A body weighing 20 g freely falls from a height of 2.5 m and falls into the center of a horizontal plate weighing

| June 2, 2021 | education

A body weighing 20 g freely falls from a height of 2.5 m and falls into the center of a horizontal plate weighing 0.2 kg, which is suspended on a vertical spring with a stiffness of 1.5 kN / m. Assuming the shock is absolutely inelastic, determine the maximum elongation of the spring.

Data: m1 (mass of the fallen body) = 20 g = 0.02 kg; h0 (initial body height) = 2.5 m; m2 (plate weight) = 0.2 kg; k (vertical spring rate) = 1.5 kN / m = 1500 N / m.

Constants: g (acceleration due to gravity) ≈ 10 m / s2.

1) The speed of the body before hitting the plate: m1 * V1 ^ 2/2 = m * g * h0, whence V1 = √ (g * h0 * 2) = √ (10 * 2.5 * 2) = 7.07 m /from.

2) Velocity of plate and body after central inelastic impact: (m1 + m2) * V = m1 * V1, whence V = m1 * V1 / (m1 + m2) = 0.02 * 7.07 / (0.02 + 0 , 2) = 0.64 m / s.

3) Maximum elongation: (m1 + m2) * V ^ 2/2 = k * Δx ^ 2/2, whence Δx = √ ((m1 + m2) * V2 / k) = √ ((0.02 + 0, 2) * 0.642 / 1500) = 7.8 * 10 ^ -3 m = 7.8 mm.



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