Answer to Question #185004 in Classical Mechanics for Alan Enrico V Tuibeo

Question

Two blocks of mass 0.30 kg and 0.20 kg are moving toward each other along a frictionless horizontal surface with velocities of 0.50 m.s–¹ and 1.00 m.s–¹, respectively. Find the final velocity and direction of each block after collision if the collision is elastic. A complete solution is one with labeled diagram showing the blocks before and after collision.

Answer 1

$\begin{cases} \frac{m_bv_b^2}{2}+\frac{m_av_a^2}{2}=\frac{m_bu_b^2}{2}+\frac{m_au _a^2}{2},\\ m_bv_b+m_av_a=m_bu_b+m_au_a; \end{cases}$

$\begin{cases} m_b(v_b^2-u_b^2)=m_a(u_a^2-v_a^2), \\ m_b(v_b-u_b)=m_a(u_a-v_a); \end{cases}$

$\begin{cases} v_b+u_b=u_a+v_a, \\ v_b-v_a=u_a-u_b; \end{cases}$

$\begin{cases} u_a=\frac{2m_bv_b+(m_a-m_b)v_a}{m_a+m_b}, \\ u_b=\frac{2m_av_a+(m_b-m_a)v_b}{m_a+m_b}; \end{cases}$

$\begin{cases} u_a=-0.7~\frac ms, \\ u_b=0.8~\frac ms. \end{cases}$