A weight of 100 kg was dropped onto a platform with a mass of 400 kg moving along a horizontal

A weight of 100 kg was dropped onto a platform with a mass of 400 kg moving along a horizontal path at a speed of 0.4 m / s. Which one became the platform speed.

Task data: m1 (mass of the platform moving along the horizontal path) = 400 kg; V1 (initial platform speed) = 0.4 m / s; m2 (mass of the cargo dropped onto the platform) = 100 kg.

The speed of the platform after the fall of the load is determined from the equality (law of conservation of momentum): (m1 + m2) * V1.2 = m1 * V1, whence V1.2 = m1 * V1 / (m1 + m2).

Calculation: V1.2 = 400 * 0.4 / (400 + 100) = 0.32 m / s.

Answer: The platform speed after dropping the load became equal to 0.32 m / s.



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