Question

A 10.1-g bullet is fired into a stationary block of wood having mass m = 5.01 kg. The bullet imbeds into the block. The speed of the bullet-plus-wood combination immediately after the collision is 0.599 m/s. What was the original speed of the bullet? (Express your answer with four significant figures.)

Answer 1

The original speed of the bullet was 297.7 m/s

Step-by-step explanation:

Step 1: Data given

mass of the bullet = 10.1 g

mass of the wood = 5.01 kg

The speed of the bullet-plus-wood combination immediately after the collision is 0.599 m/s

Step 2: Calculate the original speed of the bullet

Conserve momentum: initial p = final p

the velocity of the block + bullet after a completely inelastic impact can be given as followed:

v' = (m1v1 + m2v2) / (m1 + m2)

⇒ with m1 = mass of the bullet = 10.1 g = 0.0101 Kg

⇒ with v1 = velocity of the bullet = TO BE DETERMINED

⇒ with m2 = mass of the wood = 5.01 Kg

⇒ with v2 = speed of the wood = 0 m/s

with v' = speed of the bullet-plus-wood = 0.599 m/s

0.599 = (0.0101* v1 + 5.01*0) / ( 0.0101 + 5.01)

0.599 * (0.0101 + 5.01) = (0.0101 * v1)

v1 = 297.7 m/s

The original speed of the bullet was 297.7 m/s