Question

A 1030 kg vehicle that moves at 3.4 m / s hits when the bumper of another 1140 kg vehicle that is at rest is parked, both vehicles being engaged. Determine the speed of the assembly formed by the two vehicles immediately after the collision.

Answer 1

Given data :

m₁ = 1030 kg v₁ = 3.4 m/s  

m₂ = 1140 kg v₂ = 0 m/s

Let V be the velocity of the assembly formed by two vehicles just after the collision :

Applying law of conservation of momentum :

m₁v₁ + m₂v₂  = (m₁ + m₂) V

(1030) (3.4) + (1140 ) (0) = (1030 + 1140) V

3502 = 2170 V

3502 / 2170 = V

1. 61 m/s = V

So, the velocity of the assembly formed by the two vehicles is 1.61 m/s

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