Question

An archer shoots an arrow toward a 300-g target that is sliding in her direction at a speed of 2.10 m/s on a smooth, slippery surface. The 22.5-g arrow is shot with a speed of 37.5 m/s and passes through the target, which is stopped by the impact. What is the speed of the arrow after passing through the target

Answer 1

Answer:

The speed of the arrow after passing through the target is 30.1 meters per second.

Explanation:

The situation can be modelled by means of the Principle of Linear Momentum, let suppose that the arrow and the target are moving on the same axis, where the velocity of the first one is parallel to the velocity of the second one. The Linear Momentum model is presented below:

$m_{a}\cdot v_{a,o} + m_{t}\cdot v_{t,o} = m_{a}\cdot v_{a,f} + m_{t}\cdot v_{t,f}$

Where:

$m_{a}$, $m_{t}$ - Masses of arrow and target, measured in kilograms.

$v_{a,o}$, $v_{a,f}$ - Initial and final speeds of the arrow, measured in meters per second.

$v_{t,o}$, $v_{t,f}$ - Initial and final speeds of the target, measured in meters per second.

The final speed of the arrow is now cleared:

$m_{a} \cdot v_{a,f} = m_{a} \cdot v_{a,o} + m_{t}\cdot (v_{t,o}-v_{t,f})$

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

If $v_{a,o} = 2.1\,\frac{m}{s}$, $m_{t} = 0.3\,kg$, $m_{a} = 0.0225\,kg$, $v_{t,o} = 2.10\,\frac{m}{s}$ and $v_{t,f} = 0\,\frac{m}{s}$, the speed of the arrow after passing through the target is:

$v_{a,f} = 2.1\,\frac{m}{s} + \frac{0.3\,kg}{0.0225\,kg}\cdot (2.10\,\frac{m}{s} - 0\,\frac{m}{s} )$

$v_{a,f} = 30.1\,\frac{m}{s}$

The speed of the arrow after passing through the target is 30.1 meters per second.