Question

A 833 kg automobile is sliding on an icy street. It collides with a parked car which has a mass of 561 kg. The two cars lock up and slide together with a speed of 16.8 km/h. What was the speed of the first car just before the collision?

Answer 1

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 :

$mv+m'v'=(m+m')V$

v is the speed of the first car

v' is the speed of the second car that is at rest

So,

$mv=(m+m')V$

$v=\dfrac{(m+m')V}{m}$

$v=\dfrac{(833+561)\times 4.67}{833}$

$v=7.81\ m/s$

So, the speed of the first car just before the collision is 7.841 m/s. Hence, this is the required solution.