Question

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

Answer 1

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

Step-by-step 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:

`m_(B)\cdot v_(B,o) + m_(W)\cdot v_(W,o) = m_(B)\cdot v_(B,f) + m_(W)\cdot v_(W,f)`

Where:

`m_(B)`,
`m_(W)`- Masses of the bullet and the block of wood, measured in kilograms.

`v_(B,o)`,
`v_(W,o)` - Initial speeds of the bullet and the block of wood, measured in meters per second.

`v_(B,f)`,
`v_(W,f)`- Final speeds of the bullet and the block of wood, measured in meters per second.

The final speed of the block is cleared:

`v_(W,f) = (m_(B)\cdot (v_(B,o)-v_(B,f))+m_(W)\cdot v_(W,o))/(m_(W))`

`v_(W,f) = v_(W,o) + (m_(B))/(m_(W)) \cdot (v_(B,o)-v_(B,f))`

If
`v_(W,o) = 0\,(m)/(s)`,
`m_(B) = 0.022\,kg`,
`m_(W) = 2\,kg`,
`v_(B,o) = 210\,(m)/(s)` and
`v_(B,f) = 150\,(m)/(s)`, then the final velocity of the block after the bullet passes through is:

`v_(W,f) = 0\,(m)/(s)+\left((0.022\,kg)/(2\,kg)\right)\cdot \left(210\,(m)/(s)-150\,(m)/(s) \right)`

`v_(W,f) = 0.66\,(m)/(s)`

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