A 105 kg astronaut carrying a 16 kg tool bag finds himself separated from his spaceship by 18 m and moving away from the spaceship at 0.1 m/

Question

A 105 kg astronaut carrying a 16 kg tool bag finds himself separated from his spaceship by 18 m and moving away from the spaceship at 0.1 m/s. To get back to the spaceship, he throws the tool bag away from the spaceship at 4.5 m/s (relative to the station). How long (in s) will he take to return to the spaceship

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Bơ 3 years 2021-07-20T10:36:10+00:00 1 Answers 126 views 0

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  1. dangkhoa
    0
    2021-07-20T10:37:50+00:00
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    Answer:

    T=22.5sec

    Explanation:

    From the question we are told that:

    Mass of astronaut m_a=105kg

    Mass of tool m_t=16kg

    Distance d=18m

    Velocity of separation v_s= 0.1m/s

    Velocity of tool bag v_t=4.5m/s

    Generally the equation for momentum is mathematically given by

     P=mv

    Therefore

    Initial Momentum before drop

     P_1=0.1(105+16)

     P_1=12.1

    Initial Momentum after drop

     P_2=-16(4.5)+105V

    Therefore

    Since P_1=P_2

     -72+105V=12.1

     V=0.8m/s

    Generally the equation for Time T is mathematically given by

     T=\frac{d}{V}

     T=\frac{18}{0.8}

     T=22.5sec

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