Question:

A mass of 1 kg falls from a height of 1 m and lands on a massless platform supported by a spring with spring constant \( 15 \, \text{Nm}^{-1} \). The maximum compression of the spring is (take \( g = 10 \, \text{ms}^{-2} \))
A mass of 1 kg falls from a height of 1 m

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Use conservation of energy: \( mg(h + x) = \frac{1}{2} k x^2 \) for spring-mass impact problems.
Updated On: May 18, 2025
  • 2 m
  • \( \sqrt{2} \) m
  • \( \frac{2}{\sqrt{3}} \) m
  • \( \sqrt{4} \) m
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The Correct Option is A

Solution and Explanation

Energy conservation: Total gravitational potential energy is converted into spring potential energy at maximum compression.
\[ mg(h + x) = \frac{1}{2} k x^2 \] Given: \( m = 1 \, \text{kg}, h = 1 \, \text{m}, k = 15 \, \text{Nm}^{-1}, g = 10 \)
\[ 10(1 + x) = \frac{15}{2} x^2 \Rightarrow 20(1 + x) = 15x^2 \Rightarrow 20 + 20x = 15x^2 \Rightarrow 15x^2 - 20x - 20 = 0 \] Solving the quadratic: \( x = 2 \, \text{m} \)
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