## anonymous one year ago Block A, with a mass of 4.0 kg, is moving with a speed of 2.0 m/s while block B, with a mass of 8.0 kg, is moving in the opposite direction with a speed of 3.0 m/s. The momentum of the center of mass of the two-block system is:

1. IrishBoy123

$$\large \vec{ \bar x} = \frac{m_1 \vec x_ 1 + m_2 \vec x_ 2}{m_1 + m_2}$$ $$\large \dot{\vec{ \bar x} }= \frac{m_1 \vec {\dot x_ 1} + m_2 \vec{ \dot x_ 2} }{m_1 + m_2} = \frac{\vec p_1 + \vec p_2}{m_1 + m_2}$$ $$\large \implies \vec { \bar p} = \vec p_1 + \vec p_2$$ seems to make sense

2. anonymous

12 m/s in the same direction as B?

3. IrishBoy123

4(2) + 8(-3) = -16

4. anonymous

oh should it be in the same direction as B?

5. IrishBoy123

|dw:1439109531470:dw|

6. IrishBoy123

i "think" you are multiplying the relative speed by the total mass what i tried to show in the first post is that the momentum of the CoM is actually the momentum of the system PS units of momentum?? kg m/s

7. anonymous

oh thank you so much :)