A 0.0466-kg bullet is fired horizontally into a 3.87-kg wooden block attached to

Question

A 0.0466-kg bullet is fired horizontally into a 3.87-kg wooden block attached to one end of a massless, horizontal spring (k= 860 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 0.286 m. What is the speed of the bullet?

Explanation / Answer

To calculate the speed:the kinetic energy at the instant the bullet is pierced is 1/2(m+M)v2

the energy stored in the spring is =1/2kx2

kinetic energy and energy stored in spring are equal

1/2(m+M)V2=1/2KX2

THEREFORE v2=kx2/(m+M)

u=velocity of the bullet

mu=(m+M)v

mu=(m+M ) square root(kx2/m+m)

u=x/m square root k(m+M)

u=860/0.0466 square root 860(0.0466+3.87)=18454.93562 square root 860*3.9166=1,071,066.337m/second

speed of the bullet is 1071066.337m/sec

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