A 0.0178-kg bullet is fired horizontally into a 3.06-kg wooden block attached to

Question

A 0.0178-kg bullet is fired horizontally into a 3.06-kg wooden block attached to one end of a massless, horizontal spring (k = 880 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 0.269 m. What is the speed of the bullet?

Explanation / Answer

mass (m) of the bullet = 1.78*10-2kg mass (M) of the wooden block = 3.06kg sprong constant K = 880 N/m according to the law of conservation of momentum       mv = (m+M)V or V = mv/(m+M) here v is the initial speed of thebullet& V be thecombined speed of the bullet & block according to the law of conservation of energy 1/2(m+M)V2 = 1/2Kx2 or (m+M)V2 = Kx2 or (m+M)(mv/m+M)2 = kx2 solving finally we have        v = x/m*(m+M)K here x = 0.269m plug in the values for v we get the speed of thebullet

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