1. On the planet Microid, trains that run on air tracks much like the air track you used in lab are the principle means of transportation. A

Question

1. On the planet Microid, trains that run on air tracks much like the air track you used in lab are the principle means of transportation. A train car with a mass of 6300 kg is traveling at 12.0 m/s when it strikes a second car moving in the same direction at 2.2 m/s. The two stick together and move off with a speed of 4.00 m/s. What is the mass of the second car? Explain your reasoning. (10 pts) 2. How much kinetic energy was lost in the collision? (10 pts)

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Sapo 2 months 2021-07-29T10:24:46+00:00 1 Answers 2 views 0

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    2021-07-29T10:26:34+00:00

    Answer:

    a) the mass of the second car = 28000 kg

    b) the amount of kinetic energy that was lost in the collision = 246.9*10^3 \ J

    Explanation:

    Given that:

    mass of the train car m_1 = 6300 kg

    speed of the train car v_1 = 12.0 m/s

    mass of the second moving car m_2 = ???

    speed of the second moving car v_2 = 2.2 m/s

    After strike;

    they both move with a speed v_f = 4.00 m/s

    a)

    Using the conservation of momentum :

    m_1v_1+m_2v_2 = (m_1 + m_2)v_f

    (6300*12)+m_2(2.2) = (6300 + m_2)4

    (75600)+m_2(2.2) = 25200 + 4m_2

    75600 - 25200  = 4m_2 -2.2m_2

    50400 = 1.8m_2

    m_2 = \frac{50400}{1.8}

    m_2 = 28000 \ kg

    b)

    To determine the amount of kinetic energy that was lost in the collision;

    we will need to find the difference between the kinetic energy before the collision and after the collision;

    i.e

    K.E _{lost} = K.E_{i} - K.E _{f}

    K.E_{i}  =  \frac{1}{2} m_1v_1^2 + \frac{1}{2}m_2v_2^2

    K.E_{i}  =  \frac{1}{2} (6300)(12)^2 + \frac{1}{2}(28000)(2.2)^2

    K.E_{i}  =  521360 \ J

    K.E_{f}  = \frac{1}{2}(m_1 + m_2 ) v_f ^2

    K.E_{f}  = \frac{1}{2}(6300 + 28000 ) (4)^2

    K.E_{f}  =274400 \ J

    Now;  the  kinetic energy that was lost in the collision is calculated as follows:

    K.E _{lost} = K.E_{i} - K.E _{f}

    K.E_{lost} = (521360-274400) \ J

    K.E_{lost} =246960 \ J

    K.E_{lost} =246.9 * 10 ^3 \ J

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