Question

A railroad car of mass 2.90 ? 104 kg moving at 3.05 m/s collides and couples with two coupled railroad cars, each of the same mass as the single car and moving in the same direction at 1.20 m/s. (a) What is the speed of the three coupled cars after the collision? m/s (b) How much kinetic energy is lost in the collision? J

Answer 1

a)

m = mass of each railroad car = 2.90 x 10⁴ kg

v₁ = velocity of single railroad car before collision= 3.05 m/s

v₂ = velocity of two coupled railroad car before collision= 1.20 m/s

v = velocity of three-coupled cars after collision = ?

Using conservation of momentum

m v₁ + (2m) v₂ = (m + 2m) v

3.05 + (2) (1.20) = (3) v

v = 1.82 m/s

b)

Kinetic energy lost is given as

KE = (0.5) m v₁² + (0.5) (2m) v₂² - (0.5) (m + 2m) v²

KE = (0.5) (2.90 x 10⁴) (3.05)² + (0.5) (2(2.90 x 10⁴)) (1.20)² - (0.5) (3 (2.90 x 10⁴)) (1.82)²

KE = 3.3 x 10⁴ J