An object at rest on a flat, horizontal surface explodes into two fragments, one seven times as massive as the other. The heavier fragment s

Question

An object at rest on a flat, horizontal surface explodes into two fragments, one seven times as massive as the other. The heavier fragment slides 7.00 m before stopping. How far does the lighter fragment slide?

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Ngọc Khuê 7 hours 2021-07-22T01:46:27+00:00 1 Answers 0 views 0

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    2021-07-22T01:48:14+00:00

    Answer:

    The distance the lighter fragments slides is 343 m

    Explanation:

    Here, we have

    Let the mass of the heavier fragments be m₁

    Let the distance the heavy object slide be d₁

    Let the mass of the lighter fragments be m₂

    Let the final velocity of the heavier fragments be v₁

    Let the final velocity of the lighter fragments be v₂

    Let the distance the light object slide be d₂

    The distance traveled by  the heavy fragment = 7.00 m

    Therefore since m₁ = 7 × m₂ we have

    Initial total momentum = final total momentum

    Since the initial total momentum = 0 we have

    m₁·v₁ + m₂·v₂ = 0  or

    7·m₂·v₁  = -m₂·v₂

    ∴ v₂ = 7·v₁

    The net work done by the heavier block is

    W_{net, 7m} = μk × m₁ × g × d₁ = 1/2×m₂×v₂²

    Also v₁² = 2μk·g·d₁ and

    v₂² = 2μk·g·d₂ so that since v₂ =  7·v₁ or v₁ = v₂/7 we get

    (v₂/7)² = 2μk·g·d₁

    So v₂²/49 = 2μk·g·d₁

    or (2μk·g·d₂)/49 = 2μk·g·d₁

    ∴ d₂/49 = d₁

    d₂ = d₁×49 = 7 × 49 = 343 m

    The distance the lighter fragments slides = 343 m.

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