Ph203 Quiz 8           October 30, 2000

1. A bullet with a mass of $5\times 10^{-3}$ kg is fired horizontally into a 2.0-kg wooden block at rest on a horizontal surface. The coefficient of kinetic friction between the block and the surface is 0.2. The bullet comes to rest in the block which moves 2.0 m. Find the speed of the bullet. [5 pts]

   Solution: This is a two part problem. First you need to conserve linear momentum during the collision of the bullet and block, then you need to conserve energy as the block slides across the surface. First conservation of momentum:

   
   

   $$m_b v_o = (m_b + m_B) V$$
   
   

   where $v_0$ is the initial speed of the bullet and $V$ is the speed that the bullet plus block have right after they collide. To determine $V$ equate the kinetic energy of the bullet plus block right after collision with the work done by friction ($f_k$):

   
   

   $$\frac{1}{2} (m_b + m_B) V^2 = f_K s = \mu (m_b + m_B) g s$$
   
   
   
   
   $$V = \sqrt{2\mu g s}$$
   
   

   Plugging this in the first equation and solving for $v_0$ we get:

   
   

   $$v_0 = {m_b + m_B \over m_b}\, \sqrt{2\mu g s} = {2.005 \over ... ...imes 0.2 \times 9.8 \times 2.0} {\,\rm m/s} = 1120 {\,\rm m/s}$$
   
   

2. An automobile with a mass of 2000-kg has a wheel base of 3.0 m (the wheel base is the distance between the front and rear axles). The center of gravity of the car is located 1.75 m behind the front axle. Determine the force exerted on each of the front wheels (assumed the same) and the force exerted on each of the back wheels (assumed the same) by the level ground. [5 pts]

   Solution: You can solve this by taking torques about the front axle and then about the rear axle and, in each case, setting the sum of the torques equal to zero. About the front axle we have:

   
   

   $$\Sigma \tau_F = 0 = -l_{cg} mg + l_{wb} F_R$$
   
   
   then
   
   $$F_R = mg {l_{cg} \over l_{wb}} = 2000\times 9.8 {1.75 \over 3.0}{\,\rm N} = 11,433 {\,\rm N}$$
   
   

   About the rear axle we have

   
   

   $$\Sigma \tau_R = 0 = (l_{wb}-l_{cg}) mg - l_{wb} F_F$$
   
   
   then
   
   $$F_F = mg {l_{wb}-l_{cg} \over l_{wb}} = 2000\times 9.8 {3.0-1.75 \over 3.0}{\,\rm N} = 8,167 {\,\rm N}$$
   
   

   The force exerted by each of the front wheels is $F_F/2 = 4,083 {\,\rm N}$ and by each of the back wheels is $F_R/2 = 5,717 {\,\rm N}$.