A 0.0411-kg bullet is fired horizontally into a 2.19-kg wooden block attached to

Question

A 0.0411-kg bullet is fired horizontally into a 2.19-kg wooden block attached to one end of a massless, horizontal spring (k = 875 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 0.191 m. What is the speed of the bullet?

Explanation / Answer

m1=0.0411kg , m2=2.19kg , v2=0m/s ,  k=875 N/m , A = 0.191m

Assume initial speed of bullet = v1

When bullet hit the block and embeds in it , we will use law of conservation of momentum to find their common speed vc

Thus

m1v1+m2v2 = (m1+m2)*vc

m1v1+m2*0 = (m1+m2)*vc

m1v1 = (m1+m2)*vc----------------(1)

Using law of conservation of energy

ME of bullet + block = ME of spring

1/2(m1+m1)*vc ^2 = 1/2kA^2

1/2*(0.0411+2.19)*vc^2 = 1/2*875*0.191^2

vc  = 3.78 m/s

Plugging this value in (1)

m1v1 = (m1+m2)*vc

0.0411*v1 = (0.0411+2.19)*3.78     =>  v1 = 205.2 m/s

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

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