Two masses, \(m_{1}=2.0 \mathrm{~kg}\) and \(m_{2}=3.0 \mathrm{~kg}\), are moving in the \(x y\) -plane. The velocity of their center of mass and the velocity of mass 1 relative to mass 2 are given by the vectors \(v_{\mathrm{cm}}=(-1.0,+2.4) \mathrm{m} / \mathrm{s}\) and \(v_{\mathrm{rel}}=(+5.0,+1.0) \mathrm{m} / \mathrm{s} .\) Determine a) the total momentum of the system b) the momentum of mass 1 , and c) the momentum of mass 2 .

To find the total momentum of the system (P_total), we will first add the masses together (M) and multiply it by the given center of mass velocity (v_cm): \(P_{\text{total}} = M \times v_{\text{cm}}\) \(M = m_1 + m_2 = 2.0 kg + 3.0 kg = 5.0 kg\) \(v_{\text{cm}} = (-1.0,+2.4) \frac{m}{s}\) So, \(P_{\text{total}} = (5.0 kg) \times (-1.0,+2.4) \frac{m}{s} = (-5.0,+12.0) \frac{kg.m}{s}\)

To find the momentum of mass 1 (P_1), we need to first find its velocity (v_1). To calculate this, we will use the relative velocity of mass 1 to mass 2 (v_rel) and the center of mass velocity (v_cm). We know that: \(v_{\text{rel}} = v_1 - v_2\) \(v_1 = v_{\text{rel}} + v_2\) Now we can find the velocity of mass 2 (v_2) using the center of mass velocity: \(v_{\text{cm}} = \frac{m_1 v_1 + m_2 v_2}{m_1 + m_2}\) \(v_2 = \frac{v_{\text{cm}}(m_1 + m_2) - m_1 v_1}{m_2}\) Substitute the values: \(v_2 = \frac{(-1.0,+2.4)(5.0) - 2.0 v_1}{3.0}\) \(v_2 = \frac{(-5.0,+12.0) - 2.0 v_1}{3.0}\) Now, substitute \(v_2\) back into the equation for \(v_1\): \(v_1 = (+5.0,+1.0) + \frac{(-5.0,+12.0) - 2.0 v_1}{3.0}\) Solving for \(v_1\), we get: \(v_1 = (+3.0,-2.0) \frac{m}{s}\) Now, we can find the momentum of mass 1 by multiplying its mass with its velocity: \(P_1 = m_1 \times v_1 = (2.0 kg) \times (+3.0,-2.0) \frac{m}{s} = (+6.0,-4.0) \frac{kg.m}{s}\)

Now that we have the momentum of mass 1, we can use the total momentum of the system to determine the momentum of mass 2 (P_2): \(P_{\text{total}} = P_1 + P_2\) \(P_2 = P_{\text{total}} - P_1 = (-5.0,+12.0) \frac{kg.m}{s} - (+6.0,-4.0) \frac{kg.m}{s} = (-11.0,+16.0) \frac{kg.m}{s}\)