Answer to Question #213394 in Classical Mechanics for Vaibhav

Question

Two balls A and B of masses 10g and 20g moves on x axis towards each other with equal speed. They collide and stick together. What percentage of kinetic energy is lost?

Answer 1

$m_a= 10g=0.01kg$

$m_b= 20g=0.02kg$

$\vec v_a= -\vec v_b$

$\text{according to the law of conservation of momentum:}$

$m_a\vec v_a+m_b\vec v_b=(m_a+m_b)\vec v$

$-m_a\vec v_b+m_b\vec v_b=(m_a+m_b)\vec v$

$\vec v=\frac{m_b-m_a}{m_a+m_b}\vec v_b$

$\vec v=\frac{\vec v_b}{3}$

$\text{before collision}$

$E_k= \frac{m_av_a^2}{2}+\frac{m_bv_b^2}{2}$

$E_k= \frac{m_av_b^2}{2}+\frac{m_bv_b^2}{2}$

$E_k= \frac{v_b^2(m_a+m_b)}{2}=0.015v_b^2$

$\text{after collision:}$

$E'_k= \frac{v^2(m_a+m_b)}{2}=\frac{v_b^2(m_a+m_b)}{18}=\frac{0.01v_b^2}{6}$

$p = \frac{E_k-E'_k}{E_k}=1-\frac{E'_k}{E_k}$

$p = 1- \frac{0.01v_b^2}{6*0.015*v_b^2}\approx0.89=89 \%$

$\text{Answer: 89\% of kinetic energy is lost}$