A 5.00-g bullet moving with an initial speed of v 0 = 395 m/s is fired into and

Question

A 5.00-g bullet moving with an initial speed of v0 = 395 m/s is fired into and passes through a 1.00-kg block, as in the figure below. The block, initially at rest on a frictionless horizontal surface, is connected to a spring with a spring constant of 875 N/m.

A 5.00-g bullet moving with an initial speed of v0 = 395 m/s is fired into and passes through a 1.00-kg block, as in the figure below. The block, initially at rest on a frictionless horizontal surface, is connected to a spring with a spring constant of 875 N/m.

Explanation / Answer

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

(1/2) ( 1 kg ) V^2 = (1/2) ( 875 N/m ) ( 0.05 m )^2

V = 1.09 m/sec

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

( 0.005 kg ) ( 395 m/sec ) = ( 1 kg )( 1.09 m/sec ) + ( 0.005 kg )S

Hence S = 177 m/sec exit velocity.

The various energies are:

Initial bullet: (1/2) ( 0.005 kg ) ( 395 m/sec )^2 = 390.1 J
Final bullet: (1/2) ( 0.005 kg ) ( 177 m/sec )^2 = 78.32 J
Block/Spring system: (1/2) ( 1 kg ) ( 1.09 m/sec )^2 = 0.59 J

The energy lost is therefore:

390.1 - ( 78.32 + 0.59 ) = 311.19 J = 311.2 J

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