Answer: The momentum of the system after the collision is 0.2 kg m/s.
Explanation:
According to the conservation of momentum, the total momentum of the system before the collision is same the total momentum of the system after the collision in an isolated system.
[tex]m_{1}u_{1}+m_{2}u_{2}=m_{1}v_{1}+m_{2}v_{2}[/tex]
Here, m_{1},m_{2} are the first and second object, v_{1},v_{2} are the initial velocities of the first and second objects and u_{1},u_{2} are the final velocities of the first and the second objects.
Use the formula of the conservation of the momentum.
[tex]m_{1}u_{1}+m_{2}u_{2}=m_{1}v_{1}+m_{2}v_{2}[/tex]
Put m_{1}=0.6 kg, m_{2}=0.5 kg, v_{1}=0.5 m/s and v_{2}=-0.2 m/s.
[tex]m_{1}u_{1}+m_{2}u_{2}=(0.6)(0.5)+(0.5)(-0.2)[/tex]
[tex][tex]m_{1}u_{1}+m_{2}u_{2}[/tex]= 0.2 kg m/s[/tex]
Therefore, the momentum of the system after the collision is 0.2 kg m/s.
