Answer: 0.205 m/s
Explanation:
According the conservation of momentum law, the total momentum before the collision ([tex]p_{i}[/tex]) is equal to the total momentum after the collision ([tex]p_{f}[/tex]):
[tex]p_{i}=p_{f}[/tex] (1)
Being:
[tex]p_{i}=m_{1}V_{1}+m_{2}V_{2}[/tex] (2)
[tex]p_{f}=m_{1}U_{1}+m_{2}U_{2}[/tex] (3)
Where:
[tex]m_{1}=0.1 kg[/tex] is the mass of the first cart
[tex]V_{1}=1.20 m/s[/tex] is the initial velocity of the first cart
[tex]m_{2}=1 kg[/tex] is the mass of the second cart
[tex]V_{2}=0 m/s[/tex] is the initial velocity of the second cart
[tex]U_{1}=-0.85 m/s[/tex] is the final velocity of the first cart
[tex]U_{2}[/tex] is the final velocity of the second cart
Substituting (2) and (3) in (1):
[tex]m_{1}V_{1}+m_{2}V_{2}=m_{1}U_{1}+m_{2}U_{2}[/tex] (4)
Isolating [tex]U_{2}[/tex]:
[tex]U_{2}=\frac{m_{1}(V_{1}-U_{1})}{m_{2}}[/tex] (5)
[tex]U_{2}=\frac{0.1 kg(1.20 m/s-(-0.85 m/s))}{1 kg}[/tex] (6)
Finally:
[tex]U_{2}=0.205 m/s[/tex] This is the speed of the second cart after the collision
