A professional boxer hits his opponent with a 1025 N horizontal blow that lasts 0.150 s. The opponent’s total body mass is 116 kg and the bl

Question

A professional boxer hits his opponent with a 1025 N horizontal blow that lasts 0.150 s. The opponent’s total body mass is 116 kg and the blow strikes him near his center of mass and while he is motionless in midair. Determine the following.
(a) impulse the boxer imparts to his opponent by this blow
kg · m/s
(b) the opponent’s final velocity after the blow
m/s
(c) Calculate the recoil velocity of the opponent’s 5.0-kg head if hit in this manner, assuming the head does not initially transfer significant momentum to the boxer’s body.
m/s

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Vân Khánh 4 years 2021-08-24T16:03:13+00:00 1 Answers 8 views 0

Answers ( )

  1. thuthuy
    0
    2021-08-24T16:04:19+00:00
    Reply

    Answer:

    The impulse is  I = 153.8 \ N \cdot s

    The opponents velocity is  v= 1.33 m/s

    The opponents head recoils velocity  v_r = 30.8 \ m/s

    Explanation:

    From the question we are told that

        The force of the blow is  F= 1025 \ N

        The duration of the blow is  t = 0.150

          The mass of the opponent is  m_o = 116 \ kg

           The mass of the opponents head is  m_h = 5 \ kg

    The impulse the boxer imparts is mathematically represented as

             I = F * t

    substituting values

             I = 1025 * 0.150

              I = 153.8 \ N \cdot s

    The impulse can also be mathematically evaluated as

             I = m_o * v

    substituting values

              153.8 = 116 * v

              v= \frac{153.8}{116}

              v= 1.33 m/s

    The recoil velocity is mathematically represented as  

                  v_r = \frac{I}{m_h}

    substituting values

                    v_r = \frac{153.8}{5}

                    v_r = 30.8 \ m/s

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