Question

A 4.71-g bullet is moving horizontally with a velocity of +363 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 1244 g, and its velocity is +0.677 m/s after the bullet passes through it. The mass of the second block is 1632 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.

Step 1

(a)

According to the law of conservation of momentum, total momentum of a system before the collision is equal to the total momentum of system after the collision.

Here,

*m*_{B} is the mass of the bullet.

*v*_{B} is the velocity of the bullet.

*m*_{1} is the mass of the first block.

*m*_{2} is the mass of the second block.

*v*_{1} is the velocity of the first block after the collision.

*v*_{2} is the velocity of the second block with bullet after the bullet imbeds itself.

Step 2

Rewrite the above equation for .

Step 3

Substitute the ...

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