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?

Answer 1

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

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

                                    $u_b = 228.7 m/s$

   

Answer 2

The initial speed of the bullet as it was fired horizontally into the wooden block resting on a horizontal surface is 228.7 m/s.

Given

Mass of bullet; 5.00-g

Mass of wood; - 1.20-kg

Coefficient of kinetic friction; -  0.20

Displacement of bullet; - 0.230 m

Initial speed of the bullet; - ?

Solution:

To find the velocity of the Wood,

   

Where v = velocity of the block, μ = coefficient of kinetic friction, s = displacement of the bullet, and g = acceleration due to gravity(9.8).

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

Substituting values  

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

                                 $u_b$= 228.7 m/s        

Thus, The initial speed of the bullet as it was fired horizontally into the wooden block resting on a horizontal surface is 228.7 m/s.

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