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:

Question

irishboy123 · Answer

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

anonymous · Answer

12 m/s in the same direction as B?

irishboy123 · Answer

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

anonymous · Answer

oh should it be in the same direction as B?

irishboy123 · Answer

|dw:1439109531470:dw|

irishboy123 · Answer

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

anonymous · Answer

oh thank you so much :)