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Michele_Laino
 one year ago
A "Mechanics" challenge
Michele_Laino
 one year ago
A "Mechanics" challenge

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Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.0Let'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 neutronnucleus 2) show that with respect to the center of mass, the total momentum of the system nucleusneutron is the null vector @IrishBoy123 @Empty @Astrophysics

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.0dw:1437757810812:dw

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.0hint: 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: \[\Large {{\mathbf{v}}_{CM}} = \frac{{\sum\limits_1^N {{m_i}{{\mathbf{v}}_i}} }}{{\sum\limits_1^N {{m_i}} }}\]

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.1*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.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Using 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 nonzero momentum before the collision and exactly the same momentum after the collision. How can it be zero?

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.1for 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: \(\vec v⃗_1v⃗_2=A v⃗_f\) [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 : \(\large \vec v_f = \frac{2 \vec v_1}{1+A}\) \(\large \vec v_2 = \frac{ \vec v_1 (1A)}{1+A}\) 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: \(\vec p = A (\vec v_f  \vec v_{cm})+ (1)(\vec v_2  \vec v_{cm} )\\ = A \frac{ \vec v_1}{1+A} + (1) \frac{ \vec v_1A}{1+A} = \vec 0\)

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.1in 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: \(\large \vec p_o = (1) (\vec v_1  v_{cm_o} ) + A(\vec 0  v_{cm_o}) \) \(\large \ \ = (1) (\vec v_1 \frac{\vec v_1}{1+A} ) + A(  \frac{\vec v_1}{1 + A}) = \vec 0\)

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.0nicely done!! :) @IrishBoy123
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