Answer to Question #231787 in Physics for sule

Question

An archer shoots an arrow toward a target that is sliding toward her with a speed of 2.50 m/s on a smooth surface. The 22.5-g arrow is shot with a speed of 35.0 m/s and passes through the 300-g target, which is stopped by the impact. What is the speed of the arrow after passing through the target? 

Answer 1

Given:

$v_1=35.0\:\rm m/s$

$v_2=2.50\:\rm m/s$

$m_1=22.5\:\rm g$

$m_2=300\:\rm g$

The law of conservation of momentum says

$$m_1v_1-m_2v_2=(m_1+m_2)u$$

The speed of the arrow after passing through the target

$$u=\frac{m_1v_1-m_2v_2}{m_1+m_2}$$ $$u=\frac{22.5*35.0-300*2.50}{22.5+300}=0.12\:\rm m/s$$