Question #1a6bb

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Question #1a6bb

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Answer 1 · 2017-07-19T04:00:44

Relation is great.

Explanation:

Suppose,a particle of mass $m_{1}$ $m_1$ moving with a velocity of $u_{1}$ $u_1$ collides with a particle of mass $m_{2}$ $m_2$ moving with a velocity $u_{2}$ $u_2$ .
After collision the speed of the mass $m_{1}$ $m_1$ and $m_{2}$ $m_2$ are respectively $v_{1}$ $v_1$ and $v_{2}$ $v_2$ .

Let,the $m_{1}$ $m_1$ particle pushes the other with the force $F_{1 \to 2}$ $F_(1->2)$ and the $m_{2}$ $m_2$ particle on $m_{1}$ $m_1$ is $F_{2 \to 1}$ $F_(2->1$ .
According to Newton's third law,
$F_{1 \to 2} = - F_{2 \to 1}$ $F_(1->2)=-F_(2->1)$
As we know impulse of force is equal to change of momentum ,so we can write
$F_{1 \to 2} = m_{1} u_{1} - m_{1} v_{1}$ $F_(1->2)=m_1u_1-m_1v_1$
$F_{2 \to 1} = m_{2} u_{2} - m_{2} v_{2}$ $F_(2->1)=m_2u_2-m_2v_2$
Solving all equations,
$m_{1} u_{1} - m_{1} v_{1} = - m_{2} u_{2} + m_{2} v_{2}$ $m_1u_1-m_1v_1=-m_2u_2+m_2v_2$
$\Rightarrow m_{1} u_{1} + m_{2} u_{2} = m_{1} v_{1} + m_{2} v_{2}$ $=>m_1u_1+m_2u_2=m_1v_1+m_2v_2$
That's the all.

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Question #1a6bb

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