Blocks A (mass 2.00 kg ) and B (mass 14.00 kg , to the right of A) move on a frictionless, horizontal surface. Initially, block B is moving

Question

Blocks A (mass 2.00 kg ) and B (mass 14.00 kg , to the right of A) move on a frictionless, horizontal surface. Initially, block B is moving to the left at 0.500 m/s and block A is moving to the right at 2.00 m/s. The blocks are equipped with ideal spring bumpers. The collision is headon, so all motion before and after it is along a straight line. Let +x be the direction of the initial motion of A. Part A Find the maximum energy stored in the spring bumpers.

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Eirian 3 years 2021-08-05T23:45:24+00:00 1 Answers 75 views 0

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  1. mit
    0
    2021-08-05T23:46:36+00:00
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    Answer:

    7.72 Joules

    Explanation:

    Data:

    mass of block A = 2.00 kg

    mass of block B = 14.00 kg

    velocity of block A = 2.00 m/s

    velocity of block B = 0.500 m/s

    The fundamental assumption is that there is elastic collision, therefore, the momentum and energy of the system are conserved.

    Thus:

    U_{spring max} = KE_{total} - KE_{cm}\\ KE_{total} = \frac{1}{2}(mv_{1} ^{2} + mv_{2} ^{2} )\\ = 0.5 (3.00(2.00)^{2}+ 14.00 (0.500)^{2} \\ = 7.75 J

    V_{cm} = \frac{3(2)-14(0.5)}{3+14} \\ = -0.0588m/s

    The kinetic energy is given by the following:

    KE_{cm} = \frac{1}{2}(m_{A} + m_{B})(v_{cm})^{2} \\ = \frac{1}{2}(3.00 + 14.00) (-0.0588)^{2} \\ = 0.03 J\\

    Therefore, the maximum energy is given by the following:

    U_{max} = 7.75 - 0.03\\ = 7.72 J

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Giải phương trình 1 ẩn: x + 2 - 2(x + 1) = -x . Hỏi x = ? ( )