Question

 

 A thin block of soft wood with a mass of 0.070 kg rests on a horizontal frictionless surface. A bullet with a mass of 4.67 g is fired with a speed of 629 m/s at a block of wood and passes completely through it. The speed of the block is 25 m/s immediately after the bullet exits the block.
(a) Determine the speed of the bullet as it exits the block. 
(b) Determine the loss of kinetic energy of this system (block and bullet).

Answer 1

a) Let the speed of the bullet as it exits the block is v1.

Given the initial speed of the bullet is u1=629 m/s.

The initial speed of the block is u2=0 m/s.

The final speed of the block is v2=25 m/s.

The mass of the bullet m=4.67 g=0.00467 kg.

The mass of the block M=0.070 kg

From the principle of conservation of momentum

  

     

  

  

The speeds of the bullet as it exits the block is 254.267666 m/s or 254.27 m/s approx.

b) Let the loss of kinetic energy of the system is .

We know from the conservation of energy

The loss of kinetic energy of this system is 750.98 J.