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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:

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  1. IrishBoy123
    • one year ago
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    \(\large \vec{ \bar x} = \frac{m_1 \vec x_ 1 + m_2 \vec x_ 2}{m_1 + m_2}\) \(\large \dot{\vec{ \bar x} }= \frac{m_1 \vec {\dot x_ 1} + m_2 \vec{ \dot x_ 2} }{m_1 + m_2} = \frac{\vec p_1 + \vec p_2}{m_1 + m_2}\) \(\large \implies \vec { \bar p} = \vec p_1 + \vec p_2\) seems to make sense

  2. anonymous
    • one year ago
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    12 m/s in the same direction as B?

  3. IrishBoy123
    • one year ago
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    4(2) + 8(-3) = -16

  4. anonymous
    • one year ago
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    oh should it be in the same direction as B?

  5. IrishBoy123
    • one year ago
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    |dw:1439109531470:dw|

  6. IrishBoy123
    • one year ago
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    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
    • one year ago
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    oh thank you so much :)

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