A 640 kg automobile slides across an icy street at a speed of 63.9 km/h and collides with a parked car which has a mass of 816 kg. The two cars lock up and slide together. What is the speed of the two cars just after they collide?

V3 = 7.802 m/s

Explanation:

m1 = 640 Kg, M2 = 816 kg, V1 = 63.9 Km/h = 17.75 m/s, V2 =0 m/s

Let V3 is the combine velocity after collision.

According to the law of conservation of momentum

m1 v1 + m2 v2 = (m1 + m2) v3

⇒ V3 =( m1 v1 + m2 v2 ) / (m1 + M2)

V3 =  ( 640 Kg × 17.75 m/s + 816 kg × 0m/s) / (640 Kg + 816 kg)

V3 = 7.802 m/s

28.088 km/h or 7.802 m/s

Explanation:

From the law of conservation of momentum,

Total momentum before collision = Total momentum after collision

mu+m'u' = V(m+m') Equation 1

Where m = mass of the automobile, m' = mass of the car, u = initial velocity of the car, u' = initial velocity of the car, V = velocity of the two car after collision

Make V the the subject of the equation

V = (mu+m'u')/(m+m') Equation 2

Given: m = 640 kg, m' = 816 kg, u = 63.9 km/h, u' = 0 m/s (parked).

Substitute into equation 2

V = (640×63.9+816×0)/(640+816)

V = 40896/1456

V = 28.088 km/h = 28.088(1000/3600) m/s = 7.802 m/s

Hence the speed of the two cars after they collide = 28.088 km/h or 7.802 m/s

here we know that ball will drop down by total distance 3.2 m

so here we know that

now using kinematics

now in the same time it will cover horizontal distance of 0.75 m

so speed of the ball will be

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