A railroad car of mass 2.52 104 kg is moving with a speed of 3.86 m/s. It collides and couples with three other coupled railroad cars, each of the same mass as the single car and moving in the same direction with an initial speed of 1.93 m/s. (a) What is the speed of the four cars after the collision? (Round your answer to at least two decimal places.) m/s (b) How much mechanical energy is lost in the collision?

a)   v = 2.4125 m / s  , b)  Em_{f} / Em₀ = 0.89

Explanation:

a) This is an inelastic crash problem, the system is made up of the four carriages, so the forces during the crash are internal and the moment is conserved

Initial

          p₀ = m v₁ + 3 m v₂

Final

          = (4 m) v

        p₀ =p_{f}

        m (v₁ + 3 v₂) = 4 m v

        v = (v₁ +3 v₂) / 4

Let's calculate

       v = (3.86 + 3 1.93) / 4

       v = 2.4125 m / s

b) the initial mechanical energy is

       Em₀ = K₁ + 3 K₂

       Em₀ = ½ m v₁² + ½ 3m v₂²

The final mechanical energy

          = K

         Em_{f} = ½ 4 m v²

The fraction of energy lost is

          Em_{f} / Em₀ = ½ 4m v² / ½ m (v₁² +3 v₂²)

          Em_{f} / Em₀ = 4 v₂ / (v₁² + 3 v₂²)

          Em_{f} / Em₀ = 4 2.4125² / (3.86² + 3 1.93²)

          Em_{f} / em₀ = 23.28 / 26.07

          Em_{f} / Em₀ = 0.89

i think the answer would be reserve.

the answer is f1 and d2 because if you put equal force on those two points the beam would balance trust me on this