To solve this problem we will use the concept related to the center of mass by speed of two objects that approach each other.
According to the information our values are
[tex]m_1 = 1.67g[/tex]
[tex]v_1 = 4.58m/s[/tex]
[tex]m_2 = 9.54g[/tex]
[tex]v_2 = 0m/s \rightarrow[/tex] Stationary
The center of mass of the two particles is
[tex]v_{cm} = \frac{m_1v_1+m_2v_2}{m_1+m_2}[/tex]
Replacing,
[tex]v_{cm} = \frac{(1.67)(4.58)+(9.54)(0)}{(4.47+8.05)}[/tex]
[tex]v_{cm} = 0.6109m/s[/tex]
Therefore the speed does the heavier particle approach the center of mass of two particles is -0.6109m/s (From the reference system this speed can be taken as negative. As the lightest mass (our reference system) is receiving the impact of the heaviest mass)
