A 5.00-g bullet moving with an initial speed of vi = 380 m/s is fired into and p

Question

A 5.00-g bullet moving with an initial speed of vi = 380 m/s is fired into and passes through a 1.00-kg block as shown in the figure below. The block, initially at rest on a frictionless, horizontal surface, is connected to a spring with force constant 880 N/m. The block moves d = 5.60 cm to the right after impact before being brought to rest by the spring.

(a) Find the speed at which the bullet emerges from the block. m/s

(b) Find the amount of initial kinetic energy of the bullet that is converted into internal energy in bullet–block system during the collision. J

Explanation / Answer

The bullet imparts a velocity V to the block on impact. That kinetic energy is taken up by the spring, hence we write

(1/2) ( 1 kg ) V^2 = (1/2) ( 880 N/m ) ( 0.056 m )^2

V = 1.66 m/sec

Then, since momentum is conserved, if b is the exit speed of the bullet we must have

( 0.005 kg ) ( 380 m/sec ) = ( 1 kg )( 1.66 m/sec ) + ( 0.005 kg )b

Hence b = 48 m/sec

speed at which the bullet emerges from the block is 48m/s

The various energies are:

Initial bullet: (1/2) ( 0.005 kg ) ( 380 m/sec )^2 = 361 J
Final bullet: (1/2) ( 0.005 kg ) ( 48 m/sec )^2 = 5.76 J
Block/Spring system: (1/2) ( 1 kg ) ( 1.66 m/sec )^2 = 1.377 J

The energy lost is therefore:

361 - ( 5.76 + 1.377 ) = 353.8 J

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