Question

A 1.67 g particle is moving at 4.58 m/s toward a stationary 9.54 g particle. With what speed does the heavier particle approach the center of mass of the two particles? Answer in units of m/s.

Answer 1

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

`m_1 = 1.67g`

`v_1 = 4.58m/s`

`m_2 = 9.54g`

`v_2 = 0m/s \rightarrow` Stationary

The center of mass of the two particles is

`v_(cm) = (m_1v_1+m_2v_2)/(m_1+m_2)`

Replacing,

`v_(cm) = ((1.67)(4.58)+(9.54)(0))/((4.47+8.05))`

`v_(cm) = 0.6109m/s`

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)