Question

two equal - mass particles (a and b) are located at some distan from each other. particle a is held stationary while b is mov away at speed v. what happens to the center of mass of the two - particle system? a. it does not move b. it moves towards a with speed v c. it moves away from a with speed v d. it moves towards a with speed 1/2 v e. it moves away from a with speed 1/2 v

Explanation:

Step1: Recall the formula for the velocity of the center of mass

The velocity of the center of mass \(v_{cm}\) of a two - particle system is given by \(v_{cm}=\frac{m_1v_1 + m_2v_2}{m_1 + m_2}\). Let \(m_1=m_2 = m\), \(v_1 = 0\) (since particle \(A\) is stationary) and \(v_2=v\) (velocity of particle \(B\)).

Step2: Substitute the values into the formula

Substitute \(m_1=m\), \(v_1 = 0\), \(m_2=m\) and \(v_2 = v\) into \(v_{cm}=\frac{m_1v_1 + m_2v_2}{m_1 + m_2}\). We get \(v_{cm}=\frac{m\times0+m\times v}{m + m}=\frac{mv}{2m}=\frac{v}{2}\). But we need to consider the direction. Since particle \(B\) is moving away from \(A\), the center of mass moves towards \(A\) (because the mass of \(A\) is not moving and we have a symmetric mass distribution in the formula).

Answer:

D. It moves towards A with speed \( \frac{1}{2}v\)