A thin block of soft wood with a mass of 0.080 kg rests on a horizontal friction

Question

A thin block of soft wood with a mass of 0.080 kg rests on a horizontal frictionless surface. A bullet with a mass of 4.67 g is fired with a speed of 575 m/s at a block of wood and passes completely through it. The speed of the block is 17 m/s immediately after the bullet exits the block.

(a) Determine the speed of the bullet as it exits the block. m/s

(b) Determine if the final kinetic energy of this system (block and bullet) is equal to, less than, or greater than the initial kinetic energy.

equal to the initial kinetic energy

less than the initial kinetic energy

greater than the initial kinetic energy

(c) Verify your answer to part (b) by calculating the initial and final kinetic energies of the system.

KEi = ___J

KEf = ___J

Explanation / Answer

As there is no external force on Bullet-block system, momentum of the esystem is conserved. Initially only bullet is moving, with momentum
Mb Vo = 0.00467 x 575 = 2.68525 kg m/sec
Final momentum of Bullet = 0.00467 V ( where V is final speed of bullet)
Final momentum of block =  0.08 x17 = 1.36
Initial momentum = final momentum
2.68525 = 1.36 + 0.00467 V
V = 283.78 m/s
As final relative velocity between Bullet and Block ( 283.78 - 17 = 266.78) is less than initial relative velocity of 575 m/s , Final kinetic enrgy is less than initial kinetic energy.

Ki = 1/ ( 0.00467) 5752 = 772.01 J
Kf = 1/2 (0.00467) 283.782 + 1/2 (0.08) 172 = 199.6 J

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