...................... A bullet with a mass of 4.0 g and a speed of 611 m/s is f

Question

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A bullet with a mass of 4.0 g and a speed of 611 m/s is fired at a block of wood with a mass of 0.071 kg. The block rests on a frictionless surface, and is thin enough that the bullet passes completely through it. Immediately after the bullet exits the block, the speed of the block is 23 m/s. What is the speed of the bullet when it exits the block? m/s Is the final kinetic energy of this system equal to, less than, or greater than the initial kinetic energy? less greater same (c) Verify your answer to part (b) by calculating the initial and final kinetic energies of the system. Ki = Kf =

Explanation / Answer

Conservation of momentum:

m1u1 + m2u2 = m1v1 + m2v2

4*611 + 0 = 4v + 71*23

v = 202.75 m/s

Kinetic energy of the system will be reduced.

Ki = 0.5*0.004*611*611 = 746.642 J

Kf = 0.5*0.004*202.75*202.75 + 0.5*0.071*23*23 = 100.99 J

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