A 0.0430-kg bullet is fired horizontally into a 3.92-kg wooden block attached to

Question

A 0.0430-kg bullet is fired horizontally into a 3.92-kg wooden block attached to one end of a massless, horizontal spring (k = 884 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 0.256 m. What is the speed of the bullet?

Explanation / Answer

solution
as we know that
let v is speed of bullet

energy transfered= 884x0.2562 = 57.9J
v= (2x57.9/0.0430)= 51.9m/s

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