A railroad car weighing 40 tons moving at a speed of 2 m / s, overtakes a platform weighing 20 tons, moving at a speed of 1 m /. After the automatic coupler, the car platform continues to move. Determine their joint speed.
m1 = 40 t = 40,000 kg.
V1 = 2 m / s.
m2 = 20 t = 20,000 kg.
V2 = 1 m / s.
When coupled, railway cars interact only with each other, then they can be considered a closed system of interacting bodies. For railway cars, the law of conservation of momentum in vector form is valid: m1 * V1 + m2 * V2 = m1 * V1 ^ + m2 * V2 ^.
Since before the coupling, the cars were moving in one direction, and after that they move together with the same speed V1 ^ = V2 ^ = V, then the law of conservation of momentum will take the form: m1 * V1 + m2 * V2 = (m1 + m2) * V.
V = (m1 * V1 + m2 * V2) / (m1 + m2).
V = (40,000 kg * 2 m / s + 20,000 kg * 1 m / s) / (40,000 kg + 20,000 kg) = 1.67 m / s.
Answer: after coupling, railway cars will move at a speed of V = 1.67 m / s.
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