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

Answer:

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:

$m_{B}\cdot v_{B,o} + m_{W}\cdot v_{W,o} = m_{B}\cdot v_{B,f} + m_{W}\cdot v_{W,f}$

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:

$v_{W,f} = \frac{m_{B}\cdot (v_{B,o}-v_{B,f})+m_{W}\cdot v_{W,o}}{m_{W}}$

$v_{W,f} = v_{W,o} + \frac{m_{B}}{m_{W}} \cdot (v_{B,o}-v_{B,f})$

If $v_{W,o} = 0\,\frac{m}{s}$, $m_{B} = 0.022\,kg$, $m_{W} = 2\,kg$, $v_{B,o} = 210\,\frac{m}{s}$ and $v_{B,f} = 150\,\frac{m}{s}$, then the final velocity of the block after the bullet passes through is:

$v_{W,f} = 0\,\frac{m}{s}+\left(\frac{0.022\,kg}{2\,kg}\right)\cdot \left(210\,\frac{m}{s}-150\,\frac{m}{s} \right)$

$v_{W,f} = 0.66\,\frac{m}{s}$

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