Physics Solution: Momenta in Center of Mass Frame
The velocity of a particle in the COM frame is $\vec{v}’ = \vec{v}_{lab} – \vec{v}_{cm}$. In the COM frame, the total momentum of the system is always zero ($p’_A + p’_B = 0$).
Step-by-Step Calculation
1. Laboratory Frame Velocities
Both particles start from rest. A force $F$ acts on both for time $t$.
- Particle A ($m$): $v_A = \frac{F}{m}t$
- Particle B ($2m$): $v_B = \frac{F}{2m}t$
2. Velocity of the Center of Mass ($v_{cm}$)
$$v_{cm} = \frac{m v_A + 2m v_B}{m + 2m}$$
$$v_{cm} = \frac{m(\frac{Ft}{m}) + 2m(\frac{Ft}{2m})}{3m} = \frac{Ft + Ft}{3m} = \frac{2Ft}{3m}$$
3. Momentum in the COM Frame
For Particle A:
$$v’_A = v_A – v_{cm} = \frac{Ft}{m} – \frac{2Ft}{3m} = \frac{3Ft – 2Ft}{3m} = \frac{Ft}{3m}$$
$$\text{Momentum } p’_A = m v’_A = m \left( \frac{Ft}{3m} \right) = \frac{Ft}{3}$$
For Particle B:
$$v’_B = v_B – v_{cm} = \frac{Ft}{2m} – \frac{2Ft}{3m} = \frac{3Ft – 4Ft}{6m} = -\frac{Ft}{6m}$$
$$\text{Momentum } p’_B = 2m v’_B = 2m \left( -\frac{Ft}{6m} \right) = -\frac{Ft}{3}$$