Question

A 9.9-3 bullet is fired into a stationary block of wood having mass m = 5.02 kg. The bullet imbeds into the block. The speed of the bullet-plus-wood combination immediately after the collision is 0.606 m/s. What was the original speed of the bullet? (Express your answer with four significant figures.) 4.0

Answer 1

Given mass of bullet = 9.9 g = 0.0099 kg

mass of wood block = 5.02 kg

Speed of the block before the collision  

fo=0

The sum of the moment of bodies mV before collision is equal to the sum after the collision

miVi +m1V2 = m V' + mavz

The bullet speed equal to the speed of the block after the shock

Vi = V = V = 0.606m/s

Then we have

miVi + m2V2 = (mı + m2)

Solving for the velocity of the bullet

Vi = [(mı + m2 V - m2V21 mi

Where

mol = 0

So that

(mı + m2) Vi = ? mi

Vi = [(0.0099 + 5.02 0.606 0.0099

Vi = 307.9m/s