A 5.00-g bullet is fired horizontally into a 1.20-kg wooden block resting on a horizontal surface. The coefficient of kinetic friction betwe

Question

A 5.00-g bullet is fired horizontally into a 1.20-kg wooden block resting on a horizontal surface. The coefficient of kinetic friction between block and surface is 0.20. The bullet remains embedded in the block, which is observed to slide 0.230 m along the surface before stopping. What was the initial speed of the bullet?

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Dulcie 5 years 2021-08-08T19:33:44+00:00 1 Answers 177 views 1

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  1. Calantha
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    2021-08-08T19:35:11+00:00
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    Answer:

    The initial velocity of the bullet is  u_b = 228.7 m/s

    Explanation:

    From the question we are told that

              The mass of the bullet is  m_b = 5.00g =\frac{5}{1000} = 0.005kg

               The mass of the wooden block m_w = 1.20kg

              The coefficient of kinetic friction \mu_g = 0.20

              The  distance traveled by block and bullet d_{t } = 0.230m

    let  F_ b denote the force on the bullet in the block and this mathematically denoted as

                                F_b = \mu_g mg

    Where m is the mass of both bullet and block

      Substituting values

                                 = (0.20)* (0.005 + 1.20)(9.8)

                                 = 2.36N

    Generally workdone is mathematically represented as

                             W = F *d_t

    the  work in this cause in the kinetic energy used to move the block and bullet

                     i.e      W = \frac{1}{2}mv_f^2

    Where  v_f is the final speed of the block and bullet

               Therefore  \frac{1}{2}mv_f^2 = F * d_t

    making v_f the subject

                                  v_f = \sqrt{\frac{2Fd_t}{m} }

    Here m is the mass of both bullet and block

    Now substituting values

                                v_f = \sqrt{\frac{2(2.36)(0.230)}{(0.005 + 1.20)} }

                                     = 0.949m/s

    From the law of conservation of momentum this collision can be mathematically represented as

                   

                                 m_bu_b +m_wu_w = (m_w +m_b)v_f

    Now from the question  m_wu_w =0  this because the block was at rest

                              Therefore u_b = \frac{(m_w +m_u) v_f}{m_b}

        Substituting values

                                   u_b = \frac{(0.005 +1.2) * 0.949}{0.005}

                                        u_b = 228.7 m/s

       

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