Bullet with mass "m1" and speed "v1", hits a bag with mass "m2" and speed "v2=0", which means that the bag is standing still. find the formula that explains how much speed has the bag after the bullet hits it. sorry for my bad english, it's not my ethnic language.

IF the bag is suspended freely in mid-air and free to move
without friction, then the linear momentum is the same both
before and after the hit.

Momentum of the bullet before the hit = ( m₁ v₁ ).

Momentum of the (bullet + bag) after the hit = (m₁ + m₂) (v₂) .

Momentum is conserved, so we can say that ...

             ( m₁ v₁ ) = (m₁ + m₂) (v₂)

                 v₂ =  ( m₁ v₁ ) / (m₁ + m₂)

Usually the mass of the bag is much much more than the mass
of the bullet, so the mass of the bullet can be ignored after the
hit, and the formula is very closely ...

                 v₂ =  v₁ ( m₁ / m₂ ) .

Stop apologizing for your English.  It's better than many native
Americans writing on this site.  Be proud.
 

  the answer is don't know.