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Physics collisions question?
A thin block of soft wood with a mass of 0.080 kg rests on a horizontal frictionless surface. A bullet with a mass of 4.67 g is fired with a speed of 613 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
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
Correct: Your answer is correct.
(c)
Verify your answer to part (b) by calculating the initial and final kinetic energies of the system in joules.
KEi
= J
KEf
= J
1 Answer
- NCSLv 72 years agoFavorite Answer
(a) conserve momentum:
4.67g * 613m/s + 0 = 4.67g * v + 80g * 23m/s
solves to
v = 219 m/s
(b) Of course the final KE will be less than the bullet's initial KE -- there has been deformation.
initial KE = ½ * 0.00467kg * (613m/s)² = 877 J
final KE = ½ * (0.00467*219² + 0.080*23²) J = 133 J
Hope this helps!
