Question

A 1094 kg van, stopped at a traffic light, is hit directly in the rear by a 727 kg car traveling with a velocity of +2.41 m/s. Assume that the transmission of the van is in neutral, the brakes are not being applied, and the collision is elastic. What is the final velocity of the car?

Answer 1

From the law of conservation of momentum, we have

       m2v2 + m1v1 = 0 + m1u1

1094*v2 + 727*v1 = 727*2.41

Rewrite the equation for v2.

      v2 = [727*2.41-727*v1] / 1094 = 1.60 - (0.6645)v₁

From the law of conservation of energy, we have

1/2m1v1² + 1/2m2v2² = 0+ 1/2m2u2²

   1094*v2^2 +727v1^2 = 727*2.41^2

   1094*v2^2 +727v1^2 = 4222.4887

   1094[1.60 - (0.6645)v₁]2 + 727v1^2 = 4222.4887

   {1094[2.56 +(0.44156025)v₁²-(2.1264)v₁]} + 727v1^2 = 4222.4887

Write the equation in the form of quadratic equation:

    (1208.0669135)v₁² - (2326.2816)v1 - 1421.8487= 0

Solutions: v1 = 2.41 m/s and -0.487 m/s. In this case, the car travel after collision along the negative x-axis.

So, the final velocity of the car is, -0.487 m/s.

Note:

Final velocity of the van is, v2 = 1.60 - (0.6645)v₁= 1.60 - 0.6645*-0.487 = 1.92 m/s