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27 June, 17:13

The same ball is shot straight up a second time from the same gun, but this time the spring is compressed only half as far before firing. How far up does the ball go this time? Neglect friction. Assume that the spring is ideal and that the distance by which the spring is compressed is negligible compared to H.

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  1. 27 June, 20:49
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    The new height the ball will reach = (1/4) of the initial height it reached.

    Explanation:

    The energy stored in any spring material is given as (1/2) kx²

    This energy is converted to potential energy, mgH, of the ball at its maximum height.

    If the initial height reached is H

    And the initial compression of the spring = x

    So, mgH = (1/2) kx²

    H = kx²/2mg

    The new compression, x₁ = x/2

    New energy of loaded spring = (1/2) kx₁²

    And the new potential energy = mgH₁

    mgH₁ = (1/2) kx₁²

    But x₁ = x/2

    mgH₁ = (1/2) k (x/2) ² = kx²/8

    H₁ = kx²/8mg = H/4 (provided all the other parameters stay constant)
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