A 0.0125 kg bullet strikes a 0.240 kg block attached to a fixed horizontal spring whose spring constant is 2.25*10^3N/m and sets it into osc

Question

A 0.0125 kg bullet strikes a 0.240 kg block attached to a fixed horizontal spring whose spring constant is 2.25*10^3N/m and sets it into oscillation with amplitude of 12.4 cm. What was the initial speed of the bullet if the two objects move together after impact?

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Khang Minh 6 months 2021-07-29T16:04:57+00:00 1 Answers 6 views 0

Answers ( )

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    2021-07-29T16:06:16+00:00

    Answer:

    The  value is  u_1 =  236 \ m/s

    Explanation:

    From the question we are told that

       The  mass of bullet  is  m_b  =  0.0125 \  kg

       The  mass of the block is  M_B  =  0.240 \  kg

        The  spring constant is  k  =  2.25*10^{3} \  N/m

       The  amplitude is  A= 12.4 \ cm  =  0.124 \ m

    Generally according to the conservation of momentum is  

         m_b u_1 + M_B  u_2 =  (m_b + M_B) v

    given that the block was at rest we have that

            m_b u_1  =  (m_b + M_B) v

    Now the angular velocity of the both bodies is mathematically represented as

         w =  \sqrt{\frac{k}{M_B  + m_b} }

              w = \sqrt{\frac{ 2.25*10^{3}}{ 0.0125 + 0.240 } }

             w =  94.4 \  rad/s

    Given that the system after collision set into oscillation

    The maximum  linear velocity of the system after impact  is mathematically represented as

           v  = A * w

          v  =  94.4 *0.124

          v  =  11.7 \  m/s

    From above equation

        u_1  =  \frac{(m_b +  M_B ) * v}{m_b}

    =>    u_1 =  \frac{(0.240 + 0.0125) *  11.7}{ 0.0125}

    =>   u_1 =  236 \ m/s

       

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