Answer:
[tex]m_1=8\ kg,\ m_2=6\ kg,\ v_1=12\ m/s, v_2=4\ m/s,\ v_1'=-6\ m/s,\ v_2'=28\ m/s[/tex]
Explanation:
Conservation of Momentum
The total momentum of a system of two particles is
[tex]p=m_1v_1+m_2v_2[/tex]
Where m1,m2,v1, and v2 are the respective masses and velocities of the particles at a given time. Then, the two particles collide and change their velocities to v1' and v2'. The final momentum is now
[tex]p'=m_1v_1'+m_2v_2'[/tex]
The momentum is conserved if no external forces are acting on the system, thus
[tex]m_1v_1+m_2v_2=m_1v_1'+m_2v_2'[/tex]
Let's put some numbers in the problem and say
[tex]m_1=8\ kg,\ m_2=6\ kg,\ v_1=12\ m/s, v_2=4\ m/s,\ v_1'=-6\ m/s,\ v_2'=28\ m/s[/tex]
[tex](8)(12)+(6)(4)=(8)(-6)+(6)(28)[/tex]
[tex]96+24=-48+168[/tex]
120=120
It means that when the particles collide, the first mass returns at 6 m/s and the second continues in the same direction at 28 m/s
