will the block and bullet move after the collision? [convert g to kg]

Answer 1

Answer:

the velocity of the bullet-wood system after the collision is 2.48 m/s

Explanation:

Given;

mass of the bullet, m₀ = 20 g = 0.02 kg

velocity of the bullet, v₀ = 250 m/s

mass of the wood, m₁ = 2 kg

velocity of the wood, v₁ = 0

Let the velocity of the bullet-wood system after collision = v

Apply the principle of conservation of linear momentum to calculate the final velocity of the system;

Initial momentum = final momentum

m₀v₀ + m₁v₁ = v(m₀ + m₁)

0.02 x 250  + 2 x 0    =    v(2  + 0.02)

5 + 0 = v(2.02)

5 = 2.02v

v = 5/2.02

v = 2.48 m/s

Therefore, the velocity of the bullet-wood system after the collision is 2.48 m/s