This question can be answered with the concept of Conservation of Momentum. Because momentum is always conserved, we can write an equation that can be used for this question:
[tex]m_{1}[/tex][tex]v_{i}[/tex] + [tex]m_{2}[/tex][tex]v_{0}[/tex] = ([tex]m_{1}[/tex] + [tex]m_{2}[/tex])[tex]v_{f}[/tex]
Since the asteroid is flying down vertically, it's speed in this equation can be disregarded and is essentially 0 since we're measuring horizontal speed.
Therefore: [tex]m_{1}[/tex][tex]v_{i}[/tex] = ([tex]m_{1}[/tex] + [tex]m_{2}[/tex])[tex]v_{f}[/tex]
We can simplify this equation to make our life easier when we plug in the numbers to crunch the final value:
[tex]v_{f}[/tex] = [tex]\frac{m_{1}v_{1}}{(m_{1} + m_{2})}[/tex]
There we go. Let's plug in the numbers:
[tex]v_{f}[/tex] = [tex]\frac{m_{1}v_{1}}{(m_{1} + m_{2})}[/tex]
[tex]v_{f}[/tex] = [tex]\frac{(1200)(25)}{(1200+220)}[/tex]
[tex]v_{f}[/tex] ≈ 21 (m/s)
Answer: Option B
And there we have it! If you have any questions about my answer, just ask :)
