A thin block of soft wood with a mass of 0.072 kg rests on a horizontal friction

Question

A thin block of soft wood with a mass of 0.072 kg rests on a horizontal frictionless surface. A bullet with a mass of 4.67 g is fired with a speed of 605 m/s at a block of wood and passes completely through it. The speed of the block is 15 m/s immediately after the bullet exits the block.

(a) Determine the speed of the bullet as it exits the block.
m/s

(b) Determine if the final kinetic energy of this system (block and bullet) is equal to, less than, or greater than the initial kinetic energy.

equal to the initial kinetic energyless than the initial kinetic energy    greater than the initial kinetic energy

(c) Verify your answer to part (b) by calculating the initial and final kinetic energies of the system.

KEi =

  J

KEf =

  J

Explanation / Answer

a)

by using law of conservation momentunm

m1u1+m2u2=m1v1+m2v2

0.072*0+4.67*10^-3*605=0.072*15+4.67*10^-3*v2

v2=(4.67*10^-3*605)-(0.072*15)/(4.67*10^-3)

v2=373.736 m/sec ...is answer

b)

Ki=1/2*m1u1^2+1/2*m2u2^2

=0+1/2*4.67*10^-3*605^2

=854.668 J

kf=1/2*m1v1^2+1/2*m2v2^2

=1/2*0.072*15^2+1/2*4.67*10^-3*373.736^2

=334.249 J

c)
initial kinetic energy is greater than the final kinetic energy

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.