A 1.08 10-2 kg bullet is fired horizontally into a 2.44 kg wooden block attached

Question

A 1.08 10-2 kg bullet is fired horizontally into a 2.44 kg wooden block attached to one end of a massless, horizontal spring (k = 848 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

let the speed of bullet=v conservation of momentum 1.08*10^-2*v=(2.44+1.08*10^-2)*v1 x=0.2 conservation of energy 0.5*(2.44+1.08*10^-2)*v1^2=0.5*848*0.2^2 v1=3.72 m/s v=844.16 m/s

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