Question

IP A bullet with a mass of 4.2 g and a speed of 640 m/s is fired at a block of wood with a mass of 9.8×10−2 kg . The block rests on a frictionless surface, and is thin enough that the bullet passes completely through it. Immediately after the bullet exits the block, the speed of the block is 21 m/s . What is the speed of the bullet when it exits the block? Is the final kinetic energy of this system equal to, less than, or greater than the initial kinetic energy? Verify your answer to part B by calculating the initial and final kinetic energies of the system. Express your answer using two significant figures.

Answer 1

By law of conservation of momentum we know tha sum of initial momentum of block and bullet will he equal to the sum of final momentum of both.let final velocity if buller be v.

We know that inital masses if bullet and block is 0.0042 kg and 0.098 jg respectively.initially bullet is at rest. And initial speed of bullet us 640 m/s then. (0.0042×640)+(0.098×0)=(0.0042×v)+(0.098×21)

Solving above equation v= 150 m/s.

2) kinetic energy =1/2 (mass×velocity^2)

Initial kinetic energy of system= 0.5×0.0042×(640)^2 + 0.5×0.098×0 = 860.16

Final kinetic energy = 0.5×0.0042×(150)^2 +0.5×0.098×(21)^2 =47.25+21.609 = 68.859.

Cleary final kinetic energy is less than inital kinetic energy. (energy is lost in making hole through the block and dissipated in the form of heat and sound energy)