A locomotive of mass 6.00*10^4 kg is moving eastward with a speed of 10.8 km/h.

Question

A locomotive of mass 6.00*10^4 kg is moving eastward with a speed of 10.8 km/h. It collides and couples with two coupled railroad cars, each mass 2.00*10^4 kg, moving westward on the same track with an initial speed of 5.40 km/h. What is the speed of the coupled system (locomotive and two railroad cars) after the collision? What is the direction of motion after the collision? How much energy is transformed into internal energy?

Explanation / Answer

m = 6.00*104 kg, v = 10.8 km/h. (choose eastward aspositive direction) m' = 2*2.00*104 kg, v' = -5.40 km/h. What is the speed u of the coupled system (locomotive and tworailroad cars) after the collision? momentum conservation: mv + m'v' = (m + m')u u = (mv + m'v')/(m + m') = 4.32 km/h What is the direction of motion after the collision? u > 0, its direction is eastward How much energy is transformed into internal energy? mv2/2 + m'v'2/2 + (m + m')u2/2 =2.43*105 J

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.