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24 February, 05:03

A dart is inserted into a spring-loaded dart gun by pushing the spring in by a distance. For the next loading, the spring is compressed a distance. How much faster does the second dart leave the gun compared with the first

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  1. 24 February, 05:35
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    Complete question is;

    A dart is inserted into a spring - loaded dart gun by pushing the spring in by a distance x. For the next loading, the spring is compressed a distance 2x. How much faster does the second dart leave the gun compared to the first?

    Answer:

    The second dart leaves the gun two times faster than the first one.

    Explanation:

    If we assume there was no energy loss during the spring - dart energy transfer, we can easily apply the principle of conservation of energy. So;

    Potential energy = kinetic energy

    Thus;

    ½kx² = ½mv²

    Making velocity "v" the subject, we have;

    v = √ (kx²/m)

    Since the initial distance is "x", thus initial launching velocity is;

    v1 = √ (kx²/m)

    Since next distance is 2x, thus, second launch velocity is;

    v2 = √ (k (2x) ²/m)

    Expanding, we have;

    v2 = √ (4kx²/m)

    v2 = 2√ (kx²/m)

    Comparing this to the one gotten for v1 earlier, we can see that it is double v1.

    So, v2 = 2v1

    Hence, The second dart leaves the gun two times faster than the first one.
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