A 1.00 times 10^-2 -kg bullet is fired horizontally into a 2.50-kg wooden block

Question

A 1.00 times 10^-2 -kg bullet is fired horizontally into a 2.50-kg wooden block attached to one end of a massless horizontal spring (k = 845 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 0.200 m. What is the speed of the bullet?

Explanation / Answer

According to the given problem,

Let the bullets velocity be v

Using the conservation linear momentum fine the speed of block and bullet toghter,

pi = pf

1*10-2*v = 2.51 v`

v` = 3.984*10-3v----[1]

Now, using the Conservation of Energy principal,

K.E = S.P.E

1/2Mv`2 = 1/2kx2 where x = aplitude = 0.2m

2.51*( 3.984*10-3v)2 = 845*0.22

Calcualte for Velocity,

v = 921.075 m/s

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