In the “before” part of Fig. 9-60, car A (mass 1100 kg) is stopped at a traffic light when it is rear-ended by car B (mass 1400 kg). Both ca

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In the “before” part of Fig. 9-60, car A (mass 1100 kg) is stopped at a traffic light when it is rear-ended by car B (mass 1400 kg). Both cars then slide with locked wheels until the frictional force from the slick road (with a low

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Thanh Hà 5 months 2021-08-19T18:58:15+00:00 2 Answers 74 views 0

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    0
    2021-08-19T18:59:52+00:00

    Complete Question:

    In the “before” part of Fig. 9-60, car A (mass 1100 kg) is stopped at a traffic light when it is rear-ended by car B (mass 1400 kg). Both cars then slide with locked wheels until the frictional force from the slick road (with a low ?k of 0.15) stops them, at distances dA = 6.1 m and dB = 4.4 m. What are the speeds of (a) car A and (b) car B at the start of the sliding, just after the collision? (c) Assuming that linear momentum is conserved during the collision, find the speed of car B just before the collision.

    Answer:

    a) Speed of car A at the start of sliding = 4.23 m/s

    b) speed of car B at the start of sliding = 3.957 m/s

    c) Speed of car B before the collision = 7.28 m/s

    Explanation:

    NB: The figure is not provided but all the parameters needed to solve the question have been given.

    Let the frictional force acting on car A, f_{ra} = \mu mg\\…………(1)

    Since frictional force is a type of force, we are safe to say f_{ra} = ma…….(2)

    Equating (1) and (2)

    ma = \mu mg\\a = \mu g\\\mu = 0.15\\a = 0.15 * 9.8 = 1.47 m/s^{2}

    a) Speed of A at the start of the sliding

    d_{A} = 6.1 m\\Speed of A at the start of sliding, v_{A} = \sqrt{2ad_{A} }\\ v_{A} = \sqrt{2*1.47*6.1 } \\v_{A} = \sqrt{17.934 } \\v_{A} = 4.23 m/s

    b) Speed of B at the start of the sliding

    d_{A} = 4.4 m\\Speed of A at the start of sliding, v_{B} = \sqrt{2ad_{B} }\\ v_{B} = \sqrt{2*1.47*4.4 } \\v_{B} = \sqrt{12.936 } \\v_{B} = 3.957 m/s

    Let the speed of car B before collision = v_{B1}

    Momentum of car B before collision = m_{B} v_{B1}

    Momentum after collision = m_{A} v_{A} + m_{B} v_{B2}

    Applying the law of conservation of momentum:

    m_{B} v_{B1}  = m_{A} v_{A} +m_{B} v_{B2}

    m_{A} = 1100 kg\\m_{B} = 1400 kg

    (1400*v_{B1} ) = (1100 * 4.23) + ( 1400 * 3.957)\\(1400*v_{B1} ) = 10192.8\\v_{B1} = 10192.8/1400\\v_{B1 = 7.28 m/s

    0
    2021-08-19T19:00:08+00:00

    Answer:

    v= 6.9245 m/sec

    Explanation:

    complete question

    In the “before” part of Fig. 9-60, car A (mass 1100 kg) is stopped at a traffic light when it is rear-ended by car B (mass 1400 kg). Both cars then slide with locked wheels until the frictional force from the slick road (with a low ?k of 0.15) stops them, at distances dA = 6.1 m and dB = 4.4 m. What are the speeds of (a) car A and (b) car B at the start of the sliding, just after the collision? (c) Assuming that linear momentum is conserved during the collision, find the speed of car B just before the collision.

    solution:

    is attached below

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