完全弹性碰撞公式推导

原式:
$m_1{v}_0=m_1{v}_1+m_2{v}_2$
$\frac{1}{2}{m}_1{v}_0^2=\frac{1}{2}{m}_1{v}_1^2+\frac{1}{2}{m}_2{v}_2^2$
解:
通过1式求得$v_2$
$v_2=\frac{m_1(v_0-v_1)}{m_2}$
带入2式中,且约去$\frac{1}{2}$有:
$m_1{v}_0^2=m_1{v}_0^2+m_2(\frac{m_1(v_0-v_1)}{m_2}{)}^2$
两边同时乘以 $\frac{m_2}{m_1}$ 有:
$m_2{v}_0^2=m_2{v}_1^2+m_1{v}_0^2+m_1{v}_1^2-2m_1{v}_0{v}_1$
全部移到等号左侧有:
$(m_1+m_2)v_1^2-2m_1{v}_0{v}_1+(m_1-m_2)v_0^2=0$
由一元二次方程公式得:
$\pm x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$
有:
$\pm v_1=\frac{2m_1{v}_0\pm \sqrt{4m_1^2{v}_0^2-4(m_1+m_2)(m_1-m_2)v_0^2}}{2(m_1+m_2)}$

$=\frac{2m_1{v}_0\pm \sqrt{4m_1^2{v}_0^2-4m_1^2{v}_0^2+4m_2^2{v}_0^2}}{2(m_1+m_2)}$

$=\frac{2m_1{v}_0\pm 2m_2{v}_0}{2(m_1+m_2)}$
若取正
$v_1=v_0$即碰撞后未发生速度变化,或未发生碰撞,故舍弃
取负有:
$v_1=\frac{m_1-m_2}{m_1+m_2}{v}_0$

将$v_1$反代入1式:
$m_1{v}_0=m_1\frac{m_1-m_2}{m_1+m_2}{v}_0+m_2{v}_2$
化简:
$m_2{v}_2=m_1{v}_0-m_1\frac{m_1-m_2}{m_1+m_2}{v}_0$
$m_2{v}_2=\frac{2m_2}{m_1+m_2}{m}_1{v}_0$
同时约去$m_2$:
$v_2=\frac{2m_1}{m_1+m_2}{v}_0$