Answer:-2.61 m/s
Explanation:
This problem can be solved by the Conservation of Momentum principle, which establishes that the initial momentum [tex]p_{o}[/tex] must be equal to the final momentum [tex]p_{f}[/tex]:
[tex]p_{o}=p_{f}[/tex] (1)
Where:
[tex]p_{o}=mV_{o}+MU_{o}[/tex] (2)
[tex]p_{f}=(m+M)V_{f}[/tex] (3)
[tex]m=1200 kg[/tex] is the mass of the first car
[tex]V_{o}=20 m/s[/tex] is the velocity of the first car, to the North
[tex]M=1400 kg[/tex] is the mass of the second car
[tex]U_{o}=-22 m/s[/tex] is the mass of the second car, to the South
[tex]V_{f}[/tex] is the final velocity of both cars after the collision
[tex]mV_{o}+MU_{o}=(m+M)V_{f}[/tex] (4)
Isolating [tex]V_{f}[/tex]:
[tex]V_{f}=\frac{mV_{o}+MU_{o}}{m+M}[/tex] (5)
[tex]V_{f}=\frac{(1200 kg)(20 m/s)+(1400 kg)(-22 m/s)}{1200 kg+1400 kg}[/tex] (6)
Finally:
[tex]V_{f}=-2.61 m/s[/tex] (7) This is the resulting velocity of the wreckage, to the south
