A 7.11-g bullet is moving horizontally with a velocity of +354 m/s, where the si

Question

A 7.11-g bullet is moving horizontally with a velocity of +354 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 1232 g, and its velocity is +0.694 m/s after the bullet passes through it. The mass of the second block is 1571 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.

Explanation / Answer

You know the mass and speed of the bullet, which means you can calculate its kinetic energy before impact (E = 0.5m*v^2). You know the mass and speed of the first block, so you know how much kinetic energy the bullet lost there. You now know the remaining kinetic energy of the bullet and the mass of the second block. The bullet does not pass through the second block so it transfers all its kinetic energy to it. This is application of just a single formula.

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