Question

A 1.70 kg eagle catches a 0.600 kg sparrow from behind in flight. What is their velocity after impact if the eagle’s velocity is initially 23 m/s to the right, and the sparrow’s velocity is 5.35 m/s in the opposite direction?Group of answer choices

Answer 1

Given,

$\begin{gathered} m_1=1.70kg \\ m_2=0.600kg \\ v_1=23\text{ m/s} \\ v_2=5.35\text{ m/s} \end{gathered}$

Given the velocity of the eagle is right so take it as positive and the velocity of the sparrow is left take it negatively.

$\begin{gathered} v=\frac{m_1v_1+m_2v_2}{m_1+m_2} \\ v=\frac{(1.70\times23)-(0.600\times5.35)_{}}{1.70+0.600_{}} \\ v=\frac{35.89}{2.3} \\ v=15.60\text{ m/s} \end{gathered}$

Thus, the velocity is positive so the direction is right with the velocity of 15.60 m/s.

Therefore, the given option c is correct.