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Michele_Laino

  • one year ago

A "Mechanics" challenge

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  1. Michele_Laino
    • one year ago
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    Let's suppose a colllision between a neutron which is moving with a velocity V1, and a nucleus at rest. The unit of measure of the masses is the nucleon mass, so the neutron has mass equal to 1, whereas the mass of the nucleus is A, where A is the mass number of that nucleus. 1) find the velocity of the center of mass of the system neutron-nucleus 2) show that with respect to the center of mass, the total momentum of the system nucleus-neutron is the null vector @IrishBoy123 @Empty @Astrophysics

  2. Michele_Laino
    • one year ago
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    |dw:1437757810812:dw|

  3. Michele_Laino
    • one year ago
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    hint: the velocity of the center of mass of a system composed by N particles, whose mass are m_i and velocities are v_i, is given by the subsequent fromula: \[\Large {{\mathbf{v}}_{CM}} = \frac{{\sum\limits_1^N {{m_i}{{\mathbf{v}}_i}} }}{{\sum\limits_1^N {{m_i}} }}\]

  4. IrishBoy123
    • one year ago
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    *first* bit: |dw:1437768326539:dw| \(\huge \vec v_{cm}=\frac {∑^N_i \ m_i \vec v_i}{∑^N_1 \ m_i}\) conservation of momentum: \(\large \vec v_1 = \vec v_2 + A \vec v_f \) \(\huge \vec v_{cm}=\frac { \vec v_2 + A \vec v_f }{A + 1} = \frac {\vec v_1 }{A + 1}\) if that's totally off beam, pls advise.

  5. anonymous
    • one year ago
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    Using conservation of momentum,\[m _{neutron}v _{1} + m _{nucleus}v _{nucleus} = m _{neutron}v _{1}^{'} + m _{nucleus}v _{nucleus}^{'}\]\[\left( 1 \right) v _{1} + A \left( 0 \right)= \left( 1 \right)v{'} + A \left( v^{'} \right)\]\[v^{'} = \frac{ v_1 }{ A+1 }\]I do not understand part 2. The system has non-zero momentum before the collision and exactly the same momentum after the collision. How can it be zero?

  6. IrishBoy123
    • one year ago
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    for second part using conservation of energy: \(\vec v_1 ^2 = \vec v_2^2 + A \vec v_f^2\) \((\vec v_1 - \vec v_2)\bullet (\vec v_1 + \vec v_2) = A (\vec v_f \bullet \vec v_f)\) from conservation momentum above: \(\vec v⃗_1-v⃗_2=A v⃗_f\) [A] \(A v⃗_f \bullet (\vec v_1 + \vec v_2) = A \vec v_f \bullet \vec v_f\) \(\vec v_1 + \vec v_2 = \vec v_f\) [B] [A] and [B] give : \(\large \vec v_f = \frac{2 \vec v_1}{1+A}\) \(\large \vec v_2 = \frac{ \vec v_1 (1-A)}{1+A}\) and these follow also from \( \vec v_{cm}\) in previous post: \(\large \vec v_f - \vec v_{cm} = \frac{ \vec v_1}{1+A}\) \(\large \vec v_2 - \vec v_{cm} = \frac{ -\vec v_1A}{1+A}\) momentum \(\vec p\) of system measured from the reference frame of the centre of mass is: \(\vec p = A (\vec v_f - \vec v_{cm})+ (1)(\vec v_2 - \vec v_{cm} )\\ = A \frac{ \vec v_1}{1+A} + (1) \frac{ -\vec v_1A}{1+A} = \vec 0\)

  7. IrishBoy123
    • one year ago
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    in order to verify conservation of momentum from reference frame of centre of mass, looking at the "before" picture: \(\large \vec v_{cm_o} =\frac{(1)\vec v_1+A\vec0}{1 + A} = \frac{\vec v_1}{1 + A}\) ie same from reference frame of centre of mass, momentum \(\vec p_o\) was: \(\large \vec p_o = (1) (\vec v_1 - v_{cm_o} ) + A(\vec 0 - v_{cm_o}) \) \(\large \ \ = (1) (\vec v_1 -\frac{\vec v_1}{1+A} ) + A( - \frac{\vec v_1}{1 + A}) = \vec 0\)

  8. Michele_Laino
    • one year ago
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    nicely done!! :) @IrishBoy123

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