A 2-kg ball B traveling at 22m.s-1 Overtakes a 4-kg ball traveling in the same direction at 10m.s-1 Find the velocity of each ball after collision.
{mbvb22+mava22=mbub22+maua22,mbvb+mava=mbub+maua;\begin{cases} \frac{m_bv_b^2}{2}+\frac{m_av_a^2}{2}=\frac{m_bu_b^2}{2}+\frac{m_au _a^2}{2},\\ m_bv_b+m_av_a=m_bu_b+m_au_a; \end{cases}{2mbvb2+2mava2=2mbub2+2maua2,mbvb+mava=mbub+maua;
{mb(vb2−ub2)=ma(ua2−va2),mb(vb−ub)=ma(ua−va);\begin{cases} m_b(v_b^2-u_b^2)=m_a(u_a^2-v_a^2), \\ m_b(v_b-u_b)=m_a(u_a-v_a); \end{cases}{mb(vb2−ub2)=ma(ua2−va2),mb(vb−ub)=ma(ua−va);
{vb+ub=ua+va,vb−va=ua−ub;\begin{cases} v_b+u_b=u_a+v_a, \\ v_b-v_a=u_a-u_b; \end{cases}{vb+ub=ua+va,vb−va=ua−ub;
{ua=2mbvb+(ma−mb)vama+mb,ub=2mava+(mb−ma)vbma+mb;\begin{cases} u_a=\frac{2m_bv_b+(m_a-m_b)v_a}{m_a+m_b}, \\ u_b=\frac{2m_av_a+(m_b-m_a)v_b}{m_a+m_b}; \end{cases}{ua=ma+mb2mbvb+(ma−mb)va,ub=ma+mb2mava+(mb−ma)vb;
{ua=18 ms,ub=6 ms.\begin{cases} u_a=18~\frac ms, \\ u_b=6~\frac ms. \end{cases}{ua=18 sm,ub=6 sm.
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