A 5.00 g bullet moving with an initial speed of 475 m/s is fired into and passes

Question

A 5.00 g bullet moving with an initial speed of 475 m/s is fired into and passes through a 1.00 kg block. The block, initially at rest on a frictionless, horizontal surface, is connected to a spring of force constant 890 N/m.

(a) If the block moves 4.70 cm to the right after impact, find the speed at which the bullet emerges from the block.



Your response differs from the correct answer by more than 100%. m/s

(b) If the block moves 4.70 cm to the right after impact, find the mechanical energy converted into internal energy in the collision.
J

Explanation / Answer

                       Kblock = Uspring

                       1/2mblockv2'2= 1/2kx2

             From conservation of momentum, we have

                      mbulletv1 + 0 = mbulletv1' + mblockv2'

                      (5.00 x 10-3kg)(475m/s) = (5.00 x 10-3kg)v1' + (1.00kg)(1.4m/s), which gives

                         v1' = 195m/s.

             The initial energy of the system is the kinetic energy of the bullet:

                       Ei = Ki = 1/2mv12 = 1/2(5.00 x 10-3kg)(475m/s)2 = 564J

            The final energy of the system is the spring potential energy plus the kinetic energy of the bullet:

                       Ef = Uspring + 1/2mbulletv1'2 = 1/2kx2 + 1/2mbulletv1'2

                           = 1/2(890N/m)(0.047m)2 + 1/2(5.00 x 10-3kg)(195m/s)2

                           = 0.983J + 95J = 95.98J

            The energy lost is

                     DE = Ei - Ef = 468.02 J

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