A 1.03x10^-2 kg bullet is fired horizontally into a 2.38 kg wooden block attache

Question

A 1.03x10^-2 kg bullet is fired horizontally into a 2.38 kg wooden block attached to one end of a massless, horizontal spring (k = 830 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 with in 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

From the conservation of energy of the spring, we have

.5kx2 = .5mv2

Therefore we can find the initial speed of the bullet/block combo

(830)(.2)2 = (2.38 + 1.03 X 10-2)v2

v = 3.737 m/s

Then, by conservation of momentum,

mv(bullet) = (mv) combo

(1.03 X 10-2)(v) = (2.38 + 1.03 X 10-2)(3.737)

v = 865 m/s

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