A thin block of soft wood with a mass of 0.086 kg rests on a horizontal friction

Question

A thin block of soft wood with a mass of 0.086 kg rests on a horizontal frictionless surface. A bullet with a mass of 4.67 g is fired with a speed of 591 m/s at a block of wood and passes completely through it. The speed of the block is 15 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

a) to determine the speed of the bullet as it exits the block, we apply the principle pf conservation of momentum

mbullet vi = mbullet v1 + mblock v2

0.00467 * 591 = 0.00467 * v1 + 0.086 * 15

so v1 = 314.76m/s

b) Initial Kinetic energy of the system = Kinetic energy of the bullet = 1/2mbullet vi2 = 1/2 * 0.00467 * (591)2 = 815.57J

Final Kinetic energy of the system = Kinetic energy of the block + Kinetic energy of the bullet after leaving the block

                                                           = 1/2mbullet v12 + 1/2mblock v22 = 1/2 * 0.00467 * (314.76)2 + 1/2 * 0.086 * (15)2

                                                            = 231.34 + 9.675 = 241.02J

therefore Kf < Ki

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