A 22g bullet traveling 210 m/s penetrates a 2.0kg block of wood and emerges going 150m/s. If the block were stationary on a frictionless plane before the collision, what is the velocity of the block after the bullet passes through

The final velocity of the block after the bullet passes through is 0.66 meters per second.

Explanation:

The interaction between the bullet and the block of woods is a clear example of a perfectly inelastic collision, which can be modelled after the Principle of Momentum Conservation. There are no external forces exerted on the bullet-block system. The equation describing the collision is described below:

Where:

, - Masses of the bullet and the block of wood, measured in kilograms.

, - Initial speeds of the bullet and the block of wood, measured in meters per second.

, - Final speeds of the bullet and the block of wood, measured in meters per second.

The final speed of the block is cleared:

If , , , and , then the final velocity of the block after the bullet passes through is:

The final velocity of the block after the bullet passes through is 0.66 meters per second.

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