IP A bullet with a mass of 4.2 g and a speed of 650m/s is fired at a block of wo

Question

IP A bullet with a mass of 4.2 g and a speed of 650m/s is fired at a block of wood with a mass of 9.7×102 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 .

A. What is the speed of the bullet when it exits the block?

B. Is the final kinetic energy of this system equal to, less than, or greater than the initial kinetic energy?

C. (Please explain why) ^^^

Verify your answer to part B by calculating the initial and final kinetic energies of the system.

D. Initial

E. Final

Explanation / Answer

m1 =4.2 g, m2 = 9.7x10^-2 kg,u1 = 650 m/s, u2 =0, v2 = 21 m/s

Conservation of momentum

m1u1+m2u2 =m1v1+m2v2

(0.0042*650)+0= (0.0042*v1)+(0.097*21)

2.73 = (0.0042 v1)+2.307

v1 = 0.423/0.0042

v1 =100.7 m/s

B) Final kinetic energy of the system is less than initital kinetic energy of the system.

(c) It is inelastic collision. Some kinetic energy is used to work done on the block.

Ki = (1/2)m1u1^2 +(1/2)m2u2^2

= (1/2)(0.0042)(650*650) = 887.25 J

Kf =(1/2)m1v1^2 +(1/2)m2v2^2
= [(1/2)(0.0042)(100.7*100.7) ]+[ (1/2)(0.097)(21*21)]

Kf = 21.295 +21.39 =42.685 J

Ki -Kf = 887.25 -42.685 =844.565 J

Kf<Ki

(D) Ki = 887.25 J

(E) Kf =42.685 J

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