Answer :
Answer:
The speed of the first car just before the collision is 7.841 m/s.
Explanation:
Given that,
Mass of the automobile, m = 833 kg
Mass of the parked car, m' = 561 kg
The final speed of the cars lock up and slide together with a speed of 16.8 km/h, V = 16.8 km/h = 4.67 m/s
If two cars lock up and slide together, then it is the case of inelastic collision. The momentum will remains conserved such as :
[tex]mv+m'v'=(m+m')V[/tex]
v is the speed of the first car
v' is the speed of the second car that is at rest
So,
[tex]mv=(m+m')V[/tex]
[tex]v=\dfrac{(m+m')V}{m}[/tex]
[tex]v=\dfrac{(833+561)\times 4.67}{833}[/tex]
[tex]v=7.81\ m/s[/tex]
So, the speed of the first car just before the collision is 7.841 m/s. Hence, this is the required solution.