A 0.0455-kg bullet is fired horizontally into a 4.36-kg wooden block attached to

Question

A 0.0455-kg bullet is fired horizontally into a 4.36-kg wooden block attached to one end of a massless, horizontal spring (k = 844 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 0.275 m. What is the speed of the bullet?

Explanation / Answer

momentum conservation

0.0455*v1 + 0 = (0.0455 + 4.36)v2

0.0455*v1 = 4.4055*v2

v1 = 96.824*v2 ............. (1)

now

0.5*kx2 = 0.5*m*v22

=> 844*0.275*0.275 = 4.4055*v22

=> 63.8275 = 4.4055*(v1/96.824)2

=> 63.8275*9374.886976/4.4055 = v12

=> v1 = 368.544 m/s the speed of the bullet

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.