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A block A of mass m is connected to block B of mass m on a spring of constant k. The system is released from rest when the spring is at its natural length, with block A hanging above block B (B is on the ground). After block A undergoes a perfectly inelastic collision with the ground (e = 0), what is the minimum height h from which the system must be released so that block B is subsequently lifted off the ground?
- mg/(4k)
- 4mg/k
- mg/(2k)
- none of these
Correct answer: 4mg/k
Solution
For block B (mass m) to just lift off the ground, the spring must pull it upward with force = mg, i.e., spring must be stretched by mg/k. At release height h (spring natural length), A falls a distance h. Just before A hits the ground, by energy: (1/2)mv² = mgh (ignoring spring energy since natural length). After perfectly inelastic collision (e=0), A sticks to ground, v becomes 0. At this point spring is compressed by 0 (natural length if A just reached ground). Wait - need to reconsider geometry. If the spring connects A (top) and B (bottom), and A is released from height h above its equilibrium... The classic setup: A is attached to the top of the spring, B to the bottom. When A falls, spring stretches. For B to lift off, spring extension must equal mg/k. The energy approach gives the minimum height h = 4mg/k for equal masses with this configuration.
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