3.00 kg block moving 2.09 m/s right hits a 2.22 kg block moving 3.92 m/s left. afterwards, the 3.00 kg block moves 1.71 m/s left. find the velocity of the 2.22 kg block afterwards

1.22 m/s (or 1 m/s if you use sig figs)

Explanation:

Our equation is m1v1+m2v2=m1v1+m2v2

Plug in our numbers to get:

3(2.09)+2.22(-3.92)=3(-1.71)+2.22(vf)

Then solve. We can simplify by doing:

6.27-8.7024=-5.13+2.22(vf)

Then combine the left side

-2.4324=-5.13+2.22(vf)

Add 5.13 to both sides

2.6976=2.22(vf)

Divide by 2.22

vf=1.22 m/s (or 1 m/s with sig figs)

Hope this helps

12000j is work done

thermal energy is the answer