Question

g 0 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 0.390 m along the surface before stopping. Part A What was the initial speed of the bullet? Express your answer with the appropriate units. v = nothing

Answer 1

Answer:

vb = 298 m/s

Explanation:

Given:-

- The mass of the bullet, mb = 5.00 g

- The mass of the wooden block, mw = 1.2 kg

- The coefficient of kinetic friction between block and horizontal surface, $u_k = 0.2$

- The bullet and block combined moves a distance after impact, s = 0.390

Solution:-

- We will first consider the motion of the block and bullet combined after the bullet is embedded into the block.

- We are told that the wooden block is rested on a horizontal floor with coefficient of kinetic friction ( uk ).

- When the bullet is embedded into the block. Both combined will move with a velocity ( V ). Both will eventually loose all of their kinetic energy by doing work against the friction.

- We will apply the work - energy principle to the system ( block + bullet ) as follows:

                                $W = dE_k$

Where,

                W: Work done against friction

                dEk: The change in kinetic energy of the system

- The work-done against friction is the product of frictional force ( Ff ) and the displacement over which the system travels ( s ).

- The frictional force  ( Ff ) is proportional to the contact force ( N ) between the system and the surface.

- Apply static balance on the system in the direction normal to the surface as follows:

                              $N - ( m_b + m_w )*g = 0\\\\N = ( m_b + m_w )*g$

- The frictional force is defined as:

                              $F_f = u_k*N\\\\F_f = u_k* ( m_b + m_w ) * g$

- The work done against friction ( W ) is defined as:

                              $W = F_f*s\\\\W = u_k*(m_b + m_w )*g*s$

Where,

                         g: the gravitational acceleration constant = 9.81 m/s^2

- Now use the energy balance to determine the velocity of the system after impact ( V ):

                          $u_k*( m_b + m_w )*g*s = 0.5*( m_b + m_w )*V^2\\\\V = \sqrt{2*u_k*g*s } \\\\V = \sqrt{2*0.2*9.81*0.39 }\\\\V = 1.23707 m/s$

- Now we will again consider an independent system of bullet and the wooden block resting on the horizontal surface.

- The bullet is fired with velocity ( vb ) towards the wooden block. The system can be considered to be isolated and all other fictitious effects can be ignored. This validates the use of conservation of linear momentum for this system.

- The conservation of linear momentum denotes:

                          $P_i = P_f$  

Where,

                 Pi: the inital momentum of the system

                 Pf: the final momentum of the system

- Initially the block was at rest and bullet had a velocity ( vb ) and after striking the bullet is embedded into the block and moves with a velocity ( V ).

                      $m_b*v_b = ( m_b + m_w )*V\\\\v_b = \frac{ ( m_b + m_w )}{m_b}*V\\\\v_b = \frac{ ( 0.005 + 1.2 )}{0.005}*(1.23707)\\\\v_b = 298 m/s$

Answer: The initial speed of the bullet is equivalent to the speed of the bullet just before the impact as vb = 298 m/s. This is under the assumption that forces like ( air resistance or gravitational or impulse) have negligible effect.