A thin block of soft wood with a mass of 0.070 kg rests on a horizontal friction

Question

A thin block of soft wood with a mass of 0.070 kg rests on a horizontal frictionless surface. A bullet with a mass of 4.67 g is fired with a speed of 609 m/s at a block of wood and passes completely through it. The speed of the block is 23 m/s immediately after the bullet exits the block. (a) Determine the speed (in m/s) 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 energy less 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 in joules.

Explanation / Answer

From the law of conservation of momentum

mv(i) +mv(i) = mv(f) + mv(f)

As the block is initially at rest, mv(i) = 0

v(f) = [mv(i) - mv(f)] / m

Substitute numerical values,

v(f) = [( 0.00467 kg ) ( 609 m/s) - (0.070 kg) (23 m/s) ] / 0.00467 kg

= 264.2 m/s

If round off the result to 3 significant digits

v(f) = 264 m/s

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(b) As the final speed of bullet decreases,

the final kientic energy of the less than initial kinetic energy

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(c)

Intial kinetic energy of the system,

KEi = 0.5mv² = 0.5 ( 0.00467 kg ) (609 m/s)^2 = 866 J

Final kinetic energy of system:

KEf = 0.5 m v'2 + 0.5 M V'2

= 0.5 ( 0.00467 kg ) ( 264.2 m/s)^2 +0.5 (0.070 kg ) (23 m/s)^2

= 181.55 J

If round off the result to 3 significant digits

KEf = 182 J

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