A 4.68-g bullet is moving horizontally with a velocity of +366 m/s, where the si

Question

A 4.68-g bullet is moving horizontally with a velocity of +366 m/s, where the sign + indicates that it is moving to the right (see part a of the drawing). The bullet is approaching two blocks resting on a horizontal frictionless surface. Air resistance is negligible. The bullet passes completely through the first block (an inelastic collision) and embeds itself in the second one, as indicated in part b. Note that both blocks are moving after the collision with the bullet. The mass of the first block is 1172 g, and its velocity is +0.592 m/s after the bullet passes through it. The mass of the second block is 1526 g. (a) What is the velocity of the second block after the bullet imbeds itself? (b) Find the ratio of the total kinetic energy after the collision to that before the collision.

I can't get part B! Any help is very appreciated!

A 4.68-g bullet is moving horizontally with a velocity of +366 m/s, where the sign indicates that it is moving to the right (see part a of the drawing). The bullet is approaching two blocks resting on a horizontal frictionless surface. Air resistance is negligible. The bullet passes completely through the first block (an inelastic collision) and embeds itself in the second one, as indicated in part b. Note that both blocks are moving after the collision with the bullet. The mass of the first block is 1172 g, and its velocity is +0.592 m/s after the bullet passes through it. The mass of the second block is 1526 g. (a) What is the velocity of the second block after the bullet imbeds itself? (b) Find the ratio of the total kinetic energy after the collision to that before the collision. +366m/s Block 2 nm (a) Before collision +0.592m/s block 2 mblock 1 1172g mblock 2 15268 let 4.688 (b) After collision a) Vblock2 Number +.666 Units m/s Units No units (b) KEafter/KEbefore Number 00224

Explanation / Answer

Initial kinetic energy of the bullet: 0.00468 x 3662 / 2 = 313.46 kg-m/s

Final kinetic energy of block1: 1.172 x 0.5922/ 2 = 0.21 kg-m/s

Final kinetic energy of the bullet/block2 complex: (0.00468 + 1.526) x 0.6662 / 2 = 0.339 kg-m/s

Final kinetic energy (total): 0.21 + 0.339 = 0.549 kg-m/s

Ratio of final and initial kinetic energies: 0.549/313.46 = 0.00175

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