a .01 kg bullet is fired horizontally into a 2.5kg wooden block attached to one

Question

a .01 kg bullet is fired horizontally into a 2.5kg wooden block attached to one end or a massless, horizontalspring(k=845 N/m) the other end of the spring is fixed in place andthe spring is unstrained initially. The block rests on ahorizontal frictionless surface. the bullet strikes the blockperpendicularly and quickly comes to a halt wihtin it. as a aresultof this completely inelastic collision the spring is compressedalong its axis and causes the block/bullet to oscillate with anamplitude of 0.2 m.. What is the speed of the bullet?

Explanation / Answer

Here: [Re-check the algebra] Using the conservation of energy at the equilibriumposition: E = kA2/2 = (m+M) v2/2 =>v2 = 845 (0.2)2/(0.01+2.5) => v = 3.6696m/s Now using the conservation of momentum: mu = (m+M) v or, u = (m+M) v/m = 2.51(3.6696)/0.01 = 921.1 m/s

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