A 0.0340-kg bullet is fired horizontally into a 4.16-kg wooden block attached to

Question

A 0.0340-kg bullet is fired horizontally into a 4.16-kg wooden block attached to one end of a massless, horizontal spring (k = 877 N/m). The other end of the spring is fixed in place, and the spring is unstrained initially. The block rests on a horizontal, frictionless surface. The bullet strikes the block perpendicularly and quickly comes to a halt within it. As a result of this completely inelastic collision, the spring is compressed along its axis and causes the block/bullet to oscillate with an amplitude of 0.292 m. What is the speed of the bullet?

Explanation / Answer

Amplitude = 0.292m

x = 0.292/2 = 0.146 m

kx^2/2 = (m+M)v^2/2 - mv'^2/2 [Since all of bullet's kinetic energy is converted into spring potential energy]

877 * 0.146^2 = 4.194*v^2 + 0.034v'^2

Now, according to conservation of momentum

mv' = (m+M)v

v = 0.034v'/4.194

v = 0.008v'

Substituting this into the above energy equation and solving for v', we get

v' = 23.35 m/s

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