Conservation of Momentum EX 1

(a) The kinetic energy is related to the momentum as shown below: $K=\frac{1}{2}m{v}^2=\frac{1}{2}\frac{m}{m}m{v}^2=\frac{1}{2}\frac{m^2{v}^2}{m}=\frac{1}{2}\frac{(mv{)}^2}{m}=\frac{1}{2}\frac{p^2}{m}$

Given that $p_B=p_A$, we can compute the ratio of their kinetic energies using the above expression as

$\frac{K_B}{K_A}=\frac{\frac{1}{2}\frac{{p_B}^2}{m_B}}{\frac{1}{2}\frac{{p_A}^2}{m_A}}=\frac{\frac{1}{m_B}}{\frac{1}{m_A}}=\frac{m_A}{m_B}=\frac{60}{0.020}=3000$

(b) We can compute the momentum from the kinetic energy

$K=\frac{1}{2}\frac{p^2}{m}$

$2mK=p^2$

$p=\sqrt{2mK}$

Given that $K_B=K_A$, we can compute the ratio

$\frac{p_B}{p_A}=\frac{\sqrt{2m_B{K}_B}}{\sqrt{2m_A{K}_A}}=\sqrt{\frac{m_B}{m_A}}=\sqrt{\frac{0.020}{60}}=0.018$