A 5.00-g bullet comes out a 1.00-kg handgun at a speed of 8.00× 102 m/s when the handgun is held still. What is speed of the bullet if the handgun is held very loosely, assuming that the same amount of kinetic energy is generated each time when the handgun is fired?

The velocity is  

Explanation:

From the question we are told that

   The mass of the bullet is  

   The mass of the hand gun is  

    The speed of the gun is  

Generally from the law of momentum conservation

here are zero given that before the bullet was shot the gun was at rest

is negative given because from Newtons third law action and reaction is equal and opposite so if the bullet is moving forward the gun will be moving backward

So

=>      

Hence the velocity of the bullet with respect to the loose gun is  

=>  

=>  

f=ma a/f/m a= 480/60 a = 8m/s/s that is quick wow