A 9.20 x 10^-3 kg bullet is fired horizontally into a 2.38 kg wooden block attac

Question

A 9.20 x 10^-3 kg bullet is fired horizontally into a 2.38 kg wooden block attached to one end of a massless, horizontal spring (k = 860 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? m/s P.S. I am offering A LOT of points so I will need to see exactly how you solved this step by step. Make it very clear what each number stands as you work the problem! .

Explanation / Answer

Let speed of bullet = u

after collision let speed of combined mass = v

so 9.2*10^-3 *u = ( 2.38 + 9.2*10^-3) * v

so u = 259.696 v

so kinetic energy = (2.38 + 9.2*10^-3) * v^2 /2

at maximum compression all kinetic energy is transformed to potential energy

so ( 2.38 + 9.2*10^-3 ) * v^2/2 = k*A^2/2 = 860 * 0.2^2 /2

so v = 3.794 m/s

so u = 259.696 * 3.794 = 985.3 m/s

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