A 1.03 10-2 kg bullet is fired horizontally into a 2.55-kg wooden block attached

Question

A 1.03 10-2 kg bullet is fired horizontally into a 2.55-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 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 0.200 m. What is the speed of the bullet?

Explanation / Answer

Let the velocity of (bullet+block) is v m/s after the collision:-

=>By the law of energy conservation:-

=>KE(block+bullet) = PE(spring)

=>1/2(M+m)v^2 = 1/2kx^2

=>(1.03x 10^-2 + 2.55) x v^2 = 830 x (0.2)^2

=>v = 12.97

=>v = 3.601 m/s

Let the velocity of the bullet is u m/s before the collision:-

=>By the law of momentum conservation:-

=>mu = (M+m)v

=>1.03 x 10^-2 x u = (1.03 x 10^-2 + 2.55) x 3.601

=>u = 895.11 m/s

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