弹性碰撞证明勾股定理
预备知识
证明
如图,设\(A\),\(B\)两球质量均为\(m\)
\(B\)球静止,\(A\)球以速度\(v\)与\(B\)球作完全弹性碰撞
将\(v\)分解为连心线方向分速度\(v_2\)及与\(v_2\)垂直方向分速度\(v_1\)
\(\because\)两球作完全弹性碰撞
\(\therefore\)系统动量守恒:\(m\vec{v_1}+m\vec{v_2}=m\vec{v}\)
\(\therefore\)\(\vec{v_1}+\vec{v_2}=\vec{v}\)
即\(\vec{v_1}\)、\(\vec{v_2}\)、\(\vec{v}\)可构成以\(\vec{v}\)为斜边的\(Rt\triangle\)(\(\triangle\)\(ACD\))
\(\because\)系统动能守恒
\(\therefore\)\(\frac{1}{2}mv_1^2+\frac{1}{2}mv_2^2=\frac{1}{2}mv^2\)
\(\therefore\)\(v_1^2+v_2^2=v^2\)
即在\(Rt\triangle\)\(ACD\)中,\(AD^2=AC^2+CD^2\)
\(Q.E.D\)