Question

please someone help explain?????
A 3.72 g bullet moving at 290 m/s enters and stops in an initially stationary 3.30 kg wooden block on a horizontal frictionless surface.

What fraction of the bullet's kinetic energy was lost in this perfectly inelastic collision?

How much work was done in stopping the bullet?

i am not understanding how to find these

Answer 1

According to Law of conservation of momentum, The total inital momentum of a system must be the same as its final momentum, if no external foce acts.
So, what is the initial momentum of the bullet-block system? Find the momentum of each and add them.
What is the final momentum of the bullet-block system? After collision, the bullet and block move together with a common velocity V. Its momentum is (mass of bullet+mass of block)V.
So, equate the total initial momentum and total final momentum. Find V
Now, do u get it?

Answer 2

thanks I got it