Question

A 1250 kg car is moving down the highway with a velocity of 32.0 m/s when it bumps into the car ahead of it which has a mass of 875 kg and a velocity of 25.0 m/s. After the collision, the two cars stick together. What will be the resulting velocity of the two cars together? How much energy will be lost in this collision?

Answer 1

A. If the collision is perfectly inelastic then it follows the equation,
    
             m1v1 + m2v2 = (m1 + m2)(v3)

Substituting,

           (1250 kg)(32 m/s) + (875 kg)(25 m/s) = (1250 kg + 875 kg)(v3)

The value of v3 from the equation is 29.12 m/s. 

B. The kinetic energy is calculated through the equation,

              KE = 0.5mv²

Using this equation to solve for the total kinetic energies before and after the collision,

    Before collision:

         KE = 0.5(1250 kg)(32 m/s)² + (0.5)(875 kg)(25 m/s)²
             KE = 913437.5 J

    After collision:
           KE = (0.5)(1250 kg + 875 kg)(29.12 m/s)²
                 KE = 900972.8 J

The difference is equal to 12464.7 J